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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.

Normalize the incident plane wave to unit amplitude. The scattering asymptotics are $\psi (\mathbf{r})\sim e^{ikz}+f(\theta )e^{ikr}/r$; the outgoing coefficient $f$ is the scattering amplitude. The general axisymmetric free solution for $r>0$ isusing Spherical Bessel functions, spherical Neumann functions and Legendre polynomials.

For a real short-range central potential, the exterior radial solution in each partial wave is proportional to $\cos {\delta }_{\ell }{j}_{\ell }-\sin {\delta }_{\ell }{n}_{\ell }$, whose large-$r$ form is $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift. Multiplying this combination by $(2\ell +1)i^{\ell }{e}^{i{\delta }_{\ell }}$ matches the incoming spherical component of the incident plane wave: the factor $e^{i{\delta }_{\ell }}$ cancels the shifted incoming phase. The outgoing coefficient instead acquires $e^{2i{\delta }_{\ell }}$. Subtracting the plane wave's outgoing component therefore givesFor the barrier's $s$-wave, the regular reduced radial function inside is $u(r)=A\sinh (\kappa r)$, with $\kappa =\sqrt{{\gamma }^2-k^2}$. Outside take $u(r)=B\sin (kr+{\delta }_0)$. Continuity of $u$ and its derivative at $a$ equates logarithmic derivatives:Use the branch with ${\delta }_0\to 0$ as $k\to 0$. The matching equation gives ${\delta }_0=k(\tanh (\gamma a)/\gamma -a)+o(k)$, so the $s$-wave scattering amplitude tends toIts negative is the repulsive spherical barrier scattering length, positive for this repulsive barrier.

Label the sides by $L,R,B,T$, and parametrize the two vertical sides by $s=y\in [-1,1]$, the horizontal sides by $s=x\in [-1,1]$. Their outward normal derivatives are $q_L=-u_x(-1,s)$, $q_R=u_x(1,s)$, $q_B=-u_y(s,-1)$ and $q_T=u_y(s,1)$. Let $f_j$ denote the prescribed side values. Assume compatible, sufficiently regular boundary traces; the square's corners have zero arclength measure and do not require separate normal values.

For sine collocation of square modified Helmholtz global relations, use Legendre polynomials for the known Dirichlet boundary condition and a Fourier sine series for the unknown normal derivative:The known coefficients are $d_{j\ell }=(2\ell +1){\int }_{-1}^1{f}_j{P}_{\ell }ds/2$. The sine functions have ${\int }_{-1}^1{\varphi }_m{\varphi }_{n}ds={\delta }_{mn}$ and form a complete basis in $L^2(-1,1)$. Choosing them for the normal derivative does not impose zero flux at a corner: the expansion is an $L^2$ representation, and endpoint values are not determined by it. If preserving corner values of the approximated Dirichlet trace is necessary, subtract its endpoint-interpolating line before polynomial approximation, and add that line back. All subsequent known-data integrals can alternatively be evaluated with the exact $f_j$.

Define the entire function basis transformsAt $\tau =\pm i{p}_m$ the quotient is evaluated by its removable limit or by the defining integral. For example, ${\mathsf{P}}_0=2\sinh \tau /\tau$, and polynomial expansion of $P_{\ell }$ expresses every ${\mathsf{P}}_{\ell }$ in derivatives of ${\mathsf{P}}_0$. PutFor compactness write $A=A(\lambda )$ and $B=B(\lambda )$. Substituting these expansions into the first global relation for a linear boundary value problem givesThe signs are fixed by the outward normal vectors, rather than by a choice of traversal direction. The second approximate global relation for a linear boundary value problem replaces $A$ by $-A$, leaving $B$ unchanged:Here $\simeq 0$ records omitted boundary-expansion tails. In a finite spectral method the selected equations are set equal to zero to solve for the $4N$ unknown real coefficients $c_{jm}$. For real data the two families obey the same complex conjugation relation as their exact counterparts.

