# 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.

This page titled 8.10: Appendix I- Boltzmann Equation and Collisional Invariants is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Daniel Arovas.