# Difference between revisions of "Convection"

Plenty of redundancy to be removed here, as this was three lectures.

### Thermodynamics

\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. \\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

${\displaystyle TdS=dE+PdV.\,\!}$

We don’t really want to talk about volumes of stars though. So let’s imagine a volume within a star with mass ${\displaystyle M}$, density ${\displaystyle \rho }$, and volume ${\displaystyle V}$. 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

${\displaystyle {\frac {dV}{M}}=-{\frac {d\rho }{\rho ^{2}}}.\,\!}$

This gives for the second law, with ${\displaystyle ds=dS/M}$ and ${\displaystyle dU=dE/M}$,

${\displaystyle Tds=dU-{\frac {P}{\rho }}{\frac {d\rho }{\rho }}.\,\!}$

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

${\displaystyle Tds=0.\,\!}$

In this case, the second law reduces to

${\displaystyle dU={\frac {P}{\rho }}{d\rho }{\rho }.\,\!}$

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 ${\displaystyle \epsilon }$, with

${\displaystyle \epsilon =U\rho .\,\!}$

If we relate the pressure and energy density by some constant ${\displaystyle \phi }$,

${\displaystyle P=\phi \epsilon =\phi U\rho ,\,\!}$

then the second law in the adiabatic case becomes

${\displaystyle dU=\phi U{\frac {d\rho }{\rho }},\,\!}$
${\displaystyle {\frac {dU}{U}}=\phi {\frac {d\rho }{\rho }},\,\!}$
${\displaystyle \ln(U)=\phi \ln(\rho )+C,\,\!}$
${\displaystyle U\propto \rho ^{\phi },\,\!}$
${\displaystyle P\propto \rho ^{\phi +1}.\,\!}$

In doing this, we have shown that the adiabatic index ${\displaystyle \gamma }$ is related to ${\displaystyle \phi }$ by

${\displaystyle \gamma =\phi +1.\,\!}$

We can write then that

${\displaystyle P\propto \rho ^{\gamma },\,\!}$
${\displaystyle U\propto \rho ^{\gamma -1},\,\!}$
${\displaystyle T\propto \rho ^{\gamma -1}.\,\!}$

In the non-relativistic case, ${\displaystyle \phi =2/3}$, ${\displaystyle \gamma =5/3}$. For relativistic particles, ${\displaystyle \phi =1/3}$, ${\displaystyle \gamma =4/3}$. We can then use these relations to understand how pressure, energy, and temperature relate to density for either situation. \\Now let’s look at the entropy of an ideal gas. The second law (again) is

${\displaystyle Tds=dU-{\frac {P}{\rho }}{\frac {d\rho }{\rho }}.\,\!}$

The pressure is

${\displaystyle P=U\rho (\gamma -1).\,\!}$

Then

${\displaystyle Tds=dU-U(\gamma -1){\frac {d\rho }{\rho }},\,\!}$

and then after dividing by the energy,

${\displaystyle {\frac {T}{U}}ds={\frac {dU}{U}}-(\gamma -1){\frac {d\rho }{\rho }}.\,\!}$

Writing the energy in terms of the temperature for an ideal gas in the normal way,

${\displaystyle U={\frac {3k_{b}T}{2m}},\,\!}$

and then rearranging and writing in the general form,

${\displaystyle {\frac {T}{U}}={\frac {m}{k_{B}}}(\gamma -1).\,\!}$

We can then solve for the entropy in terms of the mass, energy, and density.

${\displaystyle {\frac {m}{k_{B}}}(\gamma -1)ds={\frac {dU}{U}}-(\gamma -1){\frac {d\rho }{\rho }}\,\!}$
${\displaystyle {\frac {m}{k_{B}}}(\gamma -1)ds=\ln(U)-(\gamma -1)\ln(\rho )\,\!}$
${\displaystyle s={\frac {k_{B}}{(\gamma -1)m}}\ln \left({\frac {U}{\rho ^{\gamma -1}}}\right).\,\!}$

We can use again the relationship between pressure, density, and energy per unit mass to write

${\displaystyle s={\frac {k_{B}}{(\gamma -1)m}}\ln \left({\frac {P}{\rho ^{\gamma }}}\right).\,\!}$

Now let’s look at the change in time of entropy, again using the second law. The time derivative is

${\displaystyle T{\frac {ds}{dt}}={\frac {dU}{dt}}-{\frac {P}{\rho ^{2}}}{d\rho }{dt}.\,\!}$

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.