A particularly simple constructive choice uses paired collocation points for a global relation. For $n=1,\dots ,N$, putFour suitable sets, one for each side, areThe braces in each set range over all $n$. For large $p_n/k$, compute $r_n^{-1}=k/({\omega }_n+p_n)$ to avoid cancellation. The corresponding $(A,B)$ pairs are, respectively,Each pair has the same exponentially growing normal direction and opposite tangential frequencies. Take the phase-weighted difference of the two collocated global relations for a linear boundary value problem. For example, the bottom pair gives the adjoint testThe other three phase-weighted differences giveThus the four spectral sets supply four real sine-tested equations per mode. We retain these $4N$ equations; requiring every uncombined complex equation as well would instead form an overdetermined finite approximation. For real traces each retained equation is the appropriate real combination of its conjugate pair.

The key advantage is that $v_{Bn},v_{Tn}$ vanish on the vertical sides, while $v_{Ln},v_{Rn}$ vanish on the horizontal sides. Orthogonality then eliminates all off-mode unknowns. Define the known right-hand sidesThe retained global relations for a linear boundary value problem areFor example, with $f_{jn}={\int }_{-1}^1{f}_j^{(M)}(s){\varphi }_n(s)ds$, the bottom and top right-hand sides are explicitlyFor completeness the left and right right-hand sides areAll these integrals use known polynomial approximations or the exact Dirichlet boundary condition.

Scale each row by $e^{-{\omega }_n}$. Every block becomesThere is just one off-diagonal entry in each row. This proves diagonal dominance of paired square global-relation collocation. Therefore the assembled $4N\times 4N$ system is strictly diagonally dominant, with row margin $1-{\delta }_n>1-e^{-\pi }$. This proof applies to the explicitly paired and scaled collocation system, rather than assuming arbitrary raw spectral rows are diagonally dominant. Each block is a positive-definite matrix and have spectral condition number of a positive-definite matrixFor instance, setting ${\rho }_{jn}=e^{-{\omega }_n}{R}_{jn}$ gives the concise answerwith the identical formula for the opposite vertical sides. If the known-data integrals use exact $f_j$, these are exactly the first $N$ sine coefficients of the true normal traces: no discarded unknown mode contributes to a retained row. With a finite Legendre polynomial approximation, the error is solely that of the known-data approximation and the final normal-trace truncation. These square modified Helmholtz Dirichlet-to-Neumann coefficients determine the unknown boundary normal derivative; Green's third identity or the separated-variable construction can then recover the interior solution.

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.

This is the Gram-Schmidt process applied degree by degree. Inductively, subtracting projections onto $p_0,\dots ,p_n$ makes $p_{n+1}$ orthogonal to every earlier polynomial. Since every subtracted polynomial has degree at most $n$, the leading coefficient of the monic $q_{n+1}$ remains one, so $p_{n+1}$ is a real monic orthogonal polynomial of degree exactly $n+1$.

Now take $q_{n+1}=x{p}_n$. For $k\le n-2$, symmetry of the inner product givesbecause $x{p}_k$ has degree at most $n-1$. Only the $p_n$ and $p_{n-1}$ projections remain, yielding the three-term recurrence for monic orthogonal polynomialswhereThe second identity uses $x{p}_{n-1}=p_n$ plus lower-degree terms.

For the symmetric Legendre inner product, parity gives ${\alpha }_n=0$. Starting with $p_0=1$, one obtains $p_1=x$, thenThus the first four monic Legendre polynomials are

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.

