Velocity distribution function and Boltzmann factor

Question

In Blundell and Blundell's Thermal physics book there is a chapter on the explanation of the Boltzmann distribution. It is explained that for canonical ensemble with a reservoir at temperature $T$ in equilibrium with the system, the probability that the system is at energy $\epsilon$ is proportional to: $$ e^{-\frac{\epsilon}{k_B T}} $$ which is called the Boltzmann factor.

Then, when considering velocity distributions in monatomic gases they consider a single particle acting as a system and the rest of the energy as a reservoir, so it should follow the Boltzmann distribution in terms of energy. However, the velocity distribution in a fixed direction itself is said to be proportional to a Boltzmann factor too. This time, since the relevant energy is $E = \frac{1}{2}mv_x^2$ the distribution becomes: $$ g(v_x) = C e^{-\frac{mv_x^2}{2k_B T}}$$ What I do not understand is why the distribution is known to follow this factor. It is supposed to apply to the energy of the particle, but why does it apply to the velocity distribution too?

Answer 1

The Boltzmann distribution applies to each component of the energy.

For instance, in your monoatomic gas can be either in the fundamental electronic state or in an excited state an energy $E_{exc}$ above the fundamental one, in addition to a kinetic energy

$$E_k=m(v_x^2+v_y^2+v_z^2)/2$$

then the distribution function for an atom in the excited state will be proportional to

$$e^{-\frac {E_{exc}+E_k} {k_B T}}$$

while that in the fundamental will just be

$$e^{-\frac {E_k} {k_B T}}$$

If you ignore the excitation state, and consider only the component of the velocity along $x$, that is, you add all the atoms with a given $v_x$ whatever the degree of excitation or the values of $v_y$ and $v_z$ are, of course then the only relevant part is

$$e^{-\frac {mv_x^2} {2k_B T}}$$