The point of a Boltzmann distribution is that it maximizes entropy given a fixed energy. The concept applies to systems with other degrees of freedom besides translational kinetic energy. The general distribution, from Wikipedia is
Thus, the simple adjustment to the Maxwell-Boltzmann distribution you cited is to replace the Newtonian kinetic energy $\frac{mv^2}{2}$ with the relativistic kinetic energy $(\gamma - 1)mc^2$ everywhere it appears in the distribution.
Pair creation is a separate issue that I'll leave to someone else.
To pass from the 1st part of your exercise to the 2nd part should not be difficult.
Imagine a 3-dimentional space in which on the $x$-axis represents the $x$ projection of the velocity vector, the $y$ axis represents the y projection, and the $z$ axis the $z$ projection.
You already calculated the probability of having velocities in a small cube of dimensions $dv_x, dv_y, dv_z$, with the cube positioned at the end of the vector $(v_x, v_y, v_z)$. You got
$$q(v_x, v_y, v_z)dv_x dv_y dv_z = C exp[(−mv^2)/2kT]dv_x dv_y dv_z.$$
where $C$ is the normalization constant.
Now, you have to distinguish between probability and DENSITY of probability, i.e. probability per UNIT VOLUME of your space. The density of probability is equal to
$$C \exp[(−mv^2)/2kT].$$
Notice that this density of probability is independent of the direction of the velocity.
Now, you are asked to calculate the probability of having the velocity between $|v|$ and $|v| + d|v|$, where $|v|$ is the length of the velocity vector.
Consider therefore a sphere of radius $|v|$ in the velocity space, and around it another sphere of radius $|v| + d|v|$.
All the velocity vectors, no matter in which direction they are directed, end-up between these two spheres if their length is between $|v|$ and $|v| + d|v|$.
So, what is the volume confined between these two spheres? It is $4\pi|v|^2 d|v|$.
The density of probability (per unit of volume in the velocity space) you already calculated, and found it independent of the direction of the velocity. Then, to find the probability $dP$ in all the volume between the two spheres you multiply the probability per unit volume, with the volume. You get
$$dP = C \exp[(−mv^2)/2kT]4\pi|v|^2 d|v|.$$
What you are requested is to calculate $dP/d|v|$.
Obviously, it is equal to
$$C \exp[(−mv^2)/2kT]4\pi|v|^2.$$
Good luck,
Sofia
Best Answer
The average speed of particles in a particular direction will always be smaller than the average speed of particles.
Imagine that you have three particles with components of their velocity $(1,1,1)\, \rm ms^{-1}$.
Their average speed is $<v>= \sqrt 3\, \rm ms^{-1}$ whilst their average speed in the x-direction is $<v_{\rm x}>=1\, \rm ms^{-1}$
So what you need to do is use the distribution of velocities in the x-direction
$$f(v_\rm{x}) = \left(\frac{m}{2 \pi k_{B} T}\right)^\frac{1}{2} \exp\left ({-\frac{m v_{\rm x}^{2}}{2 k_{B} T}}\right )$$
and do the following integration
$\displaystyle \int_0^\infty v_\rm{x}\, f(v_\rm{x}) \,dv_{\rm x}$ which will give you an answer of $\dfrac {<v>}{4}$ where $<v>$ is the avergae speed of the particles.
You may find these notes of use?