For one Fourier mode, put $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$. The collisionless equation becomesUse the unweighted Legendre polynomial expansion specified in the paper: $\Theta ={\sum }_{j\ge 0}(-i{)}^j{\Theta }_j{P}_j(\mu )$. The Legendre polynomial recurrence relation saysThe coefficient of $P_{\ell }$ in $ik\mu \Theta$ receives contributions from $j=\ell -1$ and $j=\ell +1$. Dividing by $(-i{)}^{\ell }$ gives respectively $-k\ell {\Theta }_{\ell -1}/(2\ell -1)$ and $k(\ell +1){\Theta }_{\ell +1}/(2\ell +3)$. The metric sources are $\dot{\varphi }{P}_0$ and $-ik\psi P_1$, the latter becoming $k\psi$ after division by $(-i)$. Hence the neutrino Boltzmann hierarchy isThe lower-neighbour term is absent for $\ell =0$. The low moments areThese coefficients depend on the expansion convention. In the convention with a $(2\ell +1)$ factor, the temperature multipoles are ${\Theta }_{\ell }/(2\ell +1)$; using those multipoles without converting them would give incorrect factors in the scalar neutrino anisotropic stress.

The intrinsic parity $\eta$ of a particle is the parity eigenvalue of its one-particle state at rest:For the particles in this question, $\eta =\pm 1$. A two-particle state has multiparticle parity ${\eta }_1{\eta }_2(-1{)}^{\ell }$, where $\ell$ is its relative orbital angular momentum.

The pionic-deuterium ground state has atomic $\ell =0$. Since the pion has spin zero, its total angular momentum is therefore the deuteron spin, $J=1$, and its parity is ${\eta }_d{\eta }_{\pi }={\eta }_{\pi }$. The two neutrons are identical fermions. Their antisymmetric spin-one-half singlet state would require even orbital $\ell$, but $S=0$ and $J=1$ would instead require $\ell =1$. Hence they must occupy the symmetric spin-one-half triplet state, so the orbital state is antisymmetric and $\ell$ is odd. The addition of angular momentum with $S=J=1$ permits $\ell =0,1,2$, leaving $\ell =1$. Thus parity conservation givesso the pion has intrinsic parity $\boxed{{\eta }_{\pi }=-1}$.

In ${\Delta }^{-}\to {\pi }^{-}+n$, the final total spin is $S=1/2$. Coupling it to $\ell$ to obtain $J=3/2$ givesso the allowed orbital angular momenta are $\boxed{\ell =1,2}$. The observed angular probability is ${\cos }^2\theta =|P_1(\cos \theta ){|}^2$, where $P_1$ is a Legendre polynomial; it therefore selects the $\ell =1$ partial wave rather than $\ell =2$. The parity selection rule for a two-body decay now givesTherefore the delta particle has intrinsic parity $\boxed{{\eta }_{{\Delta }^{-}}=+1}$.

For an axisymmetric separated solution of the Laplace equation whose angular factor is the Legendre polynomial $P_n(\cos \theta )$, the radial ordinary differential equation has the two powers $r^n$ and $r^{-n-1}$. Thusfor constants $A,B$.

Write $x=\cos \theta$. Since $3\cos (2\theta )=6x^2-3=4P_2(x)-P_0(x)$, only the spherical harmonics of degrees zero and two occur. The degree-zero radial coefficient satisfying its two Dirichlet boundary conditions is $1-2/r$. For degree two, write $C{r}^2+D{r}^{-3}$; thenso $C=-4/31$ and $D=128/31$. By linearity and uniqueness of this boundary value problem,Direct substitution at $r=1,2$ verifies both boundary values.

The displayed Rodrigues formula gives the Legendre polynomialA four-node Gaussian rule is exact through degree $2\cdot 4-1=7$. Solving $q_4(x)=0$ first for $x^2$ gives the four nodesThe rule is not exact for every polynomial of degree eight: its quadrature value on $q_4^2$ is zero because every node is a root of $q_4$, whereas

The Legendre polynomial $p_n$ is orthogonal on $[-1,1]$ to every polynomial of degree below $n$. Suppose it changes sign at only $r<n$ distinct interior points $x_1,\dots ,x_r$, and putchoosing its overall sign so that $p_{n}q\ge 0$ on $[-1,1]$. This product is positive except at finitely many points, socontradicting orthogonality because $\mathrm{deg}q=r<n$. Thus $p_n$ has at least $n$ sign-changing roots in $(-1,1)$; its degree is $n$, so these are all its roots and they are distinct.

