I'm trying to solve the following problem (in preparation for my exam).
Consider two one-dimensional relativistic ideal-gas particles with masses confines to a one dimensional box of length $L$. Because they are relativistic, their energies are given by $E_A=|P_A|c$ and $E_B=|P_B|c$.

Assume that the particles are in thermal equilibrium with each other, and that the total kinetic energy is $E=E_A+E_B$. Use the usual assumption that the probability density is uniform in phase space, subject to the constraints.

Calculate the probability distribution $P(E_A)$ for the energy of one of the particles.

What I've done so far
This smells like a microcanonical ensemble. To find the probability distribution I use the equation



$$\Omega(E)=\frac{1}{h^{3N}N!}\int dq\int dp \delta(E-H).$$

We can write down $$H=c(|p_A|+|p_B|).$$

Then I try evaluating

$$\Omega(E_A)=\frac{1}{h}\int dq_A\int dp_A \delta\big(E_A-c(|p_A|+|p_B|)\big)$$

where $h^{3N}$ reduces down to just $h$ because we are looking at a single particle in one dimension. $\int dq=L$, and I reason that

$$\int dp_A \delta\big(E_A-c(|p_A|+|p_B|)\big)=2$$

because there are two values for which $E_A-c(|p_A|+|p_B|)=0$ since we take the absolute value of $p_A$. We can follow the same line of thinking for $\Omega(E_B)$ and afterwards substitute $E_B=E-E_A$. Then we find


This smells like total bullshit to me, since there is no dependency of $E_A$ anywhere and we are not involving the mass of particle A vs particle B. Nevertheless I can keep on trucking and calculate $\Omega(E)$ to be

$$\Omega(E)=\frac{\pi L^2E}{2h^2c}.$$

I'm pretty confident that this one is correct. However, this results in a probability distribution of

$$P(E_A)=\frac{2c}{\pi E}.$$

Surely, that can't be right!

My problem
Am I correctly evaluating the delta functions? It seems like there should be a dependency on $E_A$ or $E_B$, but this dependency is crushed by the one dimensional-ness of the problem. What is the correc way to reason through this?


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