The purpose of this post is to outline a derivation of the classical Maxwell distribution \(\rho_{\textbf V}(\textbf v|m,T)\), i.e. the probability density function for the continuous speed random vector \(\textbf V\) in a monatomic ideal gas given the atomic mass \(m\) and temperature \(T\).
It turns out that the specific Gaussian form of the \(\textbf v\)-dependence in the Maxwell distribution, namely\(\rho_{\textbf V}(\textbf v|m,T)\propto e^{-|\textbf v|^2/2\sigma^2(m,T)}\) does not actually require any physics whatsoever, emerging instead purely from the \(SO(3)\) symmetry of the setup (valid whether or not there’s any interaction among the gas particles). Put another way, any continuous random vector like the momentum \(\textbf P\), the acceleration \(\textbf A\), the angular momentum \(\textbf L\), etc. will also be normally distributed across the gas particles! For sake of concreteness however, let us work with the velocity random vector \(\textbf V\).
To prove the magical claim above, first note that \(\rho_{\textbf V}(\textbf v)=\rho_{\textbf V}(v_x,v_y,v_z)\) is clearly the joint probability distribution formed from the probability distributions \(\rho_{V_x}(v_x),\rho_{V_y}(v_y),\rho_{V_z}\) of the particle velocities along the \(x,y\) and \(z\) directions. Mathematically, the probabilistic chain rule asserts (for instance) that \(\rho_{\textbf V}(v_x,v_y,v_z)=\rho_{V_x}(v_x)\rho_{V_y|V_x}(v_y|v_x)\rho_{V_z|V_x,V_y}(v_z|v_x,v_y)\). However, the continuous random variables \(V_x,V_y,V_z\) are independent and identically distributed by \(SO(3)\) symmetry so the conditional probability distributions all simplify to the same marginal distribution denoted \(\tilde\rho:=\rho_{V_x}=\rho_{V_y}=\rho_{V_z}\). Thus, the mathematical challenge is to solve the following functional equation simultaneously for \(\rho_{\textbf V}\) and \(\tilde\rho\):
\[\rho_{\textbf V}(v_x,v_y,v_z)=\tilde{\rho}(v_x)\tilde{\rho}(v_y)\tilde{\rho}(v_z)\]
As usual, the trick to solving functional equations is to convert them to differential equations. Taking logs to convert the product into a sum and then differentiating with respect to \(v_x,v_y\) and then \(v_z\) yields:
\[\frac{1}{\rho_{\textbf V}(v)v}\frac{d\rho_{\textbf V}}{dv}=\frac{1}{\tilde{\rho}(v_x)v_x}\frac{d\tilde{\rho}}{dv_x}=\frac{1}{\tilde{\rho}(v_y)v_y}\frac{d\tilde{\rho}}{dv_y}=\frac{1}{\tilde{\rho}(v_z)v_z}\frac{d\tilde{\rho}}{dv_z}\]
where we are viewing \(\rho_{\textbf V}=\rho_{\textbf V}(\textbf v)=\rho_{\textbf V}(|\textbf v|)=\rho_{\textbf V}(v)\) by the same \(SO(3)\) isotropy argument (and indeed the answer was already spoiled earlier!). By the usual separation of variables argument, one can set everything equal to a constant suggestively denoted \(-1/\sigma^2\) (it has to be negative or the resulting solution would not be normalizable as required for interpretation as a probability distribution). This yields a separable first-order ODE which is straightforwardly integrated (and normalized) to yield a zero-mean normal distribution:
\[\tilde{\rho}(\tilde v)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\tilde v^2/2\sigma^2}\]
Hence, the Maxwell distribution of velocity vectors is a multivariate normal distribution:
\[\rho_{\textbf V}(v)=\frac{1}{\sigma^3(2\pi)^{3/2}}e^{-v^2/2\sigma^2}\]
If one is specifically interested in the distribution \(\rho_V(v)\) of the continuous speed random variable \(V:=|\textbf V|\), then clearly \(\rho_V(v)=4\pi v^2\rho_{\textbf V}(v)\) is instead a chi distribution rather than a normal distribution.
