Worksheet 11.2, Problem 1: Radiative Energy Transport

Stars generate their energy in their cores, where nuclear fusion is taking place. The energy generated is eventually radiated out at the star's surface. Therefore, there exists a gradient in the energy density from the center (high) to the surface (low), but thermodynamic systems tend towards 'equilibrium.' In the following sections, we will determine how energy flows through the star. 

a) Inside the star, consider a mass shell of width $\Delta r$, at a radius $r$. This mass shell has an energy density $u+\Delta u$, and the next mass shell out (at a radius $r+\Delta r$) will have an energy density $u$. Both shells behave as blackbodies. 

The net outwards flow of energy, $L(r)$, must equal the total excess energy in the inner shell divided by the amount of time needed to cross the shell's width $\Delta r$. Use this to derive an expression for $L(r)$ in terms of $\frac{du}{dr}$, the energy density profile. This is the diffusion equation describing the outward flow of energy. 

Mass Shell

We need to write an expression for the outward flow of energy, L(r), which we are told is equal to the total excess energy in the inner shell divided by the time to cross the shell's width. To find the excess energy, we need to multiply the energy density of the shell by the shell's volume:

$\Delta E=V_{shell}[u-(u +\Delta u)]=-4\pi r^2\Delta r\Delta u$

And to find the time to cross the shell's width, $\Delta t$, we can use the $d=rt$ equation:

$\Delta t=\dfrac{\Delta r}{\overrightarrow{v}_{diff}}$

And $\overrightarrow{v}_{diff}$ is the diffusion velocity, which we solved for in the previous problem to get:

$\overrightarrow{v}_{diff}=\dfrac{c}{\kappa \rho (r) \Delta r}$

Where $c$ is the speed of light, $\kappa$ is the absorption coefficient, and $\rho (r)$ is the mass density. So we get a time of:

$\Delta t=\dfrac{\Delta r^2\kappa \rho (r)}{c}$

Now writing out expression for the outward flow of energy, $L(r)$, we get:

$L(r)=-\dfrac{4\pi r^2 \Delta u c}{\Delta r \kappa\rho(r)}$

Since the shell is infinitesimally small, we can write:

$L(r)=-\dfrac{du}{dr}\dfrac{4\pi r^2 c}{\kappa\rho(r)}$

We have now solved for the diffusion equation, which describes the outward flow of energy from a star.

b) From the diffusion equation, use the fact that the energy density of a blackbody is $u(T(r))=aT^4$ to derive the differential equation:

$\dfrac{dT(r)}{dr}\propto -\dfrac{L(r)\kappa \rho (r)}{\pi r^2 acT^3}$

where $a$ is the radiation constant.

Plugging our equation for $u(T(r))$ into the diffusion equation, we get:

$L(r)=-\dfrac{du(T(r))}{dr}\dfrac{4\pi r^2 c}{\kappa\rho(r)}$

         $=-\dfrac{d}{dr}aT(r)^4\dfrac{4\pi r^2 c}{\kappa\rho(r)}$

                  $=-4aT(r)^3\dfrac{dT(r)}{dr}\dfrac{4\pi r^2 c}{\kappa\rho(r)}$

Now, solving for $T(r)$, we get:

  $\dfrac{dT(r)}{dr}=-\dfrac{L(r)\kappa\rho(r)}{16\pi r^2acT^3}$

So we have the relation: 

$\dfrac{dT(r)}{dr}\propto -\dfrac{L(r)\kappa\rho(r)}{\pi r^2acT^3}$

And now we have derived the equation for radiative energy transport.

I would like to thank Charles Law and Daniel Chen for their help in solving this problem.