https:///astrobaki/api.php?action=feedcontributions&user=Jmcbride&feedformat=atom AstroBaki - User contributions [en] 2023-10-04T19:38:31Z User contributions MediaWiki 1.35.1 Hydrostatic Equilibrium 2012-04-10T21:08:16Z <p>Jmcbride: fixed typo in hydrostatic equilibrium equation</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[11pt]{article}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> \begin{document}<br /> \subsection*{Central Topics}<br /> \begin{enumerate}<br /> \item Force balance in stars: pressure vs. gravity<br /> \item Energy transport by radiation, convection, and conduction<br /> \item Energy generation by fusion<br /> \end{enumerate}<br /> These combine to give the properties of the HR diagram, the main sequence, etc. Changes in mass and composition change each of the above. These three unifying principles used repeatedly in the course. We'll look at each separately to see what we can learn, and then put them together to understand stars. Even without knowing anything about energy generation, we can understand a lot based on force balance and energy transport.<br /> \\ \\<br /> \subsection*{Force Balance in Stars: Hydrostatic Equilibrium} \\<br /> Throughout the course we will assume spherical stars, ignoring factors such as rotation that may cause a star to become oblate, for instance. \\<br /> We will start by looking at a thin layer in a star, and over a small enough area that we can approximate the layer as being flat. The layer has thickness $dr$, density $\rho$, area $A$, and is at a distance $r$ from the center of the star and encloses a mass $M_r$. The mass of the shell is <br /> $<br /> M_{\rm shell} = \rho A dr<br />$<br /> The inward gravitational force on the region is then<br /> $<br /> F_g = -\frac{G M_r M_{\rm shell}}{r^2}<br />$<br /> The competing force (which provides force balance) is the pressure gradient within the star. That is, there is a slightly greater pressure on the bottom, $P_{\rm below}$ on the shell then there is on the top, $P_{\rm above}$. We then can multiply this pressure difference by the area of the thin shell to find the total force due to the pressure gradient. The net force is the sum of these two sources, and will give the acceleration of the shell.<br /> $<br /> F_{\rm net} = M_{\rm shell} a<br />$<br /> $<br /> F_{\rm net} = (P_{\rm below} - P_{\rm above})A -\frac{G M_r M_{\rm shell}}{r^2} = M_{\rm shell} a.<br />$<br /> Since we are looking at a thin shell, we will say that <br /> $<br /> P_{\rm above} = P_{\rm below} + dP.<br />$<br /> Then, substituting above,<br /> $<br /> -dp A - \frac{G M_r}{r^2}A dr \rho = A dr \rho a.<br />$ <br /> The area element cancels out, which is good because it was sort of arbitrarily chosen. Dividing through by the differential $dr$ gives<br /> $<br /> \rho a = -\frac{dP}{dr} - \frac{G M_r}{r^2} \rho.<br />$ <br /> In the case where there is no acceleration, which is pretty common since stars generally appear to be neither contraction nor expanding, we have hydrostatic equilibrium (HE). In this case, the above reduces to<br /> $<br /> \frac{dP}{dr} = -\rho \frac{G M_r}{r^2} = -\rho g.<br />$<br /> To further justify this approximation, let's imagine that $a$ is not equal to 0. Instead, it is roughly equal to the graviational acceleration. We can then estimate the amount of time it would take for the radius of a star to change appreciably. So<br /> $<br /> a \sim -\frac{G M_r}{r^2}<br />$<br /> We can look at the acceleration in terms of the size of the star, $R$, and the timescale of change, $t$. Then<br /> $<br /> a \sim \frac{R}{t^2}.<br />$<br /> For the purposes of this estimate, we can just take $r = R$ and $M_r = M$, which<br /> gives<br /> $<br /> \frac{R}{t^2} \sim \frac{G M}{R^2},<br />$<br /> $<br /> t \sim \sqrt{\frac{R^3}{G M}} \sim \sqrt{\frac{1}{G \rho}}.<br />$<br /> This is often called the dynamical time, and applies equally well to planets or galaxies as it does to a star. It is is also called the free fall time, since it is the time it takes to move the size of the system at the free fall speed of the system. For the Sun, with an average density of $1$ g cm$^{-3}$, this is about one hour, meaning any imbalances must quickly be eliminated. There are only brief periods in the life of a star where HE does not hold, such as the collapse to form a black hole. Even during convection, which violates the assumptions that went in to HE, this approximation holds very very well. \\<br /> We should explicitly define $M_r$, which we use in HE.<br /> $<br /> M_r = \int^r_0 4 \pi r^2 \rho dr, {\rm \; or}<br />$<br /> $<br /> \frac{dM_r}{dr} = 4 \pi r^2 \rho.<br />$<br /> There are still more variables here than equations. We need to know what the source of pressure is to solve for the structure of a star, as well as how energy is transported. We can, however, use just these equations to understand the outermost layers of the Sun, or even the atmosphere of the Earth. We use the approximation of a plane parallel atmosphere, which holds very well since the thickness $z$ of the atmosphere is much smaller than the size of the star (or planet) $R$. Then we can assume that the surface gravity is set by the total mass and radius of the star, without worry about the mass or size of the atmosphere. Then<br /> $<br /> g = \frac{G M}{R^2} = {\rm constant\; at\; surface}.<br />$<br /> Then HE is <br /> $<br /> \frac{dP}{dz} = - \rho g.<br />$<br /> The pressure is due to that of an ideal gas, so<br /> $<br /> P = n k_B T, <br />$<br /> where $n$ is the number density, $k_B$ is Boltzmann's constant, and $T$ is the temperature. We need to know then the average mass per particle, $m$, so that $\rho = n m$. This gives<br /> $<br /> \frac{d(n k_B T)}{dz} = -m g n.<br />$<br /> This is still not something we can solve, so we have to make the further approximation that the atmosphere is isothermal, i.e. $T$ is constant as a function of $r$. Then<br /> $<br /> \frac{1}{n}\frac{dn}{dz} = -\frac{mg}{k_B T}.<br />$<br /> The solution of this is <br /> $<br /> \left[\ln(n)\right]^{z}_{\rm surface} = -\frac{mgz}{k_B T},<br />$<br /> $<br /> n(z) = n_{\rm surf} e^{-z/h}, {\rm \; where}<br />$<br /> $<br /> h = \frac{k_B T}{mg}.<br />$<br /> We call $h$ the scale height of the atmosphere, and it is given by the ratio of the thermal energy at the surface to the gravitational potential energy at the surface. We can think of it as the distance over which the density changes by an appreciable amount (1 / $e$). Thus the thermal energy is trying to puff up the atmoshpere, while the gravitational potential is trying to keep it close to the surface. For the Sun, the scale height is<br /> $<br /> h = 2 \times 10^7 {\rm cm} = 3 \times 10^{-4} R_\odot.<br />$<br /> This is much smaller than the radius of the Sun. There is an interesting statistical mechanics interpretation of the scale height. Particles with an energy $E$ are distributed in energy levels according to the Boltzmann factor, with<br /> $<br /> n \propto e^{-E/k_B T}.<br />$<br /> Since the gravitational potential energy for our particles at height $z$ is $mgz$, this Boltzmann argument gives the same radial dependence of number density of the atmosphere as our argument above. \\ \\<br /> \subsection*{Mean Molecular Weight} \\<br /> One of the major ways fusion affects the structure of stars is by changing the average mass per particle. If it starts out as ionized hydrogen, there are basically two particles per $m_p$, while for neutral hydrogen there is about one particle per $m_p$, and for ionized helium it is 3 particles per $4 m_p$, roughly. Each of these particles is at temperature $T$, and thus contributes to the total pressure in the star. \\ \\<br /> To account for this, we will define a quantity $\mu$ called the mean molecular weight to formally define our average mass per particle and encapsulate the summing of pressures from each element. Each element has a mass fraction $X_i$ and an atomic number $A_i$ that represents the total number of protons and neutrons. If the total mass density is $\rho$, then the number density of a given species $n_i$ is <br /> $<br /> n_i = \frac{\rho X_i}{A_i m_p}.<br />$<br /> The pressure from all ions is <br /> $<br /> P = \sum_i n_i k_B T.<br />$<br /> Substituting in the expression for ionic number density,<br /> $<br /> P = \frac{\rho k_B T}{m_p} \sum_i \frac{X_i}{A_i}.<br />$<br /> We can then define the ion mean molecular weight to be<br /> $<br /> \frac{1}{\mu_i} = \sum_i \frac{X_i}{A_i}.<br />$<br /> The ions also contribute electrons, which we also need to account for. The number density for electrons $n_e$ depends upon the charge $Z_i$ of the atom (assuming complete ionization), with<br /> $<br /> \sum_i = Z_i n_i.<br />$<br /> Then <br /> $<br /> P_e = n_e k_B T = k_b T \sum_i n_i Z_i = \frac{\rho k_B T}{m_p} \sum_i \frac{Z_i X_i}{A_i}.<br />$<br /> This allows us to define the electronic mean molecular weight as <br /> $<br /> \frac{1}{\mu_e} = \frac{Z_i X_i}{A_i}.<br />$<br /> The final mean molecular weight is then <br /> $<br /> \frac{1}{\mu} = \frac{1}{\mu_i} + \frac{1}{\mu_e}.<br />$<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Bootstrap resampling 2011-11-26T21:17:37Z <p>Jmcbride: still failing</p> <hr /> <div>==Short topical video==<br /> * (never mind, audio playback got out of sync somehow when I rendered the trimmed video)</div> Jmcbride Bootstrap resampling 2011-11-26T09:50:31Z <p>Jmcbride: created page</p> <hr /> <div>==Short topical video==<br /> *[http://www.youtube.com/watch?v=DS9VtffSnR0 Bootstrap resampling]</div> Jmcbride Radio Astronomy: Tools and Techniques 2011-11-26T09:47:19Z <p>Jmcbride: /* Topics */ added link to bootstrap resampling</p> <hr /> <div>This is course is aimed at<br /> graduate students, advanced undergraduates, and interested third<br /> parties who:<br /> * would like to understand radio astronomy better<br /> * would like to develop technical skills (programming, signal processing, instrumentation, algorithms, pedagogy, etc) to help them in their research<br /> * would like to be involved, and involve their peers, in developing concrete tools to help their research<br /> <br /> This class will follow a flexible, non-traditional format whereby each<br /> week, a group of students and I will work together to prepare<br /> public-domain pedagogical materials on a subject that will be<br /> distributed to the rest of the class in advance of each meeting.<br /> Class time will be split between discussing the subject informally,<br /> and working in groups to develop tools and address on-going research<br /> questions that each student brings to the class.<br /> <br /> My hope is that this class will be moderately time-consuming, but that<br /> the tools, collaborations, and research developed inside the class<br /> will have a broad enough scope that it can double-count as<br /> research/work time. All of our activities are aimed at creating tools<br /> (both pedagogical and research-oriented) that will have value beyond<br /> the classroom.<br /> <br /> === Topics ===<br /> <br /> Here is a (non-exhaustive) list of topics that we will consider covering in this class. Eventually, it would be nice to link in as many topics as possible and begin to organize subjects by their prerequisites and relatedness.<br /> <br /> Algorithms<br /> * [[Fast Fourier Transform]]<br /> * [[Markov-Chain Monte Carlo]]<br /> <br /> Software Development<br /> * [[Python Installation and Basic Programming]]<br /> * [[Revision Control]]<br /> * [[Programming Models]]<br /> * [[General software tools]]<br /> <br /> Computing<br /> * [[Processor Architectures]]<br /> * [[Data Representations]]<br /> * [[Network Programming]]<br /> <br /> Signal Processing / Fourier Analysis<br /> * [[Convolution Theorem]]<br /> * [[Windowing]]<br /> * [[Correlators]]<br /> * [[Deconvolution]]<br /> <br /> Interferometers <br /> * [[Basic Interferometry]]<br /> * [[Basic Interferometry II]]<br /> * Units<br /> ** [[Coordinates]]<br /> ** [[Units of radiation]] <br /> * [[Advanced Interferometry]]<br /> * [[Interferometric Imaging]]<br /> <br /> Statistics<br /> * [[Central Limit Theorem]]<br /> * [[Radiometer Equation]]<br /> * [[Bayesian Statistics]]<br /> * [[Statistics in Python]]<br /> * [[Fisher Matrices]]<br /> * [[Bootstrap resampling]]<br /> <br /> Signal Path<br /> * [[RC Filters]]<br /> * [[Transistors]]<br /> * [[Transmission Lines]]<br /> * [[Antennas and Feeds]]<br /> * [[Receivers and Amplifiers]]<br /> <br /> Pedagogy of Radio Astronomy / Meta-Information<br /> * [[Creating Short Topical Presentations]]<br /> * [[Using AstroBaki]]<br /> <br /> Science of Radio Astronomy<br /> * [[Black-Body Radiation]]<br /> * [[21cm Transition]]<br /> <br /> == Topics by Date ==<br /> * Aug 31: <br /> ** [[Convolution Theorem]]<br /> ** [[Creating Short Topical Presentations]]<br /> ** [[Python Installation and Basic Programming]]<br /> ** Brainstorming Lab Projects<br /> ** Choosing a Topic to Present<br /> ** Getting an account<br /> * Sep 07:<br /> ** [[Revision Control]]<br /> ** Choosing a Lab Project<br /> ** Begin a Python project (radioastro), revision-controlled under GIT, that<br /> *** has a convolution module (conv.py) with functions for<br /> **** performing 1D and 2D convolutions of two provided functions<br /> *** has a module (noise.py) with brightness-temperature/jansky conversions<br /> **** should take beam size and wavelength as arguments<br /> **** should predict noise levels for observations of given bandwidth, time, number of antennas, etc. <br /> * Sep 14:<br /> ** [[Radiometer Equation]]<br /> ** Discus class project: the [[Homemade Interferometer]]<br /> ** Create some software for simulating a visibility. That is, given two antennas (with x,y,z positions in equatorial coordinates) and a source (with x,y,z also in equatorial coordinates), compute the phase that you would measure as a function of frequency.<br /> * Sep 21:<br /> ** [[Basic Interferometry]]<br /> ** Discuss class project: [[General software tools|Aggregating Software Tools]]<br /> ** Extend visibility simulation software to handle many sources (with spectra), and many antennas (with passbands).<br /> * Sep 28:<br /> ** [[Basic Interferometry II]]<br /> ** Discuss class project: [[Parallel Computing]]<br /> ** Create plan of work for each class project. Catch up on class software.<br /> * Oct 05:<br /> ** [[Coordinates]]<br /> ** Working on class project.<br /> ** [[Units of radiation]]<br /> * Oct 12:<br /> ** Working on class project. Report on [[Homemade Interferometer]] deployment.<br /> * Oct 19:<br /> ** [[Data Representations]]<br /> * Oct 26:<br /> ** [[21cm Transition]]<br /> * Nov 02:<br /> ** [[Principle Component Analysis]]<br /> * Nov 09:<br /> * Nov 16:<br /> ** [[Fisher Matrices]]<br /> * Nov 23:<br /> ** [[RC Filters]]<br /> * Nov 30:<br /> ** [[Fast Fourier Transform]]<br /> ** [[Transistors]]</div> Jmcbride Statistics in Python 2011-10-26T03:45:30Z <p>Jmcbride: fixed link (removed |)</p> <hr /> <div>===Prerequisites===<br /> <br /> ===Short Topical Presentations===<br /> <br /> ===Reference Material===<br /> * [http://oneau.wordpress.com/2011/02/28/simple-statistics-with-scipy/ Simple statistics with SciPy]</div> Jmcbride Units of radiation 2011-09-27T20:26:00Z <p>Jmcbride: created page with some questions to answer, and link to nrao discussion of this topic</p> <hr /> <div>==Main Questions==<br /> * What is the point of talking about specific intensity?<br /> * Why use the Jansky? What does flux density mean?<br /> * Which quantities depend upon observing conditions and which are intrinsic to the source?<br /> <br /> ==Resources==<br /> The NRAO Essential Radio Astronomy course explains this pretty well in their chapter on [http://www.cv.nrao.edu/course/astr534/Brightness.html Brightness].</div> Jmcbride Unit Testing 2011-09-22T20:58:57Z <p>Jmcbride: added unit testing</p> <hr /> <div>There are many tools that astronomers should know about, but that are not astronomy specific. This aims to be a list of such tools, and, by its nature, will certainly be incomplete.<br /> <br /> ==Unit Testing==<br /> The basic idea of unit testing is to verify that a piece of code works the way it is supposed to work. That is, have the code try to solve the a problem with a known answer, and if the code does not give the right answer, it does not work. There is a module for doing this in python called unittest (or, alternatively, PyUnit).<br /> * [http://blip.tv/pycon-us-videos-2009-2010-2011/pycon-2010-introduction-to-unittest-a-k-a-pyunit-96-3379016 Introductory video] explaining how to use unittest, presented at PyCon<br /> * [http://pyunit.sourceforge.net/pyunit.html Tutorial] by the developer of unittest</div> Jmcbride Python Installation and Basic Programming 2011-09-22T20:45:48Z <p>Jmcbride: /* Links */ removed unnecessary pipe in link</p> <hr /> <div>Here we will assemble resources for learning Python, and for getting it and other programming-related software installed on your computer.<br /> <br /> For a scientific programmer in Python, the absolute basics you need to have installed are:<br /> * [http://www.python.org Python] 2.X (note that 3.X exists and is maturing, but a lot of scientific code and packages are not yet ported)<br /> * [http://numpy.scipy.org/ NumPy]: a package for fast numerical array processing<br /> * [http://matplotlib.sourceforge.net/ Matplotlib/Pylab]: a package for generating publication-quality plots<br /> * [http://git-scm.com/ GIT]: a revision-control program for keeping tabs on the changes you make to your code. Not just for python.<br /> <br /> ==Python==<br /> ===Topical Videos===<br /> <br /> * [http://www.khanacademy.org/video/introduction-to-programs-data-types-and-variables?playlist=Computer%20Science Introduction to Programs Data Types and Variables] by Khan Academy<br /> * [http://www.khanacademy.org/video/python-lists?playlist=Computer%20Science Python Lists] by Khan Academy<br /> * [http://www.khanacademy.org/video/for-loops-in-python?playlist=Computer%20Science For Loops in Python] by Khan Academy<br /> * [http://www.khanacademy.org/video/while-loops-in-python?playlist=Computer%20Science While Loops in Python] by Khan Academy<br /> * [http://www.khanacademy.org/video/fun-with-strings?playlist=Computer%20Science Fun with Strings] by Khan Academy<br /> * [http://www.khanacademy.org/video/writing-a-simple-factorial-program---python-2?playlist=Computer%20Science Writing a Simple Factorial Program] by Khan Academy<br /> * ... there are lots more by Khan Academy ...<br /> * [http://www.khanacademy.org/video/simpler-insertion-sort-function?playlist=Computer%20Science Simpler Insertion Sort Function]<br /> <br /> ===Links===<br /> * Josh Bloom's [http://sites.google.com/site/pythonbootcamp Python Boot Camp] offers a wealth of resources for getting started with Python. In particular, see:<br /> ** [http://sites.google.com/site/pythonbootcamp/preparation/software Instructions] for installing all of the packages mentioned above.<br /> ** [http://sites.google.com/site/pythonbootcamp/links Useful links] to Python resources<br /> * [http://fperez.org/py4science/starter_kit.html Python for Scientific Computing] has a list of resources on various topics relevant to a scientist getting started with Python<br /> * [http://www.scipy.org/Additional_Documentation/Astronomy_Tutorial?action=show Scipy astronomy tutorial] covers reading in and analyzing data in scipy, mostly focusing on FITS files, but with some other useful information for those with data in other formats<br /> * A [http://code.google.com/edu/languages/google-python-class/index.html Google Class on Python]<br /> * The [http://docs.python.org/tutorial/ Python Tutorial]<br /> <br /> ==GIT==<br /> Please see the [[Revision Control]] page,</div> Jmcbride Python Installation and Basic Programming 2011-09-22T20:45:13Z <p>Jmcbride: /* Python */ added Fernando Perez's nice list of python resources, and the astronomy tutorial</p> <hr /> <div>Here we will assemble resources for learning Python, and for getting it and other programming-related software installed on your computer.<br /> <br /> For a scientific programmer in Python, the absolute basics you need to have installed are:<br /> * [http://www.python.org Python] 2.X (note that 3.X exists and is maturing, but a lot of scientific code and packages are not yet ported)<br /> * [http://numpy.scipy.org/ NumPy]: a package for fast numerical array processing<br /> * [http://matplotlib.sourceforge.net/ Matplotlib/Pylab]: a package for generating publication-quality plots<br /> * [http://git-scm.com/ GIT]: a revision-control program for keeping tabs on the changes you make to your code. Not just for python.<br /> <br /> ==Python==<br /> ===Topical Videos===<br /> <br /> * [http://www.khanacademy.org/video/introduction-to-programs-data-types-and-variables?playlist=Computer%20Science Introduction to Programs Data Types and Variables] by Khan Academy<br /> * [http://www.khanacademy.org/video/python-lists?playlist=Computer%20Science Python Lists] by Khan Academy<br /> * [http://www.khanacademy.org/video/for-loops-in-python?playlist=Computer%20Science For Loops in Python] by Khan Academy<br /> * [http://www.khanacademy.org/video/while-loops-in-python?playlist=Computer%20Science While Loops in Python] by Khan Academy<br /> * [http://www.khanacademy.org/video/fun-with-strings?playlist=Computer%20Science Fun with Strings] by Khan Academy<br /> * [http://www.khanacademy.org/video/writing-a-simple-factorial-program---python-2?playlist=Computer%20Science Writing a Simple Factorial Program] by Khan Academy<br /> * ... there are lots more by Khan Academy ...<br /> * [http://www.khanacademy.org/video/simpler-insertion-sort-function?playlist=Computer%20Science Simpler Insertion Sort Function]<br /> <br /> ===Links===<br /> * Josh Bloom's [http://sites.google.com/site/pythonbootcamp Python Boot Camp] offers a wealth of resources for getting started with Python. In particular, see:<br /> ** [http://sites.google.com/site/pythonbootcamp/preparation/software Instructions] for installing all of the packages mentioned above.<br /> ** [http://sites.google.com/site/pythonbootcamp/links Useful links] to Python resources<br /> * [http://fperez.org/py4science/starter_kit.html Python for Scientific Computing] has a list of resources on various topics relevant to a scientist getting started with Python<br /> * [http://www.scipy.org/Additional_Documentation/Astronomy_Tutorial?action=show |Scipy astronomy tutorial] covers reading in and analyzing data in scipy, mostly focusing on FITS files, but with some other useful information for those with data in other formats<br /> * A [http://code.google.com/edu/languages/google-python-class/index.html Google Class on Python]<br /> * The [http://docs.python.org/tutorial/ Python Tutorial]<br /> <br /> ==GIT==<br /> Please see the [[Revision Control]] page,</div> Jmcbride Radio Astronomy: Tools and Techniques 2011-09-22T20:36:37Z <p>Jmcbride: broke up units in to two parts, added &quot;general tools&quot; section to software development for aggregation project</p> <hr /> <div>This is course is aimed at<br /> graduate students, advanced undergraduates, and interested third<br /> parties who:<br /> * would like to understand radio astronomy better<br /> * would like to develop technical skills (programming, signal processing, instrumentation, algorithms, pedagogy, etc) to help them in their research<br /> * would like to be involved, and involve their peers, in developing concrete tools to help their research<br /> <br /> This class will follow a flexible, non-traditional format whereby each<br /> week, a group of students and I will work together to prepare<br /> public-domain pedagogical materials on a subject that will be<br /> distributed to the rest of the class in advance of each meeting.<br /> Class time will be split between discussing the subject informally,<br /> and working in groups to develop tools and address on-going research<br /> questions that each student brings to the class.<br /> <br /> My hope is that this class will be moderately time-consuming, but that<br /> the tools, collaborations, and research developed inside the class<br /> will have a broad enough scope that it can double-count as<br /> research/work time. All of our activities are aimed at creating tools<br /> (both pedagogical and research-oriented) that will have value beyond<br /> the classroom.<br /> <br /> === Topics ===<br /> <br /> Here is a (non-exhaustive) list of topics that we will consider covering in this class. Eventually, it would be nice to link in as many topics as possible and begin to organize subjects by their prerequisites and relatedness.<br /> <br /> Algorithms<br /> * [[Fast Fourier Transform]]<br /> * [[Markov-Chain Monte Carlo]]<br /> <br /> Software Development<br /> * [[Python Installation and Basic Programming]]<br /> * [[Revision Control]]<br /> * [[Programming Models]]<br /> * [[General software tools]]<br /> <br /> Computing<br /> * [[Processor Architectures]]<br /> * [[Data Representations]]<br /> * [[Network Programming]]<br /> <br /> Signal Processing / Fourier Analysis<br /> * [[Convolution Theorem]]<br /> * [[Windowing]]<br /> * [[Correlators]]<br /> * [[Deconvolution]]<br /> <br /> Interferometers <br /> * [[Basic Interferometry]]<br /> * Units<br /> ** [[Coordinates]]<br /> ** [[Units of radiation]] <br /> * [[Advanced Interferometry]]<br /> * [[Interferometric Imaging]]<br /> <br /> Statistics<br /> * [[Central Limit Theorem]]<br /> * [[Radiometer Equation]]<br /> * [[Bayesian Statistics]]<br /> <br /> Signal Path<br /> * [[Transmission Lines]]<br /> * [[Antennas and Feeds]]<br /> * [[Receivers and Amplifiers]]<br /> <br /> Pedagogy of Radio Astronomy / Meta-Information<br /> * [[Creating Short Topical Presentations]]<br /> * [[Using AstroBaki]]<br /> <br /> Science of Radio Astronomy<br /> * [[Black-Body Radiation]]<br /> * [[21cm Transition]]<br /> <br /> == Topics by Date ==<br /> * Aug 31: <br /> ** [[Convolution Theorem]]<br /> ** [[Creating Short Topical Presentations]]<br /> ** [[Python Installation and Basic Programming]]<br /> ** Brainstorming Lab Projects<br /> ** Choosing a Topic to Present<br /> ** Getting an account<br /> * Sep 07:<br /> ** [[Revision Control]]<br /> ** [[Radiometer Equation]]<br /> ** Choosing a Lab Project<br /> ** Begin a Python project (radioastro), revision-controlled under GIT, that<br /> *** has a convolution module (conv.py) with functions for<br /> **** performing 1D and 2D convolutions of two provided functions<br /> *** has a module (noise.py) with brightness-temperature/jansky conversions<br /> **** should take beam size and wavelength as arguments<br /> **** should predict noise levels for observations of given bandwidth, time, number of antennas, etc. <br /> * Sep 14:<br /> ** [[Basic Interferometry]]<br /> ** Discus class project: the [[Homemade Interferometer]]<br /> ** Create some software for simulating a visibility. That is, given two antennas (with x,y,z positions in equatorial coordinates) and a source (with x,y,z also in equatorial coordinates), compute the phase that you would measure as a function of frequency.<br /> * Sep 21:<br /> ** [[21cm Transition]]<br /> ** [[Basic Interferometry II]]<br /> ** Discuss class project: [[General software tools|Aggregating Software Tools]]<br /> ** Extend visibility simulation software to handle many sources (with spectra), and many antennas (with passbands).<br /> * Sep 28:<br /> * Oct 05:<br /> * Oct 12:<br /> * Oct 19:<br /> * Oct 26:<br /> * Nov 02:<br /> * Nov 09:<br /> * Nov 16:<br /> * Nov 23:<br /> * Dec 03:</div> Jmcbride Radio Astronomy: Tools and Techniques 2011-09-16T18:48:48Z <p>Jmcbride: fixed formatting of homework 2 (should have previewed)</p> <hr /> <div>This is course is aimed at<br /> graduate students, advanced undergraduates, and interested third<br /> parties who:<br /> * would like to understand radio astronomy better<br /> * would like to develop technical skills (programming, signal processing, instrumentation, algorithms, pedagogy, etc) to help them in their research<br /> * would like to be involved, and involve their peers, in developing concrete tools to help their research<br /> <br /> This class will follow a flexible, non-traditional format whereby each<br /> week, a group of students and I will work together to prepare<br /> public-domain pedagogical materials on a subject that will be<br /> distributed to the rest of the class in advance of each meeting.<br /> Class time will be split between discussing the subject informally,<br /> and working in groups to develop tools and address on-going research<br /> questions that each student brings to the class.<br /> <br /> My hope is that this class will be moderately time-consuming, but that<br /> the tools, collaborations, and research developed inside the class<br /> will have a broad enough scope that it can double-count as<br /> research/work time. All of our activities are aimed at creating tools<br /> (both pedagogical and research-oriented) that will have value beyond<br /> the classroom.<br /> <br /> === Topics ===<br /> <br /> Here is a (non-exhaustive) list of topics that we will consider covering in this class. Eventually, it would be nice to link in as many topics as possible and begin to organize subjects by their prerequisites and relatedness.<br /> <br /> Algorithms<br /> * [[Fast Fourier Transform]]<br /> * [[Markov-Chain Monte Carlo]]<br /> <br /> Software Development<br /> * [[Python Installation and Basic Programming]]<br /> * [[Revision Control]]<br /> * [[Programming Models]]<br /> <br /> Computing<br /> * [[Processor Architectures]]<br /> * [[Data Representations]]<br /> * [[Network Programming]]<br /> <br /> Signal Processing / Fourier Analysis<br /> * [[Convolution Theorem]]<br /> * [[Windowing]]<br /> * [[Correlators]]<br /> * [[Deconvolution]]<br /> <br /> Interferometers<br /> * [[Basic Interferometry]]<br /> * [[Units]]<br /> * [[Advanced Interferometry]]<br /> * [[Interferometric Imaging]]<br /> <br /> Statistics<br /> * [[Central Limit Theorem]]<br /> * [[Radiometer Equation]]<br /> * [[Bayesian Statistics]]<br /> <br /> Signal Path<br /> * [[Transmission Lines]]<br /> * [[Antennas and Feeds]]<br /> * [[Receivers and Amplifiers]]<br /> <br /> Pedagogy of Radio Astronomy / Meta-Information<br /> * [[Creating Short Topical Presentations]]<br /> * [[Using AstroBaki]]<br /> <br /> Science of Radio Astronomy<br /> * [[Black-Body Radiation]]<br /> * [[21cm Transition]]<br /> <br /> == Topics by Date ==<br /> * Aug 31: <br /> ** [[Convolution Theorem]]<br /> ** [[Creating Short Topical Presentations]]<br /> ** [[Python Installation and Basic Programming]]<br /> ** Brainstorming Lab Projects<br /> ** Choosing a Topic to Present<br /> ** Getting an account<br /> * Sep 07:<br /> ** [[Revision Control]]<br /> ** [[Radiometer Equation]]<br /> ** Choosing a Lab Project<br /> ** Begin a Python project (radioastro), revision-controlled under GIT, that<br /> *** has a convolution module (conv.py) with functions for<br /> **** performing 1D and 2D convolutions of two provided functions<br /> *** has a module (noise.py) with brightness-temperature/jansky conversions<br /> **** should take beam size and wavelength as arguments<br /> **** should predict noise levels for observations of given bandwidth, time, number of antennas, etc. <br /> * Sep 14:<br /> ** Create some software for simulating a visibility. That is, given two antennas (with x,y,z positions in equatorial coordinates) and a source (with x,y,z also in equatorial coordinates), compute the phase that you would measure as a function of frequency.<br /> * Sep 21:<br /> * Sep 28:<br /> * Oct 05:<br /> * Oct 12:<br /> * Oct 19:<br /> * Oct 26:<br /> * Nov 02:<br /> * Nov 09:<br /> * Nov 16:<br /> * Nov 23:<br /> * Dec 03:</div> Jmcbride Radio Astronomy: Tools and Techniques 2011-09-16T18:48:09Z <p>Jmcbride: added homework</p> <hr /> <div>This is course is aimed at<br /> graduate students, advanced undergraduates, and interested third<br /> parties who:<br /> * would like to understand radio astronomy better<br /> * would like to develop technical skills (programming, signal processing, instrumentation, algorithms, pedagogy, etc) to help them in their research<br /> * would like to be involved, and involve their peers, in developing concrete tools to help their research<br /> <br /> This class will follow a flexible, non-traditional format whereby each<br /> week, a group of students and I will work together to prepare<br /> public-domain pedagogical materials on a subject that will be<br /> distributed to the rest of the class in advance of each meeting.<br /> Class time will be split between discussing the subject informally,<br /> and working in groups to develop tools and address on-going research<br /> questions that each student brings to the class.<br /> <br /> My hope is that this class will be moderately time-consuming, but that<br /> the tools, collaborations, and research developed inside the class<br /> will have a broad enough scope that it can double-count as<br /> research/work time. All of our activities are aimed at creating tools<br /> (both pedagogical and research-oriented) that will have value beyond<br /> the classroom.<br /> <br /> === Topics ===<br /> <br /> Here is a (non-exhaustive) list of topics that we will consider covering in this class. Eventually, it would be nice to link in as many topics as possible and begin to organize subjects by their prerequisites and relatedness.<br /> <br /> Algorithms<br /> * [[Fast Fourier Transform]]<br /> * [[Markov-Chain Monte Carlo]]<br /> <br /> Software Development<br /> * [[Python Installation and Basic Programming]]<br /> * [[Revision Control]]<br /> * [[Programming Models]]<br /> <br /> Computing<br /> * [[Processor Architectures]]<br /> * [[Data Representations]]<br /> * [[Network Programming]]<br /> <br /> Signal Processing / Fourier Analysis<br /> * [[Convolution Theorem]]<br /> * [[Windowing]]<br /> * [[Correlators]]<br /> * [[Deconvolution]]<br /> <br /> Interferometers<br /> * [[Basic Interferometry]]<br /> * [[Units]]<br /> * [[Advanced Interferometry]]<br /> * [[Interferometric Imaging]]<br /> <br /> Statistics<br /> * [[Central Limit Theorem]]<br /> * [[Radiometer Equation]]<br /> * [[Bayesian Statistics]]<br /> <br /> Signal Path<br /> * [[Transmission Lines]]<br /> * [[Antennas and Feeds]]<br /> * [[Receivers and Amplifiers]]<br /> <br /> Pedagogy of Radio Astronomy / Meta-Information<br /> * [[Creating Short Topical Presentations]]<br /> * [[Using AstroBaki]]<br /> <br /> Science of Radio Astronomy<br /> * [[Black-Body Radiation]]<br /> * [[21cm Transition]]<br /> <br /> == Topics by Date ==<br /> * Aug 31: <br /> ** [[Convolution Theorem]]<br /> ** [[Creating Short Topical Presentations]]<br /> ** [[Python Installation and Basic Programming]]<br /> ** Brainstorming Lab Projects<br /> ** Choosing a Topic to Present<br /> ** Getting an account<br /> * Sep 07:<br /> ** [[Revision Control]]<br /> ** [[Radiometer Equation]]<br /> ** Choosing a Lab Project<br /> ** Begin a Python project (radioastro), revision-controlled under GIT, that<br /> *** has a convolution module (conv.py) with functions for<br /> **** performing 1D and 2D convolutions of two provided functions<br /> *** has a module (noise.py) with brightness-temperature/jansky conversions<br /> **** should take beam size and wavelength as arguments<br /> **** should predict noise levels for observations of given bandwidth, time, number of antennas, etc. <br /> * Sep 14:<br /> ** Create some software for simulating a visibility. That is, given two<br /> antennas (with x,y,z positions in equatorial coordinates) and a source<br /> (with x,y,z also in equatorial coordinates), compute the phase that<br /> you would measure as a function of frequency.<br /> * Sep 21:<br /> * Sep 28:<br /> * Oct 05:<br /> * Oct 12:<br /> * Oct 19:<br /> * Oct 26:<br /> * Nov 02:<br /> * Nov 09:<br /> * Nov 16:<br /> * Nov 23:<br /> * Dec 03:</div> Jmcbride Radiometer Equation Applied to Telescopes 2011-09-15T00:32:47Z <p>Jmcbride: /* Reference Material */ added note about unresolved source</p> <hr /> <div>===Prerequisites===<br /> * [[Central Limit Theorem]]<br /> * [[Basic_Interferometry | Basic Interferometry]] (for Radiometer equation applied to interferometers)<br /> * [[Radio Astronomy Units]]<br /> * [[Black-Body Radiation]]<br /> <br /> ===Short Topical Videos===<br /> * [http://www.youtube.com/watch?v=dwT-tMoscsY The Radiometer Equation (by Chat Hull)]<br /> <br /> ===Reference Material===<br /> <br /> &lt;latex&gt;<br /> \documentclass[11pt]{article}<br /> \usepackage{graphicx}<br /> \usepackage{amsmath}<br /> \usepackage{fullpage}<br /> \begin{document}<br /> <br /> \section*{The Radiometer Equation}<br /> <br /> The radiometer equation is really all about signal to noise. Here's the answer:<br /> <br /> $$\frac{S}{N} = \frac{T_{src}}{T_{rms}} = \frac{T_{src}}{T_{sys}} \sqrt{\tau \Delta\nu}$$<br /> <br /> where:<br /> \begin{itemize}<br /> \item $T_{src}$ is the signal of the source you're observing<br /> \item $T_{sys}$ is your system temperature<br /> \item $T_{rms}$ is the noise in your system, or the RMS fluctuations in your system temperature<br /> \item $\Delta\nu$ is the bandwidth of your correlator (in Hz)<br /> \item $\tau$ is integration time (seconds)<br /> \end{itemize}<br /> <br /> <br /> \subsection*{What's the deal with all the temperatures?}<br /> <br /> All of the temperatures in the above equation have their origin in the Rayleigh-Jeans limit of the blackbody equation:<br /> <br /> $$I_\nu \approx \frac{2kT}{\lambda^2}$$<br /> <br /> Radio astronomers like to refer to the above temperature $T$ as<br /> the &quot;brightness temperature,&quot; $T_B$, of a source. $T_B$ is the temperature of the blackbody needed<br /> to produce the observed specific intensity at that frequency:<br /> $$T_B(\nu) = \frac{I_\nu \lambda^2}{2k}$$<br /> <br /> Now, back to specific intensity. $I_\nu$ is defined as:<br /> <br /> $$I_\nu = \frac{\Delta E}{\Delta \Omega \Delta A \Delta t \Delta \nu}$$<br /> where:<br /> \begin{itemize}<br /> \item $\Delta t$ is the exposure time<br /> \item $\Delta \nu$ is a small interval in frequency<br /> \item $\Delta E$ is the energy emitted in bandwidth $\Delta \nu$ over time $\Delta t$<br /> \item $\Delta A$ is the area of the telescope<br /> \item $\Delta \Omega$ is the observed solid angle on the source, or your beam size (steradians)<br /> \end{itemize}<br /> <br /> So, <br /> $$\boxed{I_\nu = \frac{dE}{dt d\nu dA d\Omega} = [erg\ s^{-1}\ cm^{-2}\ Hz^{-1}\ Sr^{-1}]}$$<br /> <br /> It is common to report $I_\nu$ in Jy/beam, where the beam has units of $Sr^{-1}$, and Janskys (Jy) are the units of flux density $S_\nu$, defined below.<br /> <br /> To arrive at other quantities, we must integrate over $I_\nu$:<br /> \begin{itemize}<br /> \item {\bf Flux Density:} $S_\nu = \int I_\nu d\Omega = [erg\ s^{-1}\ cm^{-2}\ Hz^{-1}]$. The fundamental unit of flux density is the Jansky. $1 \textrm{Jy} = 10^{-23} erg\ s^{-1}\ cm^{-2}\ Hz^{-1}$<br /> \item {\bf Power received:} $P_\nu = \int S_\nu dA = [erg\ s^{-1}\ Hz^{-1}]$. <br /> \end{itemize}<br /> <br /> \subsection*{Source Temperature, $T_{src}$}<br /> The source temperature is defined as the brightness temperature associated with the power<br /> received by the telescope from the source you're observing:<br /> \begin{aligned}<br /> P_\nu &amp; = I_\nu dA d\Omega \\<br /> &amp; = \frac{2kT_{src}}{\lambda^2} dA d\Omega\\<br /> &amp; = 2kT_{src}\\<br /> \end{aligned}<br /> The last step comes from the Antenna Theorem, which states that $dA d\Omega = \lambda^2$.<br /> <br /> \subsection*{K / Jy, or &quot;forward gain&quot;}<br /> Now that we've related the power $P_\nu$ received by the antenna at a given frequency, <br /> to the source temperature, $T_{src}$, and the observed intensity, $I_\nu$, we can arrive at a conversion between <br /> flux density and brightness temperature for an unresolved source:<br /> \begin{aligned}<br /> P_\nu &amp; = I_\nu A_e \Omega_a \\<br /> 2kT_A &amp; = S_\nu A_e\\<br /> T_A &amp; = \left(\frac{A_e}{2k}\right) S_\nu\\<br /> \end{aligned}<br /> Therefore, the conversion between K and Jy, also known as the &quot;forward gain&quot; of an antenna, is just <br /> $$K / Jy = \frac{A_e}{2k}$$<br /> <br /> Thus, brightness temperature is just another measure of the brightness of a source. The forward gain is a physical property of a telescope, which dictates the telescope receivers' response to a given increase in Janskys. <br /> <br /> \subsection*{The System Temperature}<br /> While the source temperature $T_{src}$ describes the energy received from the source you are interested in, <br /> the system temperature, $T_{sys}$ describes the actual power received due to both the sky ($T_{sky}$) and the receivers ($T_{Rx}$):<br /> $$T_{sys} = T_{sky} + T_{Rx}$$<br /> The main component is the receiver temperature, $T_{Rx}$, which comes from the [http://en.wikipedia.org/wiki/Johnson%E2%80%93Nyquist_noise<br /> | thermal (or Johnson) noise] from the receiver electronics. Depending on the observation waveband, receivers are sometimes cooled to reduce $T_{Rx}$.<br /> <br /> $T_{sky}$ is the power generated by everything you \textit{don't} want to be looking at on the sky: background sources, water vapor in the atmosphere (especially in mm-wave astronomy), galactic backgrounds (especially in long-wavelength astronomy), etc.<br /> <br /> \subsubsection*{Effective Area $A_e$}<br /> <br /> Above we have written the area of an antenna as $A_e$, or the &quot;effective area.&quot; <br /> <br /> $$A_e = \eta_a A_p$$ <br /> <br /> where ($\eta_a$ is the aperture efficiency and $A_p$ is the projected area of the telescope). $\eta_a$ is a number less than one, and signifies that not all radiation incident upon the telescope actually makes it to the receiver because of the dish's finite reflectivity. This loss of signal requires that the signal later be amplified. There are two scenarios that can be encountered:<br /> <br /> \begin{itemize}<br /> \item $T_{sys}$ is dominated by $T_{Rx}$: in this case aperture efficiency matters, because in amplifying the signal, noise from the receiver is also amplified.<br /> \item $T_{sys}$ is dominated by $T_{sky}$: in this case, which might happen on a cloudy day at a millimeter telescope where the noise from the atmosphere is dominant, aperture efficiency doesn't matter. This is because the dominant sky noise is first cut down by a low aperture efficiency, and then multiplied back up to its original level by the amplifiers.<br /> \end{itemize}<br /> <br /> \subsection*{Detecting the signal...}<br /> <br /> Typically $T_{sys} &gt;&gt; T_{src}$, so how do we detect <br /> the source we are interested in? Beat down the noise!<br /> <br /> In order to detect the source, we need $T_{src} &gt; T_{rms}$, where the observation-dependent quantity $T_{rms}$ is the noise in our measurement of the observation-independent quantity $T_{sys}$:<br /> <br /> $$T_{rms} = \frac{T_{sys}}{\sqrt{N}}$$<br /> <br /> where $N$ is the number of independent data points.<br /> <br /> For a telescope, the number of independent samples is $\Delta \nu \cdot \tau$, where $\Delta \nu$ is the <br /> bandwidth (Hz) and $\tau$ is the integration time (seconds). With a bandwidth $\Delta \nu$, the signal is <br /> statistically independent over a time interval $1/(2 \Delta\nu)$ (the factor of 2 comes from the Nyquist sampling rate), so that the number of independent samples is <br /> just $\tau$ divided by $1/(2 \Delta\nu)$, so $N = 2 \tau \Delta\nu$. Then, as explained in the [http://www.cv.nrao.edu/course/astr534/Radiometers.html NRAO course], the fluctuations in a given measurement of $T_{sys}$ are $\sqrt{2}~T_{sys}$, and the factor of two disappears, giving us the final result:<br /> $$T_{rms} = \frac{T_{sys}}{\sqrt{\tau \Delta\nu}}$$<br /> <br /> It is important to note that the &quot;noise&quot; (or RMS variations) in any given measurement is proportional to the uncertainty in each measurement, divided by $\sqrt{N_{samples}}$. The fact that $T_{rms} \propto T_{sys}$ is not because $T_{sys}$ represents an uncertainty in each measurement! It is because, in the Rayleigh-Jeans limit of the blackbody equation, the uncertainty in a measurement in a given mode is proportional to the number of photons $N_\gamma$ in that mode. And since $T_{sys} \propto N_\gamma$, it follows that $T_{rms} \propto T_{sys}$. Even though $T_{sys}$ is sometimes called &quot;noise temperature,&quot; we emphasize here that $T_{sys}$ is a real signal; it does not represent real &quot;noise,&quot; or fluctuations in measurements.