${\displaystyle {\rm {energy\;per\;time}}=\int dA\cdot F.\,\!}$

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.

${\displaystyle {\rm {energy\;per\;time}}=\int dV\nabla \cdot F.\,\!}$

Thus the energy per time per volume is just ${\displaystyle \nabla \cdot F}$. The energy per unit time per unit mass is then just ${\displaystyle \nabla \cdot F/\rho }$. Then the second law can be written

${\displaystyle T{\frac {ds}{dt}}={\frac {dU}{dt}}-{\frac {P}{\rho ^{2}}}{\frac {d\rho }{dt}}=\epsilon _{\rm {fusion}}-{\frac {1}{\rho }}\nabla \cdot F.\,\!}$

Here we have included energy input from fusion, as well as energy moving by photons or convection. \\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? \\The blob has some ${\displaystyle \rho _{\rm {blob}}}$, ${\displaystyle T_{\rm {blob}}}$, ${\displaystyle s_{\rm {blob}}}$, ${\displaystyle P_{\rm {blob}}}$. It then rises to some higher position, where we assume all of these quantities are set by radiative diffusion to be ${\displaystyle \rho _{\rm {star}}}$, ${\displaystyle T_{\rm {star}}}$, ${\displaystyle s_{\rm {star}}}$, ${\displaystyle P_{\rm {star}}}$. 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 ${\displaystyle 10^{7}}$ 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. \\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.

${\displaystyle s_{\rm {blob}}/=s_{\rm {star}}\,\!}$
${\displaystyle P_{\rm {blob}}=P_{\rm {star}}\,\!}$

If the entropy gradient of the star is negative (${\displaystyle {\frac {ds}{dr}}<0}$), then ${\displaystyle s_{\rm {star}}. Since ${\displaystyle S\propto \ln(P/\rho ^{\gamma })}$, 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.

### Convection

\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. \\We can compare the density of the blob to the background density of the star at some position in the star.

${\displaystyle \rho _{\rm {star}}=\rho +{\frac {d\rho }{dr}}\delta r,\,\!}$
${\displaystyle P_{\rm {star}}=P+{\frac {dP}{dr}}\delta r,\,\!}$
${\displaystyle \rho _{\rm {blob}}=\rho +\left({\frac {\delta \rho }{\delta P}}\right)_{s}\delta \rho _{\rm {blob}},\,\!}$
${\displaystyle P_{\rm {blob}}=P+\delta P_{\rm {blob}}=P_{\rm {star}},\,\!}$
${\displaystyle \rho _{\rm {blob}}=\rho +\left({\frac {d\rho }{dP}}\right)_{s}{\frac {dP}{dr}}dr,\,\!}$
${\displaystyle P\propto \rho ^{\gamma }\,\!}$
${\displaystyle \left({\frac {dP}{d\rho }}\right)_{s}={\frac {\gamma P}{\rho }},\,\!}$
${\displaystyle \rho _{\rm {blob}}=\rho +{\frac {\rho }{\gamma P}}{\frac {dP}{dr}}\delta r.\,\!}$

At some new position, the blob will feel an acceleration from its bouyancy, just by Archimedes’ Principle.

${\displaystyle a=g\left({\frac {\rho _{\rm {star}}}{\rho _{\rm {blob}}}}-1\right).\,\!}$

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

${\displaystyle a=-N^{2}\delta r,\,\!}$

where we have defined

${\displaystyle N^{2}=-g\left({\frac {d\ln \rho }{dr}}-{\frac {1}{\gamma }}{\frac {d\ln P}{dr}}\right).\,\!}$

Alternatively, we can write it in terms of the entropy gradient, with

${\displaystyle N^{2}={\frac {\gamma -1}{\gamma }}{\frac {m_{p}}{k_{b}}}g{\frac {ds}{dr}}.\,\!}$

The Brunt-Vaisala frequency is proportion then to the entropy gradient. We can write this down as a simple differential equation.

