The Chapman–Enskog Transport Coefficients for Maxwell Molecules

Derives Lemma A69.1. Take \(\varphi=(\vect{C}\otimes\vect{C})^{0}\) and write \(\vect{W}=\vect{V}-\vect{u}\). By Equation (A69.4),

the cross terms cancelling between the two molecules. Since \(\vect{W}\) and \(g\) survive the collision, only the last term changes, and

Integrate over the scattering. With \(\vect{g}'=g\vect{n}\) and Equation (A69.5), the traceless part of \(\vect{g}'\otimes\vect{g}'\) averages over the azimuth to \(g^{2}P_{2}(\cos\chi)(\hat{\vect{g}}\otimes\hat{\vect{g}})^{0} =P_{2}(\cos\chi)(\vect{g}\otimes\vect{g})^{0}\), so by Equation (A69.6)

It remains to average over \(\vect{v}_{1}\). With \(\avg{\vect{C}_{1}}=\vect{0}\) and \(\avg{\vect{C}_{1}\otimes\vect{C}_{1}} =(k_{\mathrm{B}}T/m)\identity\),

whose traceless part is \((\vect{C}\otimes\vect{C})^{0}\): the isotropic term contributes nothing. Multiplying by the number density \(n\) from \(\int f_{0}\,\dd^{3}v_{1}=n\) gives Equation (A69.7).

Derives Lemma A69.2. The term linear in \(\vect{C}\) is a collision invariant and lies in the kernel of \(L\) by Proposition 59.12, so it may be dropped: it is enough to compute \(L[C^{2}\vect{C}]\).

With \(\vect{W}=\vect{V}-\vect{u}\), expanding \(C^{2}=W^{2}+\vect{W}\cdot\vect{g}+\tfrac{1}{4}g^{2}\) and \(C_{1}^{2}=W^{2}-\vect{W}\cdot\vect{g}+\tfrac{1}{4}g^{2}\) and adding,

Only the last term changes in the collision, so

Averaging the second term over the azimuth with \(\vect{g}'=g\vect{n}\) and Equation (A69.5),

while the first term is the same expression with \(P_{2}\) replaced by \(1\). Their difference, integrated against \(F\), leaves only the traceless part, the isotropic pieces cancelling:

Now average over \(\vect{v}_{1}\). Writing \(a=k_{\mathrm{B}}T/m\) and using \(\avg{\vect{C}_{1}}=\vect{0}\), \(\avg{C_{1,i}C_{1,j}}=a\delta_{ij}\), \(\avg{\vect{C}_{1}C_{1}^{2}}=\vect{0}\) and \(\avg{\vect{C}_{1}(\vect{C}\cdot\vect{C}_{1})}=a\vect{C}\),

Subtracting Equation (A69.14) from Equation (A69.13),

and multiplying by \(A_{2}n\) gives Equation (A69.10).

Derives Lemma A69.4. From \(\ln f_{0}=\ln n+\tfrac{3}{2}\ln\!\left(m/(2\pi k_{\mathrm{B}}T)\right) -mC^{2}/(2k_{\mathrm{B}}T)\) and \(\dd\vect{C}=-\dd\vect{u}\),

with \(\mathrm{D}=\pp_{t}+\vect{v}\cdot\nabla =\mathrm{D}_{t}+\vect{C}\cdot\nabla\) and \(\mathrm{D}_{t}\) the material derivative. At zeroth order the Euler equations give \(\mathrm{D}_{t}\ln n=-\nabla\cdot\vect{u}\), \(\mathrm{D}_{t}\ln T=-\tfrac{2}{3}\nabla\cdot\vect{u}\) and \(\mathrm{D}_{t}\vect{u}=-(k_{\mathrm{B}}T/m)\nabla\ln(nT)\), the last from \(\rho\,\mathrm{D}_{t}\vect{u}=-\nabla p\) with \(p=nk_{\mathrm{B}}T\) by Phenomenon 58.3. Substituting, the terms in \(\vect{C}\cdot\nabla\ln n\) cancel identically between the first and third contributions; those in \(\vect{C}\cdot\nabla\ln T\) combine to \(\left(\mathcal{S}-\tfrac{5}{2}\right)\vect{C}\cdot\nabla\ln T\); and the terms in \(\nabla\cdot\vect{u}\) combine with \((m/k_{\mathrm{B}}T)\,\vect{C}\otimes\vect{C}:\nabla\vect{u}\) to leave its traceless part. That is Equation (A69.17).

Derives Proposition A69.5. Viscosity. The deviatoric pressure tensor is \(\int m(\vect{C}\otimes\vect{C})^{0}f_{0}\Phi\,\dd^{3}v\), and only the second term of Equation (A69.18) contributes, the first being odd in \(\vect{C}\). Using the isotropic fourth moment

whose contraction with a symmetric traceless tensor returns twice that tensor, the deviatoric stress is \(-(2nk_{\mathrm{B}}T/\nu_{\eta})(\nabla\vect{u})^{0}\). Comparison with \(-2\eta(\nabla\vect{u})^{0}\) gives the first of Equation (A69.19).

Conductivity. The heat flux is \(\vect{q}=\int\tfrac{1}{2}mC^{2}\vect{C}f_{0}\Phi\,\dd^{3}v\), and now only the first term of Equation (A69.18) contributes. Using \(\int C_{i}C_{j}h(C)f_{0}\,\dd^{3}v =\tfrac{1}{3}\delta_{ij}\int C^{2}h(C)f_{0}\,\dd^{3}v\) and \(C^{2}=(2k_{\mathrm{B}}T/m)\mathcal{S}\),

For a Maxwellian \(\mathcal{S}\) has the density proportional to \(\sqrt{\mathcal{S}}\,\ee^{-\mathcal{S}}\), so \(\avg{\mathcal{S}}=\tfrac{3}{2}\), \(\avg{\mathcal{S}^{2}}=\tfrac{15}{4}\) and \(\avg{\mathcal{S}^{3}}=\tfrac{105}{8}\) by Proposition 15.29, whence \(\avg{\mathcal{S}^{3}}-\tfrac{5}{2}\avg{\mathcal{S}^{2}} =\tfrac{105}{8}-\tfrac{75}{8}=\tfrac{15}{4}\). Substituting and using \(\nabla\ln T=\nabla T/T\) gives \(\vect{q}=-\left(5nk_{\mathrm{B}}^{2}T/(2m\nu_{\kappa})\right)\nabla T\), which against Fourier's law Equation (59.37) is the second of Equation (A69.19).

Derives Corollary A69.6. Dividing the two members of Equation (A69.19) and using Equation (A69.16),

With \(c_{V}^{(1)}=\tfrac{3}{2}k_{\mathrm{B}}\) the second member is \((15/4)/(3/2)=5/2\), and with \(c_{p}=\tfrac{5}{2}k_{\mathrm{B}}/m\) per unit mass the third is \((5/2)/(15/4)=2/3\).