# The shape of the Maxwell-Boltzmann distribution

This question is about my misunderstanding of the differences between and shapes of the Boltzmann and Maxwell-Boltzmann distributions.

I understand that the Boltzmann distribution arises as the solution to the constrained maximum entropy optimisation, and that it is commonly written:

$$p_i =\frac{e^{-E_i/kT}}{\sum_{j=1}^Me^{-E_j/kT}},$$

where $j$ and $i$ index the state of some system with energy levels $E_i$ increasing in $i$. $T$ is the temperature, which was introduced as the reciprocal of a Lagrange multiplier and $k$ is Botlzmann's constant.

Now if someone were to ask me to draw the probability density function of the states for different energies, I would draw something similar to this:

In particular, the density would take its maximum value at $E=0$. When I read the wikipedia page for the Maxwell-Boltzmann distribution, they start with the distribution I have above and then proceed to find the distribution of momenta. They then transform the distribution back to a distribution of energy, but somehow the distribution now has a different shape. In particular:

$$f(E) = 2\sqrt{\frac{E}{\pi}}\frac{1}{(kT)^{3/2}}\exp\Big(\frac{-E}{kT}\Big).$$

This expression clearly does not have the maximum value at $E=0$. Why is the new energy distribution different to the first one? Perhaps my failure to understand this is related to the condition they have: $f_E(E)\,dE=f_p(\mathbf{p})d^3\mathbf{p}.$

• $f_X(x)dx=f_Y(y)dy$ is simply conservation of probability. Commented Mar 16, 2018 at 9:38
• That’s a very nicely framed question! Commented Mar 16, 2018 at 11:11

Your plot corresponds to the probability $p_i$ of occurence of the $i$-th microscopic state whose energy is $E_i$. At the macroscopic level, you cannot distinguish the different microscopic states but you can measure macroscopic quantities as the energy E. The probability to measure an energy $E$ is $$P(E)=\sum_i p_i\delta(E-E_i)=\Omega(E){e^{-\beta E}\over{\cal Z}}$$ where $$\Omega(E)=\sum_i \delta(E-E_i)$$ is the number of microscopic states whose energy is $E$. Usually, you have very few microscopic states with a low energy but this number increases with $E$. Multiplying by the Boltzmann weight which behaves in the opposite way, you get a curve which typically increases as $\Omega(E)$ for $E$ small and then decreases exponentially.
In the particular case of the Maxwell-Boltzmann distribution, i.e. for a classical ideal gas, we have for a single particle \eqalign{ P(E)&={1\over z}\int \delta\Big(E-{p^2\over 2m}\Big) e^{-\beta{p^2\over 2m}} {d^3\vec r d^3\vec p\over h_0^3}\cr &={V\over zh_0^3}\int_0^{+\infty} \delta\Big(E-{p^2\over 2m}\Big) e^{-\beta E} 4\pi p^2dp } where $d^3\vec p$ has been written in spherical coordinates. Setting $u=p^2/2m$ and $p=\sqrt{2mu}$ so that $du=pdp/m$, we get $$P(E)={4\pi V\over zh_0^3}e^{-\beta E}\delta(E-u)\sqrt{2mu} \times mdu={4\pi V\over{\cal Z}h_0^3}m^3/2\sqrt Ee^{-\beta E}$$ Plugging now the expression of the partition function, we get $$P(E)={4\pi\over (2\pi k_BT)^{3/2}}\sqrt Ee^{-\beta E}$$
However, this is only for a single particle. The same calculation should be done for a gas of $N$ particles. $$P(E)={1\over {\cal Z}}\int \delta\Big(E-\sum_{i=1}^{3N}{p_i^2\over 2m}\Big)e^{-\beta\sum_{i=1}^{3N}{p^2\over 2m}} {d^{3N}\vec r d^{3N}\vec p\over h_0^{3N}}$$ The equivalent of spherical coordinates in a space of dimension $3N$ leads to ${\rm Cst}\ \!p^{3N-1}dp$ after integrating over the $3N-1$ angles. Therefore, I guess that we should expect something like $$P(E)={\rm Cst}\ \!E^{3N/2-1}e^{-\beta E}$$
• This is excellent, thank you! May I ask, what is $h_0$? (Sorry, I am obviously not a physicist!) Commented Mar 16, 2018 at 9:43
• In the usual derivation of the Boltzmann distribution, one introduces a discretization of the phase space. The microscopic states are then countable. $h_0$ is the surface of a cell (one degree of freedom). $e^{-\beta E}/{\cal Z}$ is the probability of a microscopic state so dividing by $h_0$, you get a probability density. You can usually safely ignore it since in the above expressions, $h_0$ also appears in the partition function so they cancel in the final expression. Commented Mar 16, 2018 at 9:50