${\displaystyle a=\delta {\dot {r}},\,\!}$
${\displaystyle \delta {\dot {r}}+N^{2}\delta r=0,\,\!}$
${\displaystyle \delta r\propto e^{\pm iNt}.\,\!}$

If ${\displaystyle N^{2}}$ is greater than zero, we have an oscillating solution, and convective stability. This corresponds to a positive entropy gradient. If ${\displaystyle N^{2}}$ is negative, then the solution of this is an exponential runaway, on a timescale of ${\displaystyle 1/|N|}$. \\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,

${\displaystyle {\frac {d\ln P}{dr}}-\gamma {\frac {d\ln \rho }{dr}}=0,\,\!}$
${\displaystyle P={\frac {\rho k_{B}T}{\mu m_{p}}},\,\!}$
${\displaystyle \ln P=\ln \rho +\ln T+{\rm {constant}},\,\!}$
${\displaystyle {\frac {d\ln P}{dr}}-\gamma {\frac {d\ln P}{dr}}+\gamma {\frac {d\ln T}{dr}}<0,\,\!}$
${\displaystyle {\frac {d\ln T}{dr}}<{\frac {\gamma -1}{\gamma }}{\frac {d\ln P}{dr}}.\,\!}$

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

${\displaystyle \left|{\frac {d\ln T}{dr}}\right|>{\frac {\gamma -1}{\gamma }}\left|{\frac {d\ln P}{dr}}\right|.\,\!}$

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

${\displaystyle \left|{\frac {d\ln T}{d\ln P}}\right|>{\frac {\gamma -1}{\gamma }}.\,\!}$

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,

${\displaystyle {\frac {dP_{\rm {rad}}}{dr}}\propto F.\,\!}$

From hydrostatic equilibrium,

${\displaystyle {\frac {dP}{dr}}=-\rho {\frac {GM_{r}}{r^{2}}}.\,\!}$

Then

${\displaystyle {\frac {P}{P_{\rm {rad}}}}{\frac {dP_{\rm {rad}}}{dP}}={\frac {L_{r}\kappa }{4\pi GM_{r}c}}{\frac {P}{P_{\rm {rad}}}},\,\!}$

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

${\displaystyle {\frac {d\ln P_{\rm {rad}}}{d\ln P}}=4{\frac {d\ln T}{d\ln P}}.\,\!}$

Plug this in above,

${\displaystyle {\frac {d\ln T}{d\ln P}}={\frac {1}{4}}{\frac {P}{P_{\rm {rad}}}}{\frac {L}{L_{edd}}}{\frac {L_{r}/L}{M_{r}/M}},\,\!}$

and finally put this in our convection criterion as

${\displaystyle {\frac {1}{4}}{\frac {P}{P_{\rm {rad}}}}{\frac {L}{L_{edd}}}{\frac {L_{r}/L}{M_{r}/M}}>{\frac {\gamma -1}{\gamma }}.\,\!}$

The Sun is convection for ${\displaystyle r>0.7R_{\odot }}$. At smaller radii, photons carry the energy. At these large radii, ${\displaystyle M_{r}\approx M}$ and ${\displaystyle L_{r}=L}$. 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 ${\displaystyle \kappa }$. At lowish temperatures (${\displaystyle \sim 10^{5}}$ 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. \\Energy Transport by Convection \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. \\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, ${\displaystyle \Delta E}$ (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 ${\displaystyle \rho _{c}v_{c}^{2}}$. Estimating the energy density in the two different ways turns out to give similar answers, so we can estimate the convective flux as

${\displaystyle F_{c}\sim \rho v_{c}^{3}.\,\!}$

We can get this quantity by looking at the work done by acceleration due to bouyancy. We define the mixing length ${\displaystyle \ell }$ 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

${\displaystyle H=\left({\frac {d\ln P}{dr}}\right)^{-1}.\,\!}$

To encapsulate our ignorance, we’ll add a fudge factor ${\displaystyle \alpha }$ and say

${\displaystyle \ell =\alpha H.\,\!}$

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. \\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,

${\displaystyle v_{c}^{2}\sim a\cdot \delta r\sim |N|^{2}\delta r^{2}.\,\!}$

We are relating then the energy gained to the effectiveness of bouyance. The distance moved, ${\displaystyle \delta r}$, we are defining as the mixing length ${\displaystyle \ell }$. We write the Brunt-Vaisala frequency in terms of the gravity and the specific heat at constant pressure as

${\displaystyle |N|^{2}={\frac {g}{C_{P}}}\left|{\frac {ds}{dr}}\right|.\,\!}$

We can write this instead in terms of a dimensionless entropy gradient by dividing and multiplying by the pressure scale height,

${\displaystyle |N|^{2}={\frac {g}{H}}\left|{\frac {H}{C_{P}}}{\frac {ds}{dr}}\right|.\,\!}$

The convective velocity is then

${\displaystyle v_{c}^{2}\sim gH\left|{\frac {H}{C_{P}}}{\frac {ds}{dr}}\right|\alpha ^{2}.\,\!}$

