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 Astronomy 16 students studying on a nice Spring day.
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 one photon take to travel a distance \(\Delta r\)?
Hint: If each step is a vector \(\overrightarrow{r_i}\), then \(\overrightarrow{D}=\sum\overrightarrow{r_i}\). However, this displacement is zero for a large number of random-walk journeys. Instead, calculate \(\overrightarrow{D}^2\) and take the square root of the result to be \(\Delta r=(\overrightarrow{D}^2)^{1/2}\). This is the root-mean-squared displacement, which is a scalar rather than vector quantity. Not that 'multiplying' two vectors isn't as simple as multiplying two scalars. You must take the dot product, which looks like \(\overrightarrow{A}\cdot \overrightarrow{B}=AB \cos{\theta}\), where \(\theta\) is the angle between the two vectors.
A random walk is a path consisting of a series of steps or movements in random directions. The motion of photons in stars as they travel from the star's center to its surface is modeled by a random walk. As they move through the star, they collide into other particles and electrons and bounce off, colliding into more particles. This process causes the random walk path that the photon travels. The image below shows an example of what the random walk of a photon in a star might look like.
For this problem, we want to find how many steps the photon has to take on average to travel a distance \(\Delta r\). The image below shows this set up, where each line is a vector \(\vec{r}_i\), with average magnitude l, and a net displacement \(\vec{D}=\sum \vec{r}_i\).
The distance \(\Delta r\) is the root-mean-squared displacement, given by:
First, calculating \(\vec{D}^2\), we must take the dot product, since \(\vec{D}\) is a vector:
A random walk is a path consisting of a series of steps or movements in random directions. The motion of photons in stars as they travel from the star's center to its surface is modeled by a random walk. As they move through the star, they collide into other particles and electrons and bounce off, colliding into more particles. This process causes the random walk path that the photon travels. The image below shows an example of what the random walk of a photon in a star might look like.
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| Random walk of a photon through a star. Image Credit: http://www.atnf.csiro.au/outreach//education/senior/astrophysics/images/stellarevolution/rndmwlkphotonsm.jpg |
For this problem, we want to find how many steps the photon has to take on average to travel a distance \(\Delta r\). The image below shows this set up, where each line is a vector \(\vec{r}_i\), with average magnitude l, and a net displacement \(\vec{D}=\sum \vec{r}_i\).
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| A random walk, where each step has an average length, l, and a net displacement vector, \(\overrightarrow{D}\). Image Credit: http://www2.ess.ucla.edu/~jewitt/images/random.gif |
The distance \(\Delta r\) is the root-mean-squared displacement, given by:
\(\Delta r=(\vec{D}^2)^{1/2}\)
\(\vec{D}\cdot\vec{D} = \left|\vec{D}\right|^2 = \left|\displaystyle\sum_{i=1}^N\vec{r}_i\right|^2 = \displaystyle\sum_{i=1}^N\left|\vec{r}_i\right|\left|\vec{r}_i\right|\cos{(0)} + \displaystyle\sum_{i,j}\left|\vec{r}_i\right|\left|\vec{r}_j\right|\cos{\theta}\)
\(|\vec{D}|^2 =Nl^2\)
\(\Delta r^2 =Nl^2\)
\(N=\left(\dfrac{\Delta r}{l}\right)^2\)
Where N is the number of steps taken by the photon to travel the distance \(\Delta r\), and \(\theta\) is the angle between vectors \(\vec{r}_i\) and \(\vec{r}_j\). Since the vectors are pointing in random directions, we can take the average of \(\cos{\theta}\), which is 0. So the second term will be zero, leaving us with the term:
\(\left|\vec{D}\right|^2 = \displaystyle\sum_{i=1}^N\left|\vec{r}_i\right|\left|\vec{r}_i\right| = \left|\vec{r}_1\right|^2 + \left| \vec{r}_2\right|^2 + \ldots + \left|\vec{r}_N\right|^2\)
We already know that the average magnitude of each vector is l, so this gives us:
We also know that \(\Delta r=(\vec{D}^2)^{1/2}\), so we can rewrite our expression and solve for N:
\(N=\left(\dfrac{\Delta r}{l}\right)^2\)
b) What is the photon's average velocity over the total displacement after many steps? Call this \(\overrightarrow{v}_{diff}\), the diffusion velocity.
To solve for the diffusion velocity, we can use the basic equation, \(d=rt\). First, we must find the time it takes to travel all N steps:
To solve for the diffusion velocity, we can use the basic equation, \(d=rt\). First, we must find the time it takes to travel all N steps:
\(t=\dfrac{Nl}{c}\)
Where c is the speed of light (the speed of each photon). Then, knowing that the photon has a displacement of \(\Delta r\), we can solve for the diffusion velocity:
\(\vec{v}_{diff}=\dfrac{\Delta r}{t}\)
From part (a), we found that \(\Delta r =\sqrt{N}l\). Plugging this in, we find:
\(\vec{v}_{diff}=\dfrac{\sqrt{N}lc}{Nl}=\dfrac{c}{\sqrt{N}}\)
Which can also be rewritten as:
\(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\)
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 \(\sigma\). Give an analytic expression for "mean free path" relating these parameters.
The image below shows a volume filled with targets, each with cross-sectional area \(\sigma\), and with number density n, through which a photon travels.
The image below shows a volume filled with targets, each with cross-sectional area \(\sigma\), and with number density n, through which a photon travels.
