(Redirected from Stars Lecture 3)

### Part one

\\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. \\Particles have an energy density ${\displaystyle U}$, a typical velocity ${\displaystyle v}$, and a mean free path ${\displaystyle \ell }$, where the mean free path represents the average distance traveled between exchanging energy. There is flux ${\displaystyle F}$ 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

${\displaystyle F={\frac {1}{6}}U(x-\ell )v-{\frac {1}{6}}U(x+\ell )v,\,\!}$
${\displaystyle F=-{\frac {1}{3}}v\ell {\frac {dU}{dx}},\,\!}$
${\displaystyle F=-{\frac {1}{3}}v\ell {\frac {dU}{dT}}{\frac {dT}{dx}},\,\!}$
${\displaystyle F=-\kappa {\frac {dT}{dx}},{\rm {\;where}}\,\!}$
${\displaystyle \kappa \sim {\frac {1}{3}}v\ell {\frac {dU}{dT}}.\,\!}$

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, ${\displaystyle \sigma }$. 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 ${\displaystyle \sigma }$:

${\displaystyle P=\sigma F.\,\!}$

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

${\displaystyle \ell ={\frac {1}{n\sigma }}.\,\!}$

Let’s estimate this for a particle in the Earth’s atmosphere. The density of air is ${\displaystyle \rho \sim 10^{-3}}$ g/cm${\displaystyle ^{3}}$, which gives a number density of roughly ${\displaystyle n\sim 10^{20}}$ particles/cm${\displaystyle ^{3}}$. 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 ${\displaystyle 10^{-8}}$ cm. The mean free path for molecules in the atmosphere is then

${\displaystyle \ell \sim 10^{-4}{\rm {\;cm,\;or\;one\;micron}}.\,\!}$

Let’s now focus on an ideal ionized gas. The energy per unit volume is ${\displaystyle {\frac {3}{2}}nk_{B}T}$. The conductivity ${\displaystyle \kappa }$ is then

${\displaystyle \kappa \sim {\frac {1}{2}}v\ell nk_{b}.\,\!}$

This is the appropriate conductivity for energy transport by electrons. For electrons in a gas,

${\displaystyle v\sim {\sqrt {\frac {kT}{m}}}.\,\!}$

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

${\displaystyle E_{e}={\frac {e^{2}}{r}}.\,\!}$

When this energy is similar to the thermal energy ${\displaystyle k_{B}T}$, an electron will scatter. Thus significant deflections occur at

${\displaystyle b={\frac {e^{2}}{kT}},\,\!}$

where ${\displaystyle b}$ is known as the impact parameter. The Coulomb cross section is then

${\displaystyle \sigma _{c}=\pi b^{2}=\pi {\frac {e^{4}}{k_{B}^{2}T^{2}}}.\,\!}$

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

${\displaystyle \kappa \sim {\frac {k_{B}^{3}vT^{2}}{\pi e^{4}}}\propto {\frac {T^{5/2}}{\sqrt {m}}}\,\!}$

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. \\We can perform a crude estimate to determine if this is important in the Sun.

${\displaystyle F_{r}=-\kappa {\frac {dT}{dr}}={\frac {L_{r}}{4\pi r^{2}}}.\,\!}$

Making some approximations to simplify (like dropping differentials),

${\displaystyle L\sim 4\pi R\kappa T_{c},\,\!}$
${\displaystyle L\sim {\frac {k_{b}^{7/2}T_{c}^{7/2}R}{e^{4}{\sqrt {m_{e}}}}},\,\!}$
${\displaystyle L\sim 10^{-4}\left({\frac {R}{R_{\odot }}}\right)\left({\frac {T_{c}}{10^{7}K}}\right)^{7/2}L_{\odot }.\,\!}$

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. \\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 ${\displaystyle v=c}$, ${\displaystyle U=aT^{4}}$, and the mean free path is different (smaller, in fact). Then the flux from photons is

