Probability distribution of outgoing energy of particle

[dark_matter_is_neat]

Homework Statement

Consider the scattering of particle 1 that is massless with particle 2 of mass m that is at rest in the lab frame. Let E and E' be the incoming and outgoing energy of particle 1 in the lab frame. Find the probability distribution of E' assuming that in the center of mass frame particle 1 is scatter in all directions with equal probability.

Relevant Equations

$p_{1} + p_{2} = p_{1}' + p_{2}'$
$v_{CM} = \frac{p_{total}}{E_{total}}$

So from particle 1 scattering in all directions with equal probability in the CM frame, I believe that that means probability of finding particle 1 in an angular range $d\theta$ is just $\frac{d \theta}{2 \pi}$. Let P(E') be the probability density of E', so from the probability of finding the particle in a range $d \theta$, $P(E')dE' = \frac{d \theta}{2 \pi}$, so $P(E') = \frac{1}{2 \pi} \frac{d \theta}{dE'}$. Since this expression is in terms of the angle in the CM frame and not the lab frame, I need to solve for E' in terms of the angle in CM frame.

In the CM frame $E'_{CM} = \frac{s - m^{2}}{2 \sqrt{s}}$ where s is just the Mandelstam variable $s = (p_{1} + p_{2})^{2}$. Since the square of a 4-vector is Lorentz invariant, $(p_{1}+p_{2})^{2} = p_{1}^{2} + p_{2}^{2} + 2 p_{1} p_{2} = m^{2} + 2mE$ (note I'm using natural units). So $E'_{CM} = \frac{mE}{\sqrt{m^{2} + 2mE}}$. Now I need to relate this expression to E' in the lab frame, so I'll use a Lorentz transform for $v_{CM} = \frac{p_{1}}{E + m} = \frac{E}{E+m}$ and take the direction that particle 1 is initially moving in to be along the z axis for simplicity, so $E' = \gamma E'_{CM} + \alpha E'_{CM}cos(\theta)$ where $\gamma = \frac{1}{\sqrt{1 - (v_{CM)^{2}}}$ and $\alpha = v_{CM} \gamma$ when using natural units.
So rearranging this $\theta = cos^{-1}(\frac{E' - \gamma E'_{CM}}{\alpha E'_{CM}})$, so plugging this back into my earlier expression for $P(E')$,
I get $P(E') = \frac{1}{2 \pi} \frac{1}{1 - (\frac{E' - \gamma E'_{CM}}{\alpha E'_{CM}})^{2} }$.

Is the right way to go about this problem or is something wrong with my thinking?

 

[dark_matter_is_neat]

So since it looks like I made a formatting error and for some reason I can't edit my post anymore, here is the section with messed up formatting fixed:
$\gamma = \frac{1}{\sqrt{1 - v_{CM}^{2}}}$ and $\alpha = v_{CM} \gamma$ when using natural units. So So rearranging the expression for E' $\theta = cos^{-1}(\frac{E' - \gamma E'_{CM}}{\alpha E'_{CM}})$ So plugging this into my expression for P(E'), I get $P(E') = \frac{1}{2 \pi} \frac{1}{1 - (\frac{E' - \gamma E'_{CM}}{\alpha E'_{CM}})^{2}}$