<br /> <br /> \subsection*{The Radiometer Equation}<br /> Now we can write down an expression for the signal to noise ratio (the radiometer equation):<br /> $$\frac{S}{N} = \frac{T_{src}}{T_{rms}} = \frac{T_{src}}{T_{sys}} \sqrt{\tau \Delta\nu}$$<br /> Typical values might be $\Delta \nu = 10$ MHz, $\tau = 1$ sec so that $\sqrt{\tau \Delta\nu} \sim 3\times10^3$.<br /> Typical values for $T_{sys}$ are 40 - 200 K.<br /> <br /> \subsubsection*{SEFD}<br /> The SEFD is the system equivalent flux density', which is the flux density equivalent of $T_{sys}$:<br /> <br /> $$\textrm{SEFD} = \frac{T_{sys}}{(K/Jy)} = \frac{T_{sys}}{A_e / 2k} = \frac{2kT_{sys}}{A_e}$$<br /> <br /> The SEFD is a useful way to compare the sensitivity of two different systems since it folds in both<br /> $T_{sys}$ and $A_e$. This also greatly simplifies the sensitivity calculation: if you know the <br /> flux in Jy of the source you want to detect, and you know the SEFD, then you can easily calculate<br /> the integration time you need to make a given S/N detection (for an unresolved source):<br /> $$\frac{S}{N} = \frac{S_\nu (Jy)}{\textrm{SEFD}} \sqrt{\tau \Delta\nu}$$<br /> <br /> After substituting in the temperature-based Radiometer Equation for S\N, we arrive at an intuitive expression for the RMS variations in flux density $S_{\nu, rms}$:<br /> <br /> $$S_{\nu, rms} = \frac{\textrm{SEFD}}{\sqrt{\tau \Delta\nu}}$$<br /> <br /> In the temperature-based Radiometer Equation, signal-to-noise \textit{increases} with increased bandwidth and integration time. In the flux-density-based Radiometer equation, RMS flux density varations {\it decrease} with increased bandwidth and integration time.<br /> <br /> \subsubsection*{Extending to interferometric arrays}<br /> <br /> You should refer to the lecture on [[Basic Interferometry]] for explanations of the multi-antenna concepts presented here.<br /> <br /> Making more independent measurements is one way to increase signal-to-noise. With an array of $N$ antennas, every baseline (involving two antennas) adds two more independent measurements: a measurement of the signal's amplitude, and its phase. Thus, for an array of antennas, the Radiometer Equation becomes:<br /> <br /> \begin{aligned}<br /> S_{\nu, rms} &amp; = \frac{\textrm{SEFD}}{\sqrt{ \frac{N(N-1)}{2} \tau (2\Delta\nu)}} \\<br /> &amp; = \frac{\textrm{SEFD}}{\sqrt{ N(N-1) \tau \Delta\nu}}<br /> \end{aligned}<br /> <br /> $S_{\nu, rms}$ are the RMS variations in the flux density that appear in an image, or in a synthesized beam. These fluctuations dictate the faintest sources you can reliably detect in an image. The factor of 2 in the denominator comes from the amplitude and phase measurements, and $N(N-1) / 2$ is the number of baselines between the $N$ antennas.<br /> <br /> \end{document}<br /> &lt;/latex&gt;<br /> <br /> ====Other Links====<br /> * Much of the preceding is based on the Radio 101 single-dish lecture, [[Single_Dish_Basics | Single Dish Basics]].<br /> * [[Media:Radiometer_equation_notes.pdf | Handwritten notes]] used to make the [http://www.youtube.com/watch?v=dwT-tMoscsY video by Chat Hull]<br /> * [http://www.cv.nrao.edu/course/astr534/Radiometers.html The NRAO Course on Radiometers]<br /> <br /> ===Related Subjects===<br /> * [[Basic Interferometry]]<br /> * [[Antennas and Feeds]]<br /> * [[Receivers and Amplifiers]]</div> Jmcbride Revision Control 2011-09-14T21:29:58Z <p>Jmcbride: /* Links */ added name to link of progit</p> <hr /> <div>= Revision Control =<br /> == Introduction ==<br /> Revision Control is broad topic, but it is primarily concerned with managing changes. One of the currently-in-vogue revision control tools is [http://git-scm.com Git]. <br /> ==GIT==<br /> ===Topical Videos===<br /> * [http://www.youtube.com/watch?v=rd-WiB_2oVQ Quick (18 min) Introduction to Revision Control using Git]<br /> * [http://www.youtube.com/watch?v=OFkgSjRnay4 Longish (1 hr) GIT Introduction]<br /> <br /> ===Links===<br /> * For an overview of common GIT commands, see [http://ktown.kde.org/~zrusin/git/git-cheat-sheet-medium.png this cheat sheet] and [http://www.sourcemage.org/Git_Guide this guide]. You can also get help on any GIT command by typing:<br /> &lt;source lang=&quot;bash&quot;&gt;<br /> git help {command}<br /> &lt;/source&gt;<br /> * [http://www.kernel.org/pub/software/scm/git/docs/gittutorial.html GIT Tutorial]<br /> * A well written tutorial on Git, with some nice figures illustrating how Git works [http://progit.org/ Professional Version Control]<br /> <br /> -----<br /> -----<br /> This page is part of [[Radio Astronomy: Tools and Techniques]]</div> Jmcbride Revision Control 2011-09-14T21:27:43Z <p>Jmcbride: /* Links */ added progit</p> <hr /> <div>= Revision Control =<br /> == Introduction ==<br /> Revision Control is broad topic, but it is primarily concerned with managing changes. One of the currently-in-vogue revision control tools is [http://git-scm.com Git]. <br /> ==GIT==<br /> ===Topical Videos===<br /> * [http://www.youtube.com/watch?v=rd-WiB_2oVQ Quick (18 min) Introduction to Revision Control using Git]<br /> * [http://www.youtube.com/watch?v=OFkgSjRnay4 Longish (1 hr) GIT Introduction]<br /> <br /> ===Links===<br /> * For an overview of common GIT commands, see [http://ktown.kde.org/~zrusin/git/git-cheat-sheet-medium.png this cheat sheet] and [http://www.sourcemage.org/Git_Guide this guide]. You can also get help on any GIT command by typing:<br /> &lt;source lang=&quot;bash&quot;&gt;<br /> git help {command}<br /> &lt;/source&gt;<br /> * [http://www.kernel.org/pub/software/scm/git/docs/gittutorial.html GIT Tutorial]<br /> * A well written tutorial on Git, with some nice figures illustrating how Git works [http://progit.org/]<br /> <br /> -----<br /> -----<br /> This page is part of [[Radio Astronomy: Tools and Techniques]]</div> Jmcbride Stellar end states 2011-09-03T02:16:18Z <p>Jmcbride: created page with multiple lectures</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \newcommand{\nic}{{}^{56}{\rm Ni}}<br /> \newcommand{\fe}{{}^{56}{\rm Fe}}<br /> <br /> \begin{document}<br /> \subsection*{End States of Massive Stars} \\<br /> We left off talking about core-collapse supernova, which are produced by stars with initial masses of greater than 10 $M_\odot$ at the end of their lives. Some small fraction of these, perhaps 1 in 1000, will behave differently due to being a rapidly rotating star, and results in what we observe as a gamma ray burst. Gamma ray bursts emit most of their energy in gamma rays, as the name implies. The total energy release is of order (slightly larger) the energy released by supernova. They are also directional, with emission emitted along cones outward, and the material that produces it is moving at highly relativistic speeds. Associated with GRBs in most cases are afterglows at longer wavelengths (x-ray, optical, radio). The interpretation of this is that the gamma ray bursts launch material out, producing gamma ray emission, and then the material runs in to surrounding material, slowing down, and producing emission at longer wavelengths. \\ \\<br /> We observed supernova as very bright optical sources. This luminosity is produced by fusion during the explosion. Fusion is ignited because the outer layers are heated very quickly by the injection of kinetic energy from the core. In order for fusion in the outer layers to fuse, the timescale for fusion needs to be shorter than the timescale for the explosion of the star. If the timescale for explosion is shorter, then the material will cool before fusion can get going. In practice, this requires temperatures in the outer layers of 2--5 $\times 10^9$ K, and, when it occurs, is called explosive nucleosynthesis. Even during the explosion, when this is satisfied, the material is in nuclear statistical equilibrium. This state favors the most bound nucleus, given the ratio of nucleons to protons in the material. This ratio is constant during the explosion because the timescale for $\beta$-decay (turning neutrons in to protons or protons plus electrons in to neutrons) is longer than the timescale of the explosion. The material in the outer layers is mostly low atomic number, meaning an equal number of protons and neutrons. When this is the case, the most bound nucleus is ${}^{56}$Ni. We might have guessed ${}^{56}$Fe, but that does not have an equal number of protons and neutrons. Despite being the most bound nucleus, $\nic$ is only stable for 6.1 days. This is fine for the explosion, since this timescale is much longer than the explosion, but means that the $\nic$ will decay not too long after the explosion. \\ \\<br /> Before getting in to the details of this explosion, remember that there are two different kinds of supernovae that arise from very different conditions. One, which we have been talking about, are supernovae that arise from the core collapse in a massive star, and are called Type Ibc or Type II supernovae. The other is a supernovae produced by a thermonuclear runaway in a degenerate white dwarf, called Type Ia. That these have confusing names is a historical result from observations, which classified supernovae by the presence of hydrogen in the spectrum.\\ \\<br /> Going back to core collapse supernovae, the total kinetic energy of the explosion is roughly $!0^{51}$ ergs. The thermal energy is roughly of order the kinetic energy, which implies a temperature on the surface of the neutron star of $10^9$ K. This is not observable, of course, because the photons need to diffuse out. The diffusion timescale is much larger than the expansion timescale though, so we only see the photons when the star has expanded enough that it becomes optically thin. As the radius increases, the thermal energy decreases. At some point, the diffusion time will be of order the expansion time, and the radiation leaks out. This point represents the peak of the supernovae emission. (I missed some stuff, so this result is coming out of nowhere) This peak occurs at<br /> $<br /> t_{\rm peak} = \left(\frac{M \kappa}{4 v c}\right)^{1/2}.<br />$<br /> This turns out to be about two weeks, or<br /> $<br /> t_{\rm peak} = 2 {\rm \; weeks} \left(\frac{M}{M_\odot}\right)^{1/2} \left(\frac{v}{10^4 {\rm \; km\;s^{-1}}}\right)^{-1/2}.<br />$<br /> The luminosity is about the thermal energy when photons can get out divided by the timescale of diffusion at the peak, which is just $t_{\rm peak}$, so<br /> $<br /> L_{\rm peak} \approx \frac{E_{th}}{t_{\rm peak}}.<br />$<br /> <br /> This is not the whole story though. There are important decays that contribute considerable energy, particularly for stars that are not large when they explode. This starts with the decay of the unstable $\nic$ atom, which releases a couple of MeV per decay.<br /> (A lot missing here. Very sneezy day.)<br /> <br /> <br /> \subsection*{Neutron Stars} \\<br /> Neutron stars are the remnants of supernovae. To first order, neutron stars are like white dwarfs, but with the degeneracy pressure provided by neutrons rather than electrons. This means they are supported by non-relativistic degeneracy pressure, and we know that $P \propto \rho^{5/3}$, meaning it is an $n = 3/2$ polytrope. We could apply our results from that discussion, and would find that the radius of a neutron star is<br /> $<br /> R \approx 15 \left(\dfrac{M}{M_\odot}\right)^{-1/3} {\rm \;km }.<br />$<br /> The reason that neutorn stars are so much smaller than white dwarfs is because the degeneracy pressure scales as <br /> $<br /> P_{\rm degen} \sim \frac{n^{5/3}}{m}.<br />$<br /> Thus for a given energy, neutrons have a lower Fermi energy, and thus a lower pressure. In order to support the same mass then, a neutron star must be smaller than a white dwarf. The central density of a neutron star, again assuming it is an $n = 3/2$ polytrope, is<br /> $<br /> \rho_c = 8 \times 10^{14} \left(\frac{M}{M_\odot}\right)^2 {\rm \; g\; cm^{-3}}.<br />$<br /> Just as in the white dwarf case, at some point, the neutrons will become relativistic, and the equation of state changes, with $P \propto \rho^{4/3}$. The implication of relativistic neutrons is the same as for electrons, with an upper limit placed on the mass of an object that can be supported by neutron degeneracy pressure. This result, written in a slightly different way, is<br /> $<br /> M_{\rm ch} = 5.8 \mu^{-2} M_\odot.<br />$<br /> For $\mu_e = 2$, this gives $1.4 M_\odot$, the result we found for white dwarfs. For a neutron star, most of the particles are neutrons, so we can take as a first approximation $\mu = 1$, giving a neutron star maximum mass of <br /> $<br /> M_{\rm ch} = 5.8 M_\odot.<br />$<br /> This result is not quite right in detail though, for two main reasons. One of these is that we have ignored general relativity. We can ignore GR when<br /> $<br /> \frac{GM}{R c^2} \sim \frac{v_{esc}^2}{c^2} &lt;&lt; 1.<br />$<br /> Using the $R(M)$ relationship for neutron stars,<br /> $<br /> \frac{GM}{R c^2} \approx 0.1 \left(\frac{M}{M_\odot}\right)^{4/3}.<br />$<br /> Clearly this is violated as the mass approaches the limit that we found. The second reason is that the neutrons are not an ideal gas. Neutrons are an ideal gas when<br /> $<br /> \rho_c &lt;&lt; \rho_{\rm nuc}.<br />$<br /> The nuclear density is about $2 \times 10^{14} {\rm g\; cm^{-3}}$. This is fairly comparable to the central density we found, so neutrons in a neutron star are not an ideal gas. \\ \\<br /> The first problem may be solved by considering hydrostatic equilibrium in general relativity, where the metric comes in to play and we have to remember that pressure gravitates:<br /> $<br /> \frac{dP}{dr} = -\dfrac{\rho \dfrac{GM_r}{r^2}\left(1 + \dfrac{P}{\rho c^2}\right) \left(1 + \dfrac{4 \pi P r^3}{c^2 M_r}\right)}{\left(1 - \dfrac{2 G M _r}{r c^2}\right)}<br />$<br /> This is the Oppenheimer-Volk (?) approximation. Solving this (but ignoring the neutron ideal gas issue), we find a maximum mass of 0.7 $M_\odot$, with a radius of 9.6 km. If this were true, there would be no neutron stars, as neutron stars form from the collapse of the iron core when it exceeds the electron degeneracy pressure maximum mass. The neutron stars do exist is because this density is much greater than the nuclear density, and the neutrons start to overlap. \\ \\<br /> Dealing with overlapping neutron stars requires nuclear physics. We'll start on the outside at low densities where things are easiest, and move in. At $\rho &lt; 4 \times 10^{11} {\rm \;g\; cm^{-3}}$, there are no free neutrons. The matter is composed of free electrons and heavy nuclei. Electron capture takes place, with nuclei becoming more neutron rich. As the density increases, the lowest energy state of matter cannot accommodate more neutrons in nuclei. This leads to free neutrons. The name neutron drip'' is applied to densities beyond this limit, as neutrons drip away from nuclei. Moving to higher and higher densities, the number of free neutrons increases, the number of neutrons in nuclei decreases, and the number of protons and electrons decreases. At $\rho &lt; 4 \times 10^{12} {\rm \;g\; cm^{-3}}$, non-relativistic layers finally overtake relativistic electrons as the dominant source of pressure. The number of protons, neutrons, and electrons also reach an equilibrium, with $n_n \approx 8 n_p$, and $n_p = n_e$. Finally, at $\rho &lt; 2 \times 10^{14} {\rm \;g\; cm^{-3}}$, neutrons interact via the strong force. We don't know what happens here. Unfortunately, 99\% of the matter in a neutron star falls in this density regime. \\ \\<br /> Observing neutron stars is hard. We already saw it was difficult in white dwarfs, because they cool so quickly after their birth. Neutron stars are born hotter, at $10^{11}$ K, but cool to $10^9$ K in a day from neutrino luminosity. The effective temperature remains at a few times $10^6$ K for some $10^5$ or $10^6$ years, after which it drops like a rock because photons are getting out on that timescale. For $T_{\rm eff} \sim 10^6$ K and a radius of 10 km, the luminosity is roughly $L_\odot$. The bulk of this luminosity comes out in the UV and X-ray region. After this $10^6$ year period, the neutron star is almost impossible to see. At the surface of a neutron star, gas pressure dominates, and the scale height is<br /> $<br /> h = \frac{kT}{mg}.<br />$<br /> This turns out to be 1 cm, because of the high gravity at the surface, and the density is about $1 {\rm \;g\; cm^{-3}}$. The conditions in the atmosphere are actually pretty comparable to the interior of the Sun then. Despite the exotic nature of neutron star interiors, the part we see is actually not exotic.<br /> <br /> <br /> \subsection*{Rotation of Neutron Stars} \\<br /> We can expect that neutron stars should rotate very rapidly from conservation of angular momentum. If we look at the Sun (which will not become a neutron star), with a period of roughly 24 days and a radius of $R_\odot \approx 7 \times 10^{10}$ cm, we can make a rough estimate of the period of a neutron star, since we know its final size. From $J \sim M R^2 \Omega$, <br /> $<br /> P_{\rm NS} \approx P_{\odot} \left(\frac{R_{\rm NS}}{R_\odot}\right)^2.<br />$<br /> This corresonds to a period of order a millisecond, which is a little on the short side of what is actually observed. This corresponds to a rotational energy of<br /> $<br /> E_{\rm rot} \approx 10^{48} \left(\frac{P_{\rm rot}}{100 {\rm \;ms}}\right)^{-2} {\rm \; ergs}.<br />$<br /> The neutron star taps in to this energy because it is strongly magnetized, which we may understand by noticing that stars are excellent conductors. The Sun has a magnetic field of 1 Gauss. By flux conservation,<br /> $<br /> B_{\rm NS} R_{\rm NS}^2 \sim B_\odot R_\odot^2.<br />$<br /> Thus we would expect that the neutron star has a magnetic field with strength $10^{10}$ Gauss. The actual range of magnetic fields observed ranges from $10^8$--$10^{15}$ Gauss. For both the rotation and magnetic field then, the values for the neutron stars are not too different from what we would expect from simple conservation of angular momentum and flux from a star like the Sun. \\ \\<br /> As a result of this high magnetic field and rapid rotation, in addition to the misalignment of the rotational and magnetic axes, pulsed emission is generated by spinning neutron stars. Or, what appears pulsed to an observer on Earth. The change in rotational energy is what powers the radio emission, sort of (we do not observe the full change in rotational energy). This change in energy is<br /> $<br /> \frac{dE_{\rm rot}}{dt} = -\frac{B^2 R^6 \Omega^4 \sin^2(\alpha)}{6 c^3}.<br />$<br /> For typical numbers, this is<br /> $<br /> \frac{dE_{\rm rot}}{dt} = 10^{35} \left(\frac{B}{10^{12} {\rm \; G}}\right)^2 \left(\frac{P}{100 {\rm \;ms}}\right)^{-4}.<br />$<br /> This leads to something called the spin down time, the timescale over which the period changes significantly, which is<br /> $<br /> t_{\rm spindown} = \frac{\Omega}{\dot{\Omega}}.<br />$<br /> By measuring the period, and rate at which the period is changing, we can infer the magnetic field strength of neutron stars, if we assume that the changing period is due to the loss of rotational energy. This simple model is not quite right though, as we have assumed that the area outside of the neutron star is effectively a vaccuum, since it has a scale height of 1 cm. It turns out that the strong magnetic and electric field of the neutron star means that charged matter on the surface has a much different scale height, and that there is a plasma surrounding the neutron star. This is what is thought to actually be happening. A lot of the energy from spinning down actually goes to ejecting material from the neutron star at relativistic speeds. This manifests itself in a bubble of material like that which is observed around the Crab Pulsar. \\ \\<br /> Average neutron stars are traveling through the galaxy at velocities of 350 ${\rm km\; s^{-1}}$ or greater. They do not participate in the rotation of the galaxy like normal stars, and many pulsars are on courses that will take them out of the galaxy. Traveling at this speed requires a significant amount of kinetic energy, something of order $10^{48}$ ergs. This energy probably comes from the explosion that formed the neutron star. If the explosion is not perfectly symmetric, then the neutron star has to be given a kick, so that momentum is conserved.<br /> <br /> \subsection*{Black Holes} \\<br /> The surface'' of a black hole is described by the event horizon, that defines the region from no information can escape. In order to detect them, indirect means are required. One method is dynamical in nature. Unseen mass may be detected based on its gravitational influence on another body. The other method relies on detecting radiation from just outside of a black hole. Radiation from deep within the potential well of a black hole will show signatures of this (ie, quasars). \\ \\<br /> There is strong evidence for the existence of black holes, even though they can only be detected indirectly. One class is supermassive black holes at the centers of galaxies, and the other is the black holes that are produced at the endpoint of stellar evolution for very massive stars. The SMBHs manifest themselves as active galactic nuclei and quasars, while we see most stellar remnant black holes as X-ray binaries, as they accrete gas from a companion star and the accreted gas emits X-rays as it falls in to the black hole. This mass exchange occurs when the companion star overflows its Roche lobe. \\ \\<br /> The result is an accretion disk around the black hole. Particles in orbit around the black hole would like to be in stable, circular orbits. However, the particles feel friction from one another, due to turbulence in the disk caused by the presence of magnetic fields. This moves angular momentum around, and as particles lose angular momentum, they fall inward on the black hole. Of course, angular momentum is globally conserved, so some particles must gain angular momentum and move out. \\ \\<br /> The energy of material on a circular orbit is <br /> $<br /> E = -\frac{GM\Delta M}{2 R}.<br />$<br /> As matter falls in closer to the black hole, it becomes more bound, going to more negative energies, and the change in energy is<br /> $<br /> \Delta E = -\frac{GM\Delta M}{2 R_{in}} - -\frac{GM\Delta M}{2 R_{out}}.<br />$<br /> For large changes in radius, this simplifies to<br /> $<br /> \Delta E = -\frac{GM\Delta M}{2 R_{in}}.<br />$<br /> If mass moves this distance over a time $\Delta t$, then <br /> $<br /> \frac{\Delta E}{\Delta t} = -\frac{GM\Delta M}{2 R_{in} \Delta t}.<br />$<br /> We can identify the mass accretion rate,<br /> $<br /> \dot{m} = \frac{\Delta M}{\Delta t}.<br />$<br /> Then <br /> $<br /> L = \frac{G M \dot{M}}{2 R_{in}}.<br />$<br /> This energy is ultimately converted to radition going out, characterized by the efficiency $\eta$ as<br /> $<br /> L = \eta \dot{M} c^2.<br />$<br /> Since matter falls so deep in to the potential for a black hole, the efficiency is very high, with<br /> $<br /> \eta = \frac{R_s}{4 R_{in}}.<br />$<br /> This is in the range $\eta \approx 0.1$--$0.4$. Compare this to fusion, where four protons of material (4 GeV) combine to release 27.3 MeV of energy. This corresponds to an $\eta \approx 0.007$. Just as for a star, there is a maximum luminosity for accreting black holes, and it is set by the same physics that set the Eddington luminosity. It turns out that most observed accreting black holes are accreting near the Eddington luminosity. This has a corresponding mass accretion rate, given by<br /> $<br /> \dot{M_{Edd}} = 10^{-8} \left(\frac{M}{M_\odot}\right) \left(\frac{\eta}{0.1}\right)^{-1} \frac{M_\odot}{{\rm yr}}.<br />$<br /> This energy is emitted in X-rays, and neutron stars and black holes may be found by looking for X-ray sources. Stellar mass black holes are identified when the companion can be seen, and from the orbit, the mass of the accreting object can be determined. If the mass is considerably above the maximum mass of the neutron star, we can be confident that the accreting object is a black hole. \\ \\<br /> The other class of black holes, of the super massive variety, manifest themselves as point sources of radiation at the centers of galaxies, and are called active galactic nuclei. The luminosity of AGN vary on very short timescales, as short as hours or days. Based on causality arguments, the size of these objects must be the size of the solar system or smaller. Yet the luminosity they produce can be as high as $10^{13} L_\odot$. This is interpreted as being due to an accreting black hole with a mass of $10^{6}$ up to $10^{10}$ $M_\odot$. For a $10^9 M_\odot$ black hole, the Eddington luminosity is about $10^{13} L_\odot$, corresponding to an accretion rate of $10 M_\odot$/year. It also has Schwarzschild radius of about 20 AU. This is all self consistent then with the idea that black holes power AGN. \\ \\<br /> In the case of our own galaxy, we can actually track orbits of stars near the center of the Milky Way, and see elliptical orbits. This is only consistent with a one over $R$ potential, and the orbit of each star is consistent with an enclosed mass of $4 \times 10^{6} M_\odot$. The region in which orbits have been seen is about 10 light days in size, about the size of the Solar System. This is again consistent with a black hole. Explaining it in any other way is very difficult (improbable).<br /> <br /> <br /> <br /> <br /> <br /> \end{document}<br /> <br /> &lt;/latex&gt;</div> Jmcbride Supernovae 2011-09-03T02:03:26Z <p>Jmcbride: created page</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Core-collapse Explosions (supernovae)} \\<br /> At the end of the fusing stage of a very massive star, the result is an onion-like structure with iron at the center and layers of less massive nuclei above, all the way out to a hydrogen exterior. After fusion stops, the star initially undergoes Kelvin-Helmholtz contraction, but accelerates and becomes a free-free collapse as pressure support is lost. There are two processes that contribute to the contraction becoming a collapse. \\ \\<br /> The first process is photodissentegration of heavy nuclei. The energetic photons in the core will break up massive nuclei into small constituents. For example, photons can break iron up in to 13 helium nuclei and 4 neutrons. At low'' temperatures ($10^9$ K), bound nuclei are favored, but at high'' temperatures ($10^{10}$), nuclear statistical equilibrium favors breaking nuclei down in to smaller parts. This means that as the core contracts and the temperature goes up, matter in the core is dissentegrated in to lighter nuclei. The star spent tens of millions of years fusing hydrogen all the way up to iron, and in this stage, the star undoes all of that work and breaks the nuclei back down. This process takes energy, which causes the core contraction to accelerate, further removing pressure support, and causing the acceleration to increase. \\ \\<br /> The second process is related to the Fermi energy of electrons. At the density of a stellar core at the end of its lifetime, the Fermi energy is of order MeV, as<br /> $<br /> E_F \approx 5 \rho_9^{1/3} {\rm \;MeV}.<br />$<br /> For a density of $10^9$ g cm$^{-3}$, this energy is greater than the difference in mass of the proton and neutron, meaning the reaction<br /> $<br /> e^- + p \rightarrow n + \nu_e<br />$<br /> is favorable. The neutrinos produced in this process take energy away, causing the collapse to accelerate. This process also converts the charged material in the star in to neutrons, and is what leads to a neutron star. That material is converted to neutrons is also important in the process of finding pressure support. The Chandrasekhar mass is<br /> $<br /> M_{ch} = 1.4 \left(\frac{\mu_e}{2}\right)^{-2} M_\odot.<br />$<br /> This decreases as electrons are consumed in the production of neutrons. At some point, the Chandrasekhar mass will drop below the mass of the core, and the star collapses in free fall. The timescale for the collapse is of order 0.1 seconds. \\ \\<br /> Once the star stars collapsing, a new source of pressure must become important in order to halt the collapse. Thermal pressure cannot help, as it drives the star to radiate away energy as neutrinos, and electron degeneracy pressure cannot help since the electrons are being removed. From hydrostatic equilibrium, we know that the pressure needed to balance gravity is<br /> $<br /> P_{HE} \approx \frac{G M^2}{R^4} \propto M^{2/3}\rho^{4/3}.<br />$<br /> The problem of pressure support actually becomes harder as the star contracts, as the density increase will drive the required pressure up. Eventually, neutron degeneracy pressure becomes important. If the neutrons are non-relativistic, then <br /> $<br /> P \sim E_F n \sim \frac{h^2 n^{5/3}}{m_n}.<br />$<br /> Initially, due to the mass of the neutron, they are unimportant compared to electrons. Their mass is useful though in that they can be non-relativistic at higher energies, and the non-relativistic degeneracy pressure increases at a higher rate with density than relativistic degeneracy pressure does. Also imporant is that the NR degeneracy pressure increases faster than the pressure needed to balance gravity. Eventually then,<br /> $<br /> P_n \approx P_{HE}.<br />$<br /> Since the radius of a degeneracy pressure supported object scales as<br /> $<br /> R \propto \frac{1}{m},<br />$<br /> and the radius of a white dwarf is roughly 10000 km, the radius of a neutron star is about 10 km. This corresponding density is then roughly $10^{14}$ g cm$^{-3}$. The separation between particles in the neutron star is about $10^{-13}$ cm, which is roughly the size of neutrons. This invalidates our assumption that the neutron star is an ideal quantum gas. The strong force is important, and affects the pressure. We will talk a bit about this a bit more later, but the important result of this for understanding core collapse is that the strong force actually makes the neutron star stiffer, as it resists compression beyond a certain point. The mass of the neutron star is about the mass of the iron core, which is of order $1 M_\odot$. All of the layers outside barely know what happened. The collapse happens in an inside out manner, since the free-fall time at the center is shorter than the free-fall time in the layers outside. Focusing on the core, its collapse releases a lot of energy from gravitational potential energy, of order<br /> $<br /> E_G \sim \frac{G M^2}{R_{NS}}.<br />$<br /> This is of order $2 \times 10^{53}$ ergs. This is more energy than the Sun releases in its lifetime, and it is released on a timescale of a tenth of a second. Mostly, this energy gets out as neutrinos, since they more easily escape than photons. Since we are concerned with explosions, how much energy would be required to unbind the outer layers? This is of order<br /> $<br /> E_G \sim \frac{G M^2}{R_{core}},<br />$<br /> where the radius of the core is about the radius of a white dwarf. This turns out to be about $10^{51}$ ergs. which is considerably smaller than the energy released by the core's collapse. Thus the core collapse only needs to deposit about 1\% of its energy in the outer layers in order to unbind it from the proto-neutron star. Even though this is a small fraction, it is not clear how the energy actually is deposited. There are a few ideas, none of which totally work. \\ \\<br /> The earliest idea was that the core overshoots in its collapse, and it bounces back outward. The bounce would then cause shocks and push outer material away from the core. As it turns out, this cannot actually blow up the star. The basic reason is that as the star starts to push back out, its kinetic energy is radiated away in neutrinos and photons, causing photodissentegration and energy to leave, and ultimately not accomplishing the goal of blowing up the star. \\ \\<br /> Another idea relies upon neutrinos. A neutron star is one of the few examples in astrophysics where the mean free path of a neutrino is of order or smaller than the size of the system. This arises because the energy and density in the centers of neutron stars are so high. The optical depth in a neutron star is<br /> $<br /> \tau \approx \frac{R}{\ell_\nu} \approx n \sigma_\nu R \approx 3 \left(\frac{E_\nu}{m_e c^2}\right)^2.<br />$<br /> In the core of the neutron star, the optical depth is about $10^5$. This will produce a blackbody spectrum of neutrinos, just like a star produces a blackbody spectrum of photons. The best guess for the cause of the explosion then is that about 10\% of the energy of neutrinos is absorbed by the infalling matter, which is enough to blow up the star. Getting a star to explode in a calculation is possible, but hard, and is very sensitive to details. A simulation which succeeds in exploding the star may not succeed if small details are changed. Simulations like this from 15 years ago succeeded, but more detailed ones today do not. As such, this remains an outstanding problem. \\ \\<br /> The time it takes for the neutrinos to get out is found in just the same way we found how long it takes for photons to get out of the Sun. This is<br /> $<br /> t_{\rm escape} \sim \frac{R^2}{\ell_\nu c}.<br />$<br /> This turns out to be of order 1--10 seconds. The neutrino luminosity is similar, but not identical, to our typical Stefan-Boltzmann expression, with<br /> $<br /> L_\nu \approx 4 \pi R^2 6 \frac{7}{16} \sigma T_{\rm eff}^4.<br />$<br /> The corrections are due to neutrinos being fermions, and there being 6 neutrino species. \\ \\<br /> This has been observed once, from SN 1987 A, a supernova in the LMC. There were 19 neutrinos observed in 13 seconds, which is consisent with predictions for the timescale and the luminosity of neutrinos. It is not enough to do anything much more quantitative than that though. With any luck, a supernova in our galaxy will help better answer questions about neutron star (and black hole) formation, since neutrino detectors are much improved in the last twenty years. \\ \\<br /> The focus today was on neutron stars, but the formation of black holes basically follows the formation of neutron stars, with the one major difference being that the material does not get exploded away, the mass on the neutron star exceeds the Chandrasekhar mass for neutron stars, and the star will collapse to a black hole. It is also possible that a black hole could form even with a supernova, if material which initially is pushed away is pushed away at less than the escape speed, and eventually falls back on the central object and leaves a black hole. We do not really understand how this happens or what this depends upon. <br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-03T01:58:43Z <p>Jmcbride: more moves to topic names</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Hydrostatic Equilibrium]]<br /> #[[Virial Theorem]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Star formation]]<br /> #[[Fusion]]<br /> #[[Main sequence]]<br /> #[[Stellar spectra]]<br /> #[[Stellar evolution]]<br /> #[[Supernovae]]<br /> #[[Stellar end states]]</div> Jmcbride Stellar evolution 2011-09-03T01:55:27Z <p>Jmcbride: created page with multiple lectures</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \newcommand{\he}{{}^4{\rm He}}<br /> \newcommand{\be}{{}^8{\rm Be}}<br /> \newcommand{\ct}{{}^{12}{\rm C}}<br /> <br /> \begin{document}<br /> \subsection*{Stellar Evolution} \\<br /> The hydrogen burning phase, or main sequence lifetime of a star, can be found by dividing the energy available in fusion of hydrogen by the luminosity of the star, with<br /> $<br /> t_{MS} = \frac{E_{\rm nuc}}{L} \propto \frac{M_{\rm core}}{L} \approx 10^{10} {\rm \; years \;} \left(\frac{M}{M_\odot}\right)^{-2}.<br />$<br /> This tells us that more massive stars are shorter lived, and stars only a bit less massive than the Sun have lifetimes longer than the age of the universe. Stellar evolution is due to changes in composition at the centers of stars where fusion is taking place. As hydrogen is consumed, and helium is produced, eventually hydrogen runs out, which drives the structure of a star to change. Even while a star is on the main sequence, its structure will change as a result of changes in composition. If we look back at the radiative transfer equation, remembering how we used this to find how luminosity scales with mass (assuming, say, Thomson scattering), we can see the compositional dependence.<br /> $<br /> F_r = \frac{L_r}{4 \pi r^2} = -\frac{4}{3} \ell \frac{d}{dr}\left(\sigma T^4\right),<br />$<br /> $<br /> kT \sim \frac{G M \mu m_p}{R},<br />$<br /> $<br /> L \propto M^3 \mu^4 \mu_e.<br />$<br /> The mean molecular weight of electrons comes in through the fact that the electrons are doing the scattering, while the total mean molecular weight comes from the virial theorem substitution. Physically, what happens is that the number of particles in the center per unit mass is changing, meaning the temperature in the center must adjust slightly in order to maintain hydrostatic equilibrium, and the rate at which energy leaks out also changes slightly. We can write the opacity in terms of the mass fractions of hydrogen and helium as<br /> $<br /> \kappa = \frac{n_e \sigma_T}{\rho} = \frac{\sigma_T}{\rho} (n_p + 2 n_\alpha) = \frac{\sigma_T}{m_p} \left(X + \frac{Y}{2}\right),<br />$ <br /> where $X$ is the mass fraction of hydrogen and $Y$ is the mass fraction of helium. For the simple case of only hydrogen and helium, this simplifies to <br /> $<br /> \kappa = \frac{\sigma_T}{2 m_p} \left(1 + X \right).<br />$<br /> Then the mean molecular weight of electrons is<br /> $<br /> \mu = \frac{2}{1 + X},<br />$<br /> and <br /> $<br /> \kappa = \frac{\sigma_T}{\mu_e m_p}.<br />$<br /> At the start of a star's lifetime, it is roughly 75\% hydrogen by mass, and the rest is helium, so $\mu = 0.6$. For a pure helium star, which is approximately the case for a very massive star at the end of its lifetime, $\mu = 4/3$. Plugging this in to the luminosity scaling above, we find that<br /> $<br /> \frac{L(t_{MS})}{L(0)} = 40.<br />$<br /> For the Sun, only about 10\% of the hydrogen fuses, and at the end of its lifetime, $\mu = 0.64$. This gives <br /> $<br /> \frac{L(t_{MS})}{L(0)} = 1.37.<br />$<br /> If we were interested, we could follow this changing composition through all the previous equations of stellar structure we have used, to see how the central temperature, effective temperature, and radius must also change as the composition changes. At the end of a star's main sequence lifetime, we are left with a situation where nearly the entire core of the star is composed of helium. Outside the core, there is a hydrogen envelope, which has a composition that is more or less the same as the star at the beginning of its lifetime. At this point, we can give a schematic overview of the stellar evolution post main sequence. \\ \\<br /> When hydrogen fusion ceases, the star loses its source of luminosity, and, as we saw for pre-main sequence stars, the core of the star must undergo Kelvin-Helmholtz contraction. As it contracts, it heats up, and it becomes possible to fuse heavier elements than was possible before, and the fusion of heavier elements will become large enough to halt contraction. Eventually, the material will run out, and with no energy source, the core must contract, and the cylce stars anew. This cycle can be halted if, during contraction, the density increases enough to the point that electron degeneracy pressure takes over and halts contraction. \\ \\<br /> The overall fate of the star thus depends on two facts. One is that a star supported by degeneracy pressure has a maximum mass, known as the Chandrasekhar mass. If this mass is exceeded, electron degeneracy pressure is incapable of halting the contraction. The other is that there is a maximum central temperature, which depends on mass, for an object that is supported by gas pressure and degeneracy pressure. This means that the end state of fusion is set by the mass of the star. If we set the mass of our object to the Chandrasekhar mass, we find that the maximum temperature of an object that can ever be electron degeneracy pressure supported is a few times $10^8$ K. Stars cannot fuse to arbitrarily highly charged elements. Some stars that become white dwarfs are able to fuse helium to carbon and oxygen, but it cannot fuse beyond that. Another way of saying this is that the ultimate fate of a star depends on its initial mass. \\ \\<br /> One fate is the become a white dwarf. For stars with core masses less than 1.4 $M_\odot$, which corresponds to stars with initial masses of 8 $M_\odot$, the end state is an electron degeneracy pressure supported object that has no need for fusion. Why the difference between core mass and initial mass? Without going in to detail, mass loss. This is an uncertain area of stellar evolution, as we do not understand in detail the mass loss history of stars. While the 1.4 $M_\odot$ limit is theoretical (with plenty of observational support), the 8 $M_\odot$ limit is empirical. <br /> The second fate, for stars with initial masses larger than 8 $M_\odot$ (meaning no white dwarf), is for fusion to continue all the way up to iron. At this point, the star cannot produce more energy from fusion, since iron is the most tightly bound nucleus. This means that there is no way to keep the star hot and maintain pressure support. This means that the iron core will collapse under its own gravity. When the collapse starts, the core will have a radius of about that of Earth. At the end of the collapse, the radius will be of order 10 km, and the star becomes either a neutron star or a black hole, and produces a supernova explosion. Some sources will say that neutron stars will be produced for stars less massive than 30 $M_\odot$, and black holes for stars more massive. This is wrong. Definitely wrong. For this reason, we'll just say neutron star or black hole. \\ \\<br /> Our overview finished, we can now talk about evolution in a bit more detail. After the main sequence fusion ends, and the core begins contracting, the density and temperature increase. In addition to the core that contracts, there is also a hydrogen envelope around the core that contracts. This envelope becomes denser and hotter, and before any fusion in the helium core begins, the hydrogen in envelope gets hot enough to fuse. Even so, the core is still contracting, which drives energy generation in the envelope up. The outer part of the star can only carry so much energy out though through radiative diffusion, and eventually the energy generated by shell fusion is larger than the energy the star can carry out. The result is that the star expands, even while the core is contracting. Eventually, the star expands until the effective temperature is around 3000--4000 K, and convection can effectively carry the energy out, and the star will actually go up the Hayashi line. \\ \\<br /> This is a key point to understanding stellar evolution. During post main sequence evolution, the core and the outer parts of the star become decoupled. Even while the core contracts, the outer part of the star expands. For stars on the main sequence, the properties and the core and the properties outside the core were not independent. \\ \\<br /> For stars more massive than 2 $M_\odot$, the core continues contracting and the star continues moving up the Hayashi line until helium ignites, and the helium main sequence begins. For stars less massive than $M_\odot$, the core actually becomes degenerate. The star still moves up the Hayashi line though. The reason for this is that the radius of the core is proportional to $M^{-1/3}$. This matters because shell fusion dumps helium on the core, increasing the mass, and causing the core to become smaller and the central temperature to increase. Eventually, helium fusion starts in a slightly different fashion, which we'll discuss later. Thus, both mass regimes lead to giants, but the details are slightly different. <br /> <br /> \subsection*{Helium fusion physics} \\<br /> The physics of helium fusion is very different from the fusion of hydrogen. If we have a core composed of hydrogen and helium, and try to fuse helium with helium or a proton with helium, we do not get anything stable. Both beryllium and lithium are unstable. What happens instead is what is called the triple $\alpha$'' reaction, where three helium fuse to form carbon. This is not quite an actual triple reaction, but involves two helium fusing and then on a timescale faster than beryllium decay, having another helium come along and fuse. At the same temperature at which this happens, helium and carbon can fuse to form oxygen. As such, this leads to a mixture of carbon and oxygen. \\ \\<br /> The first step of the triple-$\alpha$ process, the fusion of helium to beryllium, is actually endothermic, requiring an energy of 92 keV. <br /> $<br /> {}^4{\rm He} + {}^4{\rm He} \rightarrow {}^8{\rm Be}<br />$<br /> Thus, the energy of the fusing helium must be greater than 92 keV. We can write down the same sort of $E_0$ as we did before to find the energy that dominates the fusion reaction, with<br /> $<br /> E_0 = \left(\frac{E_G (kT)^2}{4}\right)^{1/3}.