# Derivation of relativistic mean free path

I am working through a short derivation found in Abramowicz 1991 regarding the mean free path of a photon.

We have a fluid moving in a particular direction with velocity $v$ and in an inertial rest frame $A$. There is an inertial frame $B$ which is comoving with the fluid. In frame $B$ the mean free path of the photons is $\ell_0$.

Abramowicz now considers many photons propagating in the medium at an angle $\theta$ with respect to $v$ and then states that the mean free path of photons as measured in frame $A$ is: $$\ell = \frac{\ell_0}{\gamma(1-\beta \cos \theta)}$$

Can anyone explain why this should be so?

Imagine a typical path.

It starts at event $A$=$(ct_A,x_A,y_A)$ and ends at event $B$=$(ct_B,x_B,y_B).$ (Note I'm using $w=ct$ as the fourth dimension so we can measure everything in meters.) In the frame of the fluid this typical path has length $l_0.$ In both frames the separation is null.

We can actually choose the origin of the fluid frame and your frame to be at $A$=$(ct_A,x_A,y_A)$=$(0,0,0).$ And the origin of the fluid frame moves in the direction $(ct,vt,0).$ Thus if the photon moves at an angle of $\theta$ and has a path length $l$ we get $B$=$(ct_B,x_B,y_B)$=$(l,l\cos(\theta),l\sin(\theta)).$ But we'd like to know what it looks like in the fluid frame.

The fluid frame has three dimensionless orthogonal vectors; $(1,v/c,0),$ which is timelike, $(0,0,1)$ which is spacelike and orthogonal to that time like vector and to the motion and the final vector $(v/c,1,0)$ which is spacelike and mutually orthogonal to the other two. To the fluid frame these are the t,y, and x vectors. We can make them unit length and get $(1,v/c,0)/\sqrt{1-(v/c)^2},$ $(0,0,1),$ and $(v/c,1,0)/\sqrt{1-(v/c)^2}.$ These are the unit orthogonal vectors for the fluid frame.

So how do you get coordinates from a vector? You dot it with the unit vector and adjust for the sign based on whether the vector you dotted with was spacelike or timelike.

So given $B$=$(l,l\cos(\theta),l\sin(\theta)),$ all we have to do is dot with $(1,v/c,0)/\sqrt{1-(v/c)^2},$ $(0,0,1),$ and $(v/c,1,0)/\sqrt{1-(v/c)^2}.$ Doing that we get $(l-lv\cos(\theta)/c)/\sqrt{1-(v/c)^2},$ $-l\sin(\theta),$ and $(-l\cos(\theta)+lv/c)/\sqrt{1-(v/c)^2}.$

All this means is that

$$B=\left[\frac{l-lv\cos(\theta)/c}{\sqrt{1-(v/c)^2}}\right]\frac{(1,v/c,0)}{\sqrt{1-(v/c)^2}}+\left[\frac{l\cos(\theta)-lv/c}{\sqrt{1-(v/c)^2}}\right]\frac{(v/c,1,0)}{\sqrt{1-(v/c)^2}}+\left[l\sin(\theta)\right](0,0,1).$$

This is just like writing $$(a,b,c)=[a](1,0,0)+[b](0,1,0)+[c](0,0,1).$$

So now since we know these events have $x'$=$\frac{l\cos(\theta)-lv/c}{\sqrt{1-(v/c)^2}}$ and $y'=l\sin(\theta).$ And since $l_0$=$\sqrt{x'^2+y'^2}$=$l\sqrt{\frac{\cos^2(\theta)+v^2/c^2-2v\cos(\theta)/c}{1-(v/c)^2}+\frac{\sin^2(\theta)(1-v^2/c^2)}{1-(v/c)^2}}$ = $l\sqrt{\frac{1+v^2/c^2-2v\cos(\theta)/c}{1-(v/c)^2}+\frac{\sin^2(\theta)(-v^2/c^2)}{1-(v/c)^2}}$ = $l\sqrt{\frac{1+v^2/c^2-2v\cos(\theta)/c}{1-(v/c)^2}+\frac{(1-\cos^2(\theta))(-v^2/c^2)}{1-(v/c)^2}}$ = $l\sqrt{\frac{1-2v\cos(\theta)/c}{1-(v/c)^2}+\frac{(-\cos^2(\theta))(-v^2/c^2)}{1-(v/c)^2}}$ = $\frac{l}{\sqrt{1-(v/c)^2}}\sqrt{1-2\beta\cos(\theta)+\beta^2\cos^2(\theta)}$ = $\frac{l}{\sqrt{1-(v/c)^2}}\sqrt{\left(1-\beta\cos(\theta)\right)^2}$ = $\frac{l(1-\beta\cos(\theta))}{\sqrt{1-(v/c)^2}}.$

So $l_0=\frac{l(1-\beta\cos(\theta))}{\sqrt{1-(v/c)^2}}.$

Now all of this was just asking how the distance of a typical path is measured in two frames. To say the mean paths are related you'd have to compare all the paths, not just the typical one.

But they have the same ratio (as long as you average over photons making that same angle). So if your measure is then related to how many you observe and they observe the same photons then everything is fine. You don't observe the same photons because of simultaneity issues but if your mean free path isn't changing in time or space then simultaneity issues won't be a problem. You'll observe different collections as simultaneous but the averages will be the same.

As for the angle you can say that you are merely considering the mean free path of photons making that angle. So the computational part was just the special relativity part.