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Write $y(t)={\sum }_{k\ge 0}{a}_k{t}^k$ at the ordinary point zero. Substituting into the Legendre differential equation and comparing coefficients givesThe two arbitrary coefficients $a_0,a_1$ generate the even and odd series separately, convergent at least for $|t|<1$. A nonzero polynomial of degree $n$ has top coefficient satisfying $[\lambda -n(n+1)]a_n=0$, so $\lambda =n(n+1)$ is necessary. For this value the recurrence in parity $n$ terminates at degree $n$, while the other parity cannot terminate unless its initial coefficient is zero. Thus a polynomial solution exists and is unique up to scale, with $\boxed{P_n(-t)=(-1{)}^n{P}_n(t)}$.

For Orthogonality of Legendre polynomials, write the equation in self-adjoint form $[(1-t^2)P_n^{\prime }{]}^{\prime }+n(n+1)P_n=0$. Multiply it by $P_m$, subtract the equation for $P_m$ multiplied by $P_n$, and integrate. The boundary term is $[(1-t^2)(P_m{P}_n^{\prime }-P_n{P}_m^{\prime }){]}_{-1}^1=0$, because the functions are polynomials. For $m\ne n$ the eigenvalues differ, giving ${\int }_{-1}^1{P}_n{P}_m=0$. For $m=n$, the integral of the square is positive. Hence $\int P_n{P}_m=k_n{\delta }_{nm}$ with $k_n>0$.

Let $G(x,t)=(1-2xt+x^2{)}^{-1/2}$. The supplied operator $x{\partial }_x^{2}x$ means $x{\partial }_x^2(xG)$, which equals ${\partial }_x(x^2{\partial }_{x}G)$. Expand $G={\sum }_n{a}_n(x)P_n(t)$ in the given identity and use the Legendre equation. Each coefficient satisfieswhose solutions are $a_n=A_n{x}^n+B_n{x}^{-n-1}$. Regularity at $x=0$ removes $B_n$. At $t=1$, normalization $P_n(1)=1$ and $G(x,1)=1/(1-x)$ give ${\sum }_n{A}_n{x}^n={\sum }_n{x}^n$, so $A_n=1$. ThusA rigorous justification for the expansion and operations is to begin with the Taylor series of $G$ in $x$. It converges uniformly for $|x|\le r<1$, $-1\le t\le 1$. Its $x^n$ coefficient is a degree-$n$ polynomial in $t$, and the differential identity makes that coefficient solve the degree-$n$ Legendre equation. Its value at one is one, so it is exactly $P_n$. This avoids assuming an unjustified endpoint convergence of an arbitrary orthogonal expansion.

Square the generating series and integrate, using uniform convergence and orthogonality:Comparing coefficients proves the generating function and norm of Legendre polynomials result

A scalar cosmological perturbation at fixed wavevector carries no helicity or distinguished transverse direction. Its metric and scalar sources are unchanged by rotations around $\mathbf{k}$, and the scalar baryon peculiar velocity is longitudinal. With scalar initial conditions, rotational invariance of the linear evolution therefore makes $\Theta$ a function only of $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$. Expand it using the stated normalization, without an extra $(2\ell +1)$ factor:In the Thomson kernel, $1+v^2=4/3+(2/3)P_2(v)$. The spherical harmonic addition theorem and Orthogonality of Legendre polynomials giveOnly the monopole and quadrupole survive; the quadrupole's $(-i{)}^2=-1$ gives the angular scattering integralThis also explains why an isotropic distribution has zero net collision term.

The streaming term $ik\mu \Theta$ and the Legendre polynomial recurrence relation yield, after division by $(-i{)}^{\ell }$, $k[(\ell +1){\Theta }_{\ell +1}/(2\ell +3)-\ell {\Theta }_{\ell -1}/(2\ell -1)]$. With ${\mathbf{v}}_b=i\hat{\mathbf{k}}{v}_b$, the velocity collision source is $+{\tau }^{\prime }{v}_b$ in the dipole equation; the gravitational gradient source is $+k\psi$. Therefore the photon Boltzmann hierarchy isSet the absent ${\Theta }_{-1}$ contribution to zero. This is algebraically the printed hierarchy. In particular,Since ${\tau }^{\prime }<0$, these collision terms damp the nonzero multipoles and drive the dipole to ${\Theta }_1=-v_b$. The latter sign follows from the paper's velocity and $(-i{)}^{\ell }$ conventions, not a reversed physical photon velocity.