<br />$<br /> This is a bit different than what we saw before, since helium has to overcome the Coulomb barrier as well as overcome the fact that the reaction is endothermic. Using $E_G$ from before, we find that<br /> $<br /> E_0 = 84 T_8^{2/3} {\rm \; keV}.<br />$<br /> Thus temperatures greater than $10^8$ K are required for helium fusion. The half-life of beryllium is <br /> $<br /> \tau = 2.6 \times 10^{-16} {\rm \;s}.<br />$<br /> Since both fusion and decay can happen on short timescales, the number of beryllium will come in to equilibrium, through the arguments we used for the Saha equatoin. In thermal equilibrium, <br /> $<br /> 2\mu(\he) = \mu(\be).<br />$<br /> Though any one beryllium will not be around for long, there is a population of beryllium that will be maintained, and can then fuse with another helium to form a carbon atom,<br /> $<br /> \be + \he \rightarrow \ct,<br />$<br /> with Q = 7.367 MeV. If this were all there was to it, the amount of carbon produced would not be enough to explain observed abundances. This led to one of the most brilliant predictions in astrophysics history, when Fred Hoyle predicted that the carbon atom had a resonant state at an energy close to the energy released by the fusion reaction. Such a resonant excited state would make fusion easier, and produce carbon. This prediction was verified, and a resonant state in carbon was observed with an energy E = 7.65 MeV above the ground state. Thus, reactions with energies of 290 keV would lead to producing an excited carbon, and we have<br /> $<br /> E_0 = 290 {\rm \;keV.}<br />$<br /> Since $E_0$ that we would find from the typical equation is<br /> $<br /> E_0 = 150 T_8^{1/3} {\rm \; keV},<br />$<br /> the required temperature is $2 \times 10^8$ K for the reaction. \\ \\<br /> The excited state of carbon (denoted with an asterisk) is itself unstable, and will break apart to form beryllium and helium, with<br /> $<br /> \ct^* \leftrightarrow \be + \he.<br />$<br /> Since this reaction goes both ways, it too will come in to chemical equilibrium. Since the same was true for the first step,<br /> $<br /> \he + \he \leftrightarrow \be,<br />$<br /> we have a total reaction with<br /> $<br /> \he + \he + \he \leftrightarrow \ct^*.<br />$<br /> Then we can write down the chemical equilibrium,<br /> $<br /> 3\mu(\he) = \mu(\ct^*).<br />$<br /> If this were all, this would still not be that helpful, since we do not observe excited carbon, and any carbon that leaks out of the center of a star will just decay and not leave us with any carbon. What saves us is that there is one other way for excited carbon to decay, with<br /> $<br /> \ct^* \rightarrow \ct + \gamma.<br />$<br /> This gives us energy and ground state carbon atoms. All of this happens by reactions that are in thermodynamic equilibrium, and only occasionally does an excited carbon make it out of the equilibrium and form a ground state carbon atom. Since it all happens in equilibrium, we know how much carbon there is in the excited state. Then, we can use the decay timescale of the excited carbon to the ground state, and the Saha equation, to find the energy produced by helium fusion. The chemical equilibriums are<br /> $<br /> \mu_{12}^* = m_{12}^* c^2 - k T \ln\left(\frac{g n_{Q, 12}}{n_{12}}\right),<br />$<br /> and <br /> $<br /> \mu_{4} = m_{4} c^2 - k T \ln\left(\frac{g n_{Q, 4}}{n_{4}}\right),<br />$<br /> The difference in energy between the masses is<br /> $<br /> \Delta E = 280 {\rm \; keV} = m_{12}^* c^2 - 3 m_{4} c^2.<br />$<br /> This is positive, unlike our usual reactions, where the energy is negative. This is a result of going to an excited state. Energy is released in the reaction when the excited carbon decays to the ground state. Then, chemical potential equilibrium gives<br /> $<br /> - k T \ln\left(\frac{g n_{Q, 4}}{n_{4}}\right)^3 = \Delta E - k T \ln\left(\frac{g n_{Q, 12}}{n_{12}}\right).<br />$<br /> Solving this equation for the number density of excited carbon, and skipping some simplification,<br /> $<br /> \frac{n_{12}^*}{n_4^3} = 3^{3/2} \left(\frac{h^2}{8 \pi m_p k T}\right)^3 e^{-\Delta E / k T}.<br />$<br /> The number density of hydrogen is<br /> $<br /> n_4 = \frac{\rho}{4 m_p} Y,<br />$<br /> where $Y$ is the mass fraction of hydrogen. The energy difference term is<br /> $<br /> \frac{\Delta E}{k T} = \frac{44}{T_8}.<br />$<br /> Then our final expression for the number density in the excited state is<br /> $<br /> n_{12}^* = \frac{Y^3 \rho^3}{(4 m_p)^3} 3^{3/2} \left(\frac{h^2}{8 \pi m_p k T}\right)^3 e^{-44 / T_8}.<br />$<br /> To get the number in the ground state, we can write the rate of ground state production as<br /> $<br /> \frac{d n_{12}}{dt} = \frac{n_{12}^*}{\tau}.<br />$<br /> This is the step that gives off energy, with an energy generation rate<br /> $<br /> \epsilon = \frac{d n_{12}}{dt}\frac{Q}{\rho},<br />$<br /> where $Q = 7.65$ MeV. A more useful final expression is<br /> $<br /> \epsilon = 5.4 \times 10^{11} \frac{\rho^2 Y^3}{T_8^3} e^{-44 / T_8} {\rm \;erg / s / g}.<br />$<br /> We can turn this in to a simple scaling relationship as before, with<br /> $<br /> \epsilon \propto \rho^\alpha T^\beta.<br />$<br /> At a temperature of $10^8$ K, $\beta sim 40$. At a temperature of $2 \times 10^8$ K, $\beta sim 20$. So this is an even stronger function of temperature than CNO is when looking at the temperatures at which the reactions actually happen in each case. \\ \\<br /> \subsection*{Helium fusion in stars} \\<br /> Now that we understand how helium fusion operates, we can talk about the helium fusing lifetime of a star. A star on the helium main sequence (horizontal branch), will have helium fusion in the core and hydrogen fusion in a shell around the core. This phase will last for a few percent of the main sequence lifetime for a star, and the luminosities are something like a factor of a few tens larger than the main sequence luminosities. \\ \\<br /> Once the helium main sequence finishes, the star has a carbon/oxygen core that, for low mass stars, is almost a white dwarf. The core will contract somewhat, and if the initial mass of the star is less than 8 $M_\odot$, then the carbon/oxygen core will be halted by electron degeneracy pressure. The rest of the material somehow leaves the star, but this is not particularly well understood. There are helium and hydrogen shells, both of which will fuse material, and produce luminosity in excess of that which can be carried out. The outer layers of the star will then puff out again, and lead to the star becoming a giant again, but this time it moves up the asymptotic giant branch. This line is more or less at the same spot as the red giant branch, the only real difference between the two being that the red giant branch stars have helium cores, and the asymptotic giant branch stars have carbon/oxygen cores. What we think happens is the outer layers basically just keep expanding until they are unbound from the carbon/oxygen core. Radiation pressure on dust probably plays a role. There are also instabilities associated with fusion in thin shells, which will produce bursts of luminosity that will help push off material. As material is blown off, the star moves across the HR diagram to hotter temperatures. This is called the planetary nebula phase. It is this process that is responsible for the fact that stars born with masses less than 8 $M_\odot$ end their lives as white dwarfs with masses less than $1.4\; M_\odot$.<br /> <br /> \subsection*{Evolution of Massive Stars} \\<br /> An initial stellar mass above 8 $M_\odot$ implies that the final core mass will be larger than 1.4 $M_\odot$, and the star cannot be supported against gravity be electron degeneracy pressure. One important consequence of this is that there is no maximum temperature for an object that cannot be supported by electron degeneracy pressure. The existence of a maximum temperature comes about because of the competition between degeneracy pressure and thermal gas pressure, but once we are above the Chandrasekhar mass, degeneracy pressure is irrelevant. Now, as the star contracts, the temperature and density both increase, but $P_{\rm degen} &lt; P_{\rm HE}$, and the core can get to arbitrarily high temperatures. This means the star can freely go through the cycle we described before, where the core contracts, temperature goes up, the core fuses heavier elements, the core runs out of the heavy element, and then the core contracts again. The difference this time is that this cycle cannot be stopped, and will proceed all the way to iron since it is the most tightly bound nuclei. Once the core is iron, the cycle is no longer useful, as fusing iron does not produce any energy. \\ \\<br /> The timescales of each step in this cycle are shorter and shorter as the cycle proceeds. This occurs primarily because of the role that neutrino radiation plays in taking energy out of the core of a star. Once the central temperature of a star is above $10^9$ K, the dominant energy loss mechanism is thermal neutrinos, meaning neutrinos that are produced because the core is so hot, not neutrinos produced by fusion. The energy loss by neutrinos drives the fusion cycling to happen very quickly, and cause the final fusion stages to last for only days or months. \\ \\<br /> Normally, we have thought about the energy from fusion balancing the energy radiated away by photons, or<br /> $<br /> L_{\rm fusion} = L_{\rm rad}.<br />$<br /> Now, we all of the energy is carried out by neutrinos, so<br /> $<br /> L_{\rm fusion} = L_{\nu}.<br />$<br /> Since the mean free path of neutrinos is large compared to the size of the star, this energy easily escapes. \\ \\<br /> One means of produces neutrinos is from an electron and positron annihilating, via<br /> $<br /> e^- + e^+ \rightarrow \nu_e + \bar{\nu_e}.<br />$<br /> Typically, what happens in annihilation is photon production, since the electromagnetic force is so much stronger than the weak force, with<br /> $<br /> e^- + e^+ \rightarrow \gamma + \gamma.<br />$<br /> Since neutrinos get out easily and photons do not, neutrino production is still important even though it happens more rarely. Where do the positrons come from? The inverse of the above step:<br /> $<br /> \gamma + \gamma \rightarrow e^- + e^+.<br />$<br /> The energy required for this reaction is the rest mass energy of the electron and positron. This limits this process to temperatures with $kT \sim 1$ MeV, corresponding to $T \sim 10^{10}$ K. This sort of process can happen with any particle, as long as things like charge and lepton number are conserved, and the energy of the photons is sufficient. It happens primarily for electrons and positrons because they have lower rest mass energies. If we could get to $T \sim 10^{13}$ K, then protons and antiprotons could be produced in this way. \\ \\<br /> These reactions are in equilibrium, meaning we can use the Saha equation. Since the chemical potential of photons is zero, we conclude that<br /> $<br /> \mu(e^+) = -\mu(e^-).<br />$<br /> If they are both non-degenerate and non-relativistic, then<br /> $<br /> m_e c^2 - kT \ln \left( \frac{g n_Q}{n_{e^-}} \right) = -m_e c^2 + kT \ln \left( \frac{g n_Q}{n_{e^+}} \right).<br />$<br /> Rearraging, we get<br /> $<br /> 2m_e c^2 = kT \ln \left( \frac{g^2 n_Q^2}{n_{e^-}n_{e^+}} \right).<br />$<br /> Solving for number densities gives<br /> $<br /> n_{e^-}n_{e^+} = g^2 n_Q^2 e^{- 2 m_e c^2 / k T}.<br />$<br /> There are two potential cases here. One is that electrons primarily come from ionized ions, and the other is that they primarily come from this process. Let's look at the first case. The number density of electrons from ions is<br /> $<br /> n_{e^-} = \frac{\rho}{\mu_e m_p} \approx 3 \times 10^{29} \rho_6 {\rm cm^{-3}},<br />$<br /> where $\rho_6$ is the density in units of $10^6$ in cgs. In this case, we can substitute this number density for the number density in our Saha solution, and<br /> $<br /> n_{e^+} = \frac{8 \times 10^{28}}{\rho_6} T_9^3 e^{-11.9 / T_9} {\rm cm^{-3}}.<br />$<br /> This gives<br /> $<br /> \frac{n_{e^+}}{n_{e^-}} = 10^{-6} {\rm \; at \; 10^9 \; K}, <br />$<br /> $<br /> \frac{n_{e^+}}{n_{e^-}} = 10^{-2} {\rm \; at \;2 \times 10^9 \;K}, <br />$<br /> $<br /> \frac{n_{e^+}}{n_{e^-}} = 1 {\rm \; at \;4 \times 10^9 \;K}. <br />$<br /> Of course, this last result violates our initial assumption that most of the electrons are coming from ions. So for temperatures above a few times $10^9$ K, we just set the number density of positrons equal to the number density of electrons, and get<br /> $<br /> n_{e^+} = 10^{29} T_9^{3/2} e^{-6 / T_9} {\rm \; cm^{-3}}.<br />$<br /> Now, we can use this to find the production rate of neutrinos. The neutrino generation rate is<br /> $<br /> r = \frac{n_{e^+}}{\tau},<br />$<br /> where $\tau$ is the timescale for collisions,<br /> $<br /> \tau = \frac{\ell}{v} = \frac{1}{n_{e^-} \sigma v}.<br />$<br /> Then the rate is<br /> $<br /> r = n_{e^+} n_{e^-} \sigma v.<br />$<br /> In order to get an energy generation rate, we need to multiply by the energy taken away by neutrinos, which is of order the rest mass energy of the positron and electron. The energy per time per mass is then<br /> $<br /> \epsilon_\nu = \frac{n_{e^+}n_{e^-}}{\rho} \sigma v (2 m_e c^2).<br />$<br /> Without worrying to much about weak physics, we'll just assume the cross section is of order $10^{-20} \sigma_T$, since the weak interaction is weaker than the electromagnetic interaction by roughly that factor. Plugging in our constants and number density expressions, we find the neutrino energy generation rate is<br /> $<br /> \epsilon_\nu = 4 \times 10^{18} \frac{T_9^3}{\rho} e^{-11.9/T_9} {\rm \; erg/s/g}.<br />$<br /> To estimate the total luminosity, we can write<br /> $<br /> L_\nu = \int \epsilon_\nu dM.<br />$<br /> Since we are concerned with the core, which has a size of order the size of a white dwarf, this is<br /> $<br /> L_\nu = \frac{4}{3} \pi R_{WD}^3 \rho \epsilon_\nu.<br />$<br /> Plugging in numbers, we find<br /> $<br /> L_\nu 10^{12} T_9^3 e^{-11.9 / T_9} \left(\frac{R_{\rm core}}{R_{WD}}\right)^3 L_\odot.<br />$<br /> How large is this? At $5 \times 10^8$ K, the luminosity from neutrinos is 10 $L_\odot$. This is fairly negligible compared to the photon luminosity of a massive star. At $10^9$ K, this becomes $10^7 L_\odot$. This is considerable. Lots of luminosity is being radiated by neutrinos now, much more than from photons. At $2 \times 10^9$ K, $L_\nu \sim 10^{10} L_\odot$. This is huge, and is what drives the short time scale of fusion in massive stars. As stars fuse more massive elements, the Coulomb barrier becomes larger, requiring higher temperatures for fusion. The higher the temperature, the larger the neutrino losses. The larger neutrino losses drive the temperature even higher so that $L_\nu = L_{\rm fusion}$. \\ \\<br /> We can roughly estimate this analytically. The timescale is, as always, the energy divided by the luminosity,<br /> $<br /> t_{\rm nuc} = \frac{E_{\rm fusion}}{L_\nu}.<br />$<br /> We can rewrite this as<br /> $<br /> t_{\rm nuc} = \frac{n_i Q}{\rho \epsilon_\nu}.<br />$<br /> For carbon fusion, $Q \sim 14$ MeV, and $T = 10^9$ K. This gives a timescale of $10^3$ years. If the temperature is increased by a factor of two, this timescale decreases by a factor of 1000, to 1 year. \\ \\<br /> For a 25 $M_\odot$ star, the timescales are summarized in the table below.<br /> \begin{table}[!htb]<br /> \begin{center}<br /> \begin{tabular}{l|l|l}<br /> Fusion Stage &amp; $t_{\rm nuc}$ &amp; $L_\nu / L_\gamma$ \\ \hline \hline<br /> C &amp; $10^3$ years &amp; 10 \\<br /> Ne &amp; 1 year &amp; 6000 \\<br /> O &amp; 1 year &amp; $2 \times 10^4$ \\<br /> Si &amp; 1 day &amp; $3 \times 10^6$ \\<br /> \end{tabular}<br /> \end{center}<br /> \end{table}<br /> All of the reactions towards iron at the very end happen in what is called nuclear statistical equilibrium, which is basically just Saha for nuclei. All of the reactions leading to iron are happening in both directions since the temperatures are so high. The reason then that iron is produced is a consequence of the most bound nucleus being favored at `low'' temperatures, where low means lower than $6 \times 10^9$ K. For equal numbers of protons and neutrons, the most bound nucleus is ${}^{56}$Ni, but in the cores of stars, there are slightly more neutrons than protons from electron capture, and the result is a core with ${}^{56}$Fe.<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Stellar spectra 2011-09-03T01:43:13Z <p>Jmcbride: oops, created page with proper content</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Stellar Spectra} \\<br /> The spectrum that we see from a star gives us information from the photosphere from the star, and contains all the information that we get from stars, aside from neutrinos. Thus, even though it is a very small fraction of the star, it is important observationally. \\ \\<br /> The surface (or photosphere) is where the mean free path is comparable to the scale height of the atmosphere. This is given by<br /> $<br /> \ell \sim H \sim \frac{P}{\rho g}<br />$<br /> Plugging in gas pressure, which dominates the photospheres of stars,<br /> $<br /> \frac{1}{\kappa \rho} \sim \frac{kT}{mg}.<br />$<br /> This gives us the pressure and density in the photosphere, with<br /> $<br /> P \approx \frac{g}{\kappa},<br />$<br /> and <br /> $<br /> \rho \approx \frac{mg}{kT\kappa}.<br />$<br /> For the Sun, $T_{\rm eff} = 5800$ K, giving a number density of $n \approx 10^{17}$ cm$^{-3}$, which is $10^{-9}$ the central number density of the Sun. The pressure is $P \approx 10^5$ erg cm$^{-3} \approx 0.1$ atm. The pressure and density, along with the effective temperature, set the observable properties of stars. To first order, we have assumed that the surface of a star is a perfect blackbody emitter. This is not accurate though, as it ignores the presence of spectral lines, the strength of which depends on the density and pressure as well. The spectra of stars are the basis of a classification system, labeling stars O, B, A, F, G, K, M, ranging from the hottest stars to the coolest stars. There are also L and T spectral types, which apply not to stars, but to brown dwarfs. The nice thing about this is that for main sequence stars, the mass maps to an effective temperature, which maps to a spectral type. Our goal is to understand the second fact (since we already understand the first fact). \\ \\<br /> The matter on the surface of a star is in thermodynamic equilibrium. This is fundamentally the reason that the temperature determines the spectral lines in a star. The ratio of number of states is just set by the Boltzmann factor, with<br /> $<br /> \frac{N_2}{N_1} = \frac{g_2}{g_1} e^{-(E_2 - E_1)/kT}.<br />$<br /> For neutral atoms and molecules, this describes the number of electrons in each state. For determining the fraction of neutral and ionized atoms, we can generalize the Boltzmann factor to an expression known as the Saha equation. In thermal equilibrium, the temperature, pressure, and chemical potential ($\mu$) will all come in to equality. These three quantities are the macroscopic properties of a gas. The implication of the chemical potential equilibrium is that for a reaction where<br /> $<br /> A + B \rightarrow C + D,<br />$<br /> and where<br /> $<br /> C + D \rightarrow A + B,<br />$<br /> then <br /> $<br /> \mu(A) + \mu(B) = \mu(C) + \mu(D).<br />$<br /> The fact that the reaction be able to go both ways is important to remember. This fact is the reason that we cannot apply chemical potential equilibrium to describe fusion in the centers of star. It is so energetically difficult to break apart a helium atom that there are many more reactions producing helium then there are reactions breaking helium apart to provide hydrogen. So, instead, we'll apply this to ionization balance, which is a problem chemical equilibrium can solve, and is what we want to understand to understand the surfaces of stars. We can start by looking at the ionization/recombination of hydrogen:<br /> $<br /> e^{-} + p \leftrightarrow H + \gamma.<br />$<br /> We can solve for the balance of this reaction by writing down the chemical potential of each side,<br /> $<br /> \mu(e^{-}) + \mu(p) \leftrightarrow \mu(H) + \mu(\gamma).<br />$<br /> The phase space distribution of particles is<br /> $<br /> n(p) = \frac{g}{h^3}\frac{1}{e^{(E_p - \mu)/kT} \pm 1}.<br />$<br /> Remember that the plus or minus depends on whether we are dealing with Fermi-Dirac or Bose-Einstein statistics. In the limit of a classical gas, this cannot matter, so we must have the exponential term dominate, leaving<br /> $<br /> n(p) = \frac{g}{h^3}e^{-(E_p - \mu)/kT}.<br />$<br /> The energy of particles is <br /> $<br /> E_p = mc^2 + \frac{p^2}{2m}.<br />$<br /> The number density of particles in real space is the integral over the momentum of the particle,<br /> $<br /> n = \int 4 \pi p^2 dp n(p) = \int 4 \pi p^2 dp \frac{g}{h^3}e^{-(mc^2 - \mu)/kT}e^{-p^2/2mkT}.<br />$<br /> Only the last term needs to stay in the integral, and it is a Gaussian integral with a nice solution. The result is<br /> $<br /> n = g e^{-(mc^2 - \mu) / kT} \left(\frac{2 \pi m k T}{h^2}\right)^{3/2}.<br />$<br /> Or, in terms of the quantum number density, <br /> $<br /> n = g n_Q e^{-(mc^2 - \mu) / kT}.<br />$<br /> This allows us to finally write down what $\mu$ is, with<br /> $<br /> \mu = mc^2 - kT \ln\left(\frac{g n_Q}{n}\right).<br />$<br /> Now, we have everything we need to solve for the ionization balance of hydrogen. We can write the chemical potential of the proton, electron, and hydrogen atom with the above expression, and the chemical potential of the photon is zero (we can create this for free, so they cannot have a chemical potential). The difference in the mass between the hydrogen atom and the sum of electon and proton masses is just the binding energy of the hydrogen atom, which we'll call $\chi$. Then<br /> $<br /> -\frac{\chi}{kT} - \ln\left(\frac{g_H n_{Q, H}}{n_H}\right) = -\ln\left(\frac{g_p n_{Q, p}}{n_p}\right) -\ln\left(\frac{g_e n_{Q, e}}{n_e}\right)<br />$<br /> Some rearranging, taking advantage of the log rules for adding/subtraction, and then exponentiating both sides, gives the Saha equation,<br /> $<br /> \frac{n_e n_p}{n_H} = \frac{g_e g_p}{g_H} n_{Q, e} e^{-\chi/kT}.<br />$<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Stellar spectra 2011-09-03T01:35:48Z <p>Jmcbride: created page</p> <hr /> <div>&lt;latex&gt;<br /> {\bf Stellar Spectra} \\<br /> The spectrum that we see from a star gives us information from the photosphere from the star, and contains all the information that we get from stars, aside from neutrinos. Thus, even though it is a very small fraction of the star, it is important observationally. \\ \\<br /> The surface (or photosphere) is where the mean free path is comparable to the scale height of the atmosphere. This is given by<br /> $<br /> \ell \sim H \sim \frac{P}{\rho g}<br />$<br /> Plugging in gas pressure, which dominates the photospheres of stars,<br /> $<br /> \frac{1}{\kappa \rho} \sim \frac{kT}{mg}.<br />$<br /> This gives us the pressure and density in the photosphere, with<br /> $<br /> P \approx \frac{g}{\kappa},<br />$<br /> and <br /> $<br /> \rho \approx \frac{mg}{kT\kappa}.<br />$<br /> For the Sun, $T_{\rm eff} = 5800$ K, giving a number density of $n \approx 10^{17}$ cm$^{-3}$, which is $10^{-9}$ the central number density of the Sun. The pressure is $P \approx 10^5$ erg cm$^{-3} \approx 0.1$ atm. The pressure and density, along with the effective temperature, set the observable properties of stars. To first order, we have assumed that the surface of a star is a perfect blackbody emitter. This is not accurate though, as it ignores the presence of spectral lines, the strength of which depends on the density and pressure as well. The spectra of stars are the basis of a classification system, labeling stars O, B, A, F, G, K, M, ranging from the hottest stars to the coolest stars. There are also L and T spectral types, which apply not to stars, but to brown dwarfs. The nice thing about this is that for main sequence stars, the mass maps to an effective temperature, which maps to a spectral type. Our goal is to understand the second fact (since we already understand the first fact). \\ \\<br /> The matter on the surface of a star is in thermodynamic equilibrium. This is fundamentally the reason that the temperature determines the spectral lines in a star. The ratio of number of states is just set by the Boltzmann factor, with<br /> $<br /> \frac{N_2}{N_1} = \frac{g_2}{g_1} e^{-(E_2 - E_1)/kT}.<br />$<br /> For neutral atoms and molecules, this describes the number of electrons in each state. For determining the fraction of neutral and ionized atoms, we can generalize the Boltzmann factor to an expression known as the Saha equation. In thermal equilibrium, the temperature, pressure, and chemical potential ($\mu$) will all come in to equality. These three quantities are the macroscopic properties of a gas. The implication of the chemical potential equilibrium is that for a reaction where<br /> $<br /> A + B \rightarrow C + D,<br />$<br /> and where<br /> $<br /> C + D \rightarrow A + B,<br />$<br /> then <br /> $<br /> \mu(A) + \mu(B) = \mu(C) + \mu(D).<br />$<br /> The fact that the reaction be able to go both ways is important to remember. This fact is the reason that we cannot apply chemical potential equilibrium to describe fusion in the centers of star. It is so energetically difficult to break apart a helium atom that there are many more reactions producing helium then there are reactions breaking helium apart to provide hydrogen. So, instead, we'll apply this to ionization balance, which is a problem chemical equilibrium can solve, and is what we want to understand to understand the surfaces of stars. We can start by looking at the ionization/recombination of hydrogen:<br /> $<br /> e^{-} + p \leftrightarrow H + \gamma.<br />$<br /> We can solve for the balance of this reaction by writing down the chemical potential of each side,<br /> $<br /> \mu(e^{-}) + \mu(p) \leftrightarrow \mu(H) + \mu(\gamma).<br />$<br /> The phase space distribution of particles is<br /> $<br /> n(p) = \frac{g}{h^3}\frac{1}{e^{(E_p - \mu)/kT} \pm 1}.<br />$<br /> Remember that the plus or minus depends on whether we are dealing with Fermi-Dirac or Bose-Einstein statistics. In the limit of a classical gas, this cannot matter, so we must have the exponential term dominate, leaving<br /> $<br /> n(p) = \frac{g}{h^3}e^{-(E_p - \mu)/kT}.<br />$<br /> The energy of particles is <br /> $<br /> E_p = mc^2 + \frac{p^2}{2m}.<br />$<br /> The number density of particles in real space is the integral over the momentum of the particle,<br /> $<br /> n = \int 4 \pi p^2 dp n(p) = \int 4 \pi p^2 dp \frac{g}{h^3}e^{-(mc^2 - \mu)/kT}e^{-p^2/2mkT}.<br />$<br /> Only the last term needs to stay in the integral, and it is a Gaussian integral with a nice solution. The result is<br /> $<br /> n = g e^{-(mc^2 - \mu) / kT} \left(\frac{2 \pi m k T}{h^2}\right)^{3/2}.<br />$<br /> Or, in terms of the quantum number density, <br /> $<br /> n = g n_Q e^{-(mc^2 - \mu) / kT}.<br />$<br /> This allows us to finally write down what $\mu$ is, with<br /> $<br /> \mu = mc^2 - kT \ln\left(\frac{g n_Q}{n}\right).<br />$<br /> Now, we have everything we need to solve for the ionization balance of hydrogen. We can write the chemical potential of the proton, electron, and hydrogen atom with the above expression, and the chemical potential of the photon is zero (we can create this for free, so they cannot have a chemical potential). The difference in the mass between the hydrogen atom and the sum of electon and proton masses is just the binding energy of the hydrogen atom, which we'll call $\chi$. Then<br /> $<br /> -\frac{\chi}{kT} - \ln\left(\frac{g_H n_{Q, H}}{n_H}\right) = -\ln\left(\frac{g_p n_{Q, p}}{n_p}\right) -\ln\left(\frac{g_e n_{Q, e}}{n_e}\right)<br />$<br /> Some rearranging, taking advantage of the log rules for adding/subtraction, and then exponentiating both sides, gives the Saha equation,<br /> $<br /> \frac{n_e n_p}{n_H} = \frac{g_e g_p}{g_H} n_{Q, e} e^{-\chi/kT}.<br />$<br /> <br /> <br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-03T01:35:02Z <p>Jmcbride: moving more to topic</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Hydrostatic Equilibrium]]<br /> #[[Virial Theorem]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Star formation]]<br /> #[[Fusion]]<br /> #[[Main sequence]]<br /> #[[Stellar spectra]]<br /> #[[Stellar evolution]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Main sequence 2011-09-03T01:29:47Z <p>Jmcbride: </p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Main Sequence} \\<br /> We understand now how to calculate the luminosity produced by fusion, in addition to understanding how luminous a star needs to be from energy transport. We have<br /> $<br /> L_{\rm fusion} = \int dM_r \epsilon(\rho, T)<br />$<br /> and <br /> $<br /> L \propto M^3 {\rm \;\; (Thomson)}<br />$<br /> $<br /> L \propto M^{5.5}R^{-0.5} {\rm \;\;(free-free)} <br />$<br /> $<br /> L \propto M^{4/7}R^2 {\rm \;\;(Convection)} <br />$<br /> When the fusion luminosity is equal to the luminosity that can leak out, then the star is in steady state. This condition, plus hydrostatic equilibrium, give the properties of stars on the main sequence. In other words, given the mass, you can solve for $L, T_c, T_{\rm eff}, R$. \\ \\<br /> We can start with an object that is not fusing, but is radiating energy away. That is, it is undergoing Kelvin Helmholtz contraction. As the object contracts, its central temperature increases, which in turn increases the luminosity due to fusion (once it is hot enough for at least some fusion to occur). This fusion<br /> luminosity will eventually increase to the point that it is equal to the luminosity being radiated away, at which point the object ceases contracting. Once this happens, the star ceases contracting, the central temperature stays fixed, the luminosity is fixed, and the star is on the main sequence. Well, almost fixed. The composition of the star will change as it fuses, which will cause very slow changes in these properties even while it is on the main sequence. This phase of hydrogen fusing lasts longer than any other stage of a star's lifetime. \\ \\<br /> For the Sun, we showed that radiative diffusion, Thomson scattering, and the virial theorem imply that $L \sim L_\odot$, independent of fusion. Then, knowing that fusion of hydrogen in to helium is the energy source, we determined that the central temperature of the Sun is $T_c \sim 10^7$ K. We then used the virial theorem to determine the radius of the star to be $R \sim 10^{11}$ cm. The way to think about this is that the central temperature is not set by the virial theorem, but fusion sets the central temperature, and this and the virial theorem set the radius. Once we know the radius and luminosity, we can use the Stefan-Boltzmann equation to determine the effective temperature to be $T_{\rm eff} \sim 6000$ K. \\ \\<br /> What we want to do now is to do this for other stars, and explain the main sequence portion of the Hertzsprung-Russell diagram. In principle, this is easy, as we can just follow the exact same procedure for other stars as we did for the Sun. What complicates this is that for other stars, the opacity, energy generation, energy transport, and dominant pressure may be different. This is because all of these properties depend on density and temperature, both of which depend on the mass of the star. Thus, in order to do this correctly, we need to know what the dominant mechanism for each of these things is for other stars. \\ \\<br /> We showed on homework that at the center of the Sun, free-free opacity and Thomson opacity are roughly of equal importance. Thomson opacity is constant with density and temperature, but $\kappa_{f-f} \propto \rho T^{-7/2}$. For stars more massive than the Sun, the central temperature is larger and the density is lower, meaning Thomson scattering dominates for stars more massive than the Sun, and free-free dominates for stars less massive than the Sun. \\ \\<br /> The energy produced by the CNO cycle is about 1\% of the energy produced by the proton-proton chain in the Sun. Since the CNO cycle scales as $T^{20}$ whereas the p-p chain scales as $T^4$, the CNO cycle is dominant for stars somewhat more massive than the Sun, and the proton-proton chain is dominant for stars the Sun's mass and lower. \\ \\<br /> Gas pressure dominantes for all but the most massive stars. Stars less massive than the Sun have convective envelopes and radiative cores. Stars more massive than the Sun have convective cores and radiative envelopes. Knowing all of this, we can look at the main sequence of mass ranges of stars. \\ \\<br /> First, we'll look at stars less massive than the Sun down to about one third of a solar mass. These have free-free opacity, photons transport the energy, proton-proton chain, and are gas pressure dominated. We know that<br /> <br /> $<br /> L_{\rm rad} \propto M^{5.5} R^{-0.5},<br />$<br /> $<br /> L_{p-p} = \int \epsilon_{p-p} dM_r = L_{\rm rad},<br />$<br /> $<br /> L_{p-p} \sim \epsilon_c M, <br />$<br /> $<br /> L_{p-p} \sim \rho T^{4.5} M, <br />$<br /> $<br /> T_c \propto \frac{M}{R}, <br />$<br /> $<br /> L_{p-p} \sim \frac{M^{6.5}}{R^{7.5}}. <br />$<br /> Equating the radiative losses to the energy generation, <br /> <br /> $<br /> L_{p-p} \sim L_{\rm rad}, <br />$<br /> $<br /> \frac{M^{6.5}}{R^{7.5}} \propto \frac{M^{5.5}}{R^{0.5}}, <br />$<br /> $<br /> R \propto M^{1/7}, <br />$<br /> $<br /> R = R_\odot \left(\frac{M}{M_\odot}\right)^{1/7} <br />$<br /> The last step is possible because we are able to normalize to the Sun. Likewise for central temperature, we find<br /> \]<br /> $<br /> T_c \propto \frac{M}{R} \propto M^{6/7},<br />$<br /> $<br /> T_c \propto (1.5 \times 10^7 K) \left(\frac{M}{M_\odot}\right)^{6/7}.<br />$<br /> The radial change with mass is tiny, so we can just say<br /> $<br /> L = L_\odot \left(\frac{M}{M_\odot}\right)^{5.5}.<br />$<br /> We can also solve for $R$ in terms of $L$, and find<br /> $<br /> R \propto L^{1/40}.<br />$<br /> This is tiny, so we can just say that<br /> $<br /> T_{\rm eff} = 5800 K \left(\frac{L}{L_\odot}\right)^{1/4}.<br />$<br /> This is everything we would want to know for a star that is between roughly 0.3 and 1 solar mass, but outside this mass range, these results do not hold. Let's do this again then for stars that are more massive than the Sun, ranging from roughly 1--50 solar masses. Here, CNO is dominant, gas pressure is dominant, and Thomson opacity is dominant. We showed in lecture that<br /> $<br /> L_{\rm rad} = L_\odot \left(\frac{M}{M_\odot}\right)^3.<br />$<br /> The energy generated by fusion is<br /> <br /> $<br /> L_{\rm fusion} = \int \epsilon dM_r \sim \epsilon_c M, <br />$<br /> $<br /> \epsilon_c \propto \rho T^{20}, <br />$<br /> $<br /> \epsilon_c \propto \frac{M^{21}}{R^{23}}. <br />$<br /> $<br /> L_{\rm fusion} \propto \frac{M^{22}}{R^{23}}.<br />$<br /> <br /> Equating the radiative luminosity to the fusion luminosity gives<br /> $<br /> \frac{M^{22}}{R^{23}} \propto M^3, <br />$<br /> $<br /> R \propto M^{19/23} \propto M^{0.8}, <br />$<br /> $<br /> R = R_\odot \left(\frac{M}{M_\odot}\right)^{0.8}.<br />$<br /> <br /> Using the virial theorem gives the central temperature dependence of<br /> $<br /> T_c = (1.5 \times 10^7 K) \left(\frac{M}{M_\odot}\right)^{0.2}.<br />$<br /> For stars on the CNO cycle, the temperature only needs to change slightly to keep up with the extra luminosity. Finally, Stefan-Boltzmann gives<br /> $<br /> T_{\rm eff} = 5800 K \left(\frac{M}{M_\odot}\right)^{3/8}.<br />$<br /> Or, in terms of luminosity, useful for plotting the HR diagram, <br /> $<br /> T_{\rm eff} = 5800 K \left(\frac{L}{L_\odot}\right)^{1/8}.<br />$<br /> Thus there is a large range of luminosities for stars on the main sequence, despite a relatively small range in effective temperatures. \\ \\<br /> We could do this again, this time for stars more massive than 50 solar masses. These only change in that the gas pressure no longer dominates, with radiation pressure taking over. This gives $L \propto M$. We will not actually do this. Likewise, we could do fully convective stars, but we'll leave that for homework. \\ \\<br /> Instead, let's explain why massive stars have convective cores. Convection sets in when there is a negative entropy gradient. A more useful condition in practice is that entropy sets in when<br /> $<br /> \frac{d \ln T}{d \ln P} &gt; \frac{\gamma - 1}{\gamma}.<br />$<br /> Remember that $\gamma$, the adiabatic index, is for the pressure providing particles. Then, for gas pressure dominated stars, we have $\gamma = 5/3$. We showed that the left hand side can be written<br /> $<br /> \frac{d \ln T}{d \ln P} = \frac{1}{4} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M}.<br />$<br /> Thus convection sets in if<br /> $<br /> \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M} &gt; \frac{8}{5}.<br />$<br /> Rearranging this in terms of the enclosed mass, convection sets in if<br /> $<br /> \frac{M_r}{M} &lt; \frac{5}{8} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r}{L}.<br />$<br /> For the CNO cycle, $L_r \approx L$ even for very small radii. We can get a rough sense of when this is important by looking at the Sun (even though it does not strictly apply for the Sun). We have $P/P_{\rm rad} \sim 3000$, and $L / L_{edd} \sim 4 \times 10^{-5}$. This gives convection for <br /> $<br /> \frac{M_r}{M} &lt; 0.1.<br />$<br /> This is why stars more massive than the Sun have convective cores. The Sun only barely avoids this, because it is dominated by the proton-proton chain, which is slightly less temperature sensitive. The convective core sets in for stars about 1.3 $M_\odot$ and above.<br /> <br /> \subsection*{Minimum and Maximum Masses of Stars} \\<br /> Not every ball of gas in hyrdostatic equilibrium is a star. This is reserved for objects that are undergoing, or at some point undergo, fusion of hydrogen in to helium. Stars only exist in between a range of masses. The lower mass limit is set by where degeneracy pressure kicks in. Masses below this limit are called brown dwarfs, and never fuse hydrogen in to helium. The upper limit is less well understood both observationally and theoretically. The upper limit has something to do with being supported by radiation pressure, but the details are unclear. \\ \\<br /> So far, we have considered gas in stars to be a classical ideal gas. To understand the minimum mass of stars, we need to consider quantum mechanical effects in a gas. The way to think about this is that the quantum mechanical nature of particles is important if the de Broglie wavelength is of order the separation between particles. The de Broglie wavelength is just the wavelength associated with the thermal momentum,<br /> $<br /> \lambda \approx \frac{h}{p_{th}}.<br />$<br /> When this wavelength is much smaller than the distance between particles, the gas does not care about quantum mechanics. So the condition for quantum mechanics to be important is<br /> $<br /> \frac{h}{p_{th}} &gt; n^{-1/3}.<br />$<br /> Or, writing in terms of the density, and putting the momentum in terms of the temperature,<br /> $<br /> n &gt; \left(\frac{m k T}{h^2}\right)^{3/2}.<br />$<br /> We will define this density as the quantum number density, $n_Q$. The particle with the lowest value of $n_Q$ is the one for which quantum mechanics is important first. Since there is an $m$ in the numerator, that means that electrons become degenerate first. The quantum number density does depend on temperature, since the thermal momentum affects the de Broglie wavelength of particles. For electrons, we find<br /> $<br /> n_q \approx 10^{26} T_7^{3/2} {\rm \; cm^{-3}}.<br />$<br /> At the center of the Sun, the temperature is about $10^7$, and the density is $n \sim n_Q$. For this reason, detailed models of the Sun always treat quantum mechanics when solving for the Sun's structure. We can see how the importance of this scales for other stars. Assuming a scaling between $M$ and $R$ of<br /> $<br /> R \propto M^{3/4},<br />$<br /> we find <br /> $<br /> T_c \propto M^{1/4}<br />$<br /> and<br /> $<br /> \rho \propto M^{-5/4}.<br />$<br /> Thus, as $M$ decreases, the central temperature decreases and the density increases, which means that quantum mechanics becomes increasingly important at lower masses. For increasing $M$, the opposite is true, and quantum mechanics is less important for stars more massive than the Sun. \\ \\<br /> When $n &gt;&gt; n_Q$, we consider the gas to be fully degenerate. This is sometimes called the low $T$ limit, where low temperature is a very relative term. The number of particles as a function of momentum is <br /> $<br /> n(p) = \frac{g / h^3}{e^{(E_p - \mu)/kT} \pm 1}.<br />$<br /> Here, $E_p$ is the total energy of the particle (mass plus momentum) and $\mu$ is the chemical potential. The chemical potential is defined as<br /> $<br /> \mu = \left(\frac{dE}{dN}\right)_{S, V}.<br />$<br /> This is a way of quantifying how changing the number of particles changes the enrgy of a system. The full thermodynamic expression is<br /> $<br /> dE = T dS - P dV + \mu dN.<br />$<br /> Finally, in the $n(p)$ expression, the $+$ is for particles which satisfy the Pauli exclusion principle, and the $-$ is for Bose-Einstein particles. In the classical limit, the quantum mechanical nature of this expression ($\pm$) is unimportant. So QM is unimportant when the exponential term is much greater than 1. Or, alternatively, the number density is much less than the quantum mechanical number density. We are actually interested in the other limit, and the fermion expression ($+$), since electrons, protons, and neutrons are fermions. Then this distribution function is <br /> $<br /> n(p) = \frac{g / h^3}{e^{(E_p - \mu)/kT} + 1}.<br />$<br /> Let's first imagine very low energies. Then $E_p$ is much smaller than $\mu$, and $kT$ is much smaller than $\mu$. This means the exponential term goes to 0, and the entire expression goes to<br /> $<br /> n(p) \approx \frac{2}{h^3}.<br />$<br /> <br /> {\bf Note: the remainder is missing, as note taker had to leave}<br /> <br /> \end{document}<br /> <br /> &lt;/latex&gt;</div> Jmcbride Fusion 2011-09-03T01:19:30Z <p>Jmcbride: added note about missing lecture</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> {\bf Note: missing an order of magnitude preamble lecture} \\ \\ <br /> <br /> \subsection*{Nuclear Reaction Rates}} \\<br /> The cross section for nuclear reaction depends upon the strong force, and it also depends upon the quantum mechanical probability for tunneling through the Coulomb barrier. These two dependences are independent of one another. The quantum mechanical part also happens to be the part that primarily sets the temperature sensitivity of fusion, and is also the part that is easier to understand. \\ \\<br /> {\bf Tunneling} \\<br /> To understand tunneling, we start by writing down the Schrodinger equation.<br /> $<br /> \left(-\frac{\hbar^2}{2m} \nabla^2 + V\right)\psi = E \psi.<br />$<br /> For a simple step like potential, outside of the step, the potential is V = 0, and the solution is a plane wave, and<br /> $<br /> E = \frac{k^2 \hbar^2}{2m}.<br />$<br /> Here, $k$ is the wavenumber. In the step, the potential is just some constant $V = V_0$, and<br /> $<br /> E - V_0 = \frac{k^2 \hbar^2}{2m}.<br />$<br /> \\<br /> {\bf Large portion missing here (notetaker had to leave lecture)}<br /> <br /> \subsection*{Fusion Reaction Rates} \\<br /> We ended last time by writing down the cross section for fusion in terms of the probability of tunneling and an intrinsic strong nuclear physics part that we will use without trying to understand the detailed physics. This cross section is<br /> $<br /> \sigma(E) = \frac{S(E)}{E} e^{-(E_G/E)^{1/2}}.<br />$<br /> Here, $S(E)$ is the nuclear physics part, $E$ is the energy of both particles, and the expontential term is the probability of tunneling through the Coulomb barrier. We wrote in the last class $E_G$ as<br /> $<br /> E_G = (1 {\rm \;MeV}) Z_1^2 Z_2^2 \frac{m_r}{m_p}.