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:

enter image description here

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: $f_E(E)\,dE=f_p(\mathbf{p})d^3\mathbf{p}.$

  • $\begingroup$ $f_X(x)dx=f_Y(y)dy$ is simply conservation of probability. $\endgroup$ – valerio Mar 16 '18 at 9:38
  • $\begingroup$ That’s a very nicely framed question! $\endgroup$ – boyfarrell Mar 16 '18 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 $$P(E)={\rm Cst}\ \!E^{3N/2-1}e^{-\beta E}$$

  • $\begingroup$ This is excellent, thank you! May I ask, what is $h_0$? (Sorry, I am obviously not a physicist!) $\endgroup$ – Student Mar 16 '18 at 9:43
  • 1
    $\begingroup$ 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. $\endgroup$ – Christophe Mar 16 '18 at 9:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.