A degree-$n$ Legendre polynomial $P_n$ is orthogonal, for integration on $[-1,1]$, to every polynomial of degree at most $n-1$, and has $n$ distinct roots in $(-1,1)$. Divide $p$ by $P_n$:The Orthogonality of Legendre polynomials gives $\int p=\int r$. At every quadrature node $x_j$, $p(x_j)=r(x_j)$. The assumed exactness for $r$ therefore proves exactness up to degree $2n-1$, the characteristic property of Gaussian quadrature.

This subpart is also present in the original PDF but missing from the local TeX. Put $p(x)=1-x^2$ and ${\lambda }_n=n(n+1)$. Multiply the Legendre differential equation for $P_n$ by $P_m$, multiply the equation for $P_m$ by $P_n$, and subtract. The product rule givesIntegrating from $-1$ to $1$, the boundary term vanishes: $p(\pm 1)=0$, and the Legendre polynomials and their derivatives are finite there. For distinct nonnegative integers $m,n$, ${\lambda }_m\ne {\lambda }_n$. Thus the Orthogonality of Legendre polynomials isThe vanishing coefficient at the endpoints causes no difficulty here because the solutions are polynomials.

Let $Q^{ij}={\hat{k}}^i{\hat{k}}^j-{\delta }^{ij}/3$. Rotations around $\hat{\mathbf{k}}$ and the vanishing trace imply that the angular integral in the scalar neutrino anisotropic stress has the form $A_{\ell }{Q}^{ij}$. Contract with ${\hat{k}}_i{\hat{k}}_j$ and use ${\mu }^2-1/3=(2/3)P_2(\mu )$. By the Orthogonality of Legendre polynomials,Only the quadrupole in the neutrino Boltzmann hierarchy contributes. Its phase factor is $(-i{)}^2=-1$, so the supplied stress definition becomesComparing with the chosen normalization $-(4/3){\bar{\rho }}_{\nu }\Pi Q^{ij}$ givesThe sign and factor here follow jointly from the stress convention and the unweighted Legendre polynomial coefficients; they should not be transferred unchanged to a differently normalized hierarchy.

Let $\Gamma =-\dot{\tau }=a{\bar{n}}_e{\sigma }_T$ and let $D={\partial }_{\eta }+\mathbf{e}\cdot \nabla$ differentiate along the unperturbed photon trajectory. Use $1+(\mathbf{e}\cdot \hat{\mathbf{m}}{)}^2=4/3+(2/3)P_2(\mathbf{e}\cdot \hat{\mathbf{m}})$ to decompose the Thomson scattering gain term:The Orthogonality of Legendre polynomials identifies $Q$ as a pure photon quadrupole. Dropping $Q$ is an approximation, justified when tight coupling makes the scattering quadrupole small; the supplied unpolarized equation already neglects polarization corrections. The Cosmic microwave background line-of-sight solution requested here omits that quadrupole source.

Set $Y=\Theta +\psi$. Adding $D\psi =\dot{\psi }+\mathbf{e}\cdot \nabla \psi$ to the Photon Boltzmann equation with Thomson scattering yieldsAlong $\mathbf{x}(\eta )={\mathbf{x}}_0-({\eta }_0-\eta )\mathbf{e}$, the integrating factor is $e^{-\tau }$ because $d{e}^{-\tau }/d\eta =\Gamma e^{-\tau }$. ConsequentlyAn early optically thick boundary makes $e^{-\tau }Y$ vanish for finite initial perturbations, and $e^{-\tau ({\eta }_0)}=1$. ThereforeHere $\dot{\varphi }$ and $\dot{\psi }$ are partial time derivatives, not total derivatives along the ray. We work at first order in scalar cosmological perturbations, evaluate sources along the background radial null geodesic in FLRW spacetime, and ignore observer peculiar velocity. Perturbed paths and lensing affect this first-order source solution only at higher perturbative order. A moving observer adds its local Doppler dipole.

