Worksheet 11.1: Photons Random Walking Out of a Star

 The problem: Consider a photon that has just been created via a nuclear reaction 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 nice Spring day.

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Part A: The photon does not travel freely from the Sun’s center to the surface. Instead it random walks, one collision at a time. Each step of the random walk traverses an average distance l, also known as the mean free path. On average, how many steps does the photon take to travel a distance ∆r? 

We know that on average, the displacement for random walk journeys would be zero, which is not very useful to us. Instead, we can look at the displacement squared: $$< \vec{ D }^2> = \sum \vec{ r_i}^2$$ and because the problem tells us $$\Delta r =  \left( < \vec{ D }^2 > \right)^{ \frac{1}{2}}$$ This means $$\Delta r^2 = \sum \vec{r_i}$$ Because r is a vector, the resultant sum would look something like: $$\sum \vec{r_N} = (\vec{r_1} + \vec{r_2} + ... \vec{r_N}) \times ( \vec{r_1} + \vec{r_2} + ... + \vec{r_N}$$ $$\Delta r^2 = \vec{r_1}^2 + \vec{r_2}^2 +...+ \vec{r_N}^2 + 2\vec{r_1}\vec{r_2}cos(\theta)+ 2\vec{r_2}\vec{r_3}cos(\theta) + ... + 2\vec{r_{N-1}}\vec{r_N}cos(\theta)$$ which is equal to: $$\Delta r^2 = nl^2 + 2l^2 \sum cos(\theta)$$ Where l is the average distance traveled defined in the problem. Due to the random nature of the photon's path of travel, $$\sum cos(\theta) = 0$$ So now, $$\Delta r^2 = Nl^2$$ and finally, $$N = \frac{\Delta r^2}{l^2}$$

Part B: What is the photon’s average velocity over the total displacement after many steps? Call this $\vec{v_{diff}}$, the diffusion velocity.

To find velocity of a photon, we need the distance it travels and the time it takes. We can get distance, or rather, displacement from above:   $$\Delta r = lN^{\frac{1}{2}}$$ to find time we can use the fact that we know photons have a speed c, and each are traveling a distance nl: $$t = \frac{d}{v} = \frac{nl}{c}$$ So $$\vec{v_{diff}} = \frac{\Delta r}{t} = \frac{lN^{\frac{1}{2}}c}{nl} = \frac{c}{N}$$ However, we do not always have N, so the formula in terms of l and r is: $$\vec{v_{diff}}= \frac{cl}{\Delta r}$$

Part C: The “mean free path” l 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 σ. Give an analytic expression for “mean free path” relating these parameters.

We can draw the situation described as shown in the diagram below, taken from our textbook because it is a much neater drawing than anything I can make in MS paint. 

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Where the volume of the cylinder is $V= \pi r^2 L$

From the described relationship, we can see that the number density multiplied by area would give us the 2D area that would result in a collision. However, we want the 3D area (well, volume?) that would result in a collision, which would give: $$n \sigma = \frac{#_{electrons}}{volume} \cdot \frac{area}{1} = \frac{#_{electrons}}{\pi r^2 L} \cdot \frac{\pi r^2}{1} = \frac{#_{electrons}}{distance}$$ For one electron, we would then have: $$n \sigma = \frac{1}{l}$$ So $$l = {1}{n \sigma}$$

Part D: The mean free path l can also be related to the mass density of absorbers ρ, and the “absorption coefficient” κ (cross-sectional area of absorbers per unit mass). How is κ related to σ? Express vdiff in terms of κ and ρ using dimensional analysis. You have now developed the tools to attack the problem of radiative diffusion.

Part 1: How is κ related to σ? 

The units for kappa and sigma would be:

• $\kappa = \frac{cm^2}{g}$
• $\sigma = \frac{cm^2}{n}$

From this we can deduce: $$\rho = n \bar{m}$$ and $$\kappa = \frac{\sigma}{\bar{m}}$$ Which allows us to find: $$n \ \frac{\rho}{\bar{m}}$$ and $$\sigma = \kappa \bar{m}$$ So plugging those last two equations into    $$l = \frac{1}{n \sigma}$$ gives   $$l = \frac{1}{\rho \kappa}$$ Now we can find velocity using: $$\vec{v_{diff}}= \frac{cl}{\Delta r}$$ $$\vec{v_{diff}}= \frac{c \left( \frac{1}{\rho \kappa} \right)}{\Delta r}$$ $$\vec{v_{diff}}= \frac{c}{ \rho \kappa \Delta r}$$

Part E: What is the diffusion timescale for a photon moving from the center of the sun to the surface? The cross section for electron scattering is $σ_T = 7 \times 10^{25} cm^2$ and you can assume pure hydrogen for the Sun’s interior. Be careful about the mass of material through which the photon travels, not just the things it scatters off of. Assume a constant density, ρ, set equal to the mean Solar density (N.B.: the subscript T is for Thomson. The scattering of photons by free (i.e. ionized) electrons where both the K.E. of the electron and λ of the photon remain constant—i.e. an elastic collision—is called Thomson scattering, and is a low-energy process appropriate if the electrons aren’t moving too fast, which is the case in the Sun.)

We know: $$t = \frac{d}{v}$$ $$t = \frac{R_{\odot}}{ \frac{c}{ \rho \kappa R_{\odot}}} = \frac{ \rho \kappa R_{\odot}^2}{c}= \frac{ \rho \left( \frac{\sigma}{\bar{m}} \right) R_{\odot}^2}{c}$$ In this case, mass would be mass of a proton, which we can look up along with the density and radius of the sun. Plugging everything in: $$t = \frac{ 1.41 \left( \frac{7 \times 10^{-25} }{1.7 \times 10^{-24}} \right) (7 \times 10^{10})^2}{3 \times 10^{10}} = 9.5 \times 10^{10} s$$