In order to determine the parameter \(\sigma\) however, it is (finally!) necessary to look at the physics. For \(N=1\) gas particle with Hamiltonian \(H\), its semi-classical canonical partition function \(Z\) is:
\[Z=\frac{1}{h^3}\int_{\textbf x,\textbf p\in\textbf R^6}e^{-\beta H}d^3\textbf xd^3\textbf p\]
For an ideal gas, the classical Hamiltonian is just \(H=|\textbf p|^2/2m\), so if the container has volume \(V=\int_{\textbf x\in\textbf R^3}d^3\textbf x\) then:
\[Z=\frac{V}{h^3}\int_{\textbf p\in\textbf R^3}e^{-\beta p^2/2m}d^3\textbf p\]
At this point the quick way to obtain \(\sigma\) would be to recognize that the semi-classical canonical partition function is the normalization factor in the Boltzmann distribution of the canonical ensemble, so the integrand must be proportional to the probability density function for the momentum random vector \(\textbf P\), from which one can directly see (using \(p^2/2m=mv^2/2\)) that \(1/2\sigma^2=\beta m/2\) so \(\sigma=1/\sqrt{\beta m}=\sqrt{kT/m}\). Alternatively, if that was too slick, one can also separate out the \(3\) Gaussian integrals in the Poisson fashion to get:
\[Z=\frac{V}{\lambda^3}\]
with \(\lambda=\sqrt{\frac{2\pi\hbar^2}{mkT}}\) the thermal de Broglie wavelength of the gas particles. Since the gas is ideal, hence non-interacting, the semi-classical canonical partition function for \(N\) ideal gas particles (denoted \(Z\) again by abuse of notation) is:
\[Z=\frac{V^N}{N!\lambda^{3N}}\]
where the \(N!\) is to avert the Gibbs paradox by accounting for identical particles (although see the article on the Gibbs paradox by E.T. Jaynes). Armed with \(Z\), everything else is at most a few derivatives away. The most immediate is the Helmholtz free energy \(F=-kT\ln Z\approx -NkT\left(\ln V-\ln N-3\ln\lambda+1\right)\) from which one deduces that the pressure \(p=-\frac{\partial F}{\partial V}=\frac{NkT}{V}\) follows the ideal gas law. The expected total energy is \(\langle E\rangle =-\partial\ln Z/\partial\beta=\frac{3}{2}NkT\) with \(C_V=\partial\langle E\rangle/\partial T=\frac{3}{2}Nk\) the heat capacity and by the fluctuation-dissipation theorem \(\sigma_E^2=kT^2C_V=3Nk^2T^2/2\) so that the relative fluctuations \(\sigma_E/\langle E\rangle=\sqrt{2/3N}\) become vanishingly small in the thermodynamic limit. The entropy \(S=k(\beta\langle E\rangle+\ln Z)\) is then given by the Sackur-Tetrode equation:
\[S=Nk\left(\ln\frac{V}{N\lambda^3}+\frac{5}{2}\right)\]
Notably, the extensivity of the entropy \(S(\xi N,\xi V, T)=\xi S(N, V, T)\) is only possible here thanks to the earlier inclusion of the \(N!\) in the denominator of the semi-classical canonical partition function \(Z\).
Anyways, back to computing \(\sigma\). We can first convert the Maxwell distribution \(\rho_{V}(v)\) for the continuous speed random variable \(V\) into a probability density function \(\rho_K(E)\) for the continuous kinetic energy random variable \(K\) by enforcing \(\rho_K(E)dE=\rho_{V}(v)dv\). Thus, one obtains a gamma distribution:
\[\rho_K(E)=\frac{dv}{dE}\rho_{V}(v)=\sqrt{\frac{2}{\pi}}\frac{v}{m\sigma^3}e^{-v^2/2\sigma^2}=\frac{2}{m^{3/2}\sigma^3}\sqrt{\frac{E}{\pi}}e^{-E/m\sigma^2}\]
Equating the average energy per ideal gas particle \(\langle E\rangle/N=\frac{3}{2}kT\) calculated earlier with the expected kinetic energy \(\text E[K]=\int_0^{\infty}E\frac{2}{m^{3/2}\sigma^3}\sqrt{\frac{E}{\pi}}e^{-E/m\sigma^2}dE=\frac{2m\sigma^2}{\sqrt{\pi}}\Gamma(5/2)=\frac{3m\sigma^2}{2}\) yields an equation for \(\sigma\) that can be solved to yield \(\sigma=\sqrt{kT/m}\) as claimed earlier.