The original PDF, unlike the transcribed TeX, also asks for the instantaneous-last-scattering limit and its interpretation. For $g(\eta )={\delta }^{(D)}(\eta -{\eta }_{\ast })$ with no later scattering, and a matter-era growing mode with constant $\varphi$ and $\psi$, the Integrated Sachs-Wolfe effect vanishes. HenceThe four terms are the intrinsic emission temperature, the gravitational potential at emission, the observer's potential, and the Doppler CMB anisotropy. The two potentials describe the gravitational redshift; the observer term is independent of direction and is absorbed into the measured temperature monopole. The intrinsic temperature and emission potential form the Sachs-Wolfe combination. On superhorizon scales for adiabatic initial conditions in matter domination, ${\Theta }_0=-2\psi /3$, giving the ordinary Sachs-Wolfe effect ${\Theta }_0+\psi =\psi /3$. The Doppler sign is positive with the paper's propagation direction $\mathbf{e}$; the viewing direction is $-\mathbf{e}$.

The asymptotic scattering wavefunction isIn each partial wave, the Partial-wave S-matrix is the ratio of its outgoing coefficient to its incoming coefficient. With the conventions in the question,For elastic scattering by a real central potential, unitarity gives $|S_l|=1$, so $S_l=e^{2i{\delta }_l}$ for a real scattering phase shift ${\delta }_l$. HenceUsing the Orthogonality of Legendre polynomials in ${\sigma }_T=\int |f(\theta ){|}^{2}d\Omega$ yields the partial-wave total scattering cross-section

For the S wave, write the reduced radial wavefunction aswhere regularity at $r=0$ selected the hyperbolic sine. Continuity of $u$ and $u^{\prime }$ at $r=a$, equivalently logarithmic-derivative matching, givesorFor $ka\ll 1$, let the scattering length be $a_s=-{\lim }_{k\to 0}\tan {\delta }_0/k$. Expanding the matching relation givesOnly the S wave contributes at leading order, soIn the hard-sphere limit $\gamma a\to \infty$, penetration is suppressed, $a_s\to a$, andThis is four times the geometric area $\pi a^2$, a wave effect associated with diffraction from an impenetrable sphere.

Insert the photon temperature multipole expansion into the Fourier-space transport equation. The Legendre polynomial recurrence relation turns multiplication by $\mu$ into nearest-neighbour multipole couplings, while Orthogonality of Legendre polynomials projects onto a fixed $\ell$. The monopole projection has no collision term because Thomson scattering conserves photon number:The dipole projection receives the electron-velocity source,For every $\ell \ge 2$, the collision term damps the anisotropic multipole and the photon Boltzmann hierarchy isThus the Free-streaming photon Boltzmann equation moves angular structure between neighbouring multipoles, whereas Thomson scattering suppresses all multipoles above the dipole in the tight-coupling approximation.

The far field is proportional to $r{P}_1(\cos \theta )$, since $P_1(\cos \theta )=\cos \theta$. After matching the growing terms, write the exterior solution asAn added constant has here been excluded by interpreting the far-field condition as $U-v_{0}r\cos \theta \to 0$. Differentiate and impose the radial Neumann boundary condition:The Orthogonality of Legendre polynomials makes each coefficient vanish separately. Thus $B_n=0$ for $n\ne 1$, and $2B_1/r_0^3=v_0$. ConsequentlyIndeed ${\partial }_{r}U=v_0(1-r_0^3/r^3)\cos \theta$, which vanishes at the boundary, while the correction decays at infinity. If the far-field condition is intended to specify only the leading growing term, an arbitrary additive constant is also allowed; its gradient has no effect on the Neumann boundary condition.

For orthonormal complex spherical harmonics, $P_{\ell }(\hat{\mathbf{n}}\cdot \hat{\mathbf{m}})=4\pi {\sum }_{m=-\ell }^{\ell }{Y}_{\ell m}(\hat{\mathbf{n}})Y_{\ell m}^{\ast }(\hat{\mathbf{m}})/(2\ell +1)$. This converts an angle between two directions into a separable sum. Together with Orthogonality of Legendre polynomials, it projects rotationally invariant scattering kernels onto angular multipoles.