Photons that are generated in the core of a star follow a random pathway to the surface, colliding with each other along the way. Below, we will examine in-depth some of the properties of a photons journey to the surface of the sun.
The figure above shows a representative path traveled by a photon to the edge of a star.
(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?
The first step to calculating the number of steps taken by a photon to a star's surface is setting up an appropriate summation. To begin, we can consider the length of each step to be l, and the overall distance to be ∆r. We can then consider each step taken as a vector, \( \vec{r_i} \), and the distance D, as the sum of these vectors, \( \vec{D} = \sum{ \vec{r_i}} \). It's important to consider, however, that this displacement will converge to zero for a large number of random-walk journeys. For this reason, we will \( \vec{D^2} \) and take the square root of the result to be \( \delta r = ( \vec{D^2)}^{\frac{1}{2}} \).
This process is equivalent to calculating the dot product of each vector step, \( \vec{r_i} \) with each other. Below is the result of this dot product multiplication: \[ \vec{D^2} = n \vec{r_0^2} + \vec{r^2} cos \theta \] Considering that the photon's random walk covers all directions and values of theta, the cosine term will favor zero, leaving us with the following relationship: \[ \vec{D^2} = n \vec{r_0^2} \] Since we are treating each step as distance of length l, we can replace \( \vec{r_0} \) with l. Additionally, since D represents the overall distance traveled, we can replace this with ∆r in the equation above. This then gives us the following relationship: \[ { \delta r}^2 = n l^2 \] By simply rearranging this, we can determine that on average, the number of steps taken by the photon to travel distance ∆r is: \[ n = \frac{ { \delta r}^2}{l^2} \]
(b) What is the photon’s average velocity over the total displacement after many steps? Call this ~vdiff, the diffusion velocity.
Calculating the diffusion velocity is a fairly straightforward process, since we know both the distance traveled by the photon, and the speed that photons (light) travel at. Knowing that our total distance is ∆r, we just need to calculate the total time to resolve the final velocity. Each step n takes \( \frac{l}{c} \) time to complete. So, \[ t_{step} = \frac{l}{c} \] To then find the total time, we multiply this by the total number of steps taken, n. \[t_{total} = n \frac{l}{c} \] From part A, we also know that \[ \delta r = l \sqrt{n} \] Dividing this total distance by the total time gives us our differential velocity. \[ \vec{v_{diff}} = \frac{ \delta r}{t_{total}} = \delta r \frac{c}{nl} = l \sqrt{n} \frac{c}{nl} = \frac{c}{ \sqrt{n}} \]
(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
Determining the "mean free path" l can be achieved through dimensional analysis. Our desired unit for this is length (cm). Let's begin with our units for number density: \( \frac{1}{cm^3} \). Furthermore, our units for cross sectional area, σ, are \( cm^2 \). By multiplying these terms together, we would get units of \( \frac{1}{cm} \). To arrive at our desired units of cm, we simply need to invert this term, to get the "mean free path." This value is expressed below: \[ l = \frac{l}{n \sigma} \]
(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.
Again, considering dimensional analysis, we can derive this expression for mean free path in terms of mass density and absorption coefficient. The desired unit of l is again cm. The units of mass density are \( \frac{g}{cm^3} \). The units of the absorption coefficient are \( \frac{cm^2}{g} \). By multiplying these terms, we again would get units of \( \frac{1}{cm} \). To arrive at our desired units, we would again need to invert this term. This final expression is shown below: \[ l = \frac{1}{ \rho \kappa} \] To see how \( \kappa \) is related to \( \sigma \), we can equate our two expressions for l: \[ \frac{1}{n \sigma} = \frac{1}{ \rho \kappa} \] Then solving for \( \sigma \), we get: \[ \sigma = \frac{ \rho \kappa }{n} \] Again applying dimensional analysis, we know that mass density has units \( \frac{g}{cm^3} \), and number density has units \( \frac{1}{cm^3} \). Seeing that these are divided in our \( \sigma \) expression, we know that these terms will cancel to give us just mass and \( \kappa \), as shown below: \[ \sigma = mass \times \kappa \] The l expression from this step can be used to also update our \( \vec{v_{diff}} \) from part b. This provides us with a new expression of \[ \vec{v_{diff}} =\delta r \frac{c}{nl} = \frac{c}{ \delta r \rho \kappa} \]
(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 σT = 7 x 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.)
To begin, let's consider our knowns:
Mass of hydrogen atom: \( 2 \times 10^{24} g \)
Solar mass density: \( \rho = 1.5 \frac{g}{cm^3} \)
Solar radius: \( 7 \times 10^{10} cm \)
Since we are trying to solve for time, which equals distance traveled divided by the velocity traveled, as shown below: \[ t = \frac{d}{v} \] The net distance, d, traveled will be equal to ∆r. Furthermore, we can apply our equation for \( \vec{v_{diff}} \) to give us a complete expression of: \[ t = \frac{ \delta r}{ \frac{c}{ \delta r \rho \kappa}} = \frac{ \delta r^2 \rho \kappa}{c} \] Additionally, we can substitute \( \kappa = \frac{ \sigma }{ m_{hydrogen}} \). This give us a final expression of \[ t = \frac{ \delta r \rho \sigma }{ m_{hydrogen} c } = 9 \times 10^{10} \, s = 3000 years \]
Acknowledgements: Team EE
Very nice.
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