Problem: Define an ideal gas (whether that be classical or quantum). Write down the single-particle canonical partition function \(Z_1\) for a classical, monatomic ideal gas, and hence the \(N\)-particle canonical partition function \(Z_N\) in terms of \(Z_1\) (assuming all \(N\) particles are identical).
Solution: The word ideal is synonymous with non-interacting \(V_{\text{int}}=0\). For the classical, monatomic ideal gas, \(Z_1\) is given by the dimensionless ratio of volumes:
\[Z_1=\int\frac{d^3\mathbf xd^3\mathbf p}{h^3}e^{-\beta|\mathbf p|^2/2m}=\frac{V}{\lambda_T^3}\]
and to avoid the Gibbs paradox, one has:
\[Z_N=\frac{Z_1^N}{N!}\]
Problem: In the canonical ensemble, the variables \(N,V,T\) are fixed. Hence, express the conjugate variables \(\mu, p, S\) of the classical, monatomic ideal gas in terms of \(N,V,T\).
Solution: From \(F=-k_BT\ln Z_N\), one has the chemical potential of the classical, monatomic ideal gas:
\[\mu=\frac{\partial F}{\partial N}=k_BT\ln n\lambda_T^3\]
the equation of state of the classical ideal gas:
\[p=-\frac{\partial F}{\partial V}=nk_BT\]
and the Sackur-Tetrode entropy of the classical, monatomic ideal gas:
\[S=-\frac{\partial F}{\partial T}=Nk_B\left(\frac{5}{2}-\ln n\lambda_T^3\right)\]
where the number density \(n:=N/V\) and dimensionless phase space density \(n\lambda_T^3\) feature.
Problem: All the results above are specific to a classical, monatomic ideal gas. Now, fixing the “classical” and “ideal” assumptions, one can first focus on generalizing the “monatomic” assumption to “polyatomic”. In particular, write down the single-particle canonical partition function \(Z_1\) of a rigid rotor, and compute the analogs of the \((\mu, p, S)\) formulas above in terms of \((N,V,T)\).
Solution: A rigid rotor is just a rigid body (i.e. averaging out vibrational perturbations via phenomenological moments of inertia) fixed at a point (i.e. stripping away translational degrees of freedom). There are \(2\) fundamentally different cases that need to be analyzed separately.
- A linear rigid rotor has principal inertia tensor \(\mathcal I=\text{diag}(I,I,0)\). Classically, its configuration space is described by just \(2\) Euler angles \((\theta,\phi)\cong S^2\) with classical Hamiltonian \(H=\frac{1}{2I}\left(p_{\theta}^2+\frac{p_{\phi}^2}{\sin^2\theta}\right)\):
\[Z_1=\frac{1}{|G_p\cap SO(3)|}\int\frac{d\theta d\phi dp_{\theta}dp_{\phi}}{h^2}e^{-\beta (p_{\theta}^2+p_{\phi}^2/\sin^2\theta)/2I}=\frac{1}{|G_p\cap SO(3)|}\frac{T}{T_I}\]
where the rotational temperature \(k_BT_I=E_I\) is associated to a quantum of rotational kinetic energy \(E_I=\hbar^2/2I\) and \(|G_p\cap SO(3)|\) is the number of proper rotational symmetries in the molecule’s point group \(G_p\).