The scale height is just ${\displaystyle {\frac {kT}{mg}}}$, which then means that ${\displaystyle gH}$ is of order the average spead of individual particles squared, or the sound speed ${\displaystyle c_{s}}$. Then

${\displaystyle v_{c}\sim \alpha c_{s}\left|{\frac {H}{C_{P}}}{\frac {ds}{dr}}\right|^{1/2}.\,\!}$

We can then use this to estimate the convective flux, with

${\displaystyle F_{c}\sim \rho v_{c}^{3}\sim \rho \alpha ^{3}c_{s}^{3}\left|{\frac {H}{C_{P}}}{\frac {ds}{dr}}\right|^{3/2}.\,\!}$

Given ${\displaystyle F_{c}}$, 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 ${\displaystyle 10^{-6}}$. 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 ${\displaystyle P\propto \rho ^{\gamma }}$.

### Convective Transport of Energy

\The main results from the end of last lecture had to do with how energy is transported via convection. That is,

${\displaystyle F_{c}\sim \rho v_{c}^{3}\sim \rho \alpha ^{3}c_{s}^{3}\left|{\frac {H}{C_{P}}}{\frac {ds}{dr}}\right|^{3/2},\,\!}$

with

${\displaystyle v_{c}\sim \alpha c_{s}\left|{\frac {H}{C_{P}}}{\frac {ds}{dr}}\right|^{1/2}.\,\!}$

As a reminder, ${\displaystyle c_{s}}$ is the sound speed, ${\displaystyle v_{c}}$ is the convective velocity, and ${\displaystyle \alpha }$ 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${\displaystyle ^{-1}}$, which implies a dimensionless entropy gradient of order ${\displaystyle 10{-6}}$. This is very much smaller than 1, meaning this is very nearly adiabatic, and we can assume that the entropy is constant. \\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

${\displaystyle P\propto \rho ^{\gamma }\,\!}$

in the convective zone (again, the profile of the star, not for the blob alone). \\Now, imagine that ${\displaystyle P\propto \rho ^{\gamma }}$ 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.

${\displaystyle {\frac {dP}{dr}}=-\rho {\frac {GM_{r}}{r^{2}}},\,\!}$
${\displaystyle {\frac {dM_{r}}{dr}}=4\pi r^{2}\rho ,\,\!}$
${\displaystyle {\frac {d}{dr}}\left[{\frac {r^{2}}{\rho }}{\frac {dP}{dr}}\right]={\frac {d}{dr}}\left[-GM_{r}\right],\,\!}$
${\displaystyle {\frac {d}{dr}}\left[{\frac {r^{2}}{\rho }}{\frac {dP}{dr}}\right]=-4\pi r^{2}\rho G.\,\!}$

Now, we use our assumption of ${\displaystyle P\propto \rho ^{\gamma }}$, and invoke polytrope solutions.

${\displaystyle P=K\rho ^{\gamma }=K\rho ^{1+1/n}.\,\!}$

For fully convective stars, ${\displaystyle \gamma =5/3}$ and ${\displaystyle n=3/2}$. 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 ${\displaystyle \gamma =4/3}$ (${\displaystyle n=3}$), 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. \\We now write down various dimensionless quantities, so we can take advantage of known polytropic solutions.

${\displaystyle \theta =\left({\frac {\rho }{\rho _{c}}}\right)^{1/n},\,\!}$
${\displaystyle \rho _{c}=\rho (r=0),\,\!}$
${\displaystyle \xi ={\frac {r}{a}},\,\!}$
${\displaystyle a=\left({\frac {(n+1)K\rho _{c}^{{\frac {1}{n}}-1}}{4\pi G}}\right)^{1/2},\,\!}$
${\displaystyle {\frac {1}{\xi ^{2}}}{\frac {d}{d\xi }}\left(\xi ^{2}{\frac {d\theta }{d\xi }}\right)=-\theta ^{n}.\,\!}$

