8.10: Appendix I- Boltzmann Equation and Collisional Invariants

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Problem : The linearized Boltzmann operator \(L\psi\) is a complicated functional. Suppose we replace \(L\) by \(\CL\), where \[\begin{split} \CL\psi&=-\gamma\,\psi(\Bv,t)+\gamma \bigg({m\over2\pi\kT}\bigg)^{\!\!3/2}\int\!\! d^3\!u\,\exp\bigg(-{m\Bu^2\over2\kT}\bigg)\\ &\qquad\times \Bigg\{ 1+{m\over\kT}\,\Bu\cdot\Bv + {2\over 3}\, \bigg( {m\Bu^2\over2\kT}-{3\over 2}\bigg) \bigg({m\Bv^2\over2\kT}-{3\over 2}\bigg) \Bigg\}\, \psi(\Bu,t)\ . \end{split}\]

Show that \(\CL\) shares all the important properties of \(L\). What is the meaning of \(\gamma\)? Expand \(\psi(\Bv,t)\) in spherical harmonics and Sonine polynomials, \[\psi(\Bv,t)=\sum_{r\ell m} a_{r\ell m}(t)\,S^r_{\ell +\half}(x)\,x^{\ell/2} \,Y^\ell_m(\nhat),\] with \(x=mv^2/2\kT\), and thus express the action of the linearized Boltzmann operator algebraically on the expansion coefficients \(a_{r\ell m}(t)\).

The Sonine polynomials \(S^n_\alpha(x)\) are a complete, orthogonal set which are convenient to use in the calculation of transport coefficients. They are defined as \[S_\alpha^n(x)=\sum_{m=0}^n {\RGamma(\alpha+n+1)\,(-x)^m\over \RGamma(\alpha+m+1)\,(n-m)!\,m!}\ ,\] and satisfy the generalized orthogonality relation \[\int\limits_0^\infty\!\!dx\,e^{-x}\,x^\alpha\,S_\alpha^n(x)\,S_\alpha^{n'}(x) = {\RGamma(\alpha+n+1)\over n!}\>\delta\nd_{nn'}\ .\]

Solution : The ‘important properties’ of \(L\) are that it annihilate the five collisional invariants, \(1\), \(\Bv\), and \(v^2\), and that all other eigenvalues are negative. That this is true for \(\CL\) can be verified by an explicit calculation.

