How can the Maxwell-Boltzmann distribution be reconciled with the Boltzmann distribution?

Question

The Boltzmann distribution asserts that a state at a lower energy is more likely to be populated than a state at a higher energy level. However, the Maxwell-Boltzmann distribution asserts that small velocities are the least likely to be observed. Since velocity is monotonically related to kinetic energy, how can the Maxwell-Boltzmann distribution be reconciled with the Boltzmann distribution?

Answer 1

When you compute the Maxwell-Boltzmann velocity distribution, you start with just the probability of getting a particular state, which in this case is parameterized by the momentum $\vec{p}$ so that $$ P(\vec{v}) \propto e^{-\beta E_\vec{p}} =e^{-\beta p^2/2m} =e^{-\beta mv^2/2}\,. $$ This distribution does show the characteristic decrease of the probability with the energy. However, this should really be thought of as function of the components of the velocity vector, i.e., $$ P(v_x,v_y,v_z) = \frac{1}{Z} e^{-\beta m(v_x^2+v_y^2+v_z^2)/2}\,, $$ where $Z$ is the partition function.

Now, this is not a function of speed directly. In order to find the probability of getting a particular speed, you have to integrate over all possible directions of the velocity. We can think of the velocity vectors as points in a "velocity"-space, and the states at a particular speed $v$ live on the surface of a sphere of radius $v$. Thus as $v$ gets larger, the number of states increases. At small $v$, this cases the probability of getting a particular $v$ to rise. At large $v$, the exponential decay in the Boltzmann distribution takes over, and the probability getting a certain $v$ decreases.

Mathematically, if you want to compute the probability of getting a particular speed $v$, we consider the normalization condition, \begin{align} 1 &= \int_0^{2\pi}d\phi\int_0^{\pi}d\theta\int_0^{\infty}v^2dv\,P(\vec{v})\\ &= \int_0^{\infty}dv\,v^2\frac{1}{Z} \int_0^{2\pi}d\phi\int_0^{\pi}d\theta e^{-\beta mv^2/2}\\ &=\int_0^{\infty}dv\,\frac{4\pi}{Z}v^2e^{-\beta mv^2/2}\,. \end{align} We can then identify the M-B distribution as $$ P(v) = \frac{4\pi}{Z}v^2e^{-\beta mv^2/2}\,, $$ and we can see how the factor of $v^2$$-$which comes mathematically from the integration measure in spherical coordinates and physically from the fact there are "more" states with larger speeds$-$causes the distribution to go to zero as $v\to0$.