<br />$<br /> This expression can be used to find the probability of a single tunneling event.We want to write down the rate of fusion for a large system of particles. We can start by writing down the rate for lots of particles that all have the same velocity. For distinguishable fusing particles that we will label 1 and 2 with number densities $n_1$ and $n_2$, we can write down a mean free path<br /> $<br /> \ell_2 = \frac{1}{n_1 \sigma}<br />$<br /> and a mean free time<br /> $<br /> \tau_2 = \frac{1}{n_1 \sigma v}.<br />$<br /> The number of reactions per unit volume per unit time is then<br /> $<br /> r_{12} = \frac{n_2}{\tau_2} = n_1 n_2 \sigma v.<br />$<br /> If we average over all possible velocities, the result is<br /> $<br /> r_{12} = n_1 n_2 &lt;\sigma(E) v&gt;.<br />$<br /> To find $&lt;\sigma(E) v&gt;$, we need to integrate over the Maxwell-Boltzmann distribution<br /> $<br /> &lt;\sigma(E) v&gt; = \int d^3v\; prob(v) \sigma(E) v.<br />$<br /> Plugging in (with normalization) the MB distribution, the result is<br /> $<br /> r_{12} = n_1 n_2 \int d^3v \sigma(E) v \left(\frac{m_r}{2 \pi k T}\right)^{3/2} e^{-\frac{\frac{1}{2}m_r v^2}{kT}}.<br />$<br /> We would like to put this entirely in terms of the energy, substituting out the velocity. We'll use<br /> $<br /> E = \frac{1}{2} m_r v^2, <br />$<br /> $<br /> dE = m_r v dv, <br />$<br /> $<br /> d^3v = 4 \pi v^2 dv = 4 \pi \frac{v^2}{v} \frac{dE}{m_r},<br />$<br /> $<br /> v d^3 v = \frac{8 \pi E}{m_r} \frac{dE}{m_r}. <br />$<br /> Now we can rewrite our velocity averaged cross section as <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE E \sigma(E) e^{-E / kT}.<br />$<br /> Now, plug in our original expression for $\sigma(E)$ to find <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE S(E) e^{-(E_G / E)^{1/2}\;-\;E / kT}.<br />$<br /> The tunneling term here favors reactions at high energies, but the Boltzmann term favors reactions at low energies where there are a large number of available particles. Compared to the energy terms in the exponential, $S(E)$ is slowly varying, and we can pull it out of the integral, writing<br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} S(E) I.<br />$<br /> with<br /> $<br /> I = \int_0^\infty e^{-(E_G / E)^{1/2}\;-\;E / kT} dE.<br />$<br /> We can draw a picture to understand roughly what this integral will give us. The MB distribution falls quickly with energy, and the probability of tunneling rises exponentially with energy. The product of the two is at a maximum at some energy $E_0$ that falls part way along the tail of each. We can find this $E_0$ by differentiating the integrand above, which we'll call $f(E)$. Then<br /> $<br /> \frac{df}{dE} = 0 = \frac{1}{kT} - \frac{E_G^{1/2}}{2 E^{3/2}}.<br />$<br /> Solving we find<br /> $<br /> E_0 = \left(\frac{1}{2} E_G^{1/2} kT\right)^{2/3}.<br />$<br /> In terms of the more fundamental quantities that define $E_G$,<br /> $<br /> E_0 = (5.7\; {\rm keV}) Z_1^{2/3} Z_2^{2/3} T_7^{2/3} \left(\frac{m_r}{m_p}\right)^{1/3}.<br />$<br /> This is the energy at which most reactions occur, even though it is far smaller than $E_G$ and far larger than $kT$, thanks to the competition between the two. In order to perform the integral above, we will treat the overlap between the two as a Gaussian (method of steepest descent).<br /> $<br /> f(E) = f(E_0) + \frac{1}{2}(E - E_0)^2 + f^{''}(E_0),<br />$<br /> where<br /> $<br /> f^{''}(E_0) = \frac{3 E_G^{1/2}}{4 E_0^{5/2}}.<br />$<br /> If we put this in our integral $I$, we get a Gaussian integral that we can perform. The result is<br /> $<br /> I = \frac{e^{-f(E_0)} \sqrt{2 \pi}}{\sqrt{f^{''}(E_0}}.<br />$<br /> The final results for our velocity averaged cross section is<br /> $<br /> &lt;\sigma(E) v&gt; = 2.6 S(E_0) \frac{E_G^{1/6}}{(kT)^{2/3} \sqrt{m_r}}e^{-3(E_G/4kT)^{1/3}}.<br />$<br /> \\<br /> Now, we can start to do something more macroscopic. Let's define the energy per unit mass per unit time from fusion as $\epsilon$. Then the total luminosity is<br /> $<br /> L = \int \epsilon dM_r = \int \epsilon 4 \pi r^2 \rho dr.<br />$<br /> Then the change in luminosity over radius is<br /> $<br /> \frac{dL_r}{dr} = 4 \pi r^2 \rho \epsilon.<br />$<br /> We can also define an energy released per reaction as $Q$. Using our rate $r_{12}$ of reactions per volume per time, we can find $\epsilon$, with<br /> $<br /> \epsilon_{12} = \frac{r_{12} Q}{\rho}.<br />$<br /> We want to put everything in terms of mass density, which we will do with<br /> $<br /> n_1 = \frac{X_1 \rho}{m_1}.<br />$<br /> Here, $X_1$ is the fraction of the total mass that is in particle 1. Together, these allow us to write the energy generation from fusion.<br /> $<br /> \epsilon_{12} = \frac{2.6 Q S(E_0) X_1 X_2}{m_1 m_2 \sqrt{m_r} (kT)^{2/3}} \rho E_G^{1/6} e^{-3(E_G / 4 kT)^{1/3}}.<br />$<br /> This expression has units ergs per second per gram, and gives the energy from reactions between particles 1 and 2. It is a function of density, temperature, and composition. Since we do not want to carry this whole expression around most of the time, we would like to just write down a simple proportionality in terms of these fundamental variables. In general, we'll approximate<br /> $<br /> \epsilon \propto \rho^\alpha T^\beta.<br />$<br /> Depending upon the reaction, $\alpha$ and $\beta$ can take different values. For any process involving collisions between two particles, $\alpha = 1$. Finding $\beta$ is a bit more complicated. Our expression for $\epsilon$ is not a simple power law in temperature. We can approximate it though by noticing that<br /> $<br /> \beta = \frac{d \ln \epsilon}{d \ln T}.<br />$<br /> Writing $\epsilon$ in terms of only the temperature dependent quantities and then performing the above differentiation, we find<br /> $<br /> \beta = -\frac{2}{3} + \left(\frac{E_G}{4 k T}\right)^{1/3}.<br />$<br /> Note that this is a function of temperature. This is only valid then when we are considering a small range of temperature. Fortunately, this is generally a fine approximation when considering fusion in stars. For the proton-proton chain at the center of a sun like star, we find $\beta \approx 4.3$. Then<br /> $<br /> \epsilon_{pp} \propto \rho T^{4.3}<br />$<br /> at temperatures near $10^7$ Kelvin. We can make a bit more detailed an estimate of energy from fusing protons to Helium. Again using $T_c \sim 10^7$ K, $\rho \sim 1$ g cm$^{-3}$, the cross section $S(E)$ is of order 1 keV barn, and $Q$ is order 10 MeV. If we put this in our expression for $\epsilon$, we find<br /> $<br /> \epsilon \sim 10^{20} {\rm \;erg/s/g}.<br />$<br /> The luminosity of the Sun is then <br /> $<br /> L = \int dM_r \epsilon \sim \epsilon M_\odot.<br />$<br /> This is<br /> $<br /> L \sim 10^{54} {\rm \; erg/s} \sim 10^{20} L_\odot.<br />$<br /> What happened in our estimate? This is WAY too big. We can think of a couple of resolutions that would give us a more reasonable estimate. One is that the weak interaction is involved somewhere, as it has a cross section that is $10^{20}$ times smaller than the strong interaction. The other possibility is that a proton is fusion with a particle that has a much larger charge, which would suppress the $E_G$ term and give much lowre energy generation rates. As it turns out, both of these solutions exist in nature. The proton-proton chain involves conversion of protons to neutrons, which requires the weak interaction. The CNO cycle produces Helium by fusion of protons with C, N, and O. Using either, we would find an energy generation rate that is in much better agreement with the observed luminosity of the Sun. \\ \\<br /> Let's look at the proton proton chain in more detail. The net reaction is <br /> $<br /> 4 p \rightarrow {}^4{\rm He} + {\rm energy}.<br />$<br /> This energy comes out as photons, kinetic energy of particles, and a small amount as neutrinos. The kinetic energy and photon energy is all shared with the star through collisions, and produces the heat inside of stars. The energy in neutrinos is not useful, as it just leaves the star without providing any heat (and, by extension, prssure support against gravity). \\ \\<br /> As implied already, the key step in the proton proton chain is the step where a proton turns in to a neutron (necessary since there are two neutrons in Helium, and we are starting with only protons). This step gives its identity as a weak interaction away by the presence of a neutrino.<br /> $<br /> p + p \rightarrow {}^2{\rm H} + e^{+} + \nu_e.<br />$<br /> The cross section for this is $S(keV) \approx 3.78 \times 10^{-22}$ keV barn. This is, as promised, much weaker than the typical strong reaction, which has a cross section of order of 1 keV barn. The rest of the reactions needed to produce Helium are easy, involving the strong reaction. First,<br /> $<br /> {}^2{\rm H} + p \rightarrow {}^3{\rm He} + \gamma,<br />$ <br /> which has a cross section of 2.5 $\times 10^{-4}$ keV barn.<br /> and then<br /> $<br /> {}^3{\rm He} + {}^3{\rm He} \rightarrow {}^4{\rm He} + 2p,<br />$<br /> with cross section 5000 keV barn. Note that the first two steps of the reaction must happen twice for the last step to happen once. That this whole process requires multiple steps is a result of the exceedlingly low probability of four protons all happening to be in the same place at the same time. Even so, determining the overall rate is made simpler by the fact that the slowest step in a cycle sets the rate for the entire cycle. Here, our rate limiting step is the first step, when two protons form a deuterium atom. Then we can calculate the energy generation rate for the entire chain with<br /> $<br /> \epsilon_{cycle} = r_{p-p\; step} Q_{cycle} / \rho.<br />$<br /> This fact will be useful in the CNO cycle as well, which has many steps, but is also rate limited by a single step, allowing us to easily write down the energy generation rate. Using our previous (complicated) expression for energy generation and applying it to the proton-proton step, we find the energy for the entire chain to be<br /> $<br /> \epsilon_{pp} \propto \rho T^{-2/3} e^{-15.7 T_7^{-1/3}}.<br />$<br /> With the constants, <br /> $<br /> \epsilon_{pp} = (5 \times 10^5) \frac{\rho X^2}{T^{2/3}} e^{-15.7 T_7^{-1/3}} {\rm erg/s/g}.<br />$<br /> We can use this to estimate the central temperature of the Sun, using the known luminosity. <br /> $<br /> L = \int \epsilon dM \sim \epsilon(center) M_\odot, <br />$<br /> $<br /> L_\odot \sim 10^7 \frac{M_\odot}{T_7^{2/3}} e^{-15.7 T_7^{-1/3}}, <br />$<br /> $<br /> T_c \approx 10^7 K.<br />$<br /> This is pretty good, and is (fortunately) self consistent.<br /> \subsection*{Hydrogen Fusion in Stars} \\<br /> One method of fusing hydrogen in the centers of stars is called the proton-proton chain.<br /> $<br /> p + p \rightarrow {}^2H + e^{+} + \nu_e<br />$<br /> $<br /> {}^2H + p \rightarrow {}^3He + \gamma <br />$<br /> $<br /> {}^3He + {}^3He \rightarrow {}^4He + 2p <br />$<br /> <br /> This process is important in lower mass stars. For higher mass stars, the CNO cycle becomes important. Carbon, nitrogen, and oxygen are not actually fused in this process. Instead, they are used as catalysts, and the reaction has the exact same input (four protons) and output (one helium atom) as the proton-proton chain does. The steps of the CNO cycle are<br /> \begin{eqnarray*}<br /> {}^{12}C + p \rightarrow {}^{13}N + \gamma <br /> {}^{13}N \rightarrow {}^{13}C + e^{+} + \nu_e <br /> {}^{13}C + p \rightarrow {}^{14}N + \gamma <br /> {}^{14}N + p \rightarrow {}^{15}O + \gamma <br /> {}^{15}O \rightarrow {}^{15}N + e^{+} + \nu_e <br /> {}^{15}N + p \rightarrow {}^{12}C + {}^4He. <br /> \]<br /> Considering our results for the Gamow Energy, which sets the probability of tunneling, how can the CNO cycle be competitive with the proton-proton chain? The charge involved is much larger, so the Gamow Energy of any of the steps are of order 50 MeV. This is five orders of magnitude larger than the typical energy at the center of a star, which is of order 1 keV. At $10^7$ K, the tunneling probability is $10^{-7}$ for the proton-proton chain, and it is a staggering $10^{-31}$ for the CNO cycle. The reason that CNO can be competitive turns out to be because the cross section in every step of the CNO cycle that requires fusing is the strong force, whereas the first step in the proton-proton chain is a weak interaction, with a significantly lower cross section. They happen to be different by roughly $10^{24}$, which roughly cancels the difference in tunneling probability. (Question: what about abundance?) The energy generation rate for the CNO cycle is set by the slowest step, as is the case for any reaction. The slowest step is the fourth step, and gives<br /> $<br /> \epsilon_{CNO} \approx (4 \times 10^27) \frac{\rho}{T_7^{2/3}} X Z e^{-70.7 T_7^{-1/3}} {\rm \;erg/g/s}.<br />$<br /> We can do the same trick as with the proton-proton chain to get a simple power law scaling, and find that <br /> $<br /> \beta = \frac{-2}{3} + \frac{23.6}{T_7^{1/3}},<br />$<br /> with <br /> $<br /> \epsilon \propto \rho T^\beta.<br />$<br /> This is phenomenally sensitive to energy, due to being way out on the tail of the Maxwell-Boltzmann distribution. Raising the temperature just a little bit will cause the number of particles at that energy to rise significantly. While the proton-proton chain was also sensitive to temperature, it is not nearly as sensitive as the CNO cycle. This means that at high temperatures, CNO dominates, and at low temperatures, the proton-proton chain dominates. It turns out that they are roughly equal for stars that are a bit more massive than the Sun. In the Sun, 99\% of the luminosity is produced by the proton-proton chain. \\ \\<br /> \subsection*{Observational Evidence for Fusion in the Sun} \\<br /> Detections of neutrinos from the Sun provide evidence that the center of the Sun is undergoing fusion. Neutrinos are very light particles, and for a long time were believed to be massless. Now, they are thought to have a mass of order (or smaller than) 0.1 eV, which is six orders of magnitude lighter than an electron. They have no charge (or color), meaning they only are affected by the weak force and gravity. There are three types of neutrinos, corresponding to the three leptons: the electron, muon, and tau particle each has a corresponding neutrino. We are only really concerned with electron neutrinos though, since that is what stars provide in fusion (which can be confirmed by looking back at the reactions we wrote down). There are multiple branches in the proton-proton chain. We talked about the dominant branch, but the other branches produce different energy neutrinos. \\<br /> The typical cross section for an interaction between neutrinos and matter is <br /> $<br /> \sigma \sim 10^{-44} \left(\frac{E_\nu}{m_e c^2}\right)^2 {\rm \; cm^2}.<br />$<br /> Note again the twenty orders of magnitude difference between this cross section and something representative of the electromagnetic force, like the Thomson cross section. We can use this to calculate the mean free path of neutrinos in the Sun.<br /> $<br /> \ell = \frac{1}{n \sigma}.<br />$<br /> Using $E_\nu \sim$ MeV, and the average number density in the Sun, we find<br /> $<br /> \ell \sim 10^9 R_\odot.<br />$<br /> This means that a neutrino typically leaves the Sun without interacting at all. A neutrino that we can detect on Earth thus tells us about the conditions at the center of the Sun, where the neutrino was produced. Most importantly, the energy of neutrinos depends on the reaction that produced them, so the energy of neutrinos that we detect tells us about fusion in the center of the Sun. Of course, since a neutrino gets out of the Sun unimpeded, detecting a neutrino on Earth is challenging. The first experiment to do this used cleaning fluid in a huge tank underground, and resulted in a Nobel Prize for Ray Davis in 2002. This cleaning fluid had chlorine, and the reaction of interest was<br /> $<br /> {}^{37}Cl + \nu_e \rightarrow {}^{37}Ar + e^{-}.<br />$<br /> Every day, $10^{22}$ neutrinos go through the tank, but on average one neutrino is detected per day. This reaction only operates for neutrinos more energetic than 0.814 MeV, so this experiment was only able to detect neutrinos from some of the less common fusion reactions, as those from the proton-proton chain were of lower energy. \\<br /> This experiment began in the 1960s, and wound up producing what is called the Solar Neutrino problem. The number of neutrinos detected was one third of the number expected. There was debate for many years between astrophysicists and particle physicists. The resolution was finally provided by Sudbury Neutrino Observatory. The SNO used heavy water, which has deuterium rather than lone protons in the water molecule. An incoming neutrino can break up the deuterium in to two protons and a positron:<br /> $<br /> \nu_e + D \rightarrow p + p + e^{-}.<br />$<br /> This experiment could also detect mu and tau neutrinos, with<br /> $<br /> \nu + D \rightarrow p + n + \nu.<br />$<br /> This measurement confirmed that the total number of neutrinos produced is the total number expected. The total was expected to be made up entirely of electron neutrinos though, and this experiment showed that only one third of them were electron neutrinos. It turns out that since neutrinos have mass, they can oscillate between different neutrino types. That means that despite the Sun producing only electron neutrinos, we can observe mu and tau neutrinos from the Sun. Thus the neutrino observation experiments provided direct confirmation of solar models, and also proved that neutrinos have mass. \\ \\<br /> <br /> &lt;/latex&gt;</div> Jmcbride Main sequence 2011-09-03T01:14:34Z <p>Jmcbride: created page</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Main Sequence} \\<br /> We understand now how to calculate the luminosity produced by fusion, in addition to understanding how luminous a star needs to be from energy transport. We have<br /> $<br /> L_{\rm fusion} = \int dM_r \epsilon(\rho, T)<br />$<br /> and <br /> $<br /> L \propto M^3 {\rm \;\; (Thomson)}<br />$<br /> $<br /> L \propto M^{5.5}R^{-0.5} {\rm \;\;(free-free)} <br />$<br /> $<br /> L \propto M^{4/7}R^2 {\rm \;\;(Convection)} <br />$<br /> When the fusion luminosity is equal to the luminosity that can leak out, then the star is in steady state. This condition, plus hydrostatic equilibrium, give the properties of stars on the main sequence. In other words, given the mass, you can solve for $L, T_c, T_{\rm eff}, R$. \\ \\<br /> We can start with an object that is not fusing, but is radiating energy away. That is, it is undergoing Kelvin Helmholtz contraction. As the object contracts, its central temperature increases, which in turn increases the luminosity due to fusion (once it is hot enough for at least some fusion to occur). This fusion<br /> luminosity will eventually increase to the point that it is equal to the luminosity being radiated away, at which point the object ceases contracting. Once this happens, the star ceases contracting, the central temperature stays fixed, the luminosity is fixed, and the star is on the main sequence. Well, almost fixed. The composition of the star will change as it fuses, which will cause very slow changes in these properties even while it is on the main sequence. This phase of hydrogen fusing lasts longer than any other stage of a star's lifetime. \\ \\<br /> For the Sun, we showed that radiative diffusion, Thomson scattering, and the virial theorem imply that $L \sim L_\odot$, independent of fusion. Then, knowing that fusion of hydrogen in to helium is the energy source, we determined that the central temperature of the Sun is $T_c \sim 10^7$ K. We then used the virial theorem to determine the radius of the star to be $R \sim 10^{11}$ cm. The way to think about this is that the central temperature is not set by the virial theorem, but fusion sets the central temperature, and this and the virial theorem set the radius. Once we know the radius and luminosity, we can use the Stefan-Boltzmann equation to determine the effective temperature to be $T_{\rm eff} \sim 6000$ K. \\ \\<br /> What we want to do now is to do this for other stars, and explain the main sequence portion of the Hertzsprung-Russell diagram. In principle, this is easy, as we can just follow the exact same procedure for other stars as we did for the Sun. What complicates this is that for other stars, the opacity, energy generation, energy transport, and dominant pressure may be different. This is because all of these properties depend on density and temperature, both of which depend on the mass of the star. Thus, in order to do this correctly, we need to know what the dominant mechanism for each of these things is for other stars. \\ \\<br /> We showed on homework that at the center of the Sun, free-free opacity and Thomson opacity are roughly of equal importance. Thomson opacity is constant with density and temperature, but $\kappa_{f-f} \propto \rho T^{-7/2}$. For stars more massive than the Sun, the central temperature is larger and the density is lower, meaning Thomson scattering dominates for stars more massive than the Sun, and free-free dominates for stars less massive than the Sun. \\ \\<br /> The energy produced by the CNO cycle is about 1\% of the energy produced by the proton-proton chain in the Sun. Since the CNO cycle scales as $T^{20}$ whereas the p-p chain scales as $T^4$, the CNO cycle is dominant for stars somewhat more massive than the Sun, and the proton-proton chain is dominant for stars the Sun's mass and lower. \\ \\<br /> Gas pressure dominantes for all but the most massive stars. Stars less massive than the Sun have convective envelopes and radiative cores. Stars more massive than the Sun have convective cores and radiative envelopes. Knowing all of this, we can look at the main sequence of mass ranges of stars. \\ \\<br /> First, we'll look at stars less massive than the Sun down to about one third of a solar mass. These have free-free opacity, photons transport the energy, proton-proton chain, and are gas pressure dominated. We know that<br /> <br /> $<br /> L_{\rm rad} \propto M^{5.5} R^{-0.5},<br />$<br /> $<br /> L_{p-p} = \int \epsilon_{p-p} dM_r = L_{\rm rad},<br />$<br /> $<br /> L_{p-p} \sim \epsilon_c M, <br />$<br /> $<br /> L_{p-p} \sim \rho T^{4.5} M, <br />$<br /> $<br /> T_c \propto \frac{M}{R}, <br />$<br /> $<br /> L_{p-p} \sim \frac{M^{6.5}}{R^{7.5}}. <br />$<br /> Equating the radiative losses to the energy generation, <br /> <br /> $<br /> L_{p-p} \sim L_{\rm rad}, <br />$<br /> $<br /> \frac{M^{6.5}}{R^{7.5}} \propto \frac{M^{5.5}}{R^{0.5}}, <br />$<br /> $<br /> R \propto M^{1/7}, <br />$<br /> $<br /> R = R_\odot \left(\frac{M}{M_\odot}\right)^{1/7} <br />$<br /> The last step is possible because we are able to normalize to the Sun. Likewise for central temperature, we find<br /> \]<br /> $<br /> T_c \propto \frac{M}{R} \propto M^{6/7},<br />$<br /> $<br /> T_c \propto (1.5 \times 10^7 K) \left(\frac{M}{M_\odot}\right)^{6/7}.<br />$<br /> The radial change with mass is tiny, so we can just say<br /> $<br /> L = L_\odot \left(\frac{M}{M_\odot}\right)^{5.5}.<br />$<br /> We can also solve for $R$ in terms of $L$, and find<br /> $<br /> R \propto L^{1/40}.<br />$<br /> This is tiny, so we can just say that<br /> $<br /> T_{\rm eff} = 5800 K \left(\frac{L}{L_\odot}\right)^{1/4}.<br />$<br /> This is everything we would want to know for a star that is between roughly 0.3 and 1 solar mass, but outside this mass range, these results do not hold. Let's do this again then for stars that are more massive than the Sun, ranging from roughly 1--50 solar masses. Here, CNO is dominant, gas pressure is dominant, and Thomson opacity is dominant. We showed in lecture that<br /> $<br /> L_{\rm rad} = L_\odot \left(\frac{M}{M_\odot}\right)^3.<br />$<br /> The energy generated by fusion is<br /> <br /> $<br /> L_{\rm fusion} = \int \epsilon dM_r \sim \epsilon_c M, <br />$<br /> $<br /> \epsilon_c \propto \rho T^{20}, <br />$<br /> $<br /> \epsilon_c \propto \frac{M^{21}}{R^{23}}. <br />$<br /> $<br /> L_{\rm fusion} \propto \frac{M^{22}}{R^{23}}.<br />$<br /> <br /> Equating the radiative luminosity to the fusion luminosity gives<br /> $<br /> \frac{M^{22}}{R^{23}} \propto M^3, <br />$<br /> $<br /> R \propto M^{19/23} \propto M^{0.8}, <br />$<br /> $<br /> R = R_\odot \left(\frac{M}{M_\odot}\right)^{0.8}.<br />$<br /> <br /> Using the virial theorem gives the central temperature dependence of<br /> $<br /> T_c = (1.5 \times 10^7 K) \left(\frac{M}{M_\odot}\right)^{0.2}.<br />$<br /> For stars on the CNO cycle, the temperature only needs to change slightly to keep up with the extra luminosity. Finally, Stefan-Boltzmann gives<br /> $<br /> T_{\rm eff} = 5800 K \left(\frac{M}{M_\odot}\right)^{3/8}.<br />$<br /> Or, in terms of luminosity, useful for plotting the HR diagram, <br /> $<br /> T_{\rm eff} = 5800 K \left(\frac{L}{L_\odot}\right)^{1/8}.<br />$<br /> Thus there is a large range of luminosities for stars on the main sequence, despite a relatively small range in effective temperatures. \\ \\<br /> We could do this again, this time for stars more massive than 50 solar masses. These only change in that the gas pressure no longer dominates, with radiation pressure taking over. This gives $L \propto M$. We will not actually do this. Likewise, we could do fully convective stars, but we'll leave that for homework. \\ \\<br /> Instead, let's explain why massive stars have convective cores. Convection sets in when there is a negative entropy gradient. A more useful condition in practice is that entropy sets in when<br /> $<br /> \frac{d \ln T}{d \ln P} &gt; \frac{\gamma - 1}{\gamma}.<br />$<br /> Remember that $\gamma$, the adiabatic index, is for the pressure providing particles. Then, for gas pressure dominated stars, we have $\gamma = 5/3$. We showed that the left hand side can be written<br /> $<br /> \frac{d \ln T}{d \ln P} = \frac{1}{4} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M}.<br />$<br /> Thus convection sets in if<br /> $<br /> \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M} &gt; \frac{8}{5}.<br />$<br /> Rearranging this in terms of the enclosed mass, convection sets in if<br /> $<br /> \frac{M_r}{M} &lt; \frac{5}{8} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r}{L}.<br />$<br /> For the CNO cycle, $L_r \approx L$ even for very small radii. We can get a rough sense of when this is important by looking at the Sun (even though it does not strictly apply for the Sun). We have $P/P_{\rm rad} \sim 3000$, and $L / L_{edd} \sim 4 \times 10^{-5}$. This gives convection for <br /> $<br /> \frac{M_r}{M} &lt; 0.1.<br />$<br /> This is why stars more massive than the Sun have convective cores. The Sun only barely avoids this, because it is dominated by the proton-proton chain, which is slightly less temperature sensitive. The convective core sets in for stars about 1.3 $M_\odot$ and above.<br /> <br /> <br /> <br /> \end{document}<br /> <br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-03T01:06:48Z <p>Jmcbride: directing to proper title</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Hydrostatic Equilibrium]]<br /> #[[Virial Theorem]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Star formation]]<br /> #[[Fusion]]<br /> #[[Main sequence]]<br /> #[[Stars Lecture 13]]<br /> #[[Stars Lecture 14]]<br /> #[[Stars Lecture 15]]<br /> #[[Stars Lecture 16]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Fusion 2011-09-03T01:03:25Z <p>Jmcbride: added more content</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Nuclear Reaction Rates}} \\<br /> The cross section for nuclear reaction depends upon the strong force, and it also depends upon the quantum mechanical probability for tunneling through the Coulomb barrier. These two dependences are independent of one another. The quantum mechanical part also happens to be the part that primarily sets the temperature sensitivity of fusion, and is also the part that is easier to understand. \\ \\<br /> {\bf Tunneling} \\<br /> To understand tunneling, we start by writing down the Schrodinger equation.<br /> $<br /> \left(-\frac{\hbar^2}{2m} \nabla^2 + V\right)\psi = E \psi.<br />$<br /> For a simple step like potential, outside of the step, the potential is V = 0, and the solution is a plane wave, and<br /> $<br /> E = \frac{k^2 \hbar^2}{2m}.<br />$<br /> Here, $k$ is the wavenumber. In the step, the potential is just some constant $V = V_0$, and<br /> $<br /> E - V_0 = \frac{k^2 \hbar^2}{2m}.<br />$<br /> \\<br /> {\bf Large portion missing here (notetaker had to leave lecture)}<br /> <br /> \subsection*{Fusion Reaction Rates} \\<br /> We ended last time by writing down the cross section for fusion in terms of the probability of tunneling and an intrinsic strong nuclear physics part that we will use without trying to understand the detailed physics. This cross section is<br /> $<br /> \sigma(E) = \frac{S(E)}{E} e^{-(E_G/E)^{1/2}}.<br />$<br /> Here, $S(E)$ is the nuclear physics part, $E$ is the energy of both particles, and the expontential term is the probability of tunneling through the Coulomb barrier. We wrote in the last class $E_G$ as<br /> $<br /> E_G = (1 {\rm \;MeV}) Z_1^2 Z_2^2 \frac{m_r}{m_p}.<br />$<br /> This expression can be used to find the probability of a single tunneling event.We want to write down the rate of fusion for a large system of particles. We can start by writing down the rate for lots of particles that all have the same velocity. For distinguishable fusing particles that we will label 1 and 2 with number densities $n_1$ and $n_2$, we can write down a mean free path<br /> $<br /> \ell_2 = \frac{1}{n_1 \sigma}<br />$<br /> and a mean free time<br /> $<br /> \tau_2 = \frac{1}{n_1 \sigma v}.<br />$<br /> The number of reactions per unit volume per unit time is then<br /> $<br /> r_{12} = \frac{n_2}{\tau_2} = n_1 n_2 \sigma v.<br />$<br /> If we average over all possible velocities, the result is<br /> $<br /> r_{12} = n_1 n_2 &lt;\sigma(E) v&gt;.<br />$<br /> To find $&lt;\sigma(E) v&gt;$, we need to integrate over the Maxwell-Boltzmann distribution<br /> $<br /> &lt;\sigma(E) v&gt; = \int d^3v\; prob(v) \sigma(E) v.<br />$<br /> Plugging in (with normalization) the MB distribution, the result is<br /> $<br /> r_{12} = n_1 n_2 \int d^3v \sigma(E) v \left(\frac{m_r}{2 \pi k T}\right)^{3/2} e^{-\frac{\frac{1}{2}m_r v^2}{kT}}.<br />$<br /> We would like to put this entirely in terms of the energy, substituting out the velocity. We'll use<br /> $<br /> E = \frac{1}{2} m_r v^2, <br />$<br /> $<br /> dE = m_r v dv, <br />$<br /> $<br /> d^3v = 4 \pi v^2 dv = 4 \pi \frac{v^2}{v} \frac{dE}{m_r},<br />$<br /> $<br /> v d^3 v = \frac{8 \pi E}{m_r} \frac{dE}{m_r}. <br />$<br /> Now we can rewrite our velocity averaged cross section as <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE E \sigma(E) e^{-E / kT}.<br />$<br /> Now, plug in our original expression for $\sigma(E)$ to find <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE S(E) e^{-(E_G / E)^{1/2}\;-\;E / kT}.<br />$<br /> The tunneling term here favors reactions at high energies, but the Boltzmann term favors reactions at low energies where there are a large number of available particles. Compared to the energy terms in the exponential, $S(E)$ is slowly varying, and we can pull it out of the integral, writing<br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} S(E) I.<br />$<br /> with<br /> $<br /> I = \int_0^\infty e^{-(E_G / E)^{1/2}\;-\;E / kT} dE.<br />$<br /> We can draw a picture to understand roughly what this integral will give us. The MB distribution falls quickly with energy, and the probability of tunneling rises exponentially with energy. The product of the two is at a maximum at some energy $E_0$ that falls part way along the tail of each. We can find this $E_0$ by differentiating the integrand above, which we'll call $f(E)$. Then<br /> $<br /> \frac{df}{dE} = 0 = \frac{1}{kT} - \frac{E_G^{1/2}}{2 E^{3/2}}.<br />$<br /> Solving we find<br /> $<br /> E_0 = \left(\frac{1}{2} E_G^{1/2} kT\right)^{2/3}.<br />$<br /> In terms of the more fundamental quantities that define $E_G$,<br /> $<br /> E_0 = (5.7\; {\rm keV}) Z_1^{2/3} Z_2^{2/3} T_7^{2/3} \left(\frac{m_r}{m_p}\right)^{1/3}.<br />$<br /> This is the energy at which most reactions occur, even though it is far smaller than $E_G$ and far larger than $kT$, thanks to the competition between the two. In order to perform the integral above, we will treat the overlap between the two as a Gaussian (method of steepest descent).<br /> $<br /> f(E) = f(E_0) + \frac{1}{2}(E - E_0)^2 + f^{''}(E_0),<br />$<br /> where<br /> $<br /> f^{''}(E_0) = \frac{3 E_G^{1/2}}{4 E_0^{5/2}}.<br />$<br /> If we put this in our integral $I$, we get a Gaussian integral that we can perform. The result is<br /> $<br /> I = \frac{e^{-f(E_0)} \sqrt{2 \pi}}{\sqrt{f^{''}(E_0}}.<br />$<br /> The final results for our velocity averaged cross section is<br /> $<br /> &lt;\sigma(E) v&gt; = 2.6 S(E_0) \frac{E_G^{1/6}}{(kT)^{2/3} \sqrt{m_r}}e^{-3(E_G/4kT)^{1/3}}.<br />$<br /> \\<br /> Now, we can start to do something more macroscopic. Let's define the energy per unit mass per unit time from fusion as $\epsilon$. Then the total luminosity is<br /> $<br /> L = \int \epsilon dM_r = \int \epsilon 4 \pi r^2 \rho dr.<br />$<br /> Then the change in luminosity over radius is<br /> $<br /> \frac{dL_r}{dr} = 4 \pi r^2 \rho \epsilon.<br />$<br /> We can also define an energy released per reaction as $Q$. Using our rate $r_{12}$ of reactions per volume per time, we can find $\epsilon$, with<br /> $<br /> \epsilon_{12} = \frac{r_{12} Q}{\rho}.<br />$<br /> We want to put everything in terms of mass density, which we will do with<br /> $<br /> n_1 = \frac{X_1 \rho}{m_1}.<br />$<br /> Here, $X_1$ is the fraction of the total mass that is in particle 1. Together, these allow us to write the energy generation from fusion.<br /> $<br /> \epsilon_{12} = \frac{2.6 Q S(E_0) X_1 X_2}{m_1 m_2 \sqrt{m_r} (kT)^{2/3}} \rho E_G^{1/6} e^{-3(E_G / 4 kT)^{1/3}}.<br />$<br /> This expression has units ergs per second per gram, and gives the energy from reactions between particles 1 and 2. It is a function of density, temperature, and composition. Since we do not want to carry this whole expression around most of the time, we would like to just write down a simple proportionality in terms of these fundamental variables. In general, we'll approximate<br /> $<br /> \epsilon \propto \rho^\alpha T^\beta.<br />$<br /> Depending upon the reaction, $\alpha$ and $\beta$ can take different values. For any process involving collisions between two particles, $\alpha = 1$. Finding $\beta$ is a bit more complicated. Our expression for $\epsilon$ is not a simple power law in temperature. We can approximate it though by noticing that<br /> $<br /> \beta = \frac{d \ln \epsilon}{d \ln T}.<br />$<br /> Writing $\epsilon$ in terms of only the temperature dependent quantities and then performing the above differentiation, we find<br /> $<br /> \beta = -\frac{2}{3} + \left(\frac{E_G}{4 k T}\right)^{1/3}.<br />$<br /> Note that this is a function of temperature. This is only valid then when we are considering a small range of temperature. Fortunately, this is generally a fine approximation when considering fusion in stars. For the proton-proton chain at the center of a sun like star, we find $\beta \approx 4.3$. Then<br /> $<br /> \epsilon_{pp} \propto \rho T^{4.3}<br />$<br /> at temperatures near $10^7$ Kelvin. We can make a bit more detailed an estimate of energy from fusing protons to Helium. Again using $T_c \sim 10^7$ K, $\rho \sim 1$ g cm$^{-3}$, the cross section $S(E)$ is of order 1 keV barn, and $Q$ is order 10 MeV. If we put this in our expression for $\epsilon$, we find<br /> $<br /> \epsilon \sim 10^{20} {\rm \;erg/s/g}.<br />$<br /> The luminosity of the Sun is then <br /> $<br /> L = \int dM_r \epsilon \sim \epsilon M_\odot.<br />$<br /> This is<br /> $<br /> L \sim 10^{54} {\rm \; erg/s} \sim 10^{20} L_\odot.<br />$<br /> What happened in our estimate? This is WAY too big. We can think of a couple of resolutions that would give us a more reasonable estimate. One is that the weak interaction is involved somewhere, as it has a cross section that is $10^{20}$ times smaller than the strong interaction. The other possibility is that a proton is fusion with a particle that has a much larger charge, which would suppress the $E_G$ term and give much lowre energy generation rates. As it turns out, both of these solutions exist in nature. The proton-proton chain involves conversion of protons to neutrons, which requires the weak interaction. The CNO cycle produces Helium by fusion of protons with C, N, and O. Using either, we would find an energy generation rate that is in much better agreement with the observed luminosity of the Sun. \\ \\<br /> Let's look at the proton proton chain in more detail. The net reaction is <br /> $<br /> 4 p \rightarrow {}^4{\rm He} + {\rm energy}.<br />$<br /> This energy comes out as photons, kinetic energy of particles, and a small amount as neutrinos. The kinetic energy and photon energy is all shared with the star through collisions, and produces the heat inside of stars. The energy in neutrinos is not useful, as it just leaves the star without providing any heat (and, by extension, prssure support against gravity). \\ \\<br /> As implied already, the key step in the proton proton chain is the step where a proton turns in to a neutron (necessary since there are two neutrons in Helium, and we are starting with only protons). This step gives its identity as a weak interaction away by the presence of a neutrino.<br /> $<br /> p + p \rightarrow {}^2{\rm H} + e^{+} + \nu_e.<br />$<br /> The cross section for this is $S(keV) \approx 3.78 \times 10^{-22}$ keV barn. This is, as promised, much weaker than the typical strong reaction, which has a cross section of order of 1 keV barn. The rest of the reactions needed to produce Helium are easy, involving the strong reaction. First,<br /> $<br /> {}^2{\rm H} + p \rightarrow {}^3{\rm He} + \gamma,<br />$ <br /> which has a cross section of 2.5 $\times 10^{-4}$ keV barn.<br /> and then<br /> $<br /> {}^3{\rm He} + {}^3{\rm He} \rightarrow {}^4{\rm He} + 2p,<br />$<br /> with cross section 5000 keV barn. Note that the first two steps of the reaction must happen twice for the last step to happen once. That this whole process requires multiple steps is a result of the exceedlingly low probability of four protons all happening to be in the same place at the same time. Even so, determining the overall rate is made simpler by the fact that the slowest step in a cycle sets the rate for the entire cycle. Here, our rate limiting step is the first step, when two protons form a deuterium atom. Then we can calculate the energy generation rate for the entire chain with<br /> $<br /> \epsilon_{cycle} = r_{p-p\; step} Q_{cycle} / \rho.<br />$<br /> This fact will be useful in the CNO cycle as well, which has many steps, but is also rate limited by a single step, allowing us to easily write down the energy generation rate. Using our previous (complicated) expression for energy generation and applying it to the proton-proton step, we find the energy for the entire chain to be<br /> $<br /> \epsilon_{pp} \propto \rho T^{-2/3} e^{-15.7 T_7^{-1/3}}.<br />$<br /> With the constants, <br /> $<br /> \epsilon_{pp} = (5 \times 10^5) \frac{\rho X^2}{T^{2/3}} e^{-15.7 T_7^{-1/3}} {\rm erg/s/g}.<br />$<br /> We can use this to estimate the central temperature of the Sun, using the known luminosity. <br /> $<br /> L = \int \epsilon dM \sim \epsilon(center) M_\odot, <br />$<br /> $<br /> L_\odot \sim 10^7 \frac{M_\odot}{T_7^{2/3}} e^{-15.7 T_7^{-1/3}}, <br />$<br /> $<br /> T_c \approx 10^7 K.<br />$<br /> This is pretty good, and is (fortunately) self consistent.<br /> \subsection*{Hydrogen Fusion in Stars} \\<br /> One method of fusing hydrogen in the centers of stars is called the proton-proton chain.<br /> $<br /> p + p \rightarrow {}^2H + e^{+} + \nu_e<br />$<br /> $<br /> {}^2H + p \rightarrow {}^3He + \gamma <br />$<br /> $<br /> {}^3He + {}^3He \rightarrow {}^4He + 2p <br />$<br /> <br /> This process is important in lower mass stars. For higher mass stars, the CNO cycle becomes important. Carbon, nitrogen, and oxygen are not actually fused in this process. Instead, they are used as catalysts, and the reaction has the exact same input (four protons) and output (one helium atom) as the proton-proton chain does. The steps of the CNO cycle are<br /> \begin{eqnarray*}<br /> {}^{12}C + p \rightarrow {}^{13}N + \gamma <br /> {}^{13}N \rightarrow {}^{13}C + e^{+} + \nu_e <br /> {}^{13}C + p \rightarrow {}^{14}N + \gamma <br /> {}^{14}N + p \rightarrow {}^{15}O + \gamma <br /> {}^{15}O \rightarrow {}^{15}N + e^{+} + \nu_e <br /> {}^{15}N + p \rightarrow {}^{12}C + {}^4He. <br /> \]<br /> Considering our results for the Gamow Energy, which sets the probability of tunneling, how can the CNO cycle be competitive with the proton-proton chain? The charge involved is much larger, so the Gamow Energy of any of the steps are of order 50 MeV. This is five orders of magnitude larger than the typical energy at the center of a star, which is of order 1 keV. At $10^7$ K, the tunneling probability is $10^{-7}$ for the proton-proton chain, and it is a staggering $10^{-31}$ for the CNO cycle. The reason that CNO can be competitive turns out to be because the cross section in every step of the CNO cycle that requires fusing is the strong force, whereas the first step in the proton-proton chain is a weak interaction, with a significantly lower cross section. They happen to be different by roughly $10^{24}$, which roughly cancels the difference in tunneling probability. (Question: what about abundance?) The energy generation rate for the CNO cycle is set by the slowest step, as is the case for any reaction. The slowest step is the fourth step, and gives<br /> $<br /> \epsilon_{CNO} \approx (4 \times 10^27) \frac{\rho}{T_7^{2/3}} X Z e^{-70.7 T_7^{-1/3}} {\rm \;erg/g/s}.<br />$<br /> We can do the same trick as with the proton-proton chain to get a simple power law scaling, and find that <br /> $<br /> \beta = \frac{-2}{3} + \frac{23.6}{T_7^{1/3}},<br />$<br /> with <br /> $<br /> \epsilon \propto \rho T^\beta.<br />$<br /> This is phenomenally sensitive to energy, due to being way out on the tail of the Maxwell-Boltzmann distribution. Raising the temperature just a little bit will cause the number of particles at that energy to rise significantly. While the proton-proton chain was also sensitive to temperature, it is not nearly as sensitive as the CNO cycle. This means that at high temperatures, CNO dominates, and at low temperatures, the proton-proton chain dominates. It turns out that they are roughly equal for stars that are a bit more massive than the Sun. In the Sun, 99\% of the luminosity is produced by the proton-proton chain. \\ \\<br /> \subsection*{Observational Evidence for Fusion in the Sun} \\<br /> Detections of neutrinos from the Sun provide evidence that the center of the Sun is undergoing fusion. Neutrinos are very light particles, and for a long time were believed to be massless. Now, they are thought to have a mass of order (or smaller than) 0.1 eV, which is six orders of magnitude lighter than an electron. They have no charge (or color), meaning they only are affected by the weak force and gravity. There are three types of neutrinos, corresponding to the three leptons: the electron, muon, and tau particle each has a corresponding neutrino. We are only really concerned with electron neutrinos though, since that is what stars provide in fusion (which can be confirmed by looking back at the reactions we wrote down). There are multiple branches in the proton-proton chain. We talked about the dominant branch, but the other branches produce different energy neutrinos. \\<br /> The typical cross section for an interaction between neutrinos and matter is <br /> $<br /> \sigma \sim 10^{-44} \left(\frac{E_\nu}{m_e c^2}\right)^2 {\rm \; cm^2}.<br />$<br /> Note again the twenty orders of magnitude difference between this cross section and something representative of the electromagnetic force, like the Thomson cross section. We can use this to calculate the mean free path of neutrinos in the Sun.<br /> $<br /> \ell = \frac{1}{n \sigma}.<br />$<br /> Using $E_\nu \sim$ MeV, and the average number density in the Sun, we find<br /> $<br /> \ell \sim 10^9 R_\odot.