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

${\displaystyle \theta (0)=1,\,\!}$

which is fairly evident. The other is that

${\displaystyle {\frac {d\theta }{d\xi }}|_{\xi =0}=0\,\!}$

This is a statement that the derivative smoothly goes to zero at the center. Basically, we run out of mass as ${\displaystyle r}$ goes to zero, which makes sense (but I skipped some justification). We also note that the radius is ${\displaystyle r=\xi a}$. These allow for the Lane-Emden equation to be solved for various polytropic indices ${\displaystyle n}$. We won’t actually do this though. There are three known analytic solutions, for ${\displaystyle n=0,1,5}$. We are interested in ${\displaystyle n=3/2,3}$. Meaning we need to solve this integral numerically, either with a computer or with a 19th century physicist by hand. \\These polytropic models in general have two free parameters. One is the constant ${\displaystyle K}$, and the other is the central density ${\displaystyle \rho _{c}}$, for a given model with set ${\displaystyle n}$. Equivalently, the two free parameters are the mass and radius of the star. That is, ${\displaystyle K}$ and ${\displaystyle \rho _{c}}$ can be written in terms of ${\displaystyle M}$ and ${\displaystyle r}$. Real stellar models are uniquely determined by the mass alone (well, composition matters too, but that is a separate issue). Aside: ${\displaystyle K}$ is actually determined from physics alone for objects dominated by degeneracy pressure. Using ${\displaystyle n=3}$, 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 ${\displaystyle n=3/2}$. \\Fully Convective Objects \The polytropes do really well for fully convective objects. This applies to stars that are less massive than about 1/3 ${\displaystyle M_{\odot }}$, 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 ${\displaystyle L\propto M^{3}}$ 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. \\We focus on objects dominated by gas pressure, meaning ${\displaystyle \gamma =5/3}$ and ${\displaystyle n=3/2}$. Using the solution of the Lane-Emden equation for such a polytrope,

${\displaystyle P_{c}=0.77{\frac {GM^{2}}{R^{4}}},\,\!}$

and

${\displaystyle kT_{c}=0.54{\frac {GM\mu m_{p}}{R}}.\,\!}$

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 ${\displaystyle P\propto \rho ^{\gamma }}$ assumption is true. \\We also know general proportionalities between the pressure, density, and temperature. Since ${\displaystyle P\propto \rho T}$ (ideal gas law),

${\displaystyle P\propto T^{5/2}.\,\!}$

Combining this with the expression at the center,

${\displaystyle {\frac {P_{ph}}{P_{c}}}=\left({\frac {T_{ph}}{T_{c}}}\right)^{5/2}.\,\!}$

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

${\displaystyle \ell ={\frac {1}{\kappa \rho }},\,\!}$

and

${\displaystyle H={\frac {kT_{eff}}{mg}}.\,\!}$

Equation, the photosphere satisfies

${\displaystyle {\frac {1}{\kappa \rho }}={\frac {kT_{eff}}{mg}},\,\!}$

or

${\displaystyle \rho _{ph}={\frac {mg}{\kappa k_{B}T_{eff}}}.\,\!}$

Then the pressure in the photosphere is

${\displaystyle P_{ph}={\frac {\rho k_{B}T_{eff}}{m}}={\frac {g}{\kappa }}.\,\!}$

Now, rearranging our temperature/pressure relationship from before,

${\displaystyle kT_{eff}=kT_{c}\left({\frac {P_{ph}}{P_{c}}}\right)^{2/5},\,\!}$
${\displaystyle kT_{eff}=kT_{c}\left({\frac {g}{\kappa _{ph}P_{c}}}\right)^{2/5},\,\!}$
${\displaystyle kT_{eff}={\frac {0.6GM\mu m_{p}}{R}}\left({\frac {R^{2}}{M\kappa _{ph}}}\right)^{2/5}.\,\!}$

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${\displaystyle ^{-}}$ ion, since the effective temperature will be of order 3000 K. The H${\displaystyle ^{-}}$ has the function form

${\displaystyle \kappa _{H^{-}}=2.5\times 10^{-31}\rho ^{1/2}T^{9}{\rm {\;cm^{2}\;g}}.\,\!}$

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

${\displaystyle T_{eff}\approx 3000\left({\frac {M}{M_{\odot }}}\right)^{1/7}\left({\frac {R}{R_{\odot }}}\right)^{1/49}.\,\!}$

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${\displaystyle -}$ will dominate, as we assumed. Writing the luminosity of a blackbody,

${\displaystyle L=4\pi R^{2}\sigma T_{eff}^{4},\,\!}$

we can plug in our result to find

${\displaystyle L\approx 0.07L_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{4/7}\left({\frac {R}{R_{\odot }}}\right)^{102/49}.\,\!}$

We can manipulate this in one more way, and write

${\displaystyle T_{eff}=3000K\left({\frac {M}{M_{\odot }}}\right)^{7/51}\left({\frac {L}{L_{\odot }}}\right)^{1/102}.\,\!}$

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.