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| Image Credit: http://press.princeton.edu/chapters/s03_8457.pdf |
The projected number of targets per unit area in the path of the photon will be ndx, and the fraction of the area that is covered by the targets is \(\sigma ndx\) (since each target has cross-sectional area \(sigma\)). So this means that the number of targets intersected by a path through the volume will be \(n\sigma dx\). Since the photons are limited to one interaction, we can set this equal to one. Here, dx will give us the typical distance a photon will travel between interactions, also known as the mean free path, l. Solving for this value, we get:
\(l=\dfrac{1}{n\sigma}\)
We can see that the units work for this: n is the number of targets per volume, so it has units of \(\text{cm}^{-3}\), and \(\sigma\) has units of area, so \(n\sigma\) has units of \(\text{cm}^{-1}\), giving l units of centemeters, which is what we wanted.
d) The mean free path, l, 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 \(\overrightarrow{v}_{diff}\) in terms of \(\kappa\) and \(\rho\) using dimensional analysis.
The absorption coefficient is given to be the cross-sectional area of the absorbers (or targets) per unit mass, so it is given by:
Where \(\overline{m}\) is the typical mass of an absorber. \(\kappa\) has units of \(\frac{\text{cm}^2}{g}\). And thus we have our relation between \(\kappa\) and \(\sigma\).
We also know that \(n=\frac{\rho}{\overline{m}}\). From parts (b) and (c) of this problem, we found that \(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\) and \(l=\dfrac{1}{n\sigma}\). We can write these expressions in terms of \(\kappa\) and \(\rho\) to get:
Plugging these relationships into the expression for the diffusion velocity, we get:
\(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\)
Checking this for dimensional analysis:
\([l]=\)m
\([c]=\)m/s
\(\Delta r=\)m
So the dimensions simplify to be m/s, which is what we wanted.
The absorption coefficient is given to be the cross-sectional area of the absorbers (or targets) per unit mass, so it is given by:
\(\kappa =\dfrac{\sigma}{\overline{m}}\)
Which can be rewritten as:
\(\sigma = \kappa\overline{m}\)
Where \(\overline{m}\) is the typical mass of an absorber. \(\kappa\) has units of \(\frac{\text{cm}^2}{g}\). And thus we have our relation between \(\kappa\) and \(\sigma\).
We also know that \(n=\frac{\rho}{\overline{m}}\). From parts (b) and (c) of this problem, we found that \(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\) and \(l=\dfrac{1}{n\sigma}\). We can write these expressions in terms of \(\kappa\) and \(\rho\) to get:
\(l=\dfrac{1}{\rho\kappa}\)
Plugging this into our expression for the diffuse velocity, we get:
\(\vec{v}_{diff}=\dfrac{c}{\rho\kappa\Delta r}\)
\(\vec{v}_{diff}=\dfrac{lc}{\Delta r}\)
Checking this for dimensional analysis:
\([l]=\)m
\([c]=\)m/s
\(\Delta r=\)m
So the dimensions simplify to be m/s, which is what we wanted.
We have now developed the tools to attack the problem of radiative diffusion.
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 \(\sigma_T =7\times 10^{-25}\text{ cm}^2\) 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, \(\rho\), set equal to the mean Solar density.
Note: 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 \(\lambda\) 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.
Here, the photons are undergoing Thomson scattering, which is when photons collide and are scattered by free, ionized elections. Since this is a low-energy process, it is only appropriate if the electrons aren't moving too fast, which is the case in the Sun. Also, since we assume that the Sun's interior is only made up of pure hydrogen, we can say that \(\overline{m}\approx m_p\) because hydrogen has only one electron and one proton.
To find the diffusion timescale for a photon, we can use the basic \(d=rt\) equation again:
I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.
Citations:
http://en.wikipedia.org/wiki/Random_walk
http://press.princeton.edu/chapters/s03_8457.pdf
Here, the photons are undergoing Thomson scattering, which is when photons collide and are scattered by free, ionized elections. Since this is a low-energy process, it is only appropriate if the electrons aren't moving too fast, which is the case in the Sun. Also, since we assume that the Sun's interior is only made up of pure hydrogen, we can say that \(\overline{m}\approx m_p\) because hydrogen has only one electron and one proton.
To find the diffusion timescale for a photon, we can use the basic \(d=rt\) equation again:
\(t_{diff}=\dfrac{\Delta r}{\vec{v}_{diff}}\)
Plugging the expressions we solved for above, we get:
\(t_{diff}=\dfrac{\kappa\rho\Delta r^2}{c}\)
In this context, we know that \(\kappa =\frac{\sigma _T}{m_p}\). Putting this into our equation:
\(t_{diff}=\dfrac{\sigma_T\rho\Delta r^2}{m_pc}\)
Now, we can plug in values. We know that \(\sigma_T =7\times 10^{-25}\text{ cm}^2\), \(\rho =1.14\text{ g cm}^{-3}\), \(\Delta r= R_\odot =6.96\times 10^{10}\text{ cm}\), \(m_p= 1.67\times 10^{-24}\text{ g}\), and \(c=3\times 10^{10}\text{ m s}^{-1}\).
\(t_{diff}=\dfrac{(7\times 10^{-25}\text{ cm}^2)(1.14\text{ g cm}^{-3})(6.96\times 10^{10}\text{ cm})^2}{(1.67\times 10^{-24}\text{ g})(3\times 10^{10}\text{ m s}^{-1})}\approx 9.54\times 10^{10}\text{ s}\approx 3000\) years
So it would take a photon approximately 3000 years to travel from the center to the surface of the Sun.
I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.
Citations:
http://en.wikipedia.org/wiki/Random_walk
http://press.princeton.edu/chapters/s03_8457.pdf
Nice post!
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