<br />$<br /> This means that a neutrino typically leaves the Sun without interacting at all. A neutrino that we can detect on Earth thus tells us about the conditions at the center of the Sun, where the neutrino was produced. Most importantly, the energy of neutrinos depends on the reaction that produced them, so the energy of neutrinos that we detect tells us about fusion in the center of the Sun. Of course, since a neutrino gets out of the Sun unimpeded, detecting a neutrino on Earth is challenging. The first experiment to do this used cleaning fluid in a huge tank underground, and resulted in a Nobel Prize for Ray Davis in 2002. This cleaning fluid had chlorine, and the reaction of interest was<br /> $<br /> {}^{37}Cl + \nu_e \rightarrow {}^{37}Ar + e^{-}.<br />$<br /> Every day, $10^{22}$ neutrinos go through the tank, but on average one neutrino is detected per day. This reaction only operates for neutrinos more energetic than 0.814 MeV, so this experiment was only able to detect neutrinos from some of the less common fusion reactions, as those from the proton-proton chain were of lower energy. \\<br /> This experiment began in the 1960s, and wound up producing what is called the Solar Neutrino problem. The number of neutrinos detected was one third of the number expected. There was debate for many years between astrophysicists and particle physicists. The resolution was finally provided by Sudbury Neutrino Observatory. The SNO used heavy water, which has deuterium rather than lone protons in the water molecule. An incoming neutrino can break up the deuterium in to two protons and a positron:<br /> $<br /> \nu_e + D \rightarrow p + p + e^{-}.<br />$<br /> This experiment could also detect mu and tau neutrinos, with<br /> $<br /> \nu + D \rightarrow p + n + \nu.<br />$<br /> This measurement confirmed that the total number of neutrinos produced is the total number expected. The total was expected to be made up entirely of electron neutrinos though, and this experiment showed that only one third of them were electron neutrinos. It turns out that since neutrinos have mass, they can oscillate between different neutrino types. That means that despite the Sun producing only electron neutrinos, we can observe mu and tau neutrinos from the Sun. Thus the neutrino observation experiments provided direct confirmation of solar models, and also proved that neutrinos have mass. \\ \\<br /> <br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-03T00:58:10Z <p>Jmcbride: fixed link</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Hydrostatic Equilibrium]]<br /> #[[Virial Theorem]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Star formation]]<br /> #[[Fusion]]<br /> #[[Stars Lecture 11]]<br /> #[[Stars Lecture 12]]<br /> #[[Stars Lecture 13]]<br /> #[[Stars Lecture 14]]<br /> #[[Stars Lecture 15]]<br /> #[[Stars Lecture 16]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Stars Lecture 9 2011-09-03T00:57:56Z <p>Jmcbride: moved Stars Lecture 9 to Fusion:&amp;#32;changed to topic</p> <hr /> <div>#REDIRECT [[Fusion]]</div> Jmcbride Fusion 2011-09-03T00:57:56Z <p>Jmcbride: moved Stars Lecture 9 to Fusion:&amp;#32;changed to topic</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Nuclear Reaction Rates}} \\<br /> The cross section for nuclear reaction depends upon the strong force, and it also depends upon the quantum mechanical probability for tunneling through the Coulomb barrier. These two dependences are independent of one another. The quantum mechanical part also happens to be the part that primarily sets the temperature sensitivity of fusion, and is also the part that is easier to understand. \\ \\<br /> {\bf Tunneling} \\<br /> To understand tunneling, we start by writing down the Schrodinger equation.<br /> $<br /> \left(-\frac{\hbar^2}{2m} \nabla^2 + V\right)\psi = E \psi.<br />$<br /> For a simple step like potential, outside of the step, the potential is V = 0, and the solution is a plane wave, and<br /> $<br /> E = \frac{k^2 \hbar^2}{2m}.<br />$<br /> Here, $k$ is the wavenumber. In the step, the potential is just some constant $V = V_0$, and<br /> $<br /> E - V_0 = \frac{k^2 \hbar^2}{2m}.<br />$<br /> \\<br /> {\bf Large portion missing here (notetaker had to leave lecture)}<br /> <br /> \subsection*{Fusion Reaction Rates} \\<br /> We ended last time by writing down the cross section for fusion in terms of the probability of tunneling and an intrinsic strong nuclear physics part that we will use without trying to understand the detailed physics. This cross section is<br /> $<br /> \sigma(E) = \frac{S(E)}{E} e^{-(E_G/E)^{1/2}}.<br />$<br /> Here, $S(E)$ is the nuclear physics part, $E$ is the energy of both particles, and the expontential term is the probability of tunneling through the Coulomb barrier. We wrote in the last class $E_G$ as<br /> $<br /> E_G = (1 {\rm \;MeV}) Z_1^2 Z_2^2 \frac{m_r}{m_p}.<br />$<br /> This expression can be used to find the probability of a single tunneling event.We want to write down the rate of fusion for a large system of particles. We can start by writing down the rate for lots of particles that all have the same velocity. For distinguishable fusing particles that we will label 1 and 2 with number densities $n_1$ and $n_2$, we can write down a mean free path<br /> $<br /> \ell_2 = \frac{1}{n_1 \sigma}<br />$<br /> and a mean free time<br /> $<br /> \tau_2 = \frac{1}{n_1 \sigma v}.<br />$<br /> The number of reactions per unit volume per unit time is then<br /> $<br /> r_{12} = \frac{n_2}{\tau_2} = n_1 n_2 \sigma v.<br />$<br /> If we average over all possible velocities, the result is<br /> $<br /> r_{12} = n_1 n_2 &lt;\sigma(E) v&gt;.<br />$<br /> To find $&lt;\sigma(E) v&gt;$, we need to integrate over the Maxwell-Boltzmann distribution<br /> $<br /> &lt;\sigma(E) v&gt; = \int d^3v\; prob(v) \sigma(E) v.<br />$<br /> Plugging in (with normalization) the MB distribution, the result is<br /> $<br /> r_{12} = n_1 n_2 \int d^3v \sigma(E) v \left(\frac{m_r}{2 \pi k T}\right)^{3/2} e^{-\frac{\frac{1}{2}m_r v^2}{kT}}.<br />$<br /> We would like to put this entirely in terms of the energy, substituting out the velocity. We'll use<br /> $<br /> E = \frac{1}{2} m_r v^2, <br />$<br /> $<br /> dE = m_r v dv, <br />$<br /> $<br /> d^3v = 4 \pi v^2 dv = 4 \pi \frac{v^2}{v} \frac{dE}{m_r},<br />$<br /> $<br /> v d^3 v = \frac{8 \pi E}{m_r} \frac{dE}{m_r}. <br />$<br /> Now we can rewrite our velocity averaged cross section as <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE E \sigma(E) e^{-E / kT}.<br />$<br /> Now, plug in our original expression for $\sigma(E)$ to find <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE S(E) e^{-(E_G / E)^{1/2}\;-\;E / kT}.<br />$<br /> The tunneling term here favors reactions at high energies, but the Boltzmann term favors reactions at low energies where there are a large number of available particles. Compared to the energy terms in the exponential, $S(E)$ is slowly varying, and we can pull it out of the integral, writing<br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} S(E) I.<br />$<br /> with<br /> $<br /> I = \int_0^\infty e^{-(E_G / E)^{1/2}\;-\;E / kT} dE.<br />$<br /> We can draw a picture to understand roughly what this integral will give us. The MB distribution falls quickly with energy, and the probability of tunneling rises exponentially with energy. The product of the two is at a maximum at some energy $E_0$ that falls part way along the tail of each. We can find this $E_0$ by differentiating the integrand above, which we'll call $f(E)$. Then<br /> $<br /> \frac{df}{dE} = 0 = \frac{1}{kT} - \frac{E_G^{1/2}}{2 E^{3/2}}.<br />$<br /> Solving we find<br /> $<br /> E_0 = \left(\frac{1}{2} E_G^{1/2} kT\right)^{2/3}.<br />$<br /> In terms of the more fundamental quantities that define $E_G$,<br /> $<br /> E_0 = (5.7\; {\rm keV}) Z_1^{2/3} Z_2^{2/3} T_7^{2/3} \left(\frac{m_r}{m_p}\right)^{1/3}.<br />$<br /> This is the energy at which most reactions occur, even though it is far smaller than $E_G$ and far larger than $kT$, thanks to the competition between the two. In order to perform the integral above, we will treat the overlap between the two as a Gaussian (method of steepest descent).<br /> $<br /> f(E) = f(E_0) + \frac{1}{2}(E - E_0)^2 + f^{''}(E_0),<br />$<br /> where<br /> $<br /> f^{''}(E_0) = \frac{3 E_G^{1/2}}{4 E_0^{5/2}}.<br />$<br /> If we put this in our integral $I$, we get a Gaussian integral that we can perform. The result is<br /> $<br /> I = \frac{e^{-f(E_0)} \sqrt{2 \pi}}{\sqrt{f^{''}(E_0}}.<br />$<br /> The final results for our velocity averaged cross section is<br /> $<br /> &lt;\sigma(E) v&gt; = 2.6 S(E_0) \frac{E_G^{1/6}}{(kT)^{2/3} \sqrt{m_r}}e^{-3(E_G/4kT)^{1/3}}.<br />$<br /> \\<br /> Now, we can start to do something more macroscopic. Let's define the energy per unit mass per unit time from fusion as $\epsilon$. Then the total luminosity is<br /> $<br /> L = \int \epsilon dM_r = \int \epsilon 4 \pi r^2 \rho dr.<br />$<br /> Then the change in luminosity over radius is<br /> $<br /> \frac{dL_r}{dr} = 4 \pi r^2 \rho \epsilon.<br />$<br /> We can also define an energy released per reaction as $Q$. Using our rate $r_{12}$ of reactions per volume per time, we can find $\epsilon$, with<br /> $<br /> \epsilon_{12} = \frac{r_{12} Q}{\rho}.<br />$<br /> We want to put everything in terms of mass density, which we will do with<br /> $<br /> n_1 = \frac{X_1 \rho}{m_1}.<br />$<br /> Here, $X_1$ is the fraction of the total mass that is in particle 1. Together, these allow us to write the energy generation from fusion.<br /> $<br /> \epsilon_{12} = \frac{2.6 Q S(E_0) X_1 X_2}{m_1 m_2 \sqrt{m_r} (kT)^{2/3}} \rho E_G^{1/6} e^{-3(E_G / 4 kT)^{1/3}}.<br />$<br /> This expression has units ergs per second per gram, and gives the energy from reactions between particles 1 and 2. It is a function of density, temperature, and composition. Since we do not want to carry this whole expression around most of the time, we would like to just write down a simple proportionality in terms of these fundamental variables. In general, we'll approximate<br /> $<br /> \epsilon \propto \rho^\alpha T^\beta.<br />$<br /> Depending upon the reaction, $\alpha$ and $\beta$ can take different values. For any process involving collisions between two particles, $\alpha = 1$. Finding $\beta$ is a bit more complicated. Our expression for $\epsilon$ is not a simple power law in temperature. We can approximate it though by noticing that<br /> $<br /> \beta = \frac{d \ln \epsilon}{d \ln T}.<br />$<br /> Writing $\epsilon$ in terms of only the temperature dependent quantities and then performing the above differentiation, we find<br /> $<br /> \beta = -\frac{2}{3} + \left(\frac{E_G}{4 k T}\right)^{1/3}.<br />$<br /> Note that this is a function of temperature. This is only valid then when we are considering a small range of temperature. Fortunately, this is generally a fine approximation when considering fusion in stars. For the proton-proton chain at the center of a sun like star, we find $\beta \approx 4.3$. Then<br /> $<br /> \epsilon_{pp} \propto \rho T^{4.3}<br />$<br /> at temperatures near $10^7$ Kelvin. We can make a bit more detailed an estimate of energy from fusing protons to Helium. Again using $T_c \sim 10^7$ K, $\rho \sim 1$ g cm$^{-3}$, the cross section $S(E)$ is of order 1 keV barn, and $Q$ is order 10 MeV. If we put this in our expression for $\epsilon$, we find<br /> $<br /> \epsilon \sim 10^{20} {\rm \;erg/s/g}.<br />$<br /> The luminosity of the Sun is then <br /> $<br /> L = \int dM_r \epsilon \sim \epsilon M_\odot.<br />$<br /> This is<br /> $<br /> L \sim 10^{54} {\rm \; erg/s} \sim 10^{20} L_\odot.<br />$<br /> What happened in our estimate? This is WAY too big. We can think of a couple of resolutions that would give us a more reasonable estimate. One is that the weak interaction is involved somewhere, as it has a cross section that is $10^{20}$ times smaller than the strong interaction. The other possibility is that a proton is fusion with a particle that has a much larger charge, which would suppress the $E_G$ term and give much lowre energy generation rates. As it turns out, both of these solutions exist in nature. The proton-proton chain involves conversion of protons to neutrons, which requires the weak interaction. The CNO cycle produces Helium by fusion of protons with C, N, and O. Using either, we would find an energy generation rate that is in much better agreement with the observed luminosity of the Sun. \\ \\<br /> Let's look at the proton proton chain in more detail. The net reaction is <br /> $<br /> 4 p \rightarrow {}^4{\rm He} + {\rm energy}.<br />$<br /> This energy comes out as photons, kinetic energy of particles, and a small amount as neutrinos. The kinetic energy and photon energy is all shared with the star through collisions, and produces the heat inside of stars. The energy in neutrinos is not useful, as it just leaves the star without providing any heat (and, by extension, prssure support against gravity). \\ \\<br /> As implied already, the key step in the proton proton chain is the step where a proton turns in to a neutron (necessary since there are two neutrons in Helium, and we are starting with only protons). This step gives its identity as a weak interaction away by the presence of a neutrino.<br /> $<br /> p + p \rightarrow {}^2{\rm H} + e^{+} + \nu_e.<br />$<br /> The cross section for this is $S(keV) \approx 3.78 \times 10^{-22}$ keV barn. This is, as promised, much weaker than the typical strong reaction, which has a cross section of order of 1 keV barn. The rest of the reactions needed to produce Helium are easy, involving the strong reaction. First,<br /> $<br /> {}^2{\rm H} + p \rightarrow {}^3{\rm He} + \gamma,<br />$ <br /> which has a cross section of 2.5 $\times 10^{-4}$ keV barn.<br /> and then<br /> $<br /> {}^3{\rm He} + {}^3{\rm He} \rightarrow {}^4{\rm He} + 2p,<br />$<br /> with cross section 5000 keV barn. Note that the first two steps of the reaction must happen twice for the last step to happen once. That this whole process requires multiple steps is a result of the exceedlingly low probability of four protons all happening to be in the same place at the same time. Even so, determining the overall rate is made simpler by the fact that the slowest step in a cycle sets the rate for the entire cycle. Here, our rate limiting step is the first step, when two protons form a deuterium atom. Then we can calculate the energy generation rate for the entire chain with<br /> $<br /> \epsilon_{cycle} = r_{p-p\; step} Q_{cycle} / \rho.<br />$<br /> This fact will be useful in the CNO cycle as well, which has many steps, but is also rate limited by a single step, allowing us to easily write down the energy generation rate. Using our previous (complicated) expression for energy generation and applying it to the proton-proton step, we find the energy for the entire chain to be<br /> $<br /> \epsilon_{pp} \propto \rho T^{-2/3} e^{-15.7 T_7^{-1/3}}.<br />$<br /> With the constants, <br /> $<br /> \epsilon_{pp} = (5 \times 10^5) \frac{\rho X^2}{T^{2/3}} e^{-15.7 T_7^{-1/3}} {\rm erg/s/g}.<br />$<br /> We can use this to estimate the central temperature of the Sun, using the known luminosity. <br /> $<br /> L = \int \epsilon dM \sim \epsilon(center) M_\odot, <br />$<br /> $<br /> L_\odot \sim 10^7 \frac{M_\odot}{T_7^{2/3}} e^{-15.7 T_7^{-1/3}}, <br />$<br /> $<br /> T_c \approx 10^7 K.<br />$<br /> This is pretty good, and is (fortunately) self consistent.<br /> <br /> &lt;/latex&gt;</div> Jmcbride Fusion 2011-09-03T00:56:34Z <p>Jmcbride: created page</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Nuclear Reaction Rates}} \\<br /> The cross section for nuclear reaction depends upon the strong force, and it also depends upon the quantum mechanical probability for tunneling through the Coulomb barrier. These two dependences are independent of one another. The quantum mechanical part also happens to be the part that primarily sets the temperature sensitivity of fusion, and is also the part that is easier to understand. \\ \\<br /> {\bf Tunneling} \\<br /> To understand tunneling, we start by writing down the Schrodinger equation.<br /> $<br /> \left(-\frac{\hbar^2}{2m} \nabla^2 + V\right)\psi = E \psi.<br />$<br /> For a simple step like potential, outside of the step, the potential is V = 0, and the solution is a plane wave, and<br /> $<br /> E = \frac{k^2 \hbar^2}{2m}.<br />$<br /> Here, $k$ is the wavenumber. In the step, the potential is just some constant $V = V_0$, and<br /> $<br /> E - V_0 = \frac{k^2 \hbar^2}{2m}.<br />$<br /> \\<br /> {\bf Large portion missing here (notetaker had to leave lecture)}<br /> <br /> \subsection*{Fusion Reaction Rates} \\<br /> We ended last time by writing down the cross section for fusion in terms of the probability of tunneling and an intrinsic strong nuclear physics part that we will use without trying to understand the detailed physics. This cross section is<br /> $<br /> \sigma(E) = \frac{S(E)}{E} e^{-(E_G/E)^{1/2}}.<br />$<br /> Here, $S(E)$ is the nuclear physics part, $E$ is the energy of both particles, and the expontential term is the probability of tunneling through the Coulomb barrier. We wrote in the last class $E_G$ as<br /> $<br /> E_G = (1 {\rm \;MeV}) Z_1^2 Z_2^2 \frac{m_r}{m_p}.<br />$<br /> This expression can be used to find the probability of a single tunneling event.We want to write down the rate of fusion for a large system of particles. We can start by writing down the rate for lots of particles that all have the same velocity. For distinguishable fusing particles that we will label 1 and 2 with number densities $n_1$ and $n_2$, we can write down a mean free path<br /> $<br /> \ell_2 = \frac{1}{n_1 \sigma}<br />$<br /> and a mean free time<br /> $<br /> \tau_2 = \frac{1}{n_1 \sigma v}.<br />$<br /> The number of reactions per unit volume per unit time is then<br /> $<br /> r_{12} = \frac{n_2}{\tau_2} = n_1 n_2 \sigma v.<br />$<br /> If we average over all possible velocities, the result is<br /> $<br /> r_{12} = n_1 n_2 &lt;\sigma(E) v&gt;.<br />$<br /> To find $&lt;\sigma(E) v&gt;$, we need to integrate over the Maxwell-Boltzmann distribution<br /> $<br /> &lt;\sigma(E) v&gt; = \int d^3v\; prob(v) \sigma(E) v.<br />$<br /> Plugging in (with normalization) the MB distribution, the result is<br /> $<br /> r_{12} = n_1 n_2 \int d^3v \sigma(E) v \left(\frac{m_r}{2 \pi k T}\right)^{3/2} e^{-\frac{\frac{1}{2}m_r v^2}{kT}}.<br />$<br /> We would like to put this entirely in terms of the energy, substituting out the velocity. We'll use<br /> $<br /> E = \frac{1}{2} m_r v^2, <br />$<br /> $<br /> dE = m_r v dv, <br />$<br /> $<br /> d^3v = 4 \pi v^2 dv = 4 \pi \frac{v^2}{v} \frac{dE}{m_r},<br />$<br /> $<br /> v d^3 v = \frac{8 \pi E}{m_r} \frac{dE}{m_r}. <br />$<br /> Now we can rewrite our velocity averaged cross section as <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE E \sigma(E) e^{-E / kT}.<br />$<br /> Now, plug in our original expression for $\sigma(E)$ to find <br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} \int dE S(E) e^{-(E_G / E)^{1/2}\;-\;E / kT}.<br />$<br /> The tunneling term here favors reactions at high energies, but the Boltzmann term favors reactions at low energies where there are a large number of available particles. Compared to the energy terms in the exponential, $S(E)$ is slowly varying, and we can pull it out of the integral, writing<br /> $<br /> &lt;\sigma(E) v&gt; = \left(\frac{2}{kT}\right)^{3/2} \frac{1}{\sqrt{\pi m_r}} S(E) I.<br />$<br /> with<br /> $<br /> I = \int_0^\infty e^{-(E_G / E)^{1/2}\;-\;E / kT} dE.<br />$<br /> We can draw a picture to understand roughly what this integral will give us. The MB distribution falls quickly with energy, and the probability of tunneling rises exponentially with energy. The product of the two is at a maximum at some energy $E_0$ that falls part way along the tail of each. We can find this $E_0$ by differentiating the integrand above, which we'll call $f(E)$. Then<br /> $<br /> \frac{df}{dE} = 0 = \frac{1}{kT} - \frac{E_G^{1/2}}{2 E^{3/2}}.<br />$<br /> Solving we find<br /> $<br /> E_0 = \left(\frac{1}{2} E_G^{1/2} kT\right)^{2/3}.<br />$<br /> In terms of the more fundamental quantities that define $E_G$,<br /> $<br /> E_0 = (5.7\; {\rm keV}) Z_1^{2/3} Z_2^{2/3} T_7^{2/3} \left(\frac{m_r}{m_p}\right)^{1/3}.<br />$<br /> This is the energy at which most reactions occur, even though it is far smaller than $E_G$ and far larger than $kT$, thanks to the competition between the two. In order to perform the integral above, we will treat the overlap between the two as a Gaussian (method of steepest descent).<br /> $<br /> f(E) = f(E_0) + \frac{1}{2}(E - E_0)^2 + f^{''}(E_0),<br />$<br /> where<br /> $<br /> f^{''}(E_0) = \frac{3 E_G^{1/2}}{4 E_0^{5/2}}.<br />$<br /> If we put this in our integral $I$, we get a Gaussian integral that we can perform. The result is<br /> $<br /> I = \frac{e^{-f(E_0)} \sqrt{2 \pi}}{\sqrt{f^{''}(E_0}}.<br />$<br /> The final results for our velocity averaged cross section is<br /> $<br /> &lt;\sigma(E) v&gt; = 2.6 S(E_0) \frac{E_G^{1/6}}{(kT)^{2/3} \sqrt{m_r}}e^{-3(E_G/4kT)^{1/3}}.<br />$<br /> \\<br /> Now, we can start to do something more macroscopic. Let's define the energy per unit mass per unit time from fusion as $\epsilon$. Then the total luminosity is<br /> $<br /> L = \int \epsilon dM_r = \int \epsilon 4 \pi r^2 \rho dr.<br />$<br /> Then the change in luminosity over radius is<br /> $<br /> \frac{dL_r}{dr} = 4 \pi r^2 \rho \epsilon.<br />$<br /> We can also define an energy released per reaction as $Q$. Using our rate $r_{12}$ of reactions per volume per time, we can find $\epsilon$, with<br /> $<br /> \epsilon_{12} = \frac{r_{12} Q}{\rho}.<br />$<br /> We want to put everything in terms of mass density, which we will do with<br /> $<br /> n_1 = \frac{X_1 \rho}{m_1}.<br />$<br /> Here, $X_1$ is the fraction of the total mass that is in particle 1. Together, these allow us to write the energy generation from fusion.<br /> $<br /> \epsilon_{12} = \frac{2.6 Q S(E_0) X_1 X_2}{m_1 m_2 \sqrt{m_r} (kT)^{2/3}} \rho E_G^{1/6} e^{-3(E_G / 4 kT)^{1/3}}.<br />$<br /> This expression has units ergs per second per gram, and gives the energy from reactions between particles 1 and 2. It is a function of density, temperature, and composition. Since we do not want to carry this whole expression around most of the time, we would like to just write down a simple proportionality in terms of these fundamental variables. In general, we'll approximate<br /> $<br /> \epsilon \propto \rho^\alpha T^\beta.<br />$<br /> Depending upon the reaction, $\alpha$ and $\beta$ can take different values. For any process involving collisions between two particles, $\alpha = 1$. Finding $\beta$ is a bit more complicated. Our expression for $\epsilon$ is not a simple power law in temperature. We can approximate it though by noticing that<br /> $<br /> \beta = \frac{d \ln \epsilon}{d \ln T}.<br />$<br /> Writing $\epsilon$ in terms of only the temperature dependent quantities and then performing the above differentiation, we find<br /> $<br /> \beta = -\frac{2}{3} + \left(\frac{E_G}{4 k T}\right)^{1/3}.<br />$<br /> Note that this is a function of temperature. This is only valid then when we are considering a small range of temperature. Fortunately, this is generally a fine approximation when considering fusion in stars. For the proton-proton chain at the center of a sun like star, we find $\beta \approx 4.3$. Then<br /> $<br /> \epsilon_{pp} \propto \rho T^{4.3}<br />$<br /> at temperatures near $10^7$ Kelvin. We can make a bit more detailed an estimate of energy from fusing protons to Helium. Again using $T_c \sim 10^7$ K, $\rho \sim 1$ g cm$^{-3}$, the cross section $S(E)$ is of order 1 keV barn, and $Q$ is order 10 MeV. If we put this in our expression for $\epsilon$, we find<br /> $<br /> \epsilon \sim 10^{20} {\rm \;erg/s/g}.<br />$<br /> The luminosity of the Sun is then <br /> $<br /> L = \int dM_r \epsilon \sim \epsilon M_\odot.<br />$<br /> This is<br /> $<br /> L \sim 10^{54} {\rm \; erg/s} \sim 10^{20} L_\odot.<br />$<br /> What happened in our estimate? This is WAY too big. We can think of a couple of resolutions that would give us a more reasonable estimate. One is that the weak interaction is involved somewhere, as it has a cross section that is $10^{20}$ times smaller than the strong interaction. The other possibility is that a proton is fusion with a particle that has a much larger charge, which would suppress the $E_G$ term and give much lowre energy generation rates. As it turns out, both of these solutions exist in nature. The proton-proton chain involves conversion of protons to neutrons, which requires the weak interaction. The CNO cycle produces Helium by fusion of protons with C, N, and O. Using either, we would find an energy generation rate that is in much better agreement with the observed luminosity of the Sun. \\ \\<br /> Let's look at the proton proton chain in more detail. The net reaction is <br /> $<br /> 4 p \rightarrow {}^4{\rm He} + {\rm energy}.<br />$<br /> This energy comes out as photons, kinetic energy of particles, and a small amount as neutrinos. The kinetic energy and photon energy is all shared with the star through collisions, and produces the heat inside of stars. The energy in neutrinos is not useful, as it just leaves the star without providing any heat (and, by extension, prssure support against gravity). \\ \\<br /> As implied already, the key step in the proton proton chain is the step where a proton turns in to a neutron (necessary since there are two neutrons in Helium, and we are starting with only protons). This step gives its identity as a weak interaction away by the presence of a neutrino.<br /> $<br /> p + p \rightarrow {}^2{\rm H} + e^{+} + \nu_e.<br />$<br /> The cross section for this is $S(keV) \approx 3.78 \times 10^{-22}$ keV barn. This is, as promised, much weaker than the typical strong reaction, which has a cross section of order of 1 keV barn. The rest of the reactions needed to produce Helium are easy, involving the strong reaction. First,<br /> $<br /> {}^2{\rm H} + p \rightarrow {}^3{\rm He} + \gamma,<br />$ <br /> which has a cross section of 2.5 $\times 10^{-4}$ keV barn.<br /> and then<br /> $<br /> {}^3{\rm He} + {}^3{\rm He} \rightarrow {}^4{\rm He} + 2p,<br />$<br /> with cross section 5000 keV barn. Note that the first two steps of the reaction must happen twice for the last step to happen once. That this whole process requires multiple steps is a result of the exceedlingly low probability of four protons all happening to be in the same place at the same time. Even so, determining the overall rate is made simpler by the fact that the slowest step in a cycle sets the rate for the entire cycle. Here, our rate limiting step is the first step, when two protons form a deuterium atom. Then we can calculate the energy generation rate for the entire chain with<br /> $<br /> \epsilon_{cycle} = r_{p-p\; step} Q_{cycle} / \rho.<br />$<br /> This fact will be useful in the CNO cycle as well, which has many steps, but is also rate limited by a single step, allowing us to easily write down the energy generation rate. Using our previous (complicated) expression for energy generation and applying it to the proton-proton step, we find the energy for the entire chain to be<br /> $<br /> \epsilon_{pp} \propto \rho T^{-2/3} e^{-15.7 T_7^{-1/3}}.<br />$<br /> With the constants, <br /> $<br /> \epsilon_{pp} = (5 \times 10^5) \frac{\rho X^2}{T^{2/3}} e^{-15.7 T_7^{-1/3}} {\rm erg/s/g}.<br />$<br /> We can use this to estimate the central temperature of the Sun, using the known luminosity. <br /> $<br /> L = \int \epsilon dM \sim \epsilon(center) M_\odot, <br />$<br /> $<br /> L_\odot \sim 10^7 \frac{M_\odot}{T_7^{2/3}} e^{-15.7 T_7^{-1/3}}, <br />$<br /> $<br /> T_c \approx 10^7 K.<br />$<br /> This is pretty good, and is (fortunately) self consistent.<br /> <br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-02T23:57:47Z <p>Jmcbride: fixed links</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Hydrostatic Equilibrium]]<br /> #[[Virial Theorem]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Star formation]]<br /> #[[Stars Lecture 9]]<br /> #[[Stars Lecture 10]]<br /> #[[Stars Lecture 11]]<br /> #[[Stars Lecture 12]]<br /> #[[Stars Lecture 13]]<br /> #[[Stars Lecture 14]]<br /> #[[Stars Lecture 15]]<br /> #[[Stars Lecture 16]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Stars Lecture 8 2011-09-02T23:57:26Z <p>Jmcbride: moved Stars Lecture 8 to Star formation:&amp;#32;changed name to topic</p> <hr /> <div>#REDIRECT [[Star formation]]</div> Jmcbride Star formation 2011-09-02T23:57:26Z <p>Jmcbride: moved Stars Lecture 8 to Star formation:&amp;#32;changed name to topic</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Star Formation} \\<br /> Stars form out of cold gas in galaxies. The gas in a galaxy like the Milky Way is in a flat sheet, not unlike the dimensions of a stack of a few pieces of paper. For the Milky Way, radial size of the disk is roughly 10 kiloparsecs, where a parsec is roughly three lightyears. The height of the disk is some hundreds of parsecs, so roughly one tenth, or less, than the radial extent. The radial extent is a trillion times larger than the Sun, which gives a hint of the problem of star formation. How does gas get taken from large scales and turned in to a (relatively) small object like a star? Another way to see this issue is to look at the mean density of the galaxy, and the mean density of the Sun. The interstellar medium (ISM) has a mean density of roughly 1 particle per cubic centimeter. The Sun has a density of one {\em gram} per cubic centimeter. Since most things are hydrogen, that means the Sun is $10^{24}$ times denser than the ISM. \\ \\<br /> The ISM actually comes in a wide range of temperatures and densities, so this single number does not tell the whole picture. All of these phases are roughly in pressure equilibrium though, as there would be forces between the phases acting to bring them to equilibrium were they not actually in equilibrium. The phases are<br /> \begin{table}<br /> \begin{tabular}{l l l l}<br /> Name &amp; Number density (n) &amp; Temperature (T) &amp; Constituent \\<br /> Hot Ionized Medium &amp; $10^{-2}$ cm$^{-3}$ &amp; $10^6$ K &amp; Ionized hydrogen\\<br /> Warm Ionized Medium &amp; $1$ cm$^{-3}$ &amp; $10^4$ K &amp; Ionized hydrogen \\<br /> Cold Neutral Medium &amp; $10^2$ cm$^{-3}$ &amp; $100$ K &amp; Neutral atomic hydrogen \\<br /> Molecular Clouds &amp; $&gt; 10^2$ cm$^{-3}$ &amp; $10$ K &amp; Molecular hydrogen \\<br /> \end{tabular}<br /> \end{table}<br /> The top two columns have most of the volume, and the bottom two most of the mass. With number densities in excess of average, the bottom two are also rare. They are where stars actually form in a galaxy, since they are slightly more dense than the rest of the ISM. Even so, they are much less dense than a star. How do we get a star then? Gravitational collapse. \\ \\<br /> Imagine a cloud with mass $M$, radius $R$, density $\rho$, and temperature $T$. The gravitational potential energy of the cloud is <br /> $<br /> U \sim -\frac{G M^2}{R}.<br />$<br /> Compare this to the thermal energy of the cloud, which is <br /> $<br /> K \sim N k T \sim \frac{M}{m_p} k T.<br />$<br /> The condition for collapse then is that the potential energy be greater than the kinetic energy,<br /> $<br /> |U| &gt; K.<br />$<br /> Plugging in our values for the cloud,<br /> $<br /> \frac{GM^2}{R} &gt; \frac{M}{m_p} k T,<br />$<br /> $<br /> M &gt; \frac{R k T}{G m_p},<br />$<br /> $<br /> \rho \sim \frac{M}{R^3},<br />$<br /> $<br /> M &gt; \frac{M^{1/3}}{\rho^{1/3}} \frac{ k T}{m_p G}, <br />$<br /> $<br /> M &gt; \left(\frac{k}{m_p G}\right)^{3/2} \frac{T^{3/2}}{\rho^{1/2}}.<br />$<br /> If we put in some numbers, we find that<br /> $<br /> M_J \approx 50 M_\odot \left(\frac{T}{10 K}\right)^{3/2} \left(\frac{n}{100 {\rm \;cm}^{-3}}\right)^{-1/2}.<br />$<br /> Only when $M &gt; M_J$ will collapse occur. This indicates that only the cool, dense gas can form stars. For the hotter components, the Jeans mass is of order the mass of all the gas in the galaxy, so it is not in danger of collapse. We can also define a Jeans length, by rearrangement of the above. We find<br /> $<br /> R_J = \left(\frac{k}{G m_p}\right)^{1/2} \frac{T^{1/2}}{\rho^{1/2}}.<br />$<br /> Again putting in numbers, we find<br /> $<br /> R_J \approx 3 {\rm \; pc} \left(\frac{T}{10 K}\right)^{1/2} \left(\frac{n}{100 {\rm \;cm}^{-3}}\right)^{-1/2}.<br />$<br /> This is much bigger than the size of a star (about one hundred million times larger). So there is a lot between collapse and the formation of a star. How long does it take for collapse to occur? Well, the free fall time is<br /> $<br /> t \sim \sqrt{\frac{1}{G \rho}}.<br />$<br /> This is roughly ten million years for the density of a molecular cloud. \\ \\<br /> What happens once collapse begins? Do we get lots of small stars, or one very massive stars? Force balance and energy losses determine the fate of a cloud. We can write down the acceleration as<br /> $<br /> \rho a = -\frac{dP}{dr} - \rho \frac{G M}{r^2}.<br />$<br /> What happens then depends on whether pressure or gravity wins during collapse. Order of magnitude, the pressure term scales as<br /> $<br /> \frac{P}{R} \sim \frac{n k T}{R} \propto \frac{T M}{R^4}.<br />$<br /> The gravity term scales as <br /> $<br /> \rho \frac{G M}{r^2} \propto \frac{M^2}{R^5}.<br />$<br /> In order to compare these, we need to know how the temperature changes as the cloud contracts. We can look at two cases here. One is when the mean free path of photons is larger than the cloud, which means they are able to effectively carry energy out of the cloud. That means the temperature is roughly constant, meaning that pressure will scale as $R^{-4}$ while gravity will scale as $R^{-5}$. Thus gravity will get stronger relative to pressure, and the collapse of the cloud will run away initially. What is happening to the Jeans' mass as this occurs? Well, the density is increasing, the temperature is constant, and the Jeans' mass decreases. Thus it will be possible for the cloud to break up in to smaller chunks as gravitational collapse proceeds, because individual clumps within the cloud can break apart in to separate clumps that are collapsing on their own. This is called gravitational fragmentation. \\ \\<br /> The second case in collapse is when the cloud becomes optically thick, or the mean free path is small relative to the size of the clump. When photons cannot escape, energy cannot easily leave the cloud, and the collapse is adiabatic. How does the temperature change in an adiabatic process? From earlier, $T \propto \rho^{\gamma - 1}$, or $T \propto \rho^{2/3} \propto M^{2/3} R^2$. The Jeans' mass has dependence $M_J \propto T^{3/2} \rho^{-1/2} \propto \rho^{1/2}$. Thus the Jeans' mass increases, and further fragmentation is halted. In terms of the pressure, we have the pressure force increasing as $R^{-6}$, which is a larger dependence than the gravity. Thus the pressure force increases more rapidly than gravity does, and the cloud will come in to hydrostatic equilibrium. At this point, the clump is much larger than a star, but it is getting close now that it is in hydrostatic equilibrium. Observationally, a typical star is about half a solar mass. For stars more massive than about a third of a solar mass, the distribution of solar masses (known as the Salpeter initial mass function) is <br /> $<br /> N(M) \propto M^{-1.35}.<br />$<br /> Our cloud in HE with no fusion will contract as it loses energy from radiating away energy at the point where it becomes optically thin. The cloud will then undergo Kelvin-Helmholtz contraction. Since it is in HE, the Virial Theorem holds. Then<br /> $<br /> E_{tot} = \frac{U}{2}.<br />$<br /> Then the luminosity is<br /> $<br /> L = -\frac{dE_{tot}}{dt} = -\frac{1}{2}\frac{dU}{dt}.<br />$<br /> This object is fully convective (which we won't fully justify, but is not surprising given how cool the cloud is). Using an $n = 3/2$ polytrope, we know then<br /> $<br /> U = -\frac{6}{7} \frac{G M^2}{R}<br />$<br /> Then the luminosity, in terms of the mass and radius, is<br /> $<br /> L = \frac{3}{7}\frac{G M^2}{R^2} \left|\frac{dR}{dt}\right|.<br />$<br /> We also know that for fully convective objects, <br /> $<br /> L = 0.2 L_\odot \left(\frac{M}{M_\odot}\right)^{4/7}\left(\frac{R}{R_\odot}\right)^2.<br />$<br /> We can solve for how the contraction proceeds by using this known convective luminosity. If we set the two expressions equal to one another, we find<br /> $<br /> \frac{3}{7}\frac{G M^2}{R^2} \left|\frac{dR}{dt}\right| = 0.2 L_\odot \left(\frac{M}{M_\odot}\right)^{4/7}\left(\frac{R}{R_\odot}\right)^2.<br />$<br /> If we solve for the radius as a function of time, we find<br /> $<br /> R \sim R_\odot \left(\frac{t}{2 \times 10^7 {\rm \;years}}\right)^{-1/3} \left(\frac{M}{M_\odot}\right)^{1/2}.<br />$<br /> We can also solve for the luminosity, finding<br /> $<br /> L \sim L_\odot \left(\frac{t}{2 \times 10^7 {\rm \;years}}\right)^{-2/3} \left(\frac{M}{M_\odot}\right)^{11/7}.<br />$<br /> Initially, the radius and luminosity of the star are both much larger than that of the Sun. As the star (or, pre-star) contracts, it will move down the Hayashi line at roughly constant temperature. At some point though, photons will take over the energy transport, at least for stars more massive than one third of a solar mass. We can find this point by equating the expressions for luminosity carried out by photons and luminosity carried out by convection. When a star crosses that point, it will cease moving down the Hayashi line, and instead move across the HR diagram until it reaches its main sequence position. We can solve for this time, and find that the time at which photons dominates is<br /> $<br /> t_{photons} \sim 10^6 {\rm \; years} \left(\frac{M}{M_\odot}\right)^{-15/7}.<br />$<br /> Thus for massive stars, photons carry out the energy for the majority of their pre main sequence lifetime. \\ \\<br /> Finally, when does contraction stop? When fusion starts. Or, more precisely, when fusion can generate the energy that leaves the star, so that it is not necessary to contract any further. This takes roughly twenty million years for a star of the mass of the Sun. This is, not coincidentally, of order the Kelvin-Helmholtz time.<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Star formation 2011-09-02T23:57:05Z <p>Jmcbride: created page</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \pagestyle{empty}<br /> \setlength{\parindent}{0in}<br /> <br /> \begin{document}<br /> \subsection*{Star Formation} \\<br /> Stars form out of cold gas in galaxies. The gas in a galaxy like the Milky Way is in a flat sheet, not unlike the dimensions of a stack of a few pieces of paper. For the Milky Way, radial size of the disk is roughly 10 kiloparsecs, where a parsec is roughly three lightyears. The height of the disk is some hundreds of parsecs, so roughly one tenth, or less, than the radial extent. The radial extent is a trillion times larger than the Sun, which gives a hint of the problem of star formation. How does gas get taken from large scales and turned in to a (relatively) small object like a star? Another way to see this issue is to look at the mean density of the galaxy, and the mean density of the Sun. The interstellar medium (ISM) has a mean density of roughly 1 particle per cubic centimeter. The Sun has a density of one {\em gram} per cubic centimeter. Since most things are hydrogen, that means the Sun is $10^{24}$ times denser than the ISM. \\ \\<br /> The ISM actually comes in a wide range of temperatures and densities, so this single number does not tell the whole picture. All of these phases are roughly in pressure equilibrium though, as there would be forces between the phases acting to bring them to equilibrium were they not actually in equilibrium. The phases are<br /> \begin{table}<br /> \begin{tabular}{l l l l}<br /> Name &amp; Number density (n) &amp; Temperature (T) &amp; Constituent \\<br /> Hot Ionized Medium &amp; $10^{-2}$ cm$^{-3}$ &amp; $10^6$ K &amp; Ionized hydrogen\\<br /> Warm Ionized Medium &amp; $1$ cm$^{-3}$ &amp; $10^4$ K &amp; Ionized hydrogen \\<br /> Cold Neutral Medium &amp; $10^2$ cm$^{-3}$ &amp; $100$ K &amp; Neutral atomic hydrogen \\<br /> Molecular Clouds &amp; $&gt; 10^2$ cm$^{-3}$ &amp; $10$ K &amp; Molecular hydrogen \\<br /> \end{tabular}<br /> \end{table}<br /> The top two columns have most of the volume, and the bottom two most of the mass. With number densities in excess of average, the bottom two are also rare. They are where stars actually form in a galaxy, since they are slightly more dense than the rest of the ISM. Even so, they are much less dense than a star. How do we get a star then? Gravitational collapse. \\ \\<br /> Imagine a cloud with mass $M$, radius $R$, density $\rho$, and temperature $T$. The gravitational potential energy of the cloud is <br /> $<br /> U \sim -\frac{G M^2}{R}.<br />$<br /> Compare this to the thermal energy of the cloud, which is <br /> $<br /> K \sim N k T \sim \frac{M}{m_p} k T.<br />$<br /> The condition for collapse then is that the potential energy be greater than the kinetic energy,<br /> $<br /> |U| &gt; K.<br />$<br /> Plugging in our values for the cloud,<br /> $<br /> \frac{GM^2}{R} &gt; \frac{M}{m_p} k T,<br />$<br /> $<br /> M &gt; \frac{R k T}{G m_p},<br />$<br /> $<br /> \rho \sim \frac{M}{R^3},<br />$<br /> $<br /> M &gt; \frac{M^{1/3}}{\rho^{1/3}} \frac{ k T}{m_p G}, <br />$<br /> $<br /> M &gt; \left(\frac{k}{m_p G}\right)^{3/2} \frac{T^{3/2}}{\rho^{1/2}}.<br />$<br /> If we put in some numbers, we find that<br /> $<br /> M_J \approx 50 M_\odot \left(\frac{T}{10 K}\right)^{3/2} \left(\frac{n}{100 {\rm \;cm}^{-3}}\right)^{-1/2}.<br />$<br /> Only when $M &gt; M_J$ will collapse occur. This indicates that only the cool, dense gas can form stars. For the hotter components, the Jeans mass is of order the mass of all the gas in the galaxy, so it is not in danger of collapse. We can also define a Jeans length, by rearrangement of the above. We find<br /> $<br /> R_J = \left(\frac{k}{G m_p}\right)^{1/2} \frac{T^{1/2}}{\rho^{1/2}}.<br />$<br /> Again putting in numbers, we find<br /> $<br /> R_J \approx 3 {\rm \; pc} \left(\frac{T}{10 K}\right)^{1/2} \left(\frac{n}{100 {\rm \;cm}^{-3}}\right)^{-1/2}.<br />$<br /> This is much bigger than the size of a star (about one hundred million times larger). So there is a lot between collapse and the formation of a star. How long does it take for collapse to occur? Well, the free fall time is<br /> $<br /> t \sim \sqrt{\frac{1}{G \rho}}.<br />$<br /> This is roughly ten million years for the density of a molecular cloud. \\ \\<br /> What happens once collapse begins? Do we get lots of small stars, or one very massive stars? Force balance and energy losses determine the fate of a cloud. We can write down the acceleration as<br /> $<br /> \rho a = -\frac{dP}{dr} - \rho \frac{G M}{r^2}.<br />$<br /> What happens then depends on whether pressure or gravity wins during collapse. Order of magnitude, the pressure term scales as<br /> $<br /> \frac{P}{R} \sim \frac{n k T}{R} \propto \frac{T M}{R^4}.<br />$<br /> The gravity term scales as <br /> $<br /> \rho \frac{G M}{r^2} \propto \frac{M^2}{R^5}.<br />$<br /> In order to compare these, we need to know how the temperature changes as the cloud contracts. We can look at two cases here. One is when the mean free path of photons is larger than the cloud, which means they are able to effectively carry energy out of the cloud. That means the temperature is roughly constant, meaning that pressure will scale as $R^{-4}$ while gravity will scale as $R^{-5}$. Thus gravity will get stronger relative to pressure, and the collapse of the cloud will run away initially. What is happening to the Jeans' mass as this occurs? Well, the density is increasing, the temperature is constant, and the Jeans' mass decreases. Thus it will be possible for the cloud to break up in to smaller chunks as gravitational collapse proceeds, because individual clumps within the cloud can break apart in to separate clumps that are collapsing on their own. This is called gravitational fragmentation. \\ \\<br /> The second case in collapse is when the cloud becomes optically thick, or the mean free path is small relative to the size of the clump. When photons cannot escape, energy cannot easily leave the cloud, and the collapse is adiabatic. How does the temperature change in an adiabatic process? From earlier, $T \propto \rho^{\gamma - 1}$, or $T \propto \rho^{2/3} \propto M^{2/3} R^2$. The Jeans' mass has dependence $M_J \propto T^{3/2} \rho^{-1/2} \propto \rho^{1/2}$. Thus the Jeans' mass increases, and further fragmentation is halted. In terms of the pressure, we have the pressure force increasing as $R^{-6}$, which is a larger dependence than the gravity. Thus the pressure force increases more rapidly than gravity does, and the cloud will come in to hydrostatic equilibrium. At this point, the clump is much larger than a star, but it is getting close now that it is in hydrostatic equilibrium. Observationally, a typical star is about half a solar mass. For stars more massive than about a third of a solar mass, the distribution of solar masses (known as the Salpeter initial mass function) is <br /> $<br /> N(M) \propto M^{-1.35}.