Plugging the conveniently parameterized form of \(\psi(\Bv,t)\) into \(\CL\), we have \[\begin{split} \CL\psi&=-\gamma\sum_{r\ell m} a_{r\ell m}(t)\>S^r_{\ell +\half}(x) \>x^{\ell/2}\>Y^\ell_m(\nhat)\ +\ {\gamma\over2\pi^{3/2}}\,\sum_{r\ell m}a_{r\ell m}(t)\!\! \int\limits_0^\infty\!\!dx\nd_1\,x_1^{1/2}\,e^{-x\nd_1}\\ &\qquad\times\int\!\!d\nhat\ns_1 \Big[1+2\,x^{1/2} x_1^{1/2}\,\nhat\!\cdot\!\nhat\ns_1+ \frac{2}{3}\big(x-\frac{3}{2}\big)\big(x\nd_1-\frac{3}{2}\big)\Big]\,S^r_{\ell +\half}(x\nd_1)\> x_1^{\ell/2}\>Y^\ell_m(\nhat_1)\ , \end{split}\] where we’ve used \[u=\sqrt{2\kT\over m}\, x_1^{1/2}\qquad,\qquad du=\sqrt{\kT\over 2m}\,x_1^{-1/2}\,dx\nd_1\ .\] Now recall \(Y^0_0(\nhat)=\frac{1}{\sqrt{4\pi}}\) and \[\begin{aligned} Y^1_1(\nhat)&=-\sqrt{3\over8\pi}\,\sin\theta\,e^{i\varphi} & Y^1_0(\nhat)&=\sqrt{3\over4\pi}\,\cos\theta & Y^1_{-1}(\nhat)&=+\sqrt{3\over8\pi}\,\sin\theta\,e^{-i\varphi} \\ \bvph S^0_{1/2}(x)&=1 & S^0_{3/2}(x)&=1 & S^1_{1/2}(x)&=\frac{3}{2}-x \ ,\end{aligned}\] which allows us to write \[\begin{aligned} \bvph 1&=4\pi\, Y^0_0(\nhat)\,{Y^0_0}^*(\nhat_1)\\ \nhat\!\cdot\!\nhat_1&={4\pi\over3} \Big[\> Y^1_0(\nhat)\,{Y^1_0}^*(\nhat_1) +Y^1_1(\nhat)\,{Y^1_1}^*(\nhat_1)+Y^1_{-1}(\nhat)\,{Y^1_{-1}}^*(\nhat_1)\>\Big]\ .\end{aligned}\] We can do the integrals by appealing to the orthogonality relations for the spherical harmonics and Sonine polynomials: \[\begin{aligned} \bvph\int\!d\nhat\,Y^\ell_m(\nhat)\,{Y^{l'}_{m'}}^*(\nhat)&=\delta_{ll'}\,\delta_{mm'}\\ \int\limits_0^\infty\!\!dx\,e^{-x}\,x^\alpha\,S^n_\alpha(x)\,S^{n'}_\alpha(x) &={\RGamma(n+\alpha+1)\over\RGamma(n+1)}\>\delta_{nn'}\ .\end{aligned}\] Integrating first over the direction vector \(\nhat_1\), \[\begin{split} \CL\psi&=-\gamma\sum_{r\ell m} a_{r\ell m}(t)\>S^r_{\ell +\half}(x)\>x^{\ell/2}\>Y^\ell_m(\nhat)\\ &\qquad+{2\gamma\over\sqrt{\pi}}\,\sum_{r\ell m}a_{r\ell m}(t)\!\int\limits_0^\infty\!\!dx\nd_1\,x_1^{1/2}\,e^{-x\nd_1}\!\int\!\!d\nhat\ns_1\> \bigg[ Y^0_0(\nhat)\,{Y^0_0}^*(\nhat_1)\,S^0_{1/2}(x)\,S^0_{1/2}(x\nd_1)\\ &\qquad\qquad+\frac{2}{3}\,x^{1/2} x_1^{1/2}\!\sum_{m'=-1}^1 Y^1_{m'}(\nhat)\, {Y^1_{m'}}^*(\nhat_1)\,S^0_{3/2}(x)\,S^0_{3/2}(x\nd_1)\\ &\qquad\qquad\qquad+\frac{2}{3} \, Y^0_0(\nhat)\,{Y^0_0}^*(\nhat_1)\,S^1_{1/2}(x)\,S^1_{1/2}(x\nd_1)\bigg] \,S^r_{\ell +\half}(x\nd_1)\>x_1^{\ell/2}\>Y^\ell_m(\nhat_1)\ , \end{split}\] we obtain the intermediate result \[\begin{split} \CL\psi&=-\gamma\sum_{r\ell m} a_{r\ell m}(t)\>S^r_{\ell +\half}(x)\>x^{\ell/2}\>Y^\ell_m(\nhat)\\ &\qquad +{2\gamma\over\sqrt{\pi}}\,\sum_{r\ell m}a_{r\ell m}(t)\!\int\limits_0^\infty\!\!dx\nd_1\,x_1^{1/2}\,e^{-x\nd_1} \bigg[Y^0_0(\nhat)\,\delta_{l0}\,\delta_{m0}\,S^0_{1/2}(x)\,S^0_{1/2}(x\nd_1)\\ &\qquad\qquad+\frac{2}{3} \, x^{1/2} x_1^{1/2}\sum_{m'=-1}^1 Y^1_{m'}(\nhat)\,\delta_{l1}\,\delta_{mm'}\,S^0_{3/2}(x)\,S^0_{3/2}(x\nd_1)\\ &\qquad\qquad\qquad+\frac{2}{3} \, Y^0_0(\nhat)\,\delta_{l0}\,\delta_{m0} \,S^1_{1/2}(x)\,S^1_{1/2}(x\nd_1)\bigg]\,S^r_{\ell +\half}(x\nd_1)\>x_1^{1/2} . \end{split}\]

