Monday, March 30, 2015

Worksheet 11.1: Photons Random Walking Out of a Star




A Diagram of the Sun showing the Photon Random Walk



The point of this problem is to show us that photons do not travel directly from the core to the surface of the sun. Instead, the photons follow a "haphazard" random walk that takes much longer to reach the surface of the sun. We can sketch a random walk ourselves to try and figure out how many steps it takes to travel a distance \(\Delta r\).
We can see that the actual displacement is the sum of the vectors of each individual step. In order to put this into a scalar quantity, we take the dot product of displacement. This dot product can be rewritten in terms of the cosine of the angle between the vectors.


 Many of the values of cosine, specifically those in which \(\theta\) is the angle between the exact same vectors, are equal to 1. For the rest of the terms, the sum of the cosine values for a large enough random walk approaches zero because all of the values cancel each other out. This leaves us with a relatively simple relationship for the number of steps needed to travel a distance \(\Delta r\):
\[N\:=\:(\frac{\Delta r}{l})^{2}\]

The diffusion velocity is how fast the photon is effectively traveling over a large displacement \(\Delta r\), usually  from the core to the surface. Although the photon is always traveling at the speed of light, its random walk path makes its velocity relative to the total displacement much less than the speed of light.

The total time it takes to travel every single step is thus equal to \(\frac{N\:\times\:l}{c}\), where \(N\)is the number of steps, \(l\) is the length of each step (mean free path), and \(c\) is the speed of light. We can then divide \(\Delta r\) by the total time in order to get the diffusion velocity:
\[v_{diff}\:=\:\frac{c}{N^{\frac{1}{2}}}\]

We can represent the photon's path over a specific region leading up to a collision with a diagram. At the beginning of the step, the photon is situated a distance \(l\) away from the electron that it will collide with. Since there is only one electron in this specific region, the number density is \(\frac{1}{volume}\). We can then express the mean free path, \(l\) in terms of the number density and the electron's cross-sectional area.



 Using our definitions for these terms and our original definition of diffusion velocity, we can do some substitutions, namely with \(l\), in order to come up with an expression for diffusion velocity in terms of \(\kappa\) and \(\rho\).


From the previous problem, we determined an expression for the mean free path in terms of \(\sigma\) and the number density. We could then substitute from our definitions of \(\sigma\) and \(\rho\). Once we had determined that \(l\:=\:\frac{1}{\kappa \rho}\), we could plug into into our known expression for N and then plug that value into our known expression for diffusion velocity:
\[N\:=\:(\frac{\Delta r}{l})^{2}\]
\[N^{\frac{1}{2}}\:=\:\frac{\Delta r}{\frac{1}{\kappa \rho}}\:=\:\Delta r \kappa \rho\]
Thus, the diffusion velocity can be written as:
\[v_{diff}\:=\:\frac{c}{\Delta r \kappa \rho}\]

Now that we have determined the diffusion velocity, we can approach easily approach this problem. The diffusion velocity helps us understand how quickly a photon can move over a total displacement. In this case, that total displacement, \(\Delta r\) is the radius of the Sun. We have already developed the expressions to solve for the total time. Now all we have to do is plug in the specific values which are listed below:

We can then use this to solve for the diffusion velocity. Finally, we can plug into our equation for total time, \(\frac{\Delta r}{v_{diff}}\) and we have solved for the time scale.


In the spirit of this class, I used some approximations for these values. Thus the time scale is not actually 3168 years but somewhere around that figure. That's a pretty long time! It's especially interesting because the light that we receive from the Sun during the day only takes about 8 minutes to reach us while traveling at the speed of light. Yet, it had to spend thousands of years escaping the sun before it could reach us!

2 comments:

  1. Very nice work, Carter! In the last step, be sure that you're using the "expert method", and solving for the time in terms of variables, only substituting at the very end.

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    1. Thanks for the advice Benjamin. I suppose the way I structured this, I solved for Vdiff first and then solved for time. I should have put my expression for time at the top and then worked from there so I would have all the variables before solving for Vdiff. I'll be sure to stick to the expert method in the future!

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