Worksheet 11.2, Problem 1

We can draw a diagram of the mass shells being described in the problem:

This shows us that the energy density of the star changes as you move farther away from the center. Since is L(r) is the net outwards flow of energy, it is expressed as \(\frac{change\:in\:energy}{time}\). The change in energy is given as \(\Delta u\) and the time, \(t\), is the time that it takes for that energy to flow, specifically how long it takes for a photon to travel the distance, \(\Delta r\). So we can express the outwards flow as;

\[L(r)\:=\:\frac{\Delta u}{t}\]

Now we know from the previous worksheet that the time it takes to travel the distance \(\Delta r\) is \(t\:=\:\frac{Nl}{c}\). We can plug this into the previous expression:

\[L(r)\:=\:\frac{\Delta u}{\frac{Nl}{c}}\]

\[L(r)\:=\:\frac{\Delta u\:\times\:c}{Nl}\]

We also know from the previous worksheet that N, the number of steps required to travel the total displacement, is equal to \((\frac{\Delta r}{l})^{2}\). Plugging this in, we get:

\[L(r)\:=\:\frac{\Delta u\:\times\:c\:\times\: l}{(\Delta r)^{2}}\]

Now this gives us the net outward flow for one infinitesimal point on the mass shell. To find the total outward flow, we multiply by the surface area of the mass shell, \(4\pi r^{2} dr\). This leaves us with:

\[L(r)\:=\:\frac{\Delta u\:\times\:c\:\times\:l}{(\Delta r)^{2}}\:\times\:4\pi  r^{2} dr\]

Rewriting the \(\frac{\Delta r}{\Delta u}\) as \(-\frac{du}{dr}, we can cancel and are left with:

\[L(r)\:=\:-4\pi r^{2}cl\frac{du}{dr}\]

We begin with the diffusion equation that we derived in the previous problem:

\[L(r)\:=\:-4\pi r^{2}cl\frac{du}{dr}\] 

Next, we find du by taking the derivative of the equation for the energy density of a blackbody:

\[u(T(r))\:=\:aT^{4}\]

\[du\:=\:4aT^{3}dT\]

We can then substitute this expression for du into the diffusion equation:

\[L(r)\:=\:-4\pi r^{2}cl\frac{4aT^{3}dT}{dr}\] 

Now, rearranging to isolate \(\frac{dT}{dr}\), we have:

\[\frac{dT}{dr}\:=\:\-frac{L(r)}{16a\pi r^{2}T^{3}cl}\]

We also know that \(l\:=\:\frac{1}{\kappa \rho (r)}\). Substituting for the \(l\) in our expression gives us:

\[\frac{dT}{dr}\:=\:-\frac{L(r) \kappa \rho (r)}{16a\pi r^{2}T^{3}c}\]

This is nearly identical to the proportional differential equation that were given at the beginning. The only difference is 16 in the denominator . The 16 can be explained because it was a proportional equation and thus left out that proportional constant. 

 Note: In our first attempt, we simply added a minus sign because we misinterpreted the conversion from \(\Delta u\) and \(\Delta r\) to du and dr. it is important that we realize that \(\frac{\Delta u}{\Delta r}\:=\:-\frac{du}{dr}\). This demonstrates the negative relationship between temperature and radius. The larger the radius, and thus the further away from the core, the lower the temperature. As you move farther out towards the surface of the star, the energy density, as well as the temperature, decreases.