Appealing now to the orthogonality of the Sonine polynomials, and recalling that \[\RGamma(\half)=\sqrt{\pi}\qquad,\qquad \RGamma(1)=1\qquad,\qquad \RGamma(z+1)=z\,\RGamma(z)\ ,\] we integrate over \(x\nd_1\). For the first term in brackets, we invoke the orthogonality relation with \(n=0\) and \(\alpha=\half\), giving \(\RGamma(\frac{3}{2})=\half\sqrt{\pi}\). For the second bracketed term, we have \(n=0\) but \(\alpha=\frac{3}{2}\), and we obtain \(\RGamma(\frac{5}{2})=\frac{3}{2}\,\RGamma(\frac{3}{2})\), while the third bracketed term involves leads to \(n=1\) and \(\alpha=\half\), also yielding \(\RGamma(\frac{5}{2})=\frac{3}{2}\,\RGamma(\frac{3}{2})\). Thus, we obtain the simple and pleasing result \[\CL\psi=-\gamma{\sum_{r\ell m}}' a_{r\ell m}(t)\>S^r_{\ell +\half}(x)\>x^{\ell/2}\>Y^\ell_m(\nhat)\] where the prime on the sum indicates that the set \[{CI}=\Big\{ (0,0,0)\ ,\quad (1,0,0)\ ,\quad (0,1,1)\ ,\quad (0,1,0)\ ,\quad (0,1,-1)\Big\}\] are to be excluded from the sum. But these are just the functions which correspond to the five collisional invariants! Thus, we learn that \[\psi_{r\ell m}(\Bv)=\CN_{r\ell m}\, S^r_{\ell +\half}(x)\,x^{\ell/2}\,Y^\ell_m(\nhat),\] is an eigenfunction of \(\CL\) with eigenvalue \(-\gamma\) if \((r,\ell,m)\) does not correspond to one of the five collisional invariants. In the latter case, the eigenvalue is zero. Thus, the algebraic action of \(\CL\) on the coefficients \(a_{r\ell m}\) is \[(\CL a)_{r\ell m}=\begin{cases} -\gamma\> a_{r\ell m} & \hbox{if $(r,\ell,m)\notin {CI}$}\\ =0 & \hbox{if $(r,\ell,m)\in {CI}$} \end{cases}\] The quantity \(\tau=\gamma^{-1}\) is the relaxation time.

It is pretty obvious that \(\CL\) is self-adjoint, since \[\begin{split} \sbraket{\phi}{\CL\psi}&\equiv\int\!\! d^3\!v \,f^0(\Bv)\,\phi(\Bv)\,\CL[\psi(\Bv)]\\ &=-\gamma\, n \left({m\over2\pi\kT}\right)^{\!\!3/2}\!\int\!\!d^3\!v\,\exp\bigg(-{m\Bv^2\over2\kT}\bigg)\phi(\Bv)\,\psi(\Bv)\\ &\qquad +\gamma\, n \bigg({m\over2\pi\kT}\bigg)^{\!\!3}\int\!\!d^3\!v\!\!\int\!\!d^3\!u\, \exp\bigg(-{m\Bu^2\over2\kT}\bigg)\exp\bigg(-{m\Bv^2\over2\kT}\bigg)\\ &\qquad\qquad\times \phi(\Bv)\,\Bigg[1+{m\over\kT}\,\Bu\cdot\Bv + {2\over 3}\, \bigg({m\Bu^2\over2\kT}-{3\over 2}\bigg)\bigg({m\Bv^2\over2\kT}-{3\over 2}\bigg)\Bigg]\,\psi(\Bu)\\ &=\sbraket{\CL\phi}{\psi}\ , \end{split}\] where \(n\) is the bulk number density and \(f^0(\Bv)\) is the Maxwellian velocity distribution.