<br />$<br /> Our cloud in HE with no fusion will contract as it loses energy from radiating away energy at the point where it becomes optically thin. The cloud will then undergo Kelvin-Helmholtz contraction. Since it is in HE, the Virial Theorem holds. Then<br /> $<br /> E_{tot} = \frac{U}{2}.<br />$<br /> Then the luminosity is<br /> $<br /> L = -\frac{dE_{tot}}{dt} = -\frac{1}{2}\frac{dU}{dt}.<br />$<br /> This object is fully convective (which we won't fully justify, but is not surprising given how cool the cloud is). Using an $n = 3/2$ polytrope, we know then<br /> $<br /> U = -\frac{6}{7} \frac{G M^2}{R}<br />$<br /> Then the luminosity, in terms of the mass and radius, is<br /> $<br /> L = \frac{3}{7}\frac{G M^2}{R^2} \left|\frac{dR}{dt}\right|.<br />$<br /> We also know that for fully convective objects, <br /> $<br /> L = 0.2 L_\odot \left(\frac{M}{M_\odot}\right)^{4/7}\left(\frac{R}{R_\odot}\right)^2.<br />$<br /> We can solve for how the contraction proceeds by using this known convective luminosity. If we set the two expressions equal to one another, we find<br /> $<br /> \frac{3}{7}\frac{G M^2}{R^2} \left|\frac{dR}{dt}\right| = 0.2 L_\odot \left(\frac{M}{M_\odot}\right)^{4/7}\left(\frac{R}{R_\odot}\right)^2.<br />$<br /> If we solve for the radius as a function of time, we find<br /> $<br /> R \sim R_\odot \left(\frac{t}{2 \times 10^7 {\rm \;years}}\right)^{-1/3} \left(\frac{M}{M_\odot}\right)^{1/2}.<br />$<br /> We can also solve for the luminosity, finding<br /> $<br /> L \sim L_\odot \left(\frac{t}{2 \times 10^7 {\rm \;years}}\right)^{-2/3} \left(\frac{M}{M_\odot}\right)^{11/7}.<br />$<br /> Initially, the radius and luminosity of the star are both much larger than that of the Sun. As the star (or, pre-star) contracts, it will move down the Hayashi line at roughly constant temperature. At some point though, photons will take over the energy transport, at least for stars more massive than one third of a solar mass. We can find this point by equating the expressions for luminosity carried out by photons and luminosity carried out by convection. When a star crosses that point, it will cease moving down the Hayashi line, and instead move across the HR diagram until it reaches its main sequence position. We can solve for this time, and find that the time at which photons dominates is<br /> $<br /> t_{photons} \sim 10^6 {\rm \; years} \left(\frac{M}{M_\odot}\right)^{-15/7}.<br />$<br /> Thus for massive stars, photons carry out the energy for the majority of their pre main sequence lifetime. \\ \\<br /> Finally, when does contraction stop? When fusion starts. Or, more precisely, when fusion can generate the energy that leaves the star, so that it is not necessary to contract any further. This takes roughly twenty million years for a star of the mass of the Sun. This is, not coincidentally, of order the Kelvin-Helmholtz time.<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-02T23:31:50Z <p>Jmcbride: fixed now broken links</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Hydrostatic Equilibrium]]<br /> #[[Virial Theorem]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Stars Lecture 8]]<br /> #[[Stars Lecture 9]]<br /> #[[Stars Lecture 10]]<br /> #[[Stars Lecture 11]]<br /> #[[Stars Lecture 12]]<br /> #[[Stars Lecture 13]]<br /> #[[Stars Lecture 14]]<br /> #[[Stars Lecture 15]]<br /> #[[Stars Lecture 16]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Stars Lecture 2 2011-09-02T23:30:52Z <p>Jmcbride: moved Stars Lecture 2 to Virial Theorem:&amp;#32;actual topic name</p> <hr /> <div>#REDIRECT [[Virial Theorem]]</div> Jmcbride Virial Theorem 2011-09-02T23:30:52Z <p>Jmcbride: moved Stars Lecture 2 to Virial Theorem:&amp;#32;actual topic name</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> <br /> \begin{document}<br /> <br /> \subsection*{ Virial Theorem} \\ \\<br /> Start out with hydrostatic equilibrium:<br /> $<br /> \frac{dP}{dr} = - \rho \frac{G M_r}{r^2}<br />$<br /> We want this to have units of energy when we integrate this equation over the entire star. In order to do this, we multiply by the factor $4 \pi r^3 dr$. Now let's look at the right hand side of this equation.<br /> $<br /> - \int^R_0 \rho \frac{G M_r}{r^2} \dot 4 \pi r^3 dr = - \int^R_0 dM_r \frac{G M_r}{r} = U.<br />$<br /> So the right side just gives us the gravitational potential energy. Now we look at the left hand side, using the same factor of $4 \pi r^3 dr$ and a little trick. That is, use the fact that <br /> $<br /> \frac{d}{dr}\left(4 \pi r^3 P\right) = 4 \pi r^3 \frac{dP}{dr} + 3 \times 4 \pi r^2 P.<br />$<br /> Now we use this on the left hand side of hydrostatic equilibrium to get<br /> $<br /> \int^R_0 4 \pi r^3 dr \frac{dP}{dr} = \int^R_0 dr \frac{d}{dr}\left(4 \pi r^3 P\right) - 3 \int^R_0 4 \pi r^2 P dr.<br />$<br /> The first factor just goes to zero, since the radius at the center is zero, and the pressure at the boundary is (nearly) zero. Thus we have<br /> $<br /> U = - 3 \int^R_0 4 \pi r^2 P dr, {\rm \; or}<br />$<br /> $<br /> U = -3 &lt;P&gt; V.<br />$<br /> The total energy in the star is <br /> $<br /> E_{tot} = K + U = -3P&lt;P&gt; + K. <br />$<br /> Here $K$ represents the total kinetic energy in the star. This kinetic energy is dominated by the random motions of particles in the gas, otherwise known as the thermal energy. Ordered motions like convection and rotation are unimportant.<br /> For an ideal gas at temperature $T$, the energy per particle is given by <br /> $<br /> \epsilon = \frac{3}{2} k_B T, {\rm \; (non-relativistic)}<br />$<br /> $<br /> \epsilon = 3 k_B T {\rm \; (relativistic).}<br />$<br /> More generally, this is <br /> $<br /> \epsilon = \frac{1}{\gamma - 1} k_B T.<br />$<br /> Thus the total thermal energy of the a star is given by <br /> $<br /> K = \int^R_0 \frac{1}{\gamma -1} k T n dV.<br />$<br /> The integrand is just the energy per volume, which is the same thing as a pressure. Thus the total thermal energy of a star is given by<br /> $<br /> K = \frac{1}{\gamma - 1} \int^R_0 P dV = \frac{1}{\gamma - 1} &lt;P&gt; V.<br />$<br /> The gravitational potential energy and kinetic energy are related by the adiabatic index, with <br /> $<br /> K = \frac{-U}{3 (\gamma - 1)}.<br />$<br /> In the non-relativistic case,<br /> $<br /> K = -\frac{U}{2},<br />$<br /> and in the relativistic case<br /> $<br /> K = -U.<br />$<br /> We can also write the total energy of the star in these terms:<br /> $<br /> E_{tot} = K (1 - 3(\gamma - 1)).<br />$<br /> This entire derivation assumes the fact that the gas is classical and ideal. Even so, the result holds much more generally, including in white dwarfs (which are degenerate) and in systems where photons dominate the pressure (high mass stars). \\ \\<br /> Now we want to find a general relationship between pressure, thermal energy, and the cases where photons or quantum effects are important. We will imagine a system with particles zooming around and bouncing off of a wall, where the wall has area $A$. Then the force on the wall from collisions is <br /> $<br /> \frac{\Delta p}{\Delta t} = P A.<br />$<br /> We could also write the change in momentum as the product of the rate at which paticles hit the wall and the change in momentum per collision. The number of particles in a length is $n A$, where $n$ is the number density of particles. Then the rate is just <br /> $<br /> {\rm rate\;} = \frac{1}{2} n A v_x.<br />$<br /> The change in momentum is use two times the x momentum of the particles, so that<br /> $<br /> \frac{\Delta p}{\Delta t} = n A v_x p_x.<br />$<br /> If we look at three dimensions rather than one, we have a factor of one third come in, which gives us<br /> $<br /> \frac{\Delta p}{\Delta t} = \frac{1}{3} n A v p = P A.<br />$<br /> The area cancels out, as it should, and we have<br /> $<br /> P = \frac{1}{3} n v p.<br />$ <br /> This is a fundamental relationship between the microscopic particles and the macroscopic, thermodynamic properties of that gas. Generally, we have a gas with a range of velocities, so we need to average over the distribution function of particles, meaning the pressure is<br /> $<br /> P = \frac{1}{3} n &lt;v p&gt;.<br />$<br /> For non-relativistic particles, the momentum and velocity are just related by the mass of the particle, meaning<br /> $<br /> P = \frac{2}{3} n &lt;\frac{1}{2} m v^2&gt;.<br />$<br /> The averaged quantity is just the kinetic energy per particle, so<br /> $<br /> P = \frac{2 K}{3 V}.<br />$<br /> This is exactly what we had before for the specific case of an ideal gas, and gives us the same result for the Virial theorem we had before. We could do the same thing for a photon, or relativistic massive particle (and Eliot did), but the procedure and result are the same. \\ \\<br /> Now let's use the Virial theorem to derive some useful results. We start by imaginging a system in HE that has no internal energy source (like fusion) that is radiating energy with power $L$. Thus the total energy, which is already negative, is decreasing, or, in an absolute value sense, the total energy is becoming {\em more} negative. That is, $|E_{tot}|$ is increasing, $|U|$ is increasing, and $K$ is increasing. The way that this happens is by contracting, specifically a process called Kelvin-Helmholtz contraction. At any stage in the life of a star, if it does not have an internal energy source, this will happen. We can estimate the timescale of this contraction (meaning the time it takes to change appreciably, say $R$ changing of order $R/2$) using the total energy and the rate at which energy is being radiatied:<br /> $<br /> t_{KH} \sim \frac{|U|}{L}.<br />$<br /> If we imagine we know nothing about the internal energy source of the Sun, and ask what the Kelvin-Helmholtz timescale is for the Sun, we would find that<br /> $<br /> t_{KH} = 3 \times 10^7 {\rm years}.<br />$<br /> This is much less than the age of the Earth or the solar system, but that did not stop Kelvin and Helmholtz from proposing that this is how the energy of the Sun was generated, which caused great debates with biologists and geologists who knew the Earth must be older. What Kelvin and Helmholtz did not know is that fusion kicked in at the center and halted this process, though the Sun did undergo this contraction during its formation. \\ \\<br /> Now let's imagine the opposite situation, where more energy is being generated than is being radiatied. Then the total energy is increasing, but $|E_{tot}|$ actually goes down, meaning $K$ decreases and $|U|$ decreases. This is pretty weird. If you add energy to the star, it gets cooler. From a thermodynamics standpoint, this means stars have a negative heat capacity. In fact, all self gravitating objects have a negative heat capacity. This is unlike anything we experience on Earth, where adding energy to a system causes it to heat up. \\ \\<br /> This is the fundamental reason that fusing stars are different from fusion bombs. A bomb is heated, which causes fusion and energy release, which further heats the bomb, leading to a runaway explosion. The star fuses in a controlled manner thanks to gravity, which acts as a safety valve. Any extra fusion in a star causes the star to expand and cool, thereby slowing the rate of fusion until it comes back in to equilibrium. Recreating this on Earth is pretty challenging, and is the reason we do not have fusion energy yet. \\ \\<br /> The Virial Theorem can also give us an estimate of the temperature in stars. Consider a star (our Sun) supported by a non-relativistic classical gas. Then<br /> $<br /> U = -2K,<br />$<br /> $<br /> \frac{G M^2}{R} \sim 3 N k_B T \sim 3 \frac{M}{m_p} k_B T.<br />$<br /> This gives<br /> $<br /> 3 k_B T \sim \frac{G M m_p}{R}.<br />$<br /> For the sun, this is about $5 \times 10^6$ K, which is much larger than the effective temperature, and a little bit smaller than the central temperature.<br /> <br /> <br /> <br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Stars Lecture 1 2011-09-02T23:30:28Z <p>Jmcbride: moved Stars Lecture 1 to Hydrostatic Equilibrium:&amp;#32;actual topic</p> <hr /> <div>#REDIRECT [[Hydrostatic Equilibrium]]</div> Jmcbride Hydrostatic Equilibrium 2011-09-02T23:30:28Z <p>Jmcbride: moved Stars Lecture 1 to Hydrostatic Equilibrium:&amp;#32;actual topic</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[11pt]{article}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> \begin{document}<br /> \subsection*{Central Topics}<br /> \begin{enumerate}<br /> \item Force balance in stars: pressure vs. gravity<br /> \item Energy transport by radiation, convection, and conduction<br /> \item Energy generation by fusion<br /> \end{enumerate}<br /> These combine to give the properties of the HR diagram, the main sequence, etc. Changes in mass and composition change each of the above. These three unifying principles used repeatedly in the course. We'll look at each separately to see what we can learn, and then put them together to understand stars. Even without knowing anything about energy generation, we can understand a lot based on force balance and energy transport.<br /> \\ \\<br /> \subsection*{Force Balance in Stars: Hydrostatic Equilibrium} \\<br /> Throughout the course we will assume spherical stars, ignoring factors such as rotation that may cause a star to become oblate, for instance. \\<br /> We will start by looking at a thin layer in a star, and over a small enough area that we can approximate the layer as being flat. The layer has thickness $dr$, density $\rho$, area $A$, and is at a distance $r$ from the center of the star and encloses a mass $M_r$. The mass of the shell is <br /> $<br /> M_{\rm shell} = \rho A dr<br />$<br /> The inward gravitational force on the region is then<br /> $<br /> F_g = -\frac{G M_r M_{\rm shell}}{r^2}<br />$<br /> The competing force (which provides force balance) is the pressure gradient within the star. That is, there is a slightly greater pressure on the bottom, $P_{\rm below}$ on the shell then there is on the top, $P_{\rm above}$. We then can multiply this pressure difference by the area of the thin shell to find the total force due to the pressure gradient. The net force is the sum of these two sources, and will give the acceleration of the shell.<br /> $<br /> F_{\rm net} = M_{\rm shell} a<br />$<br /> $<br /> F_{\rm net} = (P_{\rm below} - P_{\rm above})A -\frac{G M_r M_{\rm shell}}{r^2} = M_{\rm shell} a.<br />$<br /> Since we are looking at a thin shell, we will say that <br /> $<br /> P_{\rm above} = P_{\rm below} + dP.<br />$<br /> Then, substituting above,<br /> $<br /> -dp A - \frac{G M_r}{r^2}A dr \rho = A dr \rho a.<br />$ <br /> The area element cancels out, which is good because it was sort of arbitrarily chosen. Dividing through by the differential $dr$ gives<br /> $<br /> \rho a = -\frac{dP}{dr} - \frac{G M_r}{r^2} \rho.<br />$ <br /> In the case where there is no acceleration, which is pretty common since stars generally appear to be neither contraction nor expanding, we have hydrostatic equilibrium (HE). In this case, the above reduces to<br /> $<br /> \frac{dP}{dr} = -\rho \frac{G M_r}{r^2} = -\rho g.<br />$<br /> To further justify this approximation, let's imagine that $a$ is not equal to 0. Instead, it is roughly equal to the graviational acceleration. We can then estimate the amount of time it would take for the radius of a star to change appreciably. So<br /> $<br /> a \sim -\frac{G M_r}{r^2}<br />$<br /> We can look at the acceleration in terms of the size of the star, $R$, and the timescale of change, $t$. Then<br /> $<br /> a \sim \frac{R}{t^2}.<br />$<br /> For the purposes of this estimate, we can just take $r = R$ and $M_r = M$, which<br /> gives<br /> $<br /> \frac{R}{t^2} \sim \frac{G M}{R^2},<br />$<br /> $<br /> t \sim \sqrt{\frac{R^3}{G M}} \sim \sqrt{\frac{1}{G \rho}}.<br />$<br /> This is often called the dynamical time, and applies equally well to planets or galaxies as it does to a star. It is is also called the free fall time, since it is the time it takes to move the size of the system at the free fall speed of the system. For the Sun, with an average density of $1$ g cm$^{-3}$, this is about one hour, meaning any imbalances must quickly be eliminated. There are only brief periods in the life of a star where HE does not hold, such as the collapse to form a black hole. Even during convection, which violates the assumptions that went in to HE, this approximation holds very very well. \\<br /> We should explicitly define $M_r$, which we use in HE.<br /> $<br /> M_r = \int^r_0 4 \pi r^2 \rho dr, {\rm \; or}<br />$<br /> $<br /> \frac{dM_r}{dr} = 4 \pi r^2 \rho.<br />$<br /> There are still more variables here than equations. We need to know what the source of pressure is to solve for the structure of a star, as well as how energy is transported. We can, however, use just these equations to understand the outermost layers of the Sun, or even the atmosphere of the Earth. We use the approximation of a plane parallel atmosphere, which holds very well since the thickness $z$ of the atmosphere is much smaller than the size of the star (or planet) $R$. Then we can assume that the surface gravity is set by the total mass and radius of the star, without worry about the mass or size of the atmosphere. Then<br /> $<br /> g = \frac{G M}{R^2} = {\rm constant\; at\; surface}.<br />$<br /> Then HE is <br /> $<br /> \frac{dP}{dz} = = \rho g.<br />$<br /> The pressure is due to that of an ideal gas, so<br /> $<br /> P = n k_B T, <br />$<br /> where $n$ is the number density, $k_B$ is Boltzmann's constant, and $T$ is the temperature. We need to know then the average mass per particle, $m$, so that $\rho = n m$. This gives<br /> $<br /> \frac{d(n k_B T)}{dz} = -m g n.<br />$<br /> This is still not something we can solve, so we have to make the further approximation that the atmosphere is isothermal, i.e. $T$ is constant as a function of $r$. Then<br /> $<br /> \frac{1}{n}\frac{dn}{dz} = -\frac{mg}{k_B T}.<br />$<br /> The solution of this is <br /> $<br /> \left[\ln(n)\right]^{z}_{\rm surface} = -\frac{mgz}{k_B T},<br />$<br /> $<br /> n(z) = n_{\rm surf} e^{-z/h}, {\rm \; where}<br />$<br /> $<br /> h = \frac{k_B T}{mg}.<br />$<br /> We call $h$ the scale height of the atmosphere, and it is given by the ratio of the thermal energy at the surface to the gravitational potential energy at the surface. We can think of it as the distance over which the density changes by an appreciable amount (1 / $e$). Thus the thermal energy is trying to puff up the atmoshpere, while the gravitational potential is trying to keep it close to the surface. For the Sun, the scale height is<br /> $<br /> h = 2 \times 10^7 {\rm cm} = 3 \times 10^{-4} R_\odot.<br />$<br /> This is much smaller than the radius of the Sun. There is an interesting statistical mechanics interpretation of the scale height. Particles with an energy $E$ are distributed in energy levels according to the Boltzmann factor, with<br /> $<br /> n \propto e^{-E/k_B T}.<br />$<br /> Since the gravitational potential energy for our particles at height $z$ is $mgz$, this Boltzmann argument gives the same radial dependence of number density of the atmosphere as our argument above. \\ \\<br /> \subsection*{Mean Molecular Weight} \\<br /> One of the major ways fusion affects the structure of stars is by changing the average mass per particle. If it starts out as ionized hydrogen, there are basically two particles per $m_p$, while for neutral hydrogen there is about one particle per $m_p$, and for ionized helium it is 3 particles per $4 m_p$, roughly. Each of these particles is at temperature $T$, and thus contributes to the total pressure in the star. \\ \\<br /> To account for this, we will define a quantity $\mu$ called the mean molecular weight to formally define our average mass per particle and encapsulate the summing of pressures from each element. Each element has a mass fraction $X_i$ and an atomic number $A_i$ that represents the total number of protons and neutrons. If the total mass density is $\rho$, then the number density of a given species $n_i$ is <br /> $<br /> n_i = \frac{\rho X_i}{A_i m_p}.<br />$<br /> The pressure from all ions is <br /> $<br /> P = \sum_i n_i k_B T.<br />$<br /> Substituting in the expression for ionic number density,<br /> $<br /> P = \frac{\rho k_B T}{m_p} \sum_i \frac{X_i}{A_i}.<br />$<br /> We can then define the ion mean molecular weight to be<br /> $<br /> \frac{1}{\mu_i} = \sum_i \frac{X_i}{A_i}.<br />$<br /> The ions also contribute electrons, which we also need to account for. The number density for electrons $n_e$ depends upon the charge $Z_i$ of the atom (assuming complete ionization), with<br /> $<br /> \sum_i = Z_i n_i.<br />$<br /> Then <br /> $<br /> P_e = n_e k_B T = k_b T \sum_i n_i Z_i = \frac{\rho k_B T}{m_p} \sum_i \frac{Z_i X_i}{A_i}.<br />$<br /> This allows us to define the electronic mean molecular weight as <br /> $<br /> \frac{1}{\mu_e} = \frac{Z_i X_i}{A_i}.<br />$<br /> The final mean molecular weight is then <br /> $<br /> \frac{1}{\mu} = \frac{1}{\mu_i} + \frac{1}{\mu_e}.<br />$<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-02T06:38:35Z <p>Jmcbride: linking by subject</p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Stars Lecture 1]]<br /> #[[Stars Lecture 2]]<br /> #[[Radiative transport and conduction]]<br /> #[[Convection]]<br /> #[[Stars Lecture 8]]<br /> #[[Stars Lecture 9]]<br /> #[[Stars Lecture 10]]<br /> #[[Stars Lecture 11]]<br /> #[[Stars Lecture 12]]<br /> #[[Stars Lecture 13]]<br /> #[[Stars Lecture 14]]<br /> #[[Stars Lecture 15]]<br /> #[[Stars Lecture 16]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Stars Lecture 5 2011-09-02T06:38:12Z <p>Jmcbride: moved Stars Lecture 5 to Convection:&amp;#32;more descriptive title</p> <hr /> <div>#REDIRECT [[Convection]]</div> Jmcbride Convection 2011-09-02T06:38:12Z <p>Jmcbride: moved Stars Lecture 5 to Convection:&amp;#32;more descriptive title</p> <hr /> <div>&lt;latex&gt;<br /> <br /> \documentclass[preprint]{aastex}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> \newcommand{\rhoblob}{\rho_{\rm blob}}<br /> \newcommand{\rhostar}{\rho_{\rm star}}<br /> \newcommand{\pblob}{P_{\rm blob}}<br /> \newcommand{\pstar}{P_{\rm star}}<br /> <br /> \begin{document}<br /> Plenty of redundancy to be removed here, as this was three lectures.<br /> <br /> \subsection*{Thermodynamics} \\<br /> Convection is as important as radiative diffusion in main sequence stars. We'll make an analogy between convection and boiling, like water in a pot on the stove. At low temperatures, energy moves up through a pot through conduction. Once it gets hot enough though, the energy cannot move quickly enough through convection, causing the water to begin to boil. The basic physics that sets convection is bouyancy. Hot, underdense material at the bottom of the pot is less dense than the cooler water that sits atop it, leading to bubbles rising and depositing their energy at the top of the pot where the water was cooler. \\ \\ <br /> To understand all of this, we'll have to use some thermodynamic relationships between quantities like entropy and heat. Starting with the second law of thermodynamics, the change in entropy is given by the change in energy and volume as<br /> $<br /> T dS = dE + PdV.<br />$<br /> We don't really want to talk about volumes of stars though. So let's imagine a volume within a star with mass $M$, density $\rho$, and volume $V$. We'll focus on a this parcel with constant mass, and divide the above by the mass, to get the specific entropy in terms of a density rather than a volume. Then<br /> $<br /> \frac{dV}{M} = -\frac{d\rho}{\rho^2}.<br />$<br /> This gives for the second law, with $ds = dS / M$ and $dU = dE / M$,<br /> $<br /> T ds = dU - \frac{P}{\rho}\frac{d\rho}{\rho}.<br />$<br /> Now we imagine that the system is adiabatic, which means the heat content of the material is not exchanged with the surroundings. This is equivalent to <br /> $<br /> T ds = 0.<br />$<br /> In this case, the second law reduces to <br /> $<br /> dU = \frac{P}{\rho}{d\rho}{\rho}.<br />$<br /> The pressure and the energy per unit volume are nearly interchangeable, being related by a factor of order unit. We can define then an energy per unit volume $\epsilon$, with<br /> $<br /> \epsilon = U\rho.<br />$<br /> If we relate the pressure and energy density by some constant $\phi$,<br /> $<br /> P = \phi \epsilon = \phi U \rho,<br />$<br /> then the second law in the adiabatic case becomes<br /> $<br /> dU = \phi U \frac{d\rho}{\rho}, <br />$<br /> $<br /> \frac{dU}{U} = \phi \frac{d\rho}{\rho}, <br />$<br /> $<br /> \ln(U) = \phi \ln(\rho) + C, <br />$<br /> $<br /> U \propto \rho^\phi, <br />$<br /> $<br /> P \propto \rho^{\phi + 1}.<br />$<br /> In doing this, we have shown that the adiabatic index $\gamma$ is related to $\phi$ by<br /> $<br /> \gamma = \phi + 1.<br />$<br /> We can write then that<br /> $<br /> P \propto \rho^\gamma,<br />$<br /> $<br /> U \propto \rho^{\gamma - 1},<br />$<br /> $<br /> T \propto \rho^{\gamma - 1}.<br />$<br /> In the non-relativistic case, $\phi = 2/3$, $\gamma = 5/3$. For relativistic particles, $\phi = 1/3$, $\gamma = 4/3$. We can then use these relations to understand how pressure, energy, and temperature relate to density for either situation. \\ \\<br /> Now let's look at the entropy of an ideal gas. The second law (again) is <br /> $<br /> T ds = dU - \frac{P}{\rho}\frac{d\rho}{\rho}.<br />$<br /> The pressure is<br /> $<br /> P = U \rho (\gamma - 1).<br />$<br /> Then<br /> $<br /> T ds = dU - U(\gamma - 1) \frac{d\rho}{\rho}, <br />$<br /> <br /> and then after dividing by the energy,<br /> $<br /> \frac{T}{U} ds = \frac{dU}{U} - (\gamma - 1)\frac{d\rho}{\rho}.<br />$<br /> Writing the energy in terms of the temperature for an ideal gas in the normal way,<br /> $<br /> U = \frac{3 k_b T}{2 m},<br />$<br /> and then rearranging and writing in the general form,<br /> $<br /> \frac{T}{U} = \frac{m}{k_B} (\gamma - 1).<br />$<br /> We can then solve for the entropy in terms of the mass, energy, and density.<br /> <br /> $<br /> \frac{m}{k_B} (\gamma - 1) ds = \frac{dU}{U} - (\gamma - 1) \frac{d\rho}{\rho} <br />$<br /> $<br /> \frac{m}{k_B} (\gamma - 1) ds = \ln(U) - (\gamma - 1)\ln(\rho) <br />$<br /> $<br /> s = \frac{k_B}{(\gamma - 1)m} \ln\left(\frac{U}{\rho^{\gamma -1}}\right).<br />$<br /> <br /> We can use again the relationship between pressure, density, and energy per unit mass to write <br /> $<br /> s = \frac{k_B}{(\gamma - 1)m} \ln\left(\frac{P}{\rho^{\gamma}}\right).<br />$ %\bbox<br /> Now let's look at the change in time of entropy, again using the second law. The time derivative is<br /> $<br /> T \frac{ds}{dt} = \frac{dU}{dt} - \frac{P}{\rho^2}{d\rho}{dt}.<br />$<br /> This represents the heating per mass per time, or cooling per mass per time. We have written the radiative diffusion in terms of an energy per time per area. We would like to put the two in the same terms. To do this, we start with an integral over surface area to get energy per time.<br /> $<br /> {\rm energy\; per\; time} = \int dA \cdot F.<br />$<br /> If we use the divergence theorem, we can change the integral over area to one over volume by taking the divergence of the integrated quantity.<br /> $<br /> {\rm energy\; per\; time} = \int dV \nabla \cdot F.<br />$<br /> Thus the energy per time per volume is just $\nabla \cdot F$. The energy per unit time per unit mass is then just $\nabla \cdot F / \rho$. Then the second law can be written <br /> $<br /> T \frac{ds}{dt} = \frac{dU}{dt} - \frac{P}{\rho^2}\frac{d\rho}{dt} = \epsilon_{\rm fusion} - \frac{1}{\rho}\nabla \cdot F.<br />$<br /> Here we have included energy input from fusion, as well as energy moving by photons or convection. \\ \\<br /> Now that we have done all of this, let's determine where convection even matters. We have equations for force balance (hydrostatic equilibrium) and radiative diffusion that we can solve to find the temperature and pressure as a function of radius. That does not mean that the solution always represents the physical situation. The reason for this is that even if the solution works, it does not necessarily have to be stable. Convection is important when the solution of HE and radiative diffusion is not stable to small changes in the density. We can ask then what happens in a star when a small blob of low density starts to rise in a star, will it continue rising or will it fall back down to where it started? \\ \\<br /> The blob has some $\rho_{\rm blob}$, $T_{\rm blob}$, $s_{\rm blob}$, $P_{\rm blob}$. It then rises to some higher position, where we assume all of these quantities are set by radiative diffusion to be $\rho_{\rm star}$, $T_{\rm star}$, $s_{\rm star}$, $P_{\rm star}$. To see whether the blob moves up quickly, we'll assume the blob is adiabatic, meaning it does not exchange heat. We can say this because the boiling time for the Sun is order a month, whereas the timescale for energy to leak out is about $10^7$ years. We will also assume that the blob is in presure equilibrium, meaning the pressure of the blob is the same as the pressure of it surroundings. This is true so long as the boiling timescale is much larger than the dynamic timescale, which is the time required to maintain hydrostatic equilibrium. Since the dynamic timescale is about one hour, this is also a very good assumption. \\ \\<br /> With these assumptions, the entropy of the blob is constant as it rises, and will thus be different from the entropy of the surroundings, because in general there will be an entropy gradient in the star. Pressure equilibrium implies that the pressure of the blob is equal to the pressure of the surrounding star.<br /> <br /> $<br /> s_{\rm blob} /= s_{\rm star} <br />$<br /> $<br /> P_{\rm blob} = P_{\rm star}<br />$<br /> If the entropy gradient of the star is negative ($\frac{ds}{dr} &lt; 0$), then<br /> $s_{\rm star} &lt; s_{\rm blob}$. Since $S \propto \ln(P/\rho^\gamma)$, and the pressure of the blob is equal to the pressure of the surroundings, this implies that the density of the blob must be less than the density of its surroundings. This, in turn, means the blob is convectively unstable. If, on the other hand, there is a positive entropy gradient, the blob is convectively stable, by the exact same reason, but with the opposite sign. The condition that the entropy gradient by negative in order for convection to set in is known as the Schwarzschild criterion. Any solution of the structure of a star that gives a negative entropy gradient is an unstable solution.<br /> <br /> \subsection*{Convection} \\<br /> Quick conceptual review of last lecture: the density, pressure, temperature and entropy profile of a star all have a long time to come in to some sort of equilibrium state. A rising blob of material moves on a much smaller timescale though, and does not need to change as quickly. The blob does come in to pressure eqiulibrium with the star, but its entropy remain fixed. This is because the blob moves on a timescale much longer than the dynamical time for the star (time to come in to pressure equilibrium), but much shorter than the thermal timescale of the star. We arrived at the condition that convection takes places if and only if the entropy gradient is negative. When this is true, the density of the blob becomes less dense faster than the surrounding star becomes less dense as it rises through the star, meaning it is bouyant and the star is convectively unstable. \\ \\<br /> We can compare the density of the blob to the background density of the star at some position in the star.<br /> $<br /> \rho_{\rm star} = \rho + \frac{d\rho}{dr} \delta r,<br />$<br /> $<br /> P_{\rm star} = P + \frac{dP}{dr} \delta r,<br />$<br /> $<br /> \rho_{\rm blob} = \rho + \left(\frac{\delta \rho}{\delta P}\right)_s \delta \rho_{\rm blob}, <br />$<br /> $<br /> P_{\rm blob}= P + \delta P_{\rm blob} = P_{\rm star},<br />$<br /> $<br /> \rhoblob= \rho + \left(\frac{d\rho}{dP}\right)_s \frac{dP}{dr}dr, <br />$<br /> $<br /> P \propto \rho^\gamma<br />$<br /> $<br /> \left(\frac{dP}{d\rho}\right)_s = \frac{\gamma P}{\rho},<br />$<br /> $<br /> \rhoblob = \rho + \frac{\rho}{\gamma P} \frac{dP}{dr} \delta r.<br />$<br /> <br /> At some new position, the blob will feel an acceleration from its bouyancy, just by Archimedes' Principle.<br /> $<br /> a = g \left(\frac{\rhostar}{\rhoblob} - 1\right).<br />$<br /> If the density of the blob is less than that of the star, acceleration is positive, and the blob runs away. If the density of the blob is greater, the acceleration is negative and the blob sinks back down, so the star is convectively stable. The acceleration can be written in terms of what is called the Brunt-Vaisala frequency as <br /> $<br /> a = -N^2 \delta r,<br />$<br /> where we have defined <br /> $<br /> N^2 = -g \left(\frac{d \ln \rho}{dr} - \frac{1}{\gamma} \frac{d \ln P}{dr}\right).<br />$<br /> Alternatively, we can write it in terms of the entropy gradient, with<br /> $<br /> N^2 = \frac{\gamma - 1}{\gamma}\frac{m_p}{k_b} g \frac{ds}{dr}.<br />$<br /> The Brunt-Vaisala frequency is proportion then to the entropy gradient. We can write this down as a simple differential equation.<br /> $<br /> a = \delta \dot{r},<br />$<br /> $<br /> \delta \dot{r} + N^2 \delta r = 0, % should be double dot<br />$<br /> $<br /> \delta r \propto e^{\pm i N t}.<br />$<br /> <br /> If $N^2$ is greater than zero, we have an oscillating solution, and convective stability. This corresponds to a positive entropy gradient. If $N^2$ is negative, then the solution of this is an exponential runaway, on a timescale of $1 / |N|$. \\ \\<br /> We can write yet another convection criterion that is useful in stars. Using the negative entropy gradient condition and the relation between entropy and pressure and density, <br /> <br /> $<br /> \frac{d \ln P}{dr} - \gamma\frac{d \ln \rho}{dr} = 0,<br />$<br /> $<br /> P = \frac{\rho k_B T}{\mu m_p},<br />$<br /> $<br /> \ln P = \ln \rho + \ln T + {\rm constant},<br />$<br /> $<br /> \frac{d \ln P}{dr} - \gamma\frac{d \ln P}{dr} + \gamma \frac{d \ln T}{dr} &lt; 0, <br />$<br /> $<br /> \frac{d \ln T}{dr} &lt; \frac{\gamma - 1}{\gamma} \frac{d \ln P}{dr}.<br />$<br /> <br /> Both the temperature and pressure gradients are negative, which means we need to be careful here. It may be easier to write this in terms of the absolute values, so that the condition for convection to occur is <br /> $<br /> \left|\frac{d \ln T}{dr}\right| &gt; \frac{\gamma - 1}{\gamma} \left|\frac{d \ln P}{dr}\right|.<br />$<br /> We can interpret this then as saying that if the temperture gradient is too large for the pressure gradient, convection will set in. One more way to write this is <br /> $<br /> \left|\frac{d \ln T}{d \ln P}\right| &gt; \frac{\gamma - 1}{\gamma}.<br />$<br /> The reason to write it this way is that it is easy to calculate the quantity on the left hand side implied by hydrostatic equilibrium and radiative diffusion. In practice then, this last form is easy to use with the equations we already know. From radiative diffusion,<br /> $<br /> \frac{dP_{\rm rad}}{dr} \propto F.<br />$<br /> From hydrostatic equilibrium,<br /> $<br /> \frac{dP}{dr} = -\rho \frac{G M_r}{r^2}.<br />$<br /> Then <br /> $<br /> \frac{P}{P_{\rm rad}} \frac{dP_{\rm rad}}{dP} = \frac{L_r \kappa}{4 \pi G M_r c} \frac{P}{P_{\rm rad}},<br />$<br /> The leftmost part of the above is the log derivative of the radiation pressure with respect to the total pressure. In terms of temperature, this is<br /> $<br /> \frac{d \ln P_{\rm rad}}{d \ln P} = 4 \frac{d \ln T}{d \ln P}.<br />$<br /> Plug this in above, <br /> $<br /> \frac{d \ln T}{d \ln P} = \frac{1}{4} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M},<br />$<br /> and finally put this in our convection criterion as<br /> $<br /> \frac{1}{4} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M} &gt; \frac{\gamma - 1}{\gamma}. <br />$<br /> The Sun is convection for $r &gt; 0.7 R_\odot$. At smaller radii, photons carry the energy. At these large radii, $M_r \approx M$ and $L_r = L$. The fraction of radiation pressure relative to total pressure, and luminosity relative to Eddington luminosity are both constant, so the only variable that changes is the opacity $\kappa$. At lowish temperatures ($\sim 10^5$ K), the opacity becomes very large, radiative diffusion is inefficient at carrying energy, and convection dominates. As we move to less massive stars, the temperatures are lower and the gas is denser, and both act to increase the opacity. This means that a greater and greater fraction of the star is convective, down to about one third of a solar mass where the star is completely convective. In more massive stars, convective is less important in the atmosphere (and disappears for stars about twice as massive as the Sun). Core convection, however, becomes important. This is due to the way energy is generated in the core. Basically, there is a lot of energy in a very small region, and photons cannot get all of it out. \\ \\<br /> {\bf Energy Transport by Convection} \\<br /> Now that we have talked about convectoin criteria, we can talk about how it transports energy. The we way do this is with mixing length theory (where theory is applied generously here). We cannot predict the exact sizes of convective blobs or velocities. Rather, this is only good to about a factor of two or maybe a bit better. That we can then predict radii of stars to of order 1\% is then rather remarkable. \\ \\<br /> We can try to estimate the energy flux carried by convection by looking at the average energy difference between the hot blobs that rise and the cool blobs that sink, $\Delta E$ (where this is energy per volume). An energy density multiplied by a velocity is a flux, so we need the convective velocity to do this. We can also estimate the energy density by looking at the kinetic energy density of blobs, which is just $\rho_c v_c^2$. Estimating the energy density in the two different ways turns out to give similar answers, so we can estimate the convective flux as <br /> $<br /> F_c \sim \rho v_c^3.<br />$<br /> We can get this quantity by looking at the work done by acceleration due to bouyancy. We define the mixing length $\ell$ to be the typical distance traveled by convective blobs before exchanging energy. This is a macroscopic analogy of the mean free path, and the two quantities are very much related. We can make a guess for the mixing length to be of order the pressure scale height of the system. This is the distance over which the pressure changes by an appreciable amount, which we formally define to be<br /> $<br /> H = \left(\frac{d \ln P}{dr}\right)^{-1}.<br />$<br /> To encapsulate our ignorance, we'll add a fudge factor $\alpha$ and say<br /> $<br /> \ell = \alpha H.<br />$<br /> Our justification for (nearly) equating the mixing length and the pressure scale height has to do with the fact the blob is staying in pressure equilibrium with the surroundings. If the pressure of the surroundings changes greatly, then the density of the blob must also change, and in doing so become much larger. After traveling this pressure scale height, all of the blobs will have grown in this way, and will begin to collide, and in doing so exchange heat and lose its identity. \\ \\<br /> Now we want to take advantage of energy conservation. The energy gained by the fluid is equal to the work done by the bouyancy force. This means, in terms of our variables of interest,<br /> $<br /> v_c^2 \sim a \cdot \delta r \sim |N|^2 \delta r^2.<br />$<br /> We are relating then the energy gained to the effectiveness of bouyance. The distance moved, $\delta r$, we are defining as the mixing length $\ell$. We write the Brunt-Vaisala frequency in terms of the gravity and the specific heat at constant pressure as<br /> $<br /> |N|^2 = \frac{g}{C_P} \left|\frac{ds}{dr}\right|.<br />$<br /> We can write this instead in terms of a dimensionless entropy gradient by dividing and multiplying by the pressure scale height,<br /> $<br /> |N|^2 = \frac{g}{H} \left|\frac{H}{C_P}\frac{ds}{dr}\right|.<br />$<br /> The convective velocity is then<br /> $<br /> v_c^2 \sim g H \left|\frac{H}{C_P}\frac{ds}{dr}\right| \alpha^2.<br />$<br /> The scale height is just $\frac{kT}{mg}$, which then means that $gH$ is of order the average spead of individual particles squared, or the sound speed $c_s$. Then<br /> $<br /> v_c \sim \alpha c_s \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{1/2}.<br />$<br /> We can then use this to estimate the convective flux, with <br /> $<br /> F_c \sim \rho v_c^3 \sim \rho \alpha^3 c_s^3 \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{3/2}.<br />$<br /> Given $F_c$, by knowing how much energy is getting out, we can turn this around to estimate the convective velocity, or the entropy gradient. Plugging in some typical numbers for the convective zone in the Sun, the convective velocity is about 70 meters per second, which is about 1000 times smaller than the speed of sound. This, in turn, implies that the dimensionless entropy gradient is of order $10^{-6}$. That this is so small is a very important result. For practical purposes, the entropy is essentially constant in the convective zone. Or, in other words, barely any entropy gradient is required to get convection operating. That this number is so small is what allows us to understand the structure in the convective zone, since constant(ish) entropy gives $P \propto \rho^\gamma$.<br /> <br /> <br /> \subsection*{ Convective Transport of Energy} \\ <br /> The main results from the end of last lecture had to do with how energy is transported via convection. That is,<br /> $<br /> F_c \sim \rho v_c^3 \sim \rho \alpha^3 c_s^3 \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{3/2},<br />$<br /> with <br /> $<br /> v_c \sim \alpha c_s \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{1/2}.<br />$<br /> As a reminder, $c_s$ is the sound speed, $v_c$ is the convective velocity, and $\alpha$ is our dimensionless fudge factor that relates the pressure scale height and the convective mixing length, and basically summarizes the fact that we only really know all of this to a factor of order unity. The quantity inside the absolute value is the dimensionless entropy gradient, which tells us the strength of the bouyancy. This is actually set by the convective velocity. In a convective zone, the energy is being carried out by convection, meaning the convective velocity must have a certain value in order to be able to carry out the flux from the star, which means the convective velocity dictates the entropy gradient. For the Sun, the convective velocity is roughly 70 m s$^{-1}$, which implies a dimensionless entropy gradient of order $10{-6}$. This is very much smaller than 1, meaning this is very nearly adiabatic, and we can assume that the entropy is constant. <br /> \\ \\<br /> This is the key to solving for the structure of a star in its convective zone, or understanding a star that is entirely convective. Aside: this is a slightly different statement than the one we made before about the entropy of the individual blob. The entropy of the blob could be constant as it rises without the entropy profile of the star being constant. The constant entropy profile allows us to write<br /> $<br /> P \propto \rho^\gamma<br />$<br /> in the convective zone (again, the profile of the star, not for the blob alone). \\ \\<br /> Now, imagine that $P \propto \rho^\gamma$ is true throughout a star. If we do that, then we can combine it with hydrostatic equilibrium to solve for the structure of a star in the same way we did with radiative diffusion. First, we'll manipulate hydrostatic equilibrium to put it in more favorable form.<br /> $<br /> \frac{dP}{dr} = -\rho \frac{G M_r}{r^2},<br />$<br /> $<br /> \frac{dM_r}{dr} = 4 \pi r^2 \rho,<br />$<br /> $<br /> \frac{d}{dr}\left[\frac{r^2}{\rho}\frac{dP}{dr}\right] = \frac{d}{dr}\left[-GM_r\right],<br />$<br /> $<br /> \frac{d}{dr}\left[\frac{r^2}{\rho}\frac{dP}{dr}\right] = -4 \pi r^2 \rho G.<br />$<br /> Now, we use our assumption of $P \propto \rho^\gamma$, and invoke polytrope solutions. <br /> $<br /> P = K \rho^\gamma = K \rho^{1 + 1/n}.