WS 9.1, Problem 1: Random Walks

Consider a photon that has just been created via a nuclear reaction between two protons in the center of the Sun. The photon now starts a long and arduous journey to the Earth to be enjoyed by Ay16 students studying on a calm Spring day.

a)  The photon does not freely travel to the surface of the Sun. Instead it random walks, one collision at a time. Each step of the random walk traverses an average distance w, also known as the mean free path. On average, how many steps does the photon take to travel a distance $\Delta r$?

As always, our first step was to draw a picture illustrating a photon's random walk journey over a distance, $D=\Delta r$.

We were given the hint that if each step with length w is a vector, $\vec{r_i}$, then the vector 

$\vec{D}=\sum \vec{r_i}$

We're asked to find the number of steps it takes a photon to travel $\Delta r$, which is a scalar quantity.  So the hint also told us that we should find 

$\Delta r=\left ( \vec{D}^2 \right )^{1/2}=\left ( \sum \vec{r_i}^2 \right )^{1/2}$

To solve this, we have to find the dot products of each vector, $\vec{r_i}$ and then take the square root of the scalar sum of those dot products.  This could be a really tedious process, but the concept of randomness actually helps us out here.  The sum of these dot products is 

$\vec{D}^2=r_1^2+r_2^2+r_3^2...r_N^2$

because all of the terms that aren't a vector squared are the dot puct between two vectors.  Since each vector is completely random, there's a chance that any two vectors are perpendicular to each other, and therefore have a dot product equal to 0.  Yes, it's not at all realistic, but neither is assuming that everything in space is a spherical blackbody.  

So, now we can rewrite the last equation as 

$\vec{D}^2=\sum_{i}^{N}r_i^2$

$= \sum_{i}^{N}w^2$

because each random walk has a length, w.

$=Nw^2$

Plugging this into the equation for $\Delta r$, we find that

$\Delta r=D=\left ( Nw^2 \right )^{1/2}=w\sqrt{N}$

We can rearrange this equation to find N, the number of steps it takes the photon to travel $\Delta r$.

$N=\left ( \frac{D}{w} \right )^2$

b)  What is the photon's average velocity over this distance? Call this $v_{diff}$ , the di ffusion velocity.

We started with the general idea that velocity equals distance over time.  The distance, $\Delta r$, is stated in the problem.  The tricky part was finding the time.

We're dealing with photons--light particles, so the speed of light has to be involved.  The problem with a random walk, though, is that the photon keeps hitting things, so it's only moving at the speed of light for little bursts at a time.  More specifically, it's only moving at the speed of light for the time it takes the photon to move a distance, w.

With this knowledge, we found the time it takes a photon to move one step,

$t_i=\frac{w}{c}$

To find the total time, multiply by the total number of steps, N, so that 

$t=N\frac{w}{c}$

$=\frac{D^2w}{w^2c}=\frac{D^2}{wc}$

because we have to put N in terms of known values, like $D=\Delta r$ and w.  

This leaves us with this equation for velocity:

$v_{diff}=\frac{wc}{\Delta r}$

c)  The `mean free path' w is the characteristic (i.e. average) distance between collisions. Consider a photon moving through a cloud of electrons with a number density n. Each electron presents an effective cross-section $\sigma$ . How are these parameters related?

We started solving this problem by making a list of each quantity and their units.

$\sigma$=cm$^2$

w=cm

n=cm$^{-3}$

We originally tried to solve the problem only using dimensional analysis, but that's not the most effective method.  So, instead we stopped to think about how the parameters should effect each other.

A high number density, n, should decrease the mean free path, w, because there are more particles for the photon to hit, so it can't travel as far without hitting something.

A high effective cross section, $\sigma$, should also decrease the mean free path, w, because the particles are bigger, so they're harder to avoid and the photon can't go as far without hitting something.

These two statements led us to the idea that n and $\sigma$ should both be inversely proportional to w, so we tried the following relation:

$w=\frac{1}{n\sigma}$

We tested the units, and after finding that this relation gave us centimeters, we figured we were correct!

d)  The "mean free path" w can also be related to the mass density of absorbers, $\rho$ and the absorption coefficient, $\kappa$ (cross-sectional area of absorbers per unit mass).  How is $\kappa$ related to $\sigma$?  Express $v_{diff}$ in terms of $\kappa$ and $\rho$ using dimensional analysis. 

To solve this problem, we followed the same process as in part (c).  

w = cm

$\rho$ = $\frac{g}{cm^3}$

$\kappa = $\frac{cm^2}{g}$

$sigma$ = cm$^2$

Using dimensional analysis, we found that 

$w = \frac{1}{\rho\kappa}=\frac{1}{n\bar{m}\kappa}$

This makes sense because the photon shouldn't be able to go very far in denser regions.  

The relation between $\kappa$ and $\sigma$ can be given as 

$\sigma = \kappa\bar{m}$

if we substitute these equations into the equation for $v_{diff}$ from part (b), we find that 

$v_{diff}=\frac{c}{\rho\kappa\Delta r}$

$\frac{c}{\frac{\rho}{\bar{m}}\sigma\Delta r}$

e)  What is the diff usion timescale for a photon moving from the center of the sun to the surface?

To solve this part of the problem, I'll use the equation we found in part (b) for the time it takes a photon to move a distance, $\Delta r=D$

$t_{diff}=\frac{D^2}{wc}$

$=\frac{D^2\frac{\rho}{\bar{m}}\sigma}{c}$

By plugging in the values for the radius and density of the sun and the mass of a hydrogen atom, you get 

$t_{diff}=8\times 10^{10}$ seconds