${\displaystyle F=-{\frac {4caT^{3}}{3n\sigma }}{\frac {dT}{dx}}.\,\!}$

This is in analogy with the conduction equation from before, so we can define a photon conductivity in analogy to before of

${\displaystyle \kappa _{ph}={\frac {4caT^{3}}{n\sigma }}.\,\!}$

This is often written alternatively in terms of the opacity and density rather than cross section and number density, with

${\displaystyle F=-{\frac {4caT^{3}}{3\kappa \rho }}{\frac {dT}{dx}}.\,\!}$

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

${\displaystyle m_{e}{\frac {dv}{dt}}=-e({\overrightarrow {E}}+{\frac {\overrightarrow {v}}{c}}\times {\overrightarrow {B}}).\,\!}$

Since the magnitude of ${\displaystyle E}$ and ${\displaystyle B}$ are comparable, we need to only worry about the electric field, and can see that

${\displaystyle {\overrightarrow {a}}=-{\frac {e{\overrightarrow {E}}}{m_{e}}}.\,\!}$

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

${\displaystyle P={\frac {2e^{2}}{3c^{3}}}|a|^{2}.\,\!}$

Substituting the acceleration from above,

${\displaystyle P={\frac {2e^{4}}{3m_{e}^{2}c^{3}}}|E|^{2}.\,\!}$

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

${\displaystyle P=\sigma F.\,\!}$

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

${\displaystyle F={\frac {c|E|^{2}}{4\pi }}.\,\!}$

Solving ${\displaystyle P=\sigma F}$, we find the Thomson scattering cross section is

${\displaystyle \sigma ={\frac {8\pi }{3}}{\frac {e^{4}}{m_{e}^{2}c^{4}}}.\,\!}$

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

${\displaystyle \sigma \sim \pi r_{c}^{2},\,\!}$

where ${\displaystyle r_{c}}$ is the classical electron radius, which is the radius found from equation the rest mass energy to the electrostatic self energy of an electron.

### Part two

Our main result from last time was the radiative diffusion equation.

${\displaystyle F=-{\frac {4acT^{3}}{3n\sigma }}{\frac {dT}{dx}}.\,\!}$

We will also put this in terms of the opacity and density at times, with

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

Now, let’s use this to estimate the power that can be carried out of the Sun by photons. We write the luminosity ${\displaystyle L}$ as

${\displaystyle L=4\pi R^{2}F.\,\!}$

Then, using the radiative diffusion equation and dropping differentials,

${\displaystyle L\sim 4\pi R^{2}{\frac {acT^{4}}{n\sigma R}}.\,\!}$

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,

${\displaystyle L\sim {\frac {16\pi R^{4}T^{4}\mu _{e}m_{p}ac}{M\sigma _{T}}}.\,\!}$

We can remove the temperature from this equation by using the Virial Theorem. Specifically,

${\displaystyle kT={\frac {GM\mu m_{p}}{3R}}.\,\!}$

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:

${\displaystyle L\sim {\frac {a(\mu m_{p})^{4}(\mu _{e}m_{p})cG^{4}M^{3}}{\sigma _{T}k^{4}}}.\,\!}$

Putting in some numbers, we get

${\displaystyle L\sim 10^{35}\left({\frac {M}{M_{\odot }}}\right)^{3}{\rm {erg\;s}}^{-1}.\,\!}$

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

${\displaystyle L\propto \mu _{e}\mu ^{4}.\,\!}$

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. \\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,

${\displaystyle {\frac {F_{rad}}{F_{cond}}}={\frac {aT^{3}c\sigma _{e-e}}{n\sigma _{\gamma -e}k_{B}v_{e}}}.\,\!}$

We can rearrange in the following suggestive manner:

${\displaystyle {\frac {F_{rad}}{F_{cond}}}={\frac {aT^{4}}{nk_{B}T}}{\frac {c}{v_{e}}}{\frac {\sigma _{e-e}}{\sigma _{\gamma -e}}}.\,\!}$

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. \\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.