<br />$<br /> For fully convective stars, $\gamma = 5/3$ and $n = 3/2$. This also happens to be the case for non-relativistic fermions, which will be useful when we talk about the structure of white dwarfs. The other useful case is $\gamma = 4/3$ ($n = 3$), which describes fully relativistic fermions. We'll use these solutions in a few different types of stars during the semester. In some cases, they even work fairly well when the assumptions do not hold exactly. \\ \\<br /> We now write down various dimensionless quantities, so we can take advantage of known polytropic solutions.<br /> <br /> $<br /> \theta = \left(\frac{\rho}{\rho_c}\right)^{1/n},<br />$<br /> $<br /> \rho_c = \rho(r = 0),<br />$<br /> $<br /> \xi = \frac{r}{a}, <br />$<br /> $<br /> a = \left(\frac{(n + 1) K \rho_c^{\frac{1}{n} - 1}}{4 \pi G}\right)^{1/2}, <br />$<br /> $<br /> \frac{1}{\xi^2} \frac{d}{d\xi}\left(\xi^2 \frac{d\theta}{d\xi}\right) = -\theta^n.<br />$<br /> <br /> This final result is known as the Lane-Emden equation, and was used to understand, to some extent, the structure of stars before it was even known what the source of energy was inside of a star. This is a second order differential equation, so we need two boundary conditions. One condition is that<br /> $<br /> \theta(0) = 1,<br />$<br /> which is fairly evident. The other is that<br /> $<br /> \frac{d\theta}{d\xi}|_{\xi = 0} = 0<br />$<br /> This is a statement that the derivative smoothly goes to zero at the center. Basically, we run out of mass as $r$ goes to zero, which makes sense (but I skipped some justification). We also note that the radius is $r = \xi a$. These allow for the Lane-Emden equation to be solved for various polytropic indices $n$. We won't actually do this though. There are three known analytic solutions, for $n = 0, 1, 5$. We are interested in $n = 3/2, 3$. Meaning we need to solve this integral numerically, either with a computer or with a 19th century physicist by hand. \\ \\<br /> These polytropic models in general have two free parameters. One is the constant $K$, and the other is the central density $\rho_c$, for a given model with set $n$. Equivalently, the two free parameters are the mass and radius of the star. That is, $K$ and $\rho_c$ can be written in terms of $M$ and $r$. Real stellar models are uniquely determined by the mass alone (well, composition matters too, but that is a separate issue). Aside: $K$ is actually determined from physics alone for objects dominated by degeneracy pressure. Using $n=3$, polytropic models do quite well in interior of the Sun, as seen from comparing the polytropic solutions to the solutions of detailed models. They do a bit less well towards the outside of the Sun, where it becomes convective, and the polytropic index becomes $n=3/2$. \\ \\<br /> {\bf Fully Convective Objects} \\ <br /> The polytropes do really well for fully convective objects. This applies to stars that are less massive than about 1/3 $M_\odot$, as well as brown dwarfs and gas giant planets. Stars also go through a fully convective phase during their formation and during their post main sequence evolution. There are lots of cases then when the fully convective solution applies. We would like to derive a result for the mass and luminosity for fully convective objects that is analogous to the $L \propto M^3$ result we found for stars that transport energy by radiative diffusion and have Thomson scattering as their opacity source. We do this by using the result that convective objects have constant entropy, rather than trying to manipulate the expression for convective flux that we found previously. \\ \\<br /> We focus on objects dominated by gas pressure, meaning $\gamma = 5/3$ and $n = 3/2$. Using the solution of the Lane-Emden equation for such a polytrope,<br /> $<br /> P_c = 0.77 \frac{G M^2}{R^4},<br />$<br /> and<br /> $<br /> k T_c = 0.54 \frac{G M \mu m_p}{R}.<br />$<br /> The form of these results are familiar, as the look like the order of magnitude for each quantity when using hydrostatic equilibrium and the virial theorem. The polytrope models allow us to specify these much better than order of magnitude though. The constant will change as the polytrope index changes, and is accurate to the extent that $P \propto \rho^\gamma$ assumption is true. \\ \\<br /> We also know general proportionalities between the pressure, density, and temperature. Since $P \propto \rho T$ (ideal gas law),<br /> $<br /> P \propto T^{5/2}.<br />$<br /> Combining this with the expression at the center,<br /> $<br /> \frac{P_{ph}}{P_c} = \left(\frac{T_{ph}}{T_c}\right)^{5/2}.<br />$<br /> The temperature at the photosphere is just the effective temperature. Right at the photosphere, photons carry away the energy. This occurs when the mean free path of photons is of order the scale height in the Sun. Interior, the mean free path is smaller, and beyond the photosphere, the mean free path is much larger than the scale height, and photons are able to travel without interacting much with matter (free stream). The photons from the photosphere are the photons that we see from the Sun. The reason this is defined by the mean free path being of order the scale height is the dependence of the mean free path on density. The mean free path scales with the inverse of the density, and the scale height is the length over which the density changes appreciably. Thus the mean free path becomes much larger in this region. The mean free path is <br /> $<br /> \ell = \frac{1}{\kappa \rho},<br />$<br /> and<br /> $<br /> H = \frac{k T_{eff}}{m g}.<br />$<br /> Equation, the photosphere satisfies<br /> $<br /> \frac{1}{\kappa \rho} = \frac{k T_{eff}}{m g},<br />$<br /> or<br /> $<br /> \rho_{ph} = \frac{m g}{\kappa k_B T_{eff}}.<br />$<br /> Then the pressure in the photosphere is<br /> $<br /> P_{ph} = \frac{\rho k_B T_{eff}}{m} = \frac{g}{\kappa}.<br />$<br /> Now, rearranging our temperature/pressure relationship from before,<br /> $<br /> k T_{eff} = k T_c \left(\frac{P_{ph}}{P_c}\right)^{2/5},<br />$<br /> $<br /> k T_{eff} = k T_c \left(\frac{g}{\kappa_{ph} P_c}\right)^{2/5},<br />$<br /> $<br /> k T_{eff} = \frac{0.6 G M \mu m_p}{R} \left(\frac{R^2}{M \kappa_{ph}}\right)^{2/5}.<br />$<br /> <br /> Now we are almost there. We have the surface temperature in terms of the mass and radius, plus this opacity term. The opacity that matters in a convective star (which we could find by trial and error, but will say now and check later) is from the H$^-$ ion, since the effective temperature will be of order 3000 K. The H$^-$ has the function form<br /> $<br /> \kappa_{H^-} = 2.5 \times 10^{-31} \rho^{1/2} T^9 {\rm \;cm^2\; g}.<br />$<br /> If we substitue this in, it implies that the density at the photosphere is just a function of the effective temperature, which then means that the opacity at the photosphere is just a function of effective temperature. This just requires some algebra that we will skip, and can be used in the equation relating effective temperature, mass, and radius, which requires more algebra that we skip. The final result is<br /> $<br /> T_{eff} \approx 3000 \left(\frac{M}{M_\odot}\right)^{1/7} \left(\frac{R}{R_\odot}\right)^{1/49}.<br />$<br /> This is independent of fusion, as it tells us how quickly photons can leak out of a star. Note that this is self consistent, as at a temperature of 3000 K H$-$ will dominate, as we assumed. Writing the luminosity of a blackbody,<br /> $<br /> L = 4 \pi R^2 \sigma T_{eff}^4,<br />$<br /> we can plug in our result to find<br /> $<br /> L \approx 0.07 L_\odot \left(\frac{M}{M_\odot}\right)^{4/7} \left(\frac{R}{R_\odot}\right)^{102/49}.<br />$<br /> We can manipulate this in one more way, and write<br /> $<br /> T_{eff} = 3000 K \left(\frac{M}{M_\odot}\right)^{7/51} \left(\frac{L}{L_\odot}\right)^{1/102}.<br />$<br /> This form is useful for putting the result on an HR diagram. Convective objects lie on a basically straight line on the HR diagram. This line is called the Hayashi line. Again, no fusion is required. As stars form, they move down this Hayashi line, contracting until they become radiative and move on to the main sequence, unless they are so low massed that they are always convective. As some stars evolve off the main sequence, they become fully convective again, and move up the Hayashi line.<br /> <br /> <br /> &lt;/latex&gt;</div> Jmcbride Convection 2011-09-02T06:37:38Z <p>Jmcbride: convection!</p> <hr /> <div>&lt;latex&gt;<br /> <br /> \documentclass[preprint]{aastex}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> \newcommand{\rhoblob}{\rho_{\rm blob}}<br /> \newcommand{\rhostar}{\rho_{\rm star}}<br /> \newcommand{\pblob}{P_{\rm blob}}<br /> \newcommand{\pstar}{P_{\rm star}}<br /> <br /> \begin{document}<br /> Plenty of redundancy to be removed here, as this was three lectures.<br /> <br /> \subsection*{Thermodynamics} \\<br /> Convection is as important as radiative diffusion in main sequence stars. We'll make an analogy between convection and boiling, like water in a pot on the stove. At low temperatures, energy moves up through a pot through conduction. Once it gets hot enough though, the energy cannot move quickly enough through convection, causing the water to begin to boil. The basic physics that sets convection is bouyancy. Hot, underdense material at the bottom of the pot is less dense than the cooler water that sits atop it, leading to bubbles rising and depositing their energy at the top of the pot where the water was cooler. \\ \\ <br /> To understand all of this, we'll have to use some thermodynamic relationships between quantities like entropy and heat. Starting with the second law of thermodynamics, the change in entropy is given by the change in energy and volume as<br /> $<br /> T dS = dE + PdV.<br />$<br /> We don't really want to talk about volumes of stars though. So let's imagine a volume within a star with mass $M$, density $\rho$, and volume $V$. We'll focus on a this parcel with constant mass, and divide the above by the mass, to get the specific entropy in terms of a density rather than a volume. Then<br /> $<br /> \frac{dV}{M} = -\frac{d\rho}{\rho^2}.<br />$<br /> This gives for the second law, with $ds = dS / M$ and $dU = dE / M$,<br /> $<br /> T ds = dU - \frac{P}{\rho}\frac{d\rho}{\rho}.<br />$<br /> Now we imagine that the system is adiabatic, which means the heat content of the material is not exchanged with the surroundings. This is equivalent to <br /> $<br /> T ds = 0.<br />$<br /> In this case, the second law reduces to <br /> $<br /> dU = \frac{P}{\rho}{d\rho}{\rho}.<br />$<br /> The pressure and the energy per unit volume are nearly interchangeable, being related by a factor of order unit. We can define then an energy per unit volume $\epsilon$, with<br /> $<br /> \epsilon = U\rho.<br />$<br /> If we relate the pressure and energy density by some constant $\phi$,<br /> $<br /> P = \phi \epsilon = \phi U \rho,<br />$<br /> then the second law in the adiabatic case becomes<br /> $<br /> dU = \phi U \frac{d\rho}{\rho}, <br />$<br /> $<br /> \frac{dU}{U} = \phi \frac{d\rho}{\rho}, <br />$<br /> $<br /> \ln(U) = \phi \ln(\rho) + C, <br />$<br /> $<br /> U \propto \rho^\phi, <br />$<br /> $<br /> P \propto \rho^{\phi + 1}.<br />$<br /> In doing this, we have shown that the adiabatic index $\gamma$ is related to $\phi$ by<br /> $<br /> \gamma = \phi + 1.<br />$<br /> We can write then that<br /> $<br /> P \propto \rho^\gamma,<br />$<br /> $<br /> U \propto \rho^{\gamma - 1},<br />$<br /> $<br /> T \propto \rho^{\gamma - 1}.<br />$<br /> In the non-relativistic case, $\phi = 2/3$, $\gamma = 5/3$. For relativistic particles, $\phi = 1/3$, $\gamma = 4/3$. We can then use these relations to understand how pressure, energy, and temperature relate to density for either situation. \\ \\<br /> Now let's look at the entropy of an ideal gas. The second law (again) is <br /> $<br /> T ds = dU - \frac{P}{\rho}\frac{d\rho}{\rho}.<br />$<br /> The pressure is<br /> $<br /> P = U \rho (\gamma - 1).<br />$<br /> Then<br /> $<br /> T ds = dU - U(\gamma - 1) \frac{d\rho}{\rho}, <br />$<br /> <br /> and then after dividing by the energy,<br /> $<br /> \frac{T}{U} ds = \frac{dU}{U} - (\gamma - 1)\frac{d\rho}{\rho}.<br />$<br /> Writing the energy in terms of the temperature for an ideal gas in the normal way,<br /> $<br /> U = \frac{3 k_b T}{2 m},<br />$<br /> and then rearranging and writing in the general form,<br /> $<br /> \frac{T}{U} = \frac{m}{k_B} (\gamma - 1).<br />$<br /> We can then solve for the entropy in terms of the mass, energy, and density.<br /> <br /> $<br /> \frac{m}{k_B} (\gamma - 1) ds = \frac{dU}{U} - (\gamma - 1) \frac{d\rho}{\rho} <br />$<br /> $<br /> \frac{m}{k_B} (\gamma - 1) ds = \ln(U) - (\gamma - 1)\ln(\rho) <br />$<br /> $<br /> s = \frac{k_B}{(\gamma - 1)m} \ln\left(\frac{U}{\rho^{\gamma -1}}\right).<br />$<br /> <br /> We can use again the relationship between pressure, density, and energy per unit mass to write <br /> $<br /> s = \frac{k_B}{(\gamma - 1)m} \ln\left(\frac{P}{\rho^{\gamma}}\right).<br />$ %\bbox<br /> Now let's look at the change in time of entropy, again using the second law. The time derivative is<br /> $<br /> T \frac{ds}{dt} = \frac{dU}{dt} - \frac{P}{\rho^2}{d\rho}{dt}.<br />$<br /> This represents the heating per mass per time, or cooling per mass per time. We have written the radiative diffusion in terms of an energy per time per area. We would like to put the two in the same terms. To do this, we start with an integral over surface area to get energy per time.<br /> $<br /> {\rm energy\; per\; time} = \int dA \cdot F.<br />$<br /> If we use the divergence theorem, we can change the integral over area to one over volume by taking the divergence of the integrated quantity.<br /> $<br /> {\rm energy\; per\; time} = \int dV \nabla \cdot F.<br />$<br /> Thus the energy per time per volume is just $\nabla \cdot F$. The energy per unit time per unit mass is then just $\nabla \cdot F / \rho$. Then the second law can be written <br /> $<br /> T \frac{ds}{dt} = \frac{dU}{dt} - \frac{P}{\rho^2}\frac{d\rho}{dt} = \epsilon_{\rm fusion} - \frac{1}{\rho}\nabla \cdot F.<br />$<br /> Here we have included energy input from fusion, as well as energy moving by photons or convection. \\ \\<br /> Now that we have done all of this, let's determine where convection even matters. We have equations for force balance (hydrostatic equilibrium) and radiative diffusion that we can solve to find the temperature and pressure as a function of radius. That does not mean that the solution always represents the physical situation. The reason for this is that even if the solution works, it does not necessarily have to be stable. Convection is important when the solution of HE and radiative diffusion is not stable to small changes in the density. We can ask then what happens in a star when a small blob of low density starts to rise in a star, will it continue rising or will it fall back down to where it started? \\ \\<br /> The blob has some $\rho_{\rm blob}$, $T_{\rm blob}$, $s_{\rm blob}$, $P_{\rm blob}$. It then rises to some higher position, where we assume all of these quantities are set by radiative diffusion to be $\rho_{\rm star}$, $T_{\rm star}$, $s_{\rm star}$, $P_{\rm star}$. To see whether the blob moves up quickly, we'll assume the blob is adiabatic, meaning it does not exchange heat. We can say this because the boiling time for the Sun is order a month, whereas the timescale for energy to leak out is about $10^7$ years. We will also assume that the blob is in presure equilibrium, meaning the pressure of the blob is the same as the pressure of it surroundings. This is true so long as the boiling timescale is much larger than the dynamic timescale, which is the time required to maintain hydrostatic equilibrium. Since the dynamic timescale is about one hour, this is also a very good assumption. \\ \\<br /> With these assumptions, the entropy of the blob is constant as it rises, and will thus be different from the entropy of the surroundings, because in general there will be an entropy gradient in the star. Pressure equilibrium implies that the pressure of the blob is equal to the pressure of the surrounding star.<br /> <br /> $<br /> s_{\rm blob} /= s_{\rm star} <br />$<br /> $<br /> P_{\rm blob} = P_{\rm star}<br />$<br /> If the entropy gradient of the star is negative ($\frac{ds}{dr} &lt; 0$), then<br /> $s_{\rm star} &lt; s_{\rm blob}$. Since $S \propto \ln(P/\rho^\gamma)$, and the pressure of the blob is equal to the pressure of the surroundings, this implies that the density of the blob must be less than the density of its surroundings. This, in turn, means the blob is convectively unstable. If, on the other hand, there is a positive entropy gradient, the blob is convectively stable, by the exact same reason, but with the opposite sign. The condition that the entropy gradient by negative in order for convection to set in is known as the Schwarzschild criterion. Any solution of the structure of a star that gives a negative entropy gradient is an unstable solution.<br /> <br /> \subsection*{Convection} \\<br /> Quick conceptual review of last lecture: the density, pressure, temperature and entropy profile of a star all have a long time to come in to some sort of equilibrium state. A rising blob of material moves on a much smaller timescale though, and does not need to change as quickly. The blob does come in to pressure eqiulibrium with the star, but its entropy remain fixed. This is because the blob moves on a timescale much longer than the dynamical time for the star (time to come in to pressure equilibrium), but much shorter than the thermal timescale of the star. We arrived at the condition that convection takes places if and only if the entropy gradient is negative. When this is true, the density of the blob becomes less dense faster than the surrounding star becomes less dense as it rises through the star, meaning it is bouyant and the star is convectively unstable. \\ \\<br /> We can compare the density of the blob to the background density of the star at some position in the star.<br /> $<br /> \rho_{\rm star} = \rho + \frac{d\rho}{dr} \delta r,<br />$<br /> $<br /> P_{\rm star} = P + \frac{dP}{dr} \delta r,<br />$<br /> $<br /> \rho_{\rm blob} = \rho + \left(\frac{\delta \rho}{\delta P}\right)_s \delta \rho_{\rm blob}, <br />$<br /> $<br /> P_{\rm blob}= P + \delta P_{\rm blob} = P_{\rm star},<br />$<br /> $<br /> \rhoblob= \rho + \left(\frac{d\rho}{dP}\right)_s \frac{dP}{dr}dr, <br />$<br /> $<br /> P \propto \rho^\gamma<br />$<br /> $<br /> \left(\frac{dP}{d\rho}\right)_s = \frac{\gamma P}{\rho},<br />$<br /> $<br /> \rhoblob = \rho + \frac{\rho}{\gamma P} \frac{dP}{dr} \delta r.<br />$<br /> <br /> At some new position, the blob will feel an acceleration from its bouyancy, just by Archimedes' Principle.<br /> $<br /> a = g \left(\frac{\rhostar}{\rhoblob} - 1\right).<br />$<br /> If the density of the blob is less than that of the star, acceleration is positive, and the blob runs away. If the density of the blob is greater, the acceleration is negative and the blob sinks back down, so the star is convectively stable. The acceleration can be written in terms of what is called the Brunt-Vaisala frequency as <br /> $<br /> a = -N^2 \delta r,<br />$<br /> where we have defined <br /> $<br /> N^2 = -g \left(\frac{d \ln \rho}{dr} - \frac{1}{\gamma} \frac{d \ln P}{dr}\right).<br />$<br /> Alternatively, we can write it in terms of the entropy gradient, with<br /> $<br /> N^2 = \frac{\gamma - 1}{\gamma}\frac{m_p}{k_b} g \frac{ds}{dr}.<br />$<br /> The Brunt-Vaisala frequency is proportion then to the entropy gradient. We can write this down as a simple differential equation.<br /> $<br /> a = \delta \dot{r},<br />$<br /> $<br /> \delta \dot{r} + N^2 \delta r = 0, % should be double dot<br />$<br /> $<br /> \delta r \propto e^{\pm i N t}.<br />$<br /> <br /> If $N^2$ is greater than zero, we have an oscillating solution, and convective stability. This corresponds to a positive entropy gradient. If $N^2$ is negative, then the solution of this is an exponential runaway, on a timescale of $1 / |N|$. \\ \\<br /> We can write yet another convection criterion that is useful in stars. Using the negative entropy gradient condition and the relation between entropy and pressure and density, <br /> <br /> $<br /> \frac{d \ln P}{dr} - \gamma\frac{d \ln \rho}{dr} = 0,<br />$<br /> $<br /> P = \frac{\rho k_B T}{\mu m_p},<br />$<br /> $<br /> \ln P = \ln \rho + \ln T + {\rm constant},<br />$<br /> $<br /> \frac{d \ln P}{dr} - \gamma\frac{d \ln P}{dr} + \gamma \frac{d \ln T}{dr} &lt; 0, <br />$<br /> $<br /> \frac{d \ln T}{dr} &lt; \frac{\gamma - 1}{\gamma} \frac{d \ln P}{dr}.<br />$<br /> <br /> Both the temperature and pressure gradients are negative, which means we need to be careful here. It may be easier to write this in terms of the absolute values, so that the condition for convection to occur is <br /> $<br /> \left|\frac{d \ln T}{dr}\right| &gt; \frac{\gamma - 1}{\gamma} \left|\frac{d \ln P}{dr}\right|.<br />$<br /> We can interpret this then as saying that if the temperture gradient is too large for the pressure gradient, convection will set in. One more way to write this is <br /> $<br /> \left|\frac{d \ln T}{d \ln P}\right| &gt; \frac{\gamma - 1}{\gamma}.<br />$<br /> The reason to write it this way is that it is easy to calculate the quantity on the left hand side implied by hydrostatic equilibrium and radiative diffusion. In practice then, this last form is easy to use with the equations we already know. From radiative diffusion,<br /> $<br /> \frac{dP_{\rm rad}}{dr} \propto F.<br />$<br /> From hydrostatic equilibrium,<br /> $<br /> \frac{dP}{dr} = -\rho \frac{G M_r}{r^2}.<br />$<br /> Then <br /> $<br /> \frac{P}{P_{\rm rad}} \frac{dP_{\rm rad}}{dP} = \frac{L_r \kappa}{4 \pi G M_r c} \frac{P}{P_{\rm rad}},<br />$<br /> The leftmost part of the above is the log derivative of the radiation pressure with respect to the total pressure. In terms of temperature, this is<br /> $<br /> \frac{d \ln P_{\rm rad}}{d \ln P} = 4 \frac{d \ln T}{d \ln P}.<br />$<br /> Plug this in above, <br /> $<br /> \frac{d \ln T}{d \ln P} = \frac{1}{4} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M},<br />$<br /> and finally put this in our convection criterion as<br /> $<br /> \frac{1}{4} \frac{P}{P_{\rm rad}} \frac{L}{L_{edd}} \frac{L_r / L}{M_r / M} &gt; \frac{\gamma - 1}{\gamma}. <br />$<br /> The Sun is convection for $r &gt; 0.7 R_\odot$. At smaller radii, photons carry the energy. At these large radii, $M_r \approx M$ and $L_r = L$. The fraction of radiation pressure relative to total pressure, and luminosity relative to Eddington luminosity are both constant, so the only variable that changes is the opacity $\kappa$. At lowish temperatures ($\sim 10^5$ K), the opacity becomes very large, radiative diffusion is inefficient at carrying energy, and convection dominates. As we move to less massive stars, the temperatures are lower and the gas is denser, and both act to increase the opacity. This means that a greater and greater fraction of the star is convective, down to about one third of a solar mass where the star is completely convective. In more massive stars, convective is less important in the atmosphere (and disappears for stars about twice as massive as the Sun). Core convection, however, becomes important. This is due to the way energy is generated in the core. Basically, there is a lot of energy in a very small region, and photons cannot get all of it out. \\ \\<br /> {\bf Energy Transport by Convection} \\<br /> Now that we have talked about convectoin criteria, we can talk about how it transports energy. The we way do this is with mixing length theory (where theory is applied generously here). We cannot predict the exact sizes of convective blobs or velocities. Rather, this is only good to about a factor of two or maybe a bit better. That we can then predict radii of stars to of order 1\% is then rather remarkable. \\ \\<br /> We can try to estimate the energy flux carried by convection by looking at the average energy difference between the hot blobs that rise and the cool blobs that sink, $\Delta E$ (where this is energy per volume). An energy density multiplied by a velocity is a flux, so we need the convective velocity to do this. We can also estimate the energy density by looking at the kinetic energy density of blobs, which is just $\rho_c v_c^2$. Estimating the energy density in the two different ways turns out to give similar answers, so we can estimate the convective flux as <br /> $<br /> F_c \sim \rho v_c^3.<br />$<br /> We can get this quantity by looking at the work done by acceleration due to bouyancy. We define the mixing length $\ell$ to be the typical distance traveled by convective blobs before exchanging energy. This is a macroscopic analogy of the mean free path, and the two quantities are very much related. We can make a guess for the mixing length to be of order the pressure scale height of the system. This is the distance over which the pressure changes by an appreciable amount, which we formally define to be<br /> $<br /> H = \left(\frac{d \ln P}{dr}\right)^{-1}.<br />$<br /> To encapsulate our ignorance, we'll add a fudge factor $\alpha$ and say<br /> $<br /> \ell = \alpha H.<br />$<br /> Our justification for (nearly) equating the mixing length and the pressure scale height has to do with the fact the blob is staying in pressure equilibrium with the surroundings. If the pressure of the surroundings changes greatly, then the density of the blob must also change, and in doing so become much larger. After traveling this pressure scale height, all of the blobs will have grown in this way, and will begin to collide, and in doing so exchange heat and lose its identity. \\ \\<br /> Now we want to take advantage of energy conservation. The energy gained by the fluid is equal to the work done by the bouyancy force. This means, in terms of our variables of interest,<br /> $<br /> v_c^2 \sim a \cdot \delta r \sim |N|^2 \delta r^2.<br />$<br /> We are relating then the energy gained to the effectiveness of bouyance. The distance moved, $\delta r$, we are defining as the mixing length $\ell$. We write the Brunt-Vaisala frequency in terms of the gravity and the specific heat at constant pressure as<br /> $<br /> |N|^2 = \frac{g}{C_P} \left|\frac{ds}{dr}\right|.<br />$<br /> We can write this instead in terms of a dimensionless entropy gradient by dividing and multiplying by the pressure scale height,<br /> $<br /> |N|^2 = \frac{g}{H} \left|\frac{H}{C_P}\frac{ds}{dr}\right|.<br />$<br /> The convective velocity is then<br /> $<br /> v_c^2 \sim g H \left|\frac{H}{C_P}\frac{ds}{dr}\right| \alpha^2.<br />$<br /> The scale height is just $\frac{kT}{mg}$, which then means that $gH$ is of order the average spead of individual particles squared, or the sound speed $c_s$. Then<br /> $<br /> v_c \sim \alpha c_s \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{1/2}.<br />$<br /> We can then use this to estimate the convective flux, with <br /> $<br /> F_c \sim \rho v_c^3 \sim \rho \alpha^3 c_s^3 \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{3/2}.<br />$<br /> Given $F_c$, by knowing how much energy is getting out, we can turn this around to estimate the convective velocity, or the entropy gradient. Plugging in some typical numbers for the convective zone in the Sun, the convective velocity is about 70 meters per second, which is about 1000 times smaller than the speed of sound. This, in turn, implies that the dimensionless entropy gradient is of order $10^{-6}$. That this is so small is a very important result. For practical purposes, the entropy is essentially constant in the convective zone. Or, in other words, barely any entropy gradient is required to get convection operating. That this number is so small is what allows us to understand the structure in the convective zone, since constant(ish) entropy gives $P \propto \rho^\gamma$.<br /> <br /> <br /> \subsection*{ Convective Transport of Energy} \\ <br /> The main results from the end of last lecture had to do with how energy is transported via convection. That is,<br /> $<br /> F_c \sim \rho v_c^3 \sim \rho \alpha^3 c_s^3 \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{3/2},<br />$<br /> with <br /> $<br /> v_c \sim \alpha c_s \left|\frac{H}{C_P}\frac{ds}{dr}\right|^{1/2}.<br />$<br /> As a reminder, $c_s$ is the sound speed, $v_c$ is the convective velocity, and $\alpha$ is our dimensionless fudge factor that relates the pressure scale height and the convective mixing length, and basically summarizes the fact that we only really know all of this to a factor of order unity. The quantity inside the absolute value is the dimensionless entropy gradient, which tells us the strength of the bouyancy. This is actually set by the convective velocity. In a convective zone, the energy is being carried out by convection, meaning the convective velocity must have a certain value in order to be able to carry out the flux from the star, which means the convective velocity dictates the entropy gradient. For the Sun, the convective velocity is roughly 70 m s$^{-1}$, which implies a dimensionless entropy gradient of order $10{-6}$. This is very much smaller than 1, meaning this is very nearly adiabatic, and we can assume that the entropy is constant. <br /> \\ \\<br /> This is the key to solving for the structure of a star in its convective zone, or understanding a star that is entirely convective. Aside: this is a slightly different statement than the one we made before about the entropy of the individual blob. The entropy of the blob could be constant as it rises without the entropy profile of the star being constant. The constant entropy profile allows us to write<br /> $<br /> P \propto \rho^\gamma<br />$<br /> in the convective zone (again, the profile of the star, not for the blob alone). \\ \\<br /> Now, imagine that $P \propto \rho^\gamma$ is true throughout a star. If we do that, then we can combine it with hydrostatic equilibrium to solve for the structure of a star in the same way we did with radiative diffusion. First, we'll manipulate hydrostatic equilibrium to put it in more favorable form.<br /> $<br /> \frac{dP}{dr} = -\rho \frac{G M_r}{r^2},<br />$<br /> $<br /> \frac{dM_r}{dr} = 4 \pi r^2 \rho,<br />$<br /> $<br /> \frac{d}{dr}\left[\frac{r^2}{\rho}\frac{dP}{dr}\right] = \frac{d}{dr}\left[-GM_r\right],<br />$<br /> $<br /> \frac{d}{dr}\left[\frac{r^2}{\rho}\frac{dP}{dr}\right] = -4 \pi r^2 \rho G.<br />$<br /> Now, we use our assumption of $P \propto \rho^\gamma$, and invoke polytrope solutions. <br /> $<br /> P = K \rho^\gamma = K \rho^{1 + 1/n}.<br />$<br /> For fully convective stars, $\gamma = 5/3$ and $n = 3/2$. This also happens to be the case for non-relativistic fermions, which will be useful when we talk about the structure of white dwarfs. The other useful case is $\gamma = 4/3$ ($n = 3$), which describes fully relativistic fermions. We'll use these solutions in a few different types of stars during the semester. In some cases, they even work fairly well when the assumptions do not hold exactly. \\ \\<br /> We now write down various dimensionless quantities, so we can take advantage of known polytropic solutions.<br /> <br /> $<br /> \theta = \left(\frac{\rho}{\rho_c}\right)^{1/n},<br />$<br /> $<br /> \rho_c = \rho(r = 0),<br />$<br /> $<br /> \xi = \frac{r}{a}, <br />$<br /> $<br /> a = \left(\frac{(n + 1) K \rho_c^{\frac{1}{n} - 1}}{4 \pi G}\right)^{1/2}, <br />$<br /> $<br /> \frac{1}{\xi^2} \frac{d}{d\xi}\left(\xi^2 \frac{d\theta}{d\xi}\right) = -\theta^n.<br />$<br /> <br /> This final result is known as the Lane-Emden equation, and was used to understand, to some extent, the structure of stars before it was even known what the source of energy was inside of a star. This is a second order differential equation, so we need two boundary conditions. One condition is that<br /> $<br /> \theta(0) = 1,<br />$<br /> which is fairly evident. The other is that<br /> $<br /> \frac{d\theta}{d\xi}|_{\xi = 0} = 0<br />$<br /> This is a statement that the derivative smoothly goes to zero at the center. Basically, we run out of mass as $r$ goes to zero, which makes sense (but I skipped some justification). We also note that the radius is $r = \xi a$. These allow for the Lane-Emden equation to be solved for various polytropic indices $n$. We won't actually do this though. There are three known analytic solutions, for $n = 0, 1, 5$. We are interested in $n = 3/2, 3$. Meaning we need to solve this integral numerically, either with a computer or with a 19th century physicist by hand. \\ \\<br /> These polytropic models in general have two free parameters. One is the constant $K$, and the other is the central density $\rho_c$, for a given model with set $n$. Equivalently, the two free parameters are the mass and radius of the star. That is, $K$ and $\rho_c$ can be written in terms of $M$ and $r$. Real stellar models are uniquely determined by the mass alone (well, composition matters too, but that is a separate issue). Aside: $K$ is actually determined from physics alone for objects dominated by degeneracy pressure. Using $n=3$, polytropic models do quite well in interior of the Sun, as seen from comparing the polytropic solutions to the solutions of detailed models. They do a bit less well towards the outside of the Sun, where it becomes convective, and the polytropic index becomes $n=3/2$. \\ \\<br /> {\bf Fully Convective Objects} \\ <br /> The polytropes do really well for fully convective objects. This applies to stars that are less massive than about 1/3 $M_\odot$, as well as brown dwarfs and gas giant planets. Stars also go through a fully convective phase during their formation and during their post main sequence evolution. There are lots of cases then when the fully convective solution applies. We would like to derive a result for the mass and luminosity for fully convective objects that is analogous to the $L \propto M^3$ result we found for stars that transport energy by radiative diffusion and have Thomson scattering as their opacity source. We do this by using the result that convective objects have constant entropy, rather than trying to manipulate the expression for convective flux that we found previously. \\ \\<br /> We focus on objects dominated by gas pressure, meaning $\gamma = 5/3$ and $n = 3/2$. Using the solution of the Lane-Emden equation for such a polytrope,<br /> $<br /> P_c = 0.77 \frac{G M^2}{R^4},<br />$<br /> and<br /> $<br /> k T_c = 0.54 \frac{G M \mu m_p}{R}.<br />$<br /> The form of these results are familiar, as the look like the order of magnitude for each quantity when using hydrostatic equilibrium and the virial theorem. The polytrope models allow us to specify these much better than order of magnitude though. The constant will change as the polytrope index changes, and is accurate to the extent that $P \propto \rho^\gamma$ assumption is true. \\ \\<br /> We also know general proportionalities between the pressure, density, and temperature. Since $P \propto \rho T$ (ideal gas law),<br /> $<br /> P \propto T^{5/2}.<br />$<br /> Combining this with the expression at the center,<br /> $<br /> \frac{P_{ph}}{P_c} = \left(\frac{T_{ph}}{T_c}\right)^{5/2}.<br />$<br /> The temperature at the photosphere is just the effective temperature. Right at the photosphere, photons carry away the energy. This occurs when the mean free path of photons is of order the scale height in the Sun. Interior, the mean free path is smaller, and beyond the photosphere, the mean free path is much larger than the scale height, and photons are able to travel without interacting much with matter (free stream). The photons from the photosphere are the photons that we see from the Sun. The reason this is defined by the mean free path being of order the scale height is the dependence of the mean free path on density. The mean free path scales with the inverse of the density, and the scale height is the length over which the density changes appreciably. Thus the mean free path becomes much larger in this region. The mean free path is <br /> $<br /> \ell = \frac{1}{\kappa \rho},<br />$<br /> and<br /> $<br /> H = \frac{k T_{eff}}{m g}.<br />$<br /> Equation, the photosphere satisfies<br /> $<br /> \frac{1}{\kappa \rho} = \frac{k T_{eff}}{m g},<br />$<br /> or<br /> $<br /> \rho_{ph} = \frac{m g}{\kappa k_B T_{eff}}.<br />$<br /> Then the pressure in the photosphere is<br /> $<br /> P_{ph} = \frac{\rho k_B T_{eff}}{m} = \frac{g}{\kappa}.<br />$<br /> Now, rearranging our temperature/pressure relationship from before,<br /> $<br /> k T_{eff} = k T_c \left(\frac{P_{ph}}{P_c}\right)^{2/5},<br />$<br /> $<br /> k T_{eff} = k T_c \left(\frac{g}{\kappa_{ph} P_c}\right)^{2/5},<br />$<br /> $<br /> k T_{eff} = \frac{0.6 G M \mu m_p}{R} \left(\frac{R^2}{M \kappa_{ph}}\right)^{2/5}.<br />$<br /> <br /> Now we are almost there. We have the surface temperature in terms of the mass and radius, plus this opacity term. The opacity that matters in a convective star (which we could find by trial and error, but will say now and check later) is from the H$^-$ ion, since the effective temperature will be of order 3000 K. The H$^-$ has the function form<br /> $<br /> \kappa_{H^-} = 2.5 \times 10^{-31} \rho^{1/2} T^9 {\rm \;cm^2\; g}.<br />$<br /> If we substitue this in, it implies that the density at the photosphere is just a function of the effective temperature, which then means that the opacity at the photosphere is just a function of effective temperature. This just requires some algebra that we will skip, and can be used in the equation relating effective temperature, mass, and radius, which requires more algebra that we skip. The final result is<br /> $<br /> T_{eff} \approx 3000 \left(\frac{M}{M_\odot}\right)^{1/7} \left(\frac{R}{R_\odot}\right)^{1/49}.<br />$<br /> This is independent of fusion, as it tells us how quickly photons can leak out of a star. Note that this is self consistent, as at a temperature of 3000 K H$-$ will dominate, as we assumed. Writing the luminosity of a blackbody,<br /> $<br /> L = 4 \pi R^2 \sigma T_{eff}^4,<br />$<br /> we can plug in our result to find<br /> $<br /> L \approx 0.07 L_\odot \left(\frac{M}{M_\odot}\right)^{4/7} \left(\frac{R}{R_\odot}\right)^{102/49}.<br />$<br /> We can manipulate this in one more way, and write<br /> $<br /> T_{eff} = 3000 K \left(\frac{M}{M_\odot}\right)^{7/51} \left(\frac{L}{L_\odot}\right)^{1/102}.<br />$<br /> This form is useful for putting the result on an HR diagram. Convective objects lie on a basically straight line on the HR diagram. This line is called the Hayashi line. Again, no fusion is required. As stars form, they move down this Hayashi line, contracting until they become radiative and move on to the main sequence, unless they are so low massed that they are always convective. As some stars evolve off the main sequence, they become fully convective again, and move up the Hayashi line.<br /> <br /> <br /> &lt;/latex&gt;</div> Jmcbride Stellar Structure 2011-09-02T06:10:55Z <p>Jmcbride: </p> <hr /> <div>==Description==<br /> These are lecture notes from an upper division undergraduate course in stellar structure, taught by Eliot Quataert. For most topics, the level of these is similar to that in the graduate version of the course.<br /> <br /> ==Lecture==<br /> #[[Stars Lecture 1]]<br /> #[[Stars Lecture 2]]<br /> #[[Radiative transport and conduction]]<br /> #[[Stars Lecture 5]]<br /> #[[Stars Lecture 6]]<br /> #[[Stars Lecture 7]]<br /> #[[Stars Lecture 8]]<br /> #[[Stars Lecture 9]]<br /> #[[Stars Lecture 10]]<br /> #[[Stars Lecture 11]]<br /> #[[Stars Lecture 12]]<br /> #[[Stars Lecture 13]]<br /> #[[Stars Lecture 14]]<br /> #[[Stars Lecture 15]]<br /> #[[Stars Lecture 16]]<br /> #[[Stars Lecture 17]]<br /> #[[Stars Lecture 18]]<br /> #[[Stars Lecture 19]]<br /> #[[Stars Lecture 20]]<br /> #[[Stars Lecture 21]]<br /> #[[Stars Lecture 22]]<br /> #[[Stars Lecture 23]]</div> Jmcbride Stars Lecture 3 2011-09-02T06:10:46Z <p>Jmcbride: moved Stars Lecture 3 to Radiative transport and conduction:&amp;#32;actual topic of lecture</p> <hr /> <div>#REDIRECT [[Radiative transport and conduction]]</div> Jmcbride Radiative transport and conduction 2011-09-02T06:10:46Z <p>Jmcbride: moved Stars Lecture 3 to Radiative transport and conduction:&amp;#32;actual topic of lecture</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> <br /> <br /> \begin{document}<br /> \subsection*{Part one} \\ \\<br /> The picture to have in mind is that there are lots of particles moving around and colliding, which deflects their motion and exchanges energy with other particles and causes a flow of energy from more energetic areas to less energetic areas. The same basic physics applies for energy transport by both photons and electrons. \\ \\<br /> Particles have an energy density $U$, a typical velocity $v$, and a mean free path $\ell$, where the mean free path represents the average distance traveled between exchanging energy. There is flux $F$ in some direction that is just the energy density multiplied by the velocity. Eliot drew a picture as a basis for the argument that the net flux is<br /> $<br /> F = \frac{1}{6} U(x - \ell) v - \frac{1}{6} U(x + \ell) v,<br />$<br /> $<br /> F = -\frac{1}{3} v \ell \frac{dU}{dx}, <br />$<br /> $<br /> F = -\frac{1}{3} v \ell \frac{dU}{dT} \frac{dT}{dx},<br />$<br /> $<br /> F = -\kappa \frac{dT}{dx}, {\rm \; where}<br />$<br /> $<br /> \kappa \sim \frac{1}{3} v \ell \frac{dU}{dT}.<br />$<br /> <br /> Now that we have gone through this, let's define the mean free path a bit more concretely, specifically in terms of the cross section for collisions, $\sigma$. For a flux of incoming particles, (an energy/area/time), we define the energy/time absorbed by the particles as the product of the flux and the cross section $\sigma$:<br /> $<br /> P = \sigma F.<br />$<br /> Once we have defined a cross section, we can determine how far a particle is likely to travel before interacting with another particle. This depends on the number density of particles as well, with<br /> $<br /> \ell = \frac{1}{n \sigma}.<br />$<br /> Let's estimate this for a particle in the Earth's atmosphere. The density of air is $\rho \sim 10^{-3}$ g/cm$^3$, which gives a number density of roughly $n \sim 10^{20}$ particles/cm$^3$. The cross section for two neutral atoms or molecules is determined simply by the size of the constituents. For a typical molecule, the size is of order an Angstrom, or $10^{-8}$ cm. The mean free path for molecules in the atmosphere is then<br /> $<br /> \ell \sim 10^{-4} {\rm \; cm,\; or\; one\; micron}.<br />$<br /> Let's now focus on an ideal ionized gas. The energy per unit volume is $\frac{3}{2} n k_B T$. The conductivity $\kappa$ is then <br /> $<br /> \kappa \sim \frac{1}{2} v \ell n k_b.<br />$<br /> This is the appropriate conductivity for energy transport by electrons. For electrons in a gas, <br /> $<br /> v \sim \sqrt{\frac{kT}{m}}.<br />$<br /> For the cross section, the Coulomb cross section is what is relevant. We define this in terms of the distance between an electron and a proton at which the proton causes a large deflection of the motion of the electron. This depends on the energy of the electron from thermal motion, and the electrostatic energy that the electron feels from a proton, which is<br /> $<br /> E_e = \frac{e^2}{r}.<br />$<br /> When this energy is similar to the thermal energy $k_B T$, an electron will scatter. Thus significant deflections occur at<br /> $<br /> b = \frac{e^2}{kT},<br />$<br /> where $b$ is known as the impact parameter. The Coulomb cross section is then<br /> $<br /> \sigma_c = \pi b^2 = \pi \frac{e^4}{k_B^2 T^2}.<br />$<br /> Note that this is independent of the mass of the particle. When this calculation is done more carefully, the right hand side is multiplied by a factor called the Coulomb logarithm, which typically is of order 10. This accounts for the cumulative effect of many small angle deflections. Though this does not depend on mass, the overall conductivity does, with<br /> $<br /> \kappa \sim \frac{k_B^3 v T^2}{\pi e^4} \propto \frac{T^{5/2}}{\sqrt{m}}<br />$<br /> However, electrons move more quickly in a gas, so they are able to carry the energy further between scatterings, and are thus more efficient at energy transport. \\ \\<br /> We can perform a crude estimate to determine if this is important in the Sun.<br /> $<br /> F_r = -\kappa \frac{dT}{dr} = \frac{L_r}{4 \pi r^2}.