For arbitrary distinct nodes $x_1,\dots ,x_n$, letbe the Lagrange basis polynomials. Every polynomial $P$ of degree below $n$ satisfies $P(t)={\sum }_{i}P(x_i)L_i(t)$, so the required weights exist and areThey are unique because applying any such formula to $P=L_i$ recovers $A_i$.

Now assume exactness for every $P$ of degree below $2n$ and define $Q(t)={\prod }_i(t-x_i)$. For every polynomial $R$ of degree below $n$, the product $QR$ has degree below $2n$ and vanishes at every node. ThereforeThere is a unique monic polynomial of degree $n$ orthogonal to all lower-degree polynomials, so $Q$ is the monic normalization of $p_n$ and the $x_i$ are precisely the roots of $p_n$. Taking $P=1$ givesFinally $L_i^2$ has degree $2n-2$, so exactness givesThese are the positive weights of Gaussian quadrature.

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrodinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

Part (a) makes ${\lambda }_k<0$ for all sufficiently large $k$, while part (b) would make ${\lambda }_k\ge 0$ whenever $P^{(k)}$ were nonzero. Hence some derivative of $P$ vanishes identically, so $P$ is a polynomial.

Let its degree be $n$. Then $P^{(n)}$ is a nonzero constant. Substituting $k=n$ into the differentiated equation annihilates both derivative terms and givesThus ${\lambda }_n=0$, andThe nonzero regular solutions are therefore constant multiples of the Legendre polynomial of degree $n$.

The scattering amplitude is defined byFor a central potential its partial-wave expansion isUsing orthogonality of the Legendre polynomials in ${\sigma }_T=\int |f{|}^{2}d\Omega$ gives

Let $\theta$ be the angle between $r$ and the positive $z$-axis. By superposition of the electric potential of point charges,For $r>d$, the Legendre polynomial expansion givesThereforeThe three displayed terms are respectively the monopole, dipole, and quadrupole terms of the electric multipole expansion. In particular, the total charge is $Q(N-M-1)$ and the electric dipole moment is $Qd(M-1)\hat{z}$. This is the multipole expansion of three collinear charges.

Project the photon Boltzmann equation onto Legendre polynomials. The angular average uses $⟨\Theta ⟩={\Theta }_0$ and $⟨\mu ⟩=0$, so the collision term has no monopole andFor the dipole, the recurrence relation makes free streaming couple ${\Theta }_1$ to ${\Theta }_0$ and ${\Theta }_2$. The gravitational term $-ik\mu \Psi$ contributes $k\Psi /3$, while projection of $i\mu v_b$ gives the baryon-velocity source. ThusNeglecting the photon quadrupole gives

Expanding the direction-dependent photon temperature in Legendre polynomials turns its transport equation into an infinite hierarchy. Free streaming couples each multipole ${\Theta }_{\ell }$ only to its neighbours ${\Theta }_{\ell -1}$ and ${\Theta }_{\ell +1}$.

A direction-dependent fractional photon temperature change. For a thermal spectrum, $f(ϵ)=\bar{f}(ϵ)-\Theta ϵ{\bar{f}}^{\prime }(ϵ)$ is its linear expansion at fixed comoving energy. For a nonthermal spectrum the same ansatz is a brightness dilation, rather than necessarily a thermodynamic temperature. In unweighted Legendre polynomial multipoles, $\Theta ={\sum }_{\ell }(-i{)}^{\ell }{\Theta }_{\ell }{P}_{\ell }$, the photon density, velocity and anisotropic stress are ${\delta }_{\gamma }=4{\Theta }_0$, $v_{\gamma }=-{\Theta }_1$ and ${\Pi }_{\gamma }=-3{\Theta }_2/5$.