${\displaystyle t_{KH}={\frac {E}{L}}\,\!}$
${\displaystyle t_{KH}={\frac {nk_{B}TR^{3}Rn\sigma }{aT^{4}cR^{2}}}\,\!}$
${\displaystyle t_{KH}={\frac {nk_{B}T}{aT^{4}}}{\frac {R^{2}}{c\ell }}.\,\!}$

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

${\displaystyle t\sim {\frac {R}{\ell }}{\frac {R}{c}}\,\!}$

is of order ${\displaystyle 10^{4}}$ 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

${\displaystyle {\sqrt {<|D|^{2}>}}=N^{1/2}\ell ,\,\!}$

where ${\displaystyle N}$ is the number of steps taken and ${\displaystyle \ell }$ is as always the mean free pat. The total number of steps to get out of the Sun is then

${\displaystyle N\sim \left({\frac {R}{\ell }}\right)^{2}.\,\!}$

The time to random walk out is then the product of the number of steps and the time per step:

${\displaystyle t_{RW}=N{\frac {\ell }{c}}={\frac {R^{2}}{\ell c}}.\,\!}$

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

${\displaystyle P_{rad}={\frac {1}{3}}aT^{4},\,\!}$

or in differential form,

${\displaystyle {\frac {dP_{rad}}{dr}}={\frac {4}{3}}aT^{3}{\frac {dT}{dr}}.\,\!}$

This looks the same as part of the radiative diffusion equation, and by examination, we write

${\displaystyle F_{r}=-{\frac {c}{\kappa \rho }}{\frac {d}{dr}}\left(P_{rad}\right).\,\!}$

In terms of the luminosity,

${\displaystyle {\frac {dP_{rad}}{dr}}=-{\frac {L_{r}\kappa \rho }{4\pi r^{2}c}}.\,\!}$

We can compare this to hydrostatic equilibrium, which looks somewhat similar.

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

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

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

This tells us the fractional contribution of radiation pressure to total pressure in a star. We now define the Eddington luminosity as

${\displaystyle L_{edd}(r)={\frac {4\pi GM_{r}c}{\kappa }}.\,\!}$

This gives the very simple expression

${\displaystyle {\frac {dP_{rad}}{dP}}={\frac {L_{r}}{L_{edd}(r)}}.\,\!}$

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 ${\displaystyle L/c}$, since each photon has momentum of ${\displaystyle E/c}$. The force then due to absorption of radiation is

${\displaystyle {\frac {dp}{dt}}={\frac {L}{c4\pi r^{2}}}\sigma .\,\!}$

Now we can ask when the radiation force is equal to the force of gravity. Equating the two,

${\displaystyle {\frac {L\sigma }{c4\pi r^{2}}}={\frac {GMm}{r^{2}}}.\,\!}$

Canceling the radius terms and rearranging, we find the same expression for the Eddington luminosity (or limit) of

${\displaystyle L={\frac {4\pi GMc}{\kappa }}.\,\!}$

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

${\displaystyle L_{edd}\sim 10^{38}\left({\frac {M}{M_{\odot }}}\right){\rm {erg\;s}}^{-1}.\,\!}$

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

${\displaystyle L=L_{\odot }\left({\frac {M}{M_{\odot }}}\right)^{3}=L_{edd}.\,\!}$

Evaluating this gives an estimate for where radiation pressure is important of

${\displaystyle M\sim 100M_{\odot }.\,\!}$

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. \\

 Temperature Regime Dominant Opacity ${\displaystyle T>10^{7}}$ K Thomson scattering ${\displaystyle T\sim 10^{4-6}}$ K photo-ionization of atoms, free-free ${\displaystyle T\sim 10^{3-4}}$ K H${\displaystyle ^{-}}$ ${\displaystyle T<10^{3}}$ K Molecules and dust