<br />$<br /> Making some approximations to simplify (like dropping differentials),<br /> $<br /> L \sim 4 \pi R \kappa T_c, <br />$<br /> $<br /> L \sim \frac{k_b^{7/2} T_c^{7/2} R}{e^4 \sqrt{m_e}}, <br />$<br /> $<br /> L \sim 10^{-4} \left(\frac{R}{R_\odot}\right) \left(\frac{T_c}{10^7 K}\right)^{7/2} L_\odot.<br />$<br /> This is much too small to carry out the energy we observe, despite evaluating this under the most favorable of conditions for electron conduction to be important. We can then confidently say that conduction is not important to carrying out energy for a star on the main sequence.<br /> \\<br /> \\<br /> While this exercise showed we did not need to consider conduction, we can use a very similar derivation for photons. For photons, what changes is that $v = c$, $U = a T^4$, and the mean free path is different (smaller, in fact). Then the flux from photons is <br /> $<br /> F = -\frac{4caT^3}{3n\sigma}{dT}{dx}.<br />$<br /> This is in analogy with the conduction equation from before, so we can define a photon conductivity in analogy to before of <br /> $<br /> \kappa_{ph} = \frac{4 c a T^3}{n \sigma}.<br />$<br /> This is often written alternatively in terms of the opacity and density rather than cross section and number density, with <br /> $<br /> F = -\frac{4caT^3}{3\kappa \rho}{dT}{dx}.<br />$<br /> It is important to keep in mind that the density is the density of the scatters, not the photons. It turns out that electrons are what dominate scattering. The cross section for radiation for electrons can be derived in a nice way. We can start by writing down the Lorentz force for an electron<br /> $<br /> m_e \frac{dv}{dt} = -e(\overrightarrow{E} + \frac{\overrightarrow{v}}{c} \times \overrightarrow{B}).<br />$<br /> Since the magnitude of $E$ and $B$ are comparable, we need to only worry about the electric field, and can see that <br /> $<br /> \overrightarrow{a} = -\frac{e\overrightarrow{E}}{m_e}.<br />$<br /> We now need to see what the power radiation by an electron that is being accelerated by the presence of an electromagnetic wave. This is given by the Larmor formula as<br /> $<br /> P = \frac{2e^2}{3c^3} |a|^2.<br />$<br /> Substituting the acceleration from above,<br /> $<br /> P = \frac{2e^4}{3m_e^2c^3} |E|^2.<br />$<br /> If we go back to the picture before for determing the cross section, we see that the flux and the power absorbed or scattered are related by the cross section, with<br /> $<br /> P = \sigma F.<br />$<br /> The Larmor power represents then reradiated power due to the photon electron interaction. We know the reradiated power, and we also know that the incoming flux is just the Poynting flux, given as<br /> $<br /> F = \frac{c|E|^2}{4 \pi}.<br />$<br /> Solving $P = \sigma F$, we find the Thomson scattering cross section is<br /> $<br /> \sigma = \frac{8 \pi}{3}\frac{e^4}{m_e^2 c^4}.<br />$<br /> This is independent of the electric field, as it should be since it is a property of the particle. It is also independent of frequency (except at high frequency, where some implicit assumptions break down). This is also sometimes written as<br /> $<br /> \sigma \sim \pi r_c^2,<br />$<br /> where $r_c$ is the classical electron radius, which is the radius found from equation the rest mass energy to the electrostatic self energy of an electron.<br /> \subsection*{Part two}<br /> Our main result from last time was the radiative diffusion equation.<br /> $<br /> F = -\frac{4acT^3}{3n\sigma} \frac{dT}{dx}.<br />$<br /> We will also put this in terms of the opacity and density at times, with<br /> $<br /> \ell = \frac{1}{n\sigma} = \frac{1}{\kappa \rho}.<br />$<br /> Now, let's use this to estimate the power that can be carried out of the Sun by photons. We write the luminosity $L$ as<br /> $<br /> L = 4 \pi R^2 F.<br />$<br /> Then, using the radiative diffusion equation and dropping differentials,<br /> $<br /> L \sim 4 \pi R^2 \frac{a c T^4}{n \sigma R}.<br />$<br /> Assuming an ionized plasma where Thomson scattering is dominant, we know the cross section. We can then put the density in terms of the mass and radius,<br /> $<br /> L \sim \frac{16 \pi R^4 T^4 \mu_e m_p a c}{M \sigma_T}.<br />$<br /> We can remove the temperature from this equation by using the Virial Theorem. Specifically,<br /> $<br /> kT = \frac{G M \mu m_p}{3 R}.<br />$<br /> Putting this in the above expression for luminosity, the radius conveniently cancels out, leaving an expression for the luminosity carried by radiative diffusion as a function of mass:<br /> $<br /> L \sim \frac{a (\mu m_p)^4 (\mu_e m_p) c G^4 M^3}{\sigma_T k^4}.<br />$<br /> Putting in some numbers, we get<br /> $<br /> L \sim 10^{35} \left(\frac{M}{M_\odot}\right)^3 {\rm erg\; s}^{-1}.<br />$<br /> It is clear that radiation is sufficient to carry out energy, while electron conduction was not. Perhaps the most remarkable fact about this is that the luminosity of a star is independent of its energy generation mechanism. The luminosity is instead set by how much energy can leak out of a self gravitating object (so long as photons are carrying the energy out). Thus we can explain the observed relationship between luminosity and mass for stars on the main sequence without knowing how they generate energy. The only other property of the star that remains in this epression is the composition of the star, with<br /> $<br /> L \propto \mu_e \mu^4.<br />$<br /> This means that the as the composition of a star changes during its main sequence life, its luminosity will change. Since the mean weight of a particle in a star increases through conversion of four hydrogen to one helium, this means that stars grow brighter during their lifetime on the main sequence. For the Sun, the luminosity when it is was formed was only 75\% of what it is today, meaning the temperature on the Earth would be 15 K cooler than it is today, not accounting for the atmosphere. Some theorize for this reason, in conjunction with the observed life on Earth, that the atmosphere was thicker early in Earth's history, allowing it to trap more of the Sun's heat. \\ \\<br /> We can compare in a different way energy transport by conduction and radiation, since we wrote down a diffusion equation for each. Taking the ratio of the diffusion equations,<br /> $<br /> \frac{F_{rad}}{F_{cond}} = \frac{a T^3 c \sigma_{e - e}}{n \sigma_{\gamma - e} k_B v_e}.<br />$<br /> We can rearrange in the following suggestive manner:<br /> $<br /> \frac{F_{rad}}{F_{cond}} = \frac{aT^4}{n k_B T} \frac{c}{v_e} \frac{\sigma_{e -e}}{\sigma_{\gamma - e}}.<br />$<br /> Electrons are more important in the first ratio on the right hand side, which is the ratio of radiation pressure to gas pressure. Radiation travels faster though and has a much larger mean free path, both of which lead to photons dominating the transport of energy in stars. \\ \\<br /> Now, let's calculate the timescale for energy to leave a star via radiation. We'll find the Kelvin-Helmholtz timescale again, but in a slightly different way.<br /> $<br /> t_{KH} = \frac{E}{L}<br />$<br /> $<br /> t_{KH} = \frac{n k_B T R^3 R n \sigma}{a T^4 c R^2}<br />$<br /> $<br /> t_{KH} = \frac{nk_B T}{aT^4} \frac{R^2}{c \ell}.<br />$<br /> The second fraction on the right hand side of the last line is the timescale for a photone ot leave the Sun. If photons traveled on straight lines, they would take only two seconds to leave, but the time we find from <br /> $<br /> t \sim \frac{R}{\ell}\frac{R}{c}<br />$<br /> is of order $10^4$ years, much longer than the free streaming time. This time is actually the timescale for the photon to get out on a random walk out of the Sun. The root mean square distance that a photon travels on a random walk is<br /> $<br /> \sqrt{&lt;|D|^2&gt;} = N^{1/2} \ell,<br />$<br /> where $N$ is the number of steps taken and $\ell$ is as always the mean free pat. The total number of steps to get out of the Sun is then<br /> $<br /> N \sim \left(\frac{R}{\ell}\right)^2.<br />$<br /> The time to random walk out is then the product of the number of steps and the time per step:<br /> $<br /> t_{RW} = N \frac{\ell}{c} = \frac{R^2}{\ell c}.<br />$<br /> We can do one more thing right now with the radiative diffusion equation. If we write it in terms of the radiation pressure, which is<br /> $<br /> P_{rad} = \frac{1}{3} a T^4,<br />$<br /> or in differential form,<br /> $<br /> \frac{dP_{rad}}{dr} = \frac{4}{3} a T^3 \frac{dT}{dr}.<br />$<br /> This looks the same as part of the radiative diffusion equation, and by examination, we write<br /> $<br /> F_r = -\frac{c}{\kappa \rho} \frac{d}{dr}\left(P_{rad}\right).<br />$<br /> In terms of the luminosity,<br /> $<br /> \frac{dP_{rad}}{dr} = -\frac{L_r \kappa \rho}{4 \pi r^2 c}.<br />$<br /> We can compare this to hydrostatic equilibrium, which looks somewhat similar.<br /> $<br /> \frac{dP}{dr} = - \rho \frac{G M_r}{r^2}.<br />$<br /> The pressure in HE is generally the total pressure, but is dominated by gas pressure for the Sun, and dominated by radiation for very massive stars. If we divide these two equations by one another, we get a ratio of radiation and total pressure<br /> $<br /> \frac{dP_{rad}}{dP} = \frac{L_r \kappa}{4 \pi G M_r c}.<br />$<br /> This tells us the fractional contribution of radiation pressure to total pressure in a star. We now define the Eddington luminosity as<br /> $<br /> L_{edd}(r) = \frac{4 \pi G M_r c}{\kappa}.<br />$<br /> This gives the very simple expression<br /> $<br /> \frac{dP_{rad}}{dP} = \frac{L_r}{L_{edd}(r)}.<br />$<br /> This tells us that if the radiation pressure is comparable to the total pressure, the luminosity is close to the Eddington luminosity. There is a somewhat simpler way of deriving the Eddington luminosity. The total momentum in photons radiatied by a star is just $L/c$, since each photon has momentum of $E/c$. The force then due to absorption of radiation is <br /> $<br /> \frac{dp}{dt} = \frac{L}{c 4 \pi r^2} \sigma.<br />$<br /> Now we can ask when the radiation force is equal to the force of gravity. Equating the two,<br /> $<br /> \frac{L \sigma}{c 4 \pi r^2} = \frac{G M m}{r^2}.<br />$<br /> Canceling the radius terms and rearranging, we find the same expression for the Eddington luminosity (or limit) of<br /> $<br /> L = \frac{4 \pi G M c}{\kappa}.<br />$<br /> We see that if the luminosity of a star exceeds the Eddington luminosity, matter is blown away from a star, since the force outwards exceeds the inward force due to gravity. Now, let's put in some numbers to see what this actually is. We'll use the Thomson cross section (and remember that even though the electrons are what do the scattering, the protons have to get dragged along too, so we need to use the proton mass, and not the electron mass) and find<br /> $<br /> L_{edd} \sim 10^{38} \left(\frac{M}{M_\odot}\right) {\rm erg\; s}^{-1}.<br />$<br /> This far exceeds the luminosity of the Sun, so clearly the Sun is not radiation pressure dominated. We can find an estimate of when radiation pressure does matter by comparing how the luminosity of a star scales with mass, and comparing that to how the Eddington luminosity scales with mass. That is, where does<br /> $<br /> L = L_\odot \left(\frac{M}{M_\odot}\right)^3 = L_{edd}.<br />$<br /> Evaluating this gives an estimate for where radiation pressure is important of <br /> $<br /> M \sim 100 M_\odot.<br />$<br /> We have assumed all along that the opacity is due to Thomson scattering, but that is not always the case. Since opacity influences the rate of energy leakage, which in turn sets the structure of the star, we should note other important opacity sources now (and in more detail later). The table shows the dominant opacity for various temperature regimes. \\ \\<br /> \begin{table}<br /> \begin{center}<br /> \begin{tabular}{l|l}<br /> {\bf Temperature Regime} &amp; {\bf Dominant Opacity} \\ \hline \hline<br /> $T &gt; 10^7$ K &amp; Thomson scattering \\ \hline<br /> $T \sim 10^{4-6}$ K &amp; photo-ionization of atoms, free-free \\ \hline<br /> $T \sim 10^{3-4}$ K &amp; H$^{-}$ \\ \hline<br /> $T &lt; 10^3$ K &amp; Molecules and dust<br /> \end{tabular}<br /> \end{center}<br /> \end{table}<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Radiative transport and conduction 2011-09-02T06:10:13Z <p>Jmcbride: decided to go thematically here</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> <br /> <br /> \begin{document}<br /> \subsection*{Part one} \\ \\<br /> The picture to have in mind is that there are lots of particles moving around and colliding, which deflects their motion and exchanges energy with other particles and causes a flow of energy from more energetic areas to less energetic areas. The same basic physics applies for energy transport by both photons and electrons. \\ \\<br /> Particles have an energy density $U$, a typical velocity $v$, and a mean free path $\ell$, where the mean free path represents the average distance traveled between exchanging energy. There is flux $F$ in some direction that is just the energy density multiplied by the velocity. Eliot drew a picture as a basis for the argument that the net flux is<br /> $<br /> F = \frac{1}{6} U(x - \ell) v - \frac{1}{6} U(x + \ell) v,<br />$<br /> $<br /> F = -\frac{1}{3} v \ell \frac{dU}{dx}, <br />$<br /> $<br /> F = -\frac{1}{3} v \ell \frac{dU}{dT} \frac{dT}{dx},<br />$<br /> $<br /> F = -\kappa \frac{dT}{dx}, {\rm \; where}<br />$<br /> $<br /> \kappa \sim \frac{1}{3} v \ell \frac{dU}{dT}.<br />$<br /> <br /> Now that we have gone through this, let's define the mean free path a bit more concretely, specifically in terms of the cross section for collisions, $\sigma$. For a flux of incoming particles, (an energy/area/time), we define the energy/time absorbed by the particles as the product of the flux and the cross section $\sigma$:<br /> $<br /> P = \sigma F.<br />$<br /> Once we have defined a cross section, we can determine how far a particle is likely to travel before interacting with another particle. This depends on the number density of particles as well, with<br /> $<br /> \ell = \frac{1}{n \sigma}.<br />$<br /> Let's estimate this for a particle in the Earth's atmosphere. The density of air is $\rho \sim 10^{-3}$ g/cm$^3$, which gives a number density of roughly $n \sim 10^{20}$ particles/cm$^3$. The cross section for two neutral atoms or molecules is determined simply by the size of the constituents. For a typical molecule, the size is of order an Angstrom, or $10^{-8}$ cm. The mean free path for molecules in the atmosphere is then<br /> $<br /> \ell \sim 10^{-4} {\rm \; cm,\; or\; one\; micron}.<br />$<br /> Let's now focus on an ideal ionized gas. The energy per unit volume is $\frac{3}{2} n k_B T$. The conductivity $\kappa$ is then <br /> $<br /> \kappa \sim \frac{1}{2} v \ell n k_b.<br />$<br /> This is the appropriate conductivity for energy transport by electrons. For electrons in a gas, <br /> $<br /> v \sim \sqrt{\frac{kT}{m}}.<br />$<br /> For the cross section, the Coulomb cross section is what is relevant. We define this in terms of the distance between an electron and a proton at which the proton causes a large deflection of the motion of the electron. This depends on the energy of the electron from thermal motion, and the electrostatic energy that the electron feels from a proton, which is<br /> $<br /> E_e = \frac{e^2}{r}.<br />$<br /> When this energy is similar to the thermal energy $k_B T$, an electron will scatter. Thus significant deflections occur at<br /> $<br /> b = \frac{e^2}{kT},<br />$<br /> where $b$ is known as the impact parameter. The Coulomb cross section is then<br /> $<br /> \sigma_c = \pi b^2 = \pi \frac{e^4}{k_B^2 T^2}.<br />$<br /> Note that this is independent of the mass of the particle. When this calculation is done more carefully, the right hand side is multiplied by a factor called the Coulomb logarithm, which typically is of order 10. This accounts for the cumulative effect of many small angle deflections. Though this does not depend on mass, the overall conductivity does, with<br /> $<br /> \kappa \sim \frac{k_B^3 v T^2}{\pi e^4} \propto \frac{T^{5/2}}{\sqrt{m}}<br />$<br /> However, electrons move more quickly in a gas, so they are able to carry the energy further between scatterings, and are thus more efficient at energy transport. \\ \\<br /> We can perform a crude estimate to determine if this is important in the Sun.<br /> $<br /> F_r = -\kappa \frac{dT}{dr} = \frac{L_r}{4 \pi r^2}.<br />$<br /> Making some approximations to simplify (like dropping differentials),<br /> $<br /> L \sim 4 \pi R \kappa T_c, <br />$<br /> $<br /> L \sim \frac{k_b^{7/2} T_c^{7/2} R}{e^4 \sqrt{m_e}}, <br />$<br /> $<br /> L \sim 10^{-4} \left(\frac{R}{R_\odot}\right) \left(\frac{T_c}{10^7 K}\right)^{7/2} L_\odot.<br />$<br /> This is much too small to carry out the energy we observe, despite evaluating this under the most favorable of conditions for electron conduction to be important. We can then confidently say that conduction is not important to carrying out energy for a star on the main sequence.<br /> \\<br /> \\<br /> While this exercise showed we did not need to consider conduction, we can use a very similar derivation for photons. For photons, what changes is that $v = c$, $U = a T^4$, and the mean free path is different (smaller, in fact). Then the flux from photons is <br /> $<br /> F = -\frac{4caT^3}{3n\sigma}{dT}{dx}.<br />$<br /> This is in analogy with the conduction equation from before, so we can define a photon conductivity in analogy to before of <br /> $<br /> \kappa_{ph} = \frac{4 c a T^3}{n \sigma}.<br />$<br /> This is often written alternatively in terms of the opacity and density rather than cross section and number density, with <br /> $<br /> F = -\frac{4caT^3}{3\kappa \rho}{dT}{dx}.<br />$<br /> It is important to keep in mind that the density is the density of the scatters, not the photons. It turns out that electrons are what dominate scattering. The cross section for radiation for electrons can be derived in a nice way. We can start by writing down the Lorentz force for an electron<br /> $<br /> m_e \frac{dv}{dt} = -e(\overrightarrow{E} + \frac{\overrightarrow{v}}{c} \times \overrightarrow{B}).<br />$<br /> Since the magnitude of $E$ and $B$ are comparable, we need to only worry about the electric field, and can see that <br /> $<br /> \overrightarrow{a} = -\frac{e\overrightarrow{E}}{m_e}.<br />$<br /> We now need to see what the power radiation by an electron that is being accelerated by the presence of an electromagnetic wave. This is given by the Larmor formula as<br /> $<br /> P = \frac{2e^2}{3c^3} |a|^2.<br />$<br /> Substituting the acceleration from above,<br /> $<br /> P = \frac{2e^4}{3m_e^2c^3} |E|^2.<br />$<br /> If we go back to the picture before for determing the cross section, we see that the flux and the power absorbed or scattered are related by the cross section, with<br /> $<br /> P = \sigma F.<br />$<br /> The Larmor power represents then reradiated power due to the photon electron interaction. We know the reradiated power, and we also know that the incoming flux is just the Poynting flux, given as<br /> $<br /> F = \frac{c|E|^2}{4 \pi}.<br />$<br /> Solving $P = \sigma F$, we find the Thomson scattering cross section is<br /> $<br /> \sigma = \frac{8 \pi}{3}\frac{e^4}{m_e^2 c^4}.<br />$<br /> This is independent of the electric field, as it should be since it is a property of the particle. It is also independent of frequency (except at high frequency, where some implicit assumptions break down). This is also sometimes written as<br /> $<br /> \sigma \sim \pi r_c^2,<br />$<br /> where $r_c$ is the classical electron radius, which is the radius found from equation the rest mass energy to the electrostatic self energy of an electron.<br /> \subsection*{Part two}<br /> Our main result from last time was the radiative diffusion equation.<br /> $<br /> F = -\frac{4acT^3}{3n\sigma} \frac{dT}{dx}.<br />$<br /> We will also put this in terms of the opacity and density at times, with<br /> $<br /> \ell = \frac{1}{n\sigma} = \frac{1}{\kappa \rho}.<br />$<br /> Now, let's use this to estimate the power that can be carried out of the Sun by photons. We write the luminosity $L$ as<br /> $<br /> L = 4 \pi R^2 F.<br />$<br /> Then, using the radiative diffusion equation and dropping differentials,<br /> $<br /> L \sim 4 \pi R^2 \frac{a c T^4}{n \sigma R}.<br />$<br /> Assuming an ionized plasma where Thomson scattering is dominant, we know the cross section. We can then put the density in terms of the mass and radius,<br /> $<br /> L \sim \frac{16 \pi R^4 T^4 \mu_e m_p a c}{M \sigma_T}.<br />$<br /> We can remove the temperature from this equation by using the Virial Theorem. Specifically,<br /> $<br /> kT = \frac{G M \mu m_p}{3 R}.<br />$<br /> Putting this in the above expression for luminosity, the radius conveniently cancels out, leaving an expression for the luminosity carried by radiative diffusion as a function of mass:<br /> $<br /> L \sim \frac{a (\mu m_p)^4 (\mu_e m_p) c G^4 M^3}{\sigma_T k^4}.<br />$<br /> Putting in some numbers, we get<br /> $<br /> L \sim 10^{35} \left(\frac{M}{M_\odot}\right)^3 {\rm erg\; s}^{-1}.<br />$<br /> It is clear that radiation is sufficient to carry out energy, while electron conduction was not. Perhaps the most remarkable fact about this is that the luminosity of a star is independent of its energy generation mechanism. The luminosity is instead set by how much energy can leak out of a self gravitating object (so long as photons are carrying the energy out). Thus we can explain the observed relationship between luminosity and mass for stars on the main sequence without knowing how they generate energy. The only other property of the star that remains in this epression is the composition of the star, with<br /> $<br /> L \propto \mu_e \mu^4.<br />$<br /> This means that the as the composition of a star changes during its main sequence life, its luminosity will change. Since the mean weight of a particle in a star increases through conversion of four hydrogen to one helium, this means that stars grow brighter during their lifetime on the main sequence. For the Sun, the luminosity when it is was formed was only 75\% of what it is today, meaning the temperature on the Earth would be 15 K cooler than it is today, not accounting for the atmosphere. Some theorize for this reason, in conjunction with the observed life on Earth, that the atmosphere was thicker early in Earth's history, allowing it to trap more of the Sun's heat. \\ \\<br /> We can compare in a different way energy transport by conduction and radiation, since we wrote down a diffusion equation for each. Taking the ratio of the diffusion equations,<br /> $<br /> \frac{F_{rad}}{F_{cond}} = \frac{a T^3 c \sigma_{e - e}}{n \sigma_{\gamma - e} k_B v_e}.<br />$<br /> We can rearrange in the following suggestive manner:<br /> $<br /> \frac{F_{rad}}{F_{cond}} = \frac{aT^4}{n k_B T} \frac{c}{v_e} \frac{\sigma_{e -e}}{\sigma_{\gamma - e}}.<br />$<br /> Electrons are more important in the first ratio on the right hand side, which is the ratio of radiation pressure to gas pressure. Radiation travels faster though and has a much larger mean free path, both of which lead to photons dominating the transport of energy in stars. \\ \\<br /> Now, let's calculate the timescale for energy to leave a star via radiation. We'll find the Kelvin-Helmholtz timescale again, but in a slightly different way.<br /> $<br /> t_{KH} = \frac{E}{L}<br />$<br /> $<br /> t_{KH} = \frac{n k_B T R^3 R n \sigma}{a T^4 c R^2}<br />$<br /> $<br /> t_{KH} = \frac{nk_B T}{aT^4} \frac{R^2}{c \ell}.<br />$<br /> The second fraction on the right hand side of the last line is the timescale for a photone ot leave the Sun. If photons traveled on straight lines, they would take only two seconds to leave, but the time we find from <br /> $<br /> t \sim \frac{R}{\ell}\frac{R}{c}<br />$<br /> is of order $10^4$ years, much longer than the free streaming time. This time is actually the timescale for the photon to get out on a random walk out of the Sun. The root mean square distance that a photon travels on a random walk is<br /> $<br /> \sqrt{&lt;|D|^2&gt;} = N^{1/2} \ell,<br />$<br /> where $N$ is the number of steps taken and $\ell$ is as always the mean free pat. The total number of steps to get out of the Sun is then<br /> $<br /> N \sim \left(\frac{R}{\ell}\right)^2.<br />$<br /> The time to random walk out is then the product of the number of steps and the time per step:<br /> $<br /> t_{RW} = N \frac{\ell}{c} = \frac{R^2}{\ell c}.<br />$<br /> We can do one more thing right now with the radiative diffusion equation. If we write it in terms of the radiation pressure, which is<br /> $<br /> P_{rad} = \frac{1}{3} a T^4,<br />$<br /> or in differential form,<br /> $<br /> \frac{dP_{rad}}{dr} = \frac{4}{3} a T^3 \frac{dT}{dr}.<br />$<br /> This looks the same as part of the radiative diffusion equation, and by examination, we write<br /> $<br /> F_r = -\frac{c}{\kappa \rho} \frac{d}{dr}\left(P_{rad}\right).<br />$<br /> In terms of the luminosity,<br /> $<br /> \frac{dP_{rad}}{dr} = -\frac{L_r \kappa \rho}{4 \pi r^2 c}.<br />$<br /> We can compare this to hydrostatic equilibrium, which looks somewhat similar.<br /> $<br /> \frac{dP}{dr} = - \rho \frac{G M_r}{r^2}.<br />$<br /> The pressure in HE is generally the total pressure, but is dominated by gas pressure for the Sun, and dominated by radiation for very massive stars. If we divide these two equations by one another, we get a ratio of radiation and total pressure<br /> $<br /> \frac{dP_{rad}}{dP} = \frac{L_r \kappa}{4 \pi G M_r c}.<br />$<br /> This tells us the fractional contribution of radiation pressure to total pressure in a star. We now define the Eddington luminosity as<br /> $<br /> L_{edd}(r) = \frac{4 \pi G M_r c}{\kappa}.<br />$<br /> This gives the very simple expression<br /> $<br /> \frac{dP_{rad}}{dP} = \frac{L_r}{L_{edd}(r)}.<br />$<br /> This tells us that if the radiation pressure is comparable to the total pressure, the luminosity is close to the Eddington luminosity. There is a somewhat simpler way of deriving the Eddington luminosity. The total momentum in photons radiatied by a star is just $L/c$, since each photon has momentum of $E/c$. The force then due to absorption of radiation is <br /> $<br /> \frac{dp}{dt} = \frac{L}{c 4 \pi r^2} \sigma.<br />$<br /> Now we can ask when the radiation force is equal to the force of gravity. Equating the two,<br /> $<br /> \frac{L \sigma}{c 4 \pi r^2} = \frac{G M m}{r^2}.<br />$<br /> Canceling the radius terms and rearranging, we find the same expression for the Eddington luminosity (or limit) of<br /> $<br /> L = \frac{4 \pi G M c}{\kappa}.<br />$<br /> We see that if the luminosity of a star exceeds the Eddington luminosity, matter is blown away from a star, since the force outwards exceeds the inward force due to gravity. Now, let's put in some numbers to see what this actually is. We'll use the Thomson cross section (and remember that even though the electrons are what do the scattering, the protons have to get dragged along too, so we need to use the proton mass, and not the electron mass) and find<br /> $<br /> L_{edd} \sim 10^{38} \left(\frac{M}{M_\odot}\right) {\rm erg\; s}^{-1}.<br />$<br /> This far exceeds the luminosity of the Sun, so clearly the Sun is not radiation pressure dominated. We can find an estimate of when radiation pressure does matter by comparing how the luminosity of a star scales with mass, and comparing that to how the Eddington luminosity scales with mass. That is, where does<br /> $<br /> L = L_\odot \left(\frac{M}{M_\odot}\right)^3 = L_{edd}.<br />$<br /> Evaluating this gives an estimate for where radiation pressure is important of <br /> $<br /> M \sim 100 M_\odot.<br />$<br /> We have assumed all along that the opacity is due to Thomson scattering, but that is not always the case. Since opacity influences the rate of energy leakage, which in turn sets the structure of the star, we should note other important opacity sources now (and in more detail later). The table shows the dominant opacity for various temperature regimes. \\ \\<br /> \begin{table}<br /> \begin{center}<br /> \begin{tabular}{l|l}<br /> {\bf Temperature Regime} &amp; {\bf Dominant Opacity} \\ \hline \hline<br /> $T &gt; 10^7$ K &amp; Thomson scattering \\ \hline<br /> $T \sim 10^{4-6}$ K &amp; photo-ionization of atoms, free-free \\ \hline<br /> $T \sim 10^{3-4}$ K &amp; H$^{-}$ \\ \hline<br /> $T &lt; 10^3$ K &amp; Molecules and dust<br /> \end{tabular}<br /> \end{center}<br /> \end{table}<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Radiative transport and conduction 2011-09-02T06:05:32Z <p>Jmcbride: Created page with '&lt;latex&gt; \documentclass[preprint]{aastex} \setlength{\parindent}{0in} \pagestyle{empty} \begin{document} \subsection*{Radiation Transport and Conduction} \\ \\ The picture to …'</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> <br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> <br /> <br /> \begin{document}<br /> \subsection*{Radiation Transport and Conduction} \\ \\<br /> The picture to have in mind is that there are lots of particles moving around and colliding, which deflects their motion and exchanges energy with other particles and causes a flow of energy from more energetic areas to less energetic areas. The same basic physics applies for energy transport by both photons and electrons. \\ \\<br /> Particles have an energy density $U$, a typical velocity $v$, and a mean free path $\ell$, where the mean free path represents the average distance traveled between exchanging energy. There is flux $F$ in some direction that is just the energy density multiplied by the velocity. Eliot drew a picture as a basis for the argument that the net flux is<br /> $<br /> F = \frac{1}{6} U(x - \ell) v - \frac{1}{6} U(x + \ell) v,<br />$<br /> $<br /> F = -\frac{1}{3} v \ell \frac{dU}{dx}, <br />$<br /> $<br /> F = -\frac{1}{3} v \ell \frac{dU}{dT} \frac{dT}{dx},<br />$<br /> $<br /> F = -\kappa \frac{dT}{dx}, {\rm \; where}<br />$<br /> $<br /> \kappa \sim \frac{1}{3} v \ell \frac{dU}{dT}.<br />$<br /> <br /> Now that we have gone through this, let's define the mean free path a bit more concretely, specifically in terms of the cross section for collisions, $\sigma$. For a flux of incoming particles, (an energy/area/time), we define the energy/time absorbed by the particles as the product of the flux and the cross section $\sigma$:<br /> $<br /> P = \sigma F.<br />$<br /> Once we have defined a cross section, we can determine how far a particle is likely to travel before interacting with another particle. This depends on the number density of particles as well, with<br /> $<br /> \ell = \frac{1}{n \sigma}.<br />$<br /> Let's estimate this for a particle in the Earth's atmosphere. The density of air is $\rho \sim 10^{-3}$ g/cm$^3$, which gives a number density of roughly $n \sim 10^{20}$ particles/cm$^3$. The cross section for two neutral atoms or molecules is determined simply by the size of the constituents. For a typical molecule, the size is of order an Angstrom, or $10^{-8}$ cm. The mean free path for molecules in the atmosphere is then<br /> $<br /> \ell \sim 10^{-4} {\rm \; cm,\; or\; one\; micron}.<br />$<br /> Let's now focus on an ideal ionized gas. The energy per unit volume is $\frac{3}{2} n k_B T$. The conductivity $\kappa$ is then <br /> $<br /> \kappa \sim \frac{1}{2} v \ell n k_b.<br />$<br /> This is the appropriate conductivity for energy transport by electrons. For electrons in a gas, <br /> $<br /> v \sim \sqrt{\frac{kT}{m}}.<br />$<br /> For the cross section, the Coulomb cross section is what is relevant. We define this in terms of the distance between an electron and a proton at which the proton causes a large deflection of the motion of the electron. This depends on the energy of the electron from thermal motion, and the electrostatic energy that the electron feels from a proton, which is<br /> $<br /> E_e = \frac{e^2}{r}.<br />$<br /> When this energy is similar to the thermal energy $k_B T$, an electron will scatter. Thus significant deflections occur at<br /> $<br /> b = \frac{e^2}{kT},<br />$<br /> where $b$ is known as the impact parameter. The Coulomb cross section is then<br /> $<br /> \sigma_c = \pi b^2 = \pi \frac{e^4}{k_B^2 T^2}.<br />$<br /> Note that this is independent of the mass of the particle. When this calculation is done more carefully, the right hand side is multiplied by a factor called the Coulomb logarithm, which typically is of order 10. This accounts for the cumulative effect of many small angle deflections. Though this does not depend on mass, the overall conductivity does, with<br /> $<br /> \kappa \sim \frac{k_B^3 v T^2}{\pi e^4} \propto \frac{T^{5/2}}{\sqrt{m}}<br />$<br /> However, electrons move more quickly in a gas, so they are able to carry the energy further between scatterings, and are thus more efficient at energy transport. \\ \\<br /> We can perform a crude estimate to determine if this is important in the Sun.<br /> $<br /> F_r = -\kappa \frac{dT}{dr} = \frac{L_r}{4 \pi r^2}.<br />$<br /> Making some approximations to simplify (like dropping differentials),<br /> $<br /> L \sim 4 \pi R \kappa T_c, <br />$<br /> $<br /> L \sim \frac{k_b^{7/2} T_c^{7/2} R}{e^4 \sqrt{m_e}}, <br />$<br /> $<br /> L \sim 10^{-4} \left(\frac{R}{R_\odot}\right) \left(\frac{T_c}{10^7 K}\right)^{7/2} L_\odot.<br />$<br /> This is much too small to carry out the energy we observe, despite evaluating this under the most favorable of conditions for electron conduction to be important. We can then confidently say that conduction is not important to carrying out energy for a star on the main sequence.<br /> \\<br /> \\<br /> While this exercise showed we did not need to consider conduction, we can use a very similar derivation for photons. For photons, what changes is that $v = c$, $U = a T^4$, and the mean free path is different (smaller, in fact). Then the flux from photons is <br /> $<br /> F = -\frac{4caT^3}{3n\sigma}{dT}{dx}.<br />$<br /> This is in analogy with the conduction equation from before, so we can define a photon conductivity in analogy to before of <br /> $<br /> \kappa_{ph} = \frac{4 c a T^3}{n \sigma}.<br />$<br /> This is often written alternatively in terms of the opacity and density rather than cross section and number density, with <br /> $<br /> F = -\frac{4caT^3}{3\kappa \rho}{dT}{dx}.<br />$<br /> It is important to keep in mind that the density is the density of the scatters, not the photons. It turns out that electrons are what dominate scattering. The cross section for radiation for electrons can be derived in a nice way. We can start by writing down the Lorentz force for an electron<br /> $<br /> m_e \frac{dv}{dt} = -e(\overrightarrow{E} + \frac{\overrightarrow{v}}{c} \times \overrightarrow{B}).<br />$<br /> Since the magnitude of $E$ and $B$ are comparable, we need to only worry about the electric field, and can see that <br /> $<br /> \overrightarrow{a} = -\frac{e\overrightarrow{E}}{m_e}.<br />$<br /> We now need to see what the power radiation by an electron that is being accelerated by the presence of an electromagnetic wave. This is given by the Larmor formula as<br /> $<br /> P = \frac{2e^2}{3c^3} |a|^2.<br />$<br /> Substituting the acceleration from above,<br /> $<br /> P = \frac{2e^4}{3m_e^2c^3} |E|^2.<br />$<br /> If we go back to the picture before for determing the cross section, we see that the flux and the power absorbed or scattered are related by the cross section, with<br /> $<br /> P = \sigma F.<br />$<br /> The Larmor power represents then reradiated power due to the photon electron interaction. We know the reradiated power, and we also know that the incoming flux is just the Poynting flux, given as<br /> $<br /> F = \frac{c|E|^2}{4 \pi}.<br />$<br /> Solving $P = \sigma F$, we find the Thomson scattering cross section is<br /> $<br /> \sigma = \frac{8 \pi}{3}\frac{e^4}{m_e^2 c^4}.<br />$<br /> This is independent of the electric field, as it should be since it is a property of the particle. It is also independent of frequency (except at high frequency, where some implicit assumptions break down). This is also sometimes written as<br /> $<br /> \sigma \sim \pi r_c^2,<br />$<br /> where $r_c$ is the classical electron radius, which is the radius found from equation the rest mass energy to the electrostatic self energy of an electron.<br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride Virial Theorem 2011-09-02T05:59:06Z <p>Jmcbride: fixed brackets</p> <hr /> <div>&lt;latex&gt;<br /> \documentclass[preprint]{aastex}<br /> \setlength{\parindent}{0in}<br /> \pagestyle{empty}<br /> <br /> <br /> \begin{document}<br /> <br /> \subsection*{ Virial Theorem} \\ \\<br /> Start out with hydrostatic equilibrium:<br /> $<br /> \frac{dP}{dr} = - \rho \frac{G M_r}{r^2}<br />$<br /> We want this to have units of energy when we integrate this equation over the entire star. In order to do this, we multiply by the factor $4 \pi r^3 dr$. Now let's look at the right hand side of this equation.<br /> $<br /> - \int^R_0 \rho \frac{G M_r}{r^2} \dot 4 \pi r^3 dr = - \int^R_0 dM_r \frac{G M_r}{r} = U.<br />$<br /> So the right side just gives us the gravitational potential energy. Now we look at the left hand side, using the same factor of $4 \pi r^3 dr$ and a little trick. That is, use the fact that <br /> $<br /> \frac{d}{dr}\left(4 \pi r^3 P\right) = 4 \pi r^3 \frac{dP}{dr} + 3 \times 4 \pi r^2 P.<br />$<br /> Now we use this on the left hand side of hydrostatic equilibrium to get<br /> $<br /> \int^R_0 4 \pi r^3 dr \frac{dP}{dr} = \int^R_0 dr \frac{d}{dr}\left(4 \pi r^3 P\right) - 3 \int^R_0 4 \pi r^2 P dr.<br />$<br /> The first factor just goes to zero, since the radius at the center is zero, and the pressure at the boundary is (nearly) zero. Thus we have<br /> $<br /> U = - 3 \int^R_0 4 \pi r^2 P dr, {\rm \; or}<br />$<br /> $<br /> U = -3 &lt;P&gt; V.<br />$<br /> The total energy in the star is <br /> $<br /> E_{tot} = K + U = -3P&lt;P&gt; + K. <br />$<br /> Here $K$ represents the total kinetic energy in the star. This kinetic energy is dominated by the random motions of particles in the gas, otherwise known as the thermal energy. Ordered motions like convection and rotation are unimportant.<br /> For an ideal gas at temperature $T$, the energy per particle is given by <br /> $<br /> \epsilon = \frac{3}{2} k_B T, {\rm \; (non-relativistic)}<br />$<br /> $<br /> \epsilon = 3 k_B T {\rm \; (relativistic).}<br />$<br /> More generally, this is <br /> $<br /> \epsilon = \frac{1}{\gamma - 1} k_B T.<br />$<br /> Thus the total thermal energy of the a star is given by <br /> $<br /> K = \int^R_0 \frac{1}{\gamma -1} k T n dV.<br />$<br /> The integrand is just the energy per volume, which is the same thing as a pressure. Thus the total thermal energy of a star is given by<br /> $<br /> K = \frac{1}{\gamma - 1} \int^R_0 P dV = \frac{1}{\gamma - 1} &lt;P&gt; V.<br />$<br /> The gravitational potential energy and kinetic energy are related by the adiabatic index, with <br /> $<br /> K = \frac{-U}{3 (\gamma - 1)}.<br />$<br /> In the non-relativistic case,<br /> $<br /> K = -\frac{U}{2},<br />$<br /> and in the relativistic case<br /> $<br /> K = -U.<br />$<br /> We can also write the total energy of the star in these terms:<br /> $<br /> E_{tot} = K (1 - 3(\gamma - 1)).<br />$<br /> This entire derivation assumes the fact that the gas is classical and ideal. Even so, the result holds much more generally, including in white dwarfs (which are degenerate) and in systems where photons dominate the pressure (high mass stars). \\ \\<br /> Now we want to find a general relationship between pressure, thermal energy, and the cases where photons or quantum effects are important. We will imagine a system with particles zooming around and bouncing off of a wall, where the wall has area $A$. Then the force on the wall from collisions is <br /> $<br /> \frac{\Delta p}{\Delta t} = P A.<br />$<br /> We could also write the change in momentum as the product of the rate at which paticles hit the wall and the change in momentum per collision. The number of particles in a length is $n A$, where $n$ is the number density of particles. Then the rate is just <br /> $<br /> {\rm rate\;} = \frac{1}{2} n A v_x.<br />$<br /> The change in momentum is use two times the x momentum of the particles, so that<br /> $<br /> \frac{\Delta p}{\Delta t} = n A v_x p_x.<br />$<br /> If we look at three dimensions rather than one, we have a factor of one third come in, which gives us<br /> $<br /> \frac{\Delta p}{\Delta t} = \frac{1}{3} n A v p = P A.<br />$<br /> The area cancels out, as it should, and we have<br /> $<br /> P = \frac{1}{3} n v p.<br />$ <br /> This is a fundamental relationship between the microscopic particles and the macroscopic, thermodynamic properties of that gas. Generally, we have a gas with a range of velocities, so we need to average over the distribution function of particles, meaning the pressure is<br /> $<br /> P = \frac{1}{3} n &lt;v p&gt;.<br />$<br /> For non-relativistic particles, the momentum and velocity are just related by the mass of the particle, meaning<br /> $<br /> P = \frac{2}{3} n &lt;\frac{1}{2} m v^2&gt;.<br />$<br /> The averaged quantity is just the kinetic energy per particle, so<br /> $<br /> P = \frac{2 K}{3 V}.<br />$<br /> This is exactly what we had before for the specific case of an ideal gas, and gives us the same result for the Virial theorem we had before. We could do the same thing for a photon, or relativistic massive particle (and Eliot did), but the procedure and result are the same. \\ \\<br /> Now let's use the Virial theorem to derive some useful results. We start by imaginging a system in HE that has no internal energy source (like fusion) that is radiating energy with power $L$. Thus the total energy, which is already negative, is decreasing, or, in an absolute value sense, the total energy is becoming {\em more} negative. That is, $|E_{tot}|$ is increasing, $|U|$ is increasing, and $K$ is increasing. The way that this happens is by contracting, specifically a process called Kelvin-Helmholtz contraction. At any stage in the life of a star, if it does not have an internal energy source, this will happen. We can estimate the timescale of this contraction (meaning the time it takes to change appreciably, say $R$ changing of order $R/2$) using the total energy and the rate at which energy is being radiatied:<br /> $<br /> t_{KH} \sim \frac{|U|}{L}.<br />$<br /> If we imagine we know nothing about the internal energy source of the Sun, and ask what the Kelvin-Helmholtz timescale is for the Sun, we would find that<br /> $<br /> t_{KH} = 3 \times 10^7 {\rm years}.<br />$<br /> This is much less than the age of the Earth or the solar system, but that did not stop Kelvin and Helmholtz from proposing that this is how the energy of the Sun was generated, which caused great debates with biologists and geologists who knew the Earth must be older. What Kelvin and Helmholtz did not know is that fusion kicked in at the center and halted this process, though the Sun did undergo this contraction during its formation. \\ \\<br /> Now let's imagine the opposite situation, where more energy is being generated than is being radiatied. Then the total energy is increasing, but $|E_{tot}|$ actually goes down, meaning $K$ decreases and $|U|$ decreases. This is pretty weird. If you add energy to the star, it gets cooler. From a thermodynamics standpoint, this means stars have a negative heat capacity. In fact, all self gravitating objects have a negative heat capacity. This is unlike anything we experience on Earth, where adding energy to a system causes it to heat up. \\ \\<br /> This is the fundamental reason that fusing stars are different from fusion bombs. A bomb is heated, which causes fusion and energy release, which further heats the bomb, leading to a runaway explosion. The star fuses in a controlled manner thanks to gravity, which acts as a safety valve. Any extra fusion in a star causes the star to expand and cool, thereby slowing the rate of fusion until it comes back in to equilibrium. Recreating this on Earth is pretty challenging, and is the reason we do not have fusion energy yet. \\ \\<br /> The Virial Theorem can also give us an estimate of the temperature in stars. Consider a star (our Sun) supported by a non-relativistic classical gas. Then<br /> $<br /> U = -2K,<br />$<br /> $<br /> \frac{G M^2}{R} \sim 3 N k_B T \sim 3 \frac{M}{m_p} k_B T.<br />$<br /> This gives<br /> $<br /> 3 k_B T \sim \frac{G M m_p}{R}.<br />$<br /> For the sun, this is about $5 \times 10^6$ K, which is much larger than the effective temperature, and a little bit smaller than the central temperature.<br /> <br /> <br /> <br /> <br /> \end{document}<br /> &lt;/latex&gt;</div> Jmcbride