Quantum mechanically, the symmetric rigid rotor Hamiltonian \(H=\frac{|\mathbf L|^2-L_z^2}{2I}+\frac{L_z^2}{2I_z}\) is diagonalized not in the usual \(|\ell,m_{\ell}\rangle\) space-frame angular momentum eigenbasis, but rather (because \(L_z\) is the principal body-frame component of \(\mathbf L\) which commutes with all its space-frame components) in a basis \(|\ell,m_{\ell},\tilde{m}_{\ell}\rangle\) where both projections \(-\ell\leq m_{\ell},\tilde{m}_{\ell}\leq\ell\). From the \(m_{\ell}\)-independent spectrum \(E_{|\ell,m_{\ell},\tilde m_{\ell}\rangle}=\ell(\ell+1)E_I+\tilde m_{\ell}^2(E_{I_z}-E_I)\), one has:
\[Z_1=\frac{1}{|G_p\cap SO(3)|}\sum_{\ell=0}^{\infty}\sum_{m_{\ell}=-\ell}^{\ell}\sum_{\tilde{m}_{\ell}=-\ell}^{\ell}e^{-\beta E_{|\ell,m_{\ell},\tilde m_{\ell}\rangle}}\]
\[=\frac{1}{|G_p\cap SO(3)|}\sum_{\ell=0}^{\infty}e^{-\beta E_I\ell(\ell+1)}(2\ell+1)\sum_{\tilde{m}_{\ell}=-\ell}^{\ell}e^{-\beta (E_{I_z}-E_I)\tilde{m}_{\ell}^2}\]
At this point, recall the rigid rotor is linear so \(I_z\to 0\Leftrightarrow E_{I_z}\to\infty\) and only the \(\tilde{m}_{\ell}=0\) term in the inner sum \(\sum_{\tilde{m}_{\ell}=-\ell}^{\ell}\) survives. Furthermore, working in the high-temperature/classical limit \(\beta E_I\ll 1\), one can apply the zeroth-order Euler-Maclaurin formula:
\[Z_1\approx\frac{1}{|G_p\cap SO(3)|}\int_0^{\infty}d\ell e^{-\beta E_I\ell(\ell+1)}(2\ell+1)=\frac{1}{|G_p\cap SO(3)|}\frac{T}{T_I}\]
2. For a nonlinear rigid rotor, the configuration space is now fundamentally higher-dimensional, being described by \(3\) Euler angles \((\theta,\phi,\psi)\cong SO(3)\) rather than just \(2\). The classical Hamiltonian is also more complicated:
\[H=\frac{1}{2I_x} \left( \frac{\sin \psi}{\sin \theta} (p_\phi – p_\psi \cos \theta) + p_\theta \cos \psi \right)^2 + \frac{1}{2I_y} \left( \frac{\cos \psi}{\sin \theta} (p_\phi – p_\psi \cos \theta) – p_\theta \sin \psi \right)^2 + \frac{p_\psi^2}{2I_z}\]
but the canonical partition function turns out to be straightforward:
\[Z_1=\frac{1}{|G_p\cap SO(3)|}\int\frac{d\theta d\phi d\psi dp_{\theta} dp_{\phi} dp_{\psi}}{h^3}e^{-\beta H}\]
\[=\frac{\sqrt{\pi}}{|G_p\cap SO(3)|}\sqrt{\frac{T^3}{T_{I_x}T_{I_y}T_{I_z}}}\]
As a sanity check, in the earlier symmetric \(I:=I_x=I_y\) case with finite \(I_z<\infty\) but still operating in the high-temperature limit \(\beta E_I,\beta E_{I_z}\ll 1\), the Gaussian integrals can be evaluated analytically:
\[Z_1=\frac{\sqrt{\pi}}{|G_p\cap SO(3)|}\sqrt{\frac{T^3}{T_I^2T_{I_z}}}e^{\beta E_I^2/4E_{I_z}}\text{erfc}\left(\sqrt{\frac{\beta}{E_{I_z}}}\frac{E_I}{2}\right)\]
which reduces to the expected result \(Z_1=\frac{\sqrt{\pi}}{|G_p\cap SO(3)|}\sqrt{\frac{T^3}{T_I^2T_{I_z}}}\) as \(\beta E_I,\beta E_{I_z}\to 0\) since \(e^{\beta E_I^2/4E_{I_z}}\text{erfc}\left(\sqrt{\frac{\beta}{E_{I_z}}}\frac{E_I}{2}\right)\to 1\).
Problem: For both the linear and nonlinear rigid rotors, compute their contributions to \(\mu,S\) and \(p\) in terms of \((N,V,T)\).
Solution: Now, \(Z=Z_1^N\) and \(F=-Nk_BT\ln Z_1\) (note: the \(1/N!\) only applies to the translational partition function). For the linear rigid rotor:
\[\mu=-k_BT\ln\frac{T}{|G_p\cap SO(3)|T_I}\]
\[S=Nk_B\left(1+\ln\frac{T}{|G_p\cap SO(3)|T_I}\right)\]
\[p=0\]
For the nonlinear rigid rotor:
\[\mu=-k_BT\ln\frac{\sqrt{\pi}}{|G_p\cap SO(3)|}\sqrt{\frac{T^3}{T_{I_1}T_{I_2}T_{I_3}}}\]
\[S=Nk_B\left(\frac{3}{2}+\ln\frac{\sqrt{\pi}}{|G_p\cap SO(3)|}\sqrt{\frac{T^3}{T_{I_1}T_{I_2}T_{I_3}}}\right)\]
\[p=0\]
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.