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A scalar Fourier mode selects one spatial direction, $\hat{\mathbf{k}}$, and has no transverse vector or tensor polarization. Rotations about $\mathbf{k}$ therefore leave its temperature and polarization patterns unchanged. Only azimuthal number $m=0$ occurs in the corresponding spherical harmonic expansion.

Let $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$ and use the paper's conventionwhich omits the $(2\ell +1)$ factor used in some other definitions of the photon temperature multipole. In Fourier space the transport term is $ik\mu \Theta$. The Legendre polynomial recurrence relation makes its coefficient at multipole $\ell$ equal toThe gravitational time derivative is purely monopolar, and the potential gradient and baryon velocity are dipolar. Thomson scattering removes the monopole from its relaxation term and adds the quadrupole polarization correction. Thus the photon Boltzmann hierarchy isThe $\ell =0$ term containing ${\Theta }_{-1}$ is understood as zero.

Write $m_{ij}=m_{ij}^{(\pm 2)}$ and $h=h^{(\pm 2)}$. Since $m_{ij}{e}^i{e}^j=\frac{1}{2}(1-{\mu }^2)e^{\pm 2i\phi }$, the line-of-sight solution isThe angular pattern $m_{ij}{e}^i{e}^j$ is proportional to $Y_{2,\pm 2}$. Projecting onto this spherical harmonic multiplies its coefficient bywhere $x=k(\eta -{\eta }^{\prime })$ and $j_2$ is a Spherical Bessel function. Thus the quadrupole isUse the isotropic tensor integralBecause $m$ is symmetric and trace-free, contracting this with $m_{ab}$ gives $2m^{ij}/15$. Substitution into the paper's signed neutrino tensor anisotropic stress therefore givesThe neutrino tensor free-streaming kernel is finite at the origin, with $K(0)=1/15$.

The material derivative follows the interfacial surfactant. The term $-C{\nabla }_s\cdot {\mathbf{u}}_s$ describes dilution by tangential expansion and concentration by compression. The term $-C(\mathbf{u}\cdot \mathbf{n}){\nabla }_s\cdot \mathbf{n}$ accounts for changing surface area due to normal motion of a curved interface. Finally, $D_s{\nabla }_s^{2}C$ describes surface diffusion. Together these give conservation of insoluble surfactant on a moving interface.

The relevant surface Péclet number isWrite $S=\mathbf{n}\cdot \mathbf{A}\cdot \mathbf{n}$ and ${\mathbf{u}}_s=\mathbf{Ax}-S\mathbf{x}$. Since $\mathbf{A}$ has zero trace and ${\nabla }_s\mathbf{n}={\mathbf{I}}_s/a$, the surface divergence isThere is no normal motion. To first order in ${\mathrm{Pe}}_s$, the steady surfactant balance becomes $D_s{\nabla }_s^2{C}^{\prime }=C_0{\nabla }_s\cdot {\mathbf{u}}_s=-3C_{0}S$. The traceless quadratic $S$ is a degree-two spherical harmonic, so ${\nabla }_s^{2}S=-6S/a^2$. The mean of $C^{\prime }$ is zero by total surfactant conservation; consequentlyThe discarded advective term ${\mathbf{u}}_s\cdot {\nabla }_s{C}^{\prime }$ is smaller by ${\mathrm{Pe}}_s$. This is the quadrupolar surfactant distribution on a spherical interface.

With the unit normal directed from the bubble into the exterior, the interfacial stress balance with variable surface tension isUsing the specified first-order curvature and $\gamma ={\gamma }_0-K{\gamma }_{1}S$, together with ${\nabla }_{s}S=2{\mathbf{I}}_s\mathbf{An}/a$, givesThe first term is the spherical capillary pressure; the last contains the normal tension correction and the tangential Marangoni stress.

The ambient Stokes flow has the symmetry of the symmetric traceless tensor $\mathbf{E}$. In the Unscaled Papkovich–Neuber representation, ${\chi }_{\infty }=\mathbf{x}\cdot \mathbf{E}\cdot \mathbf{x}/2$ is harmonic and yields $\mathbf{Ex}$. The decaying vector potential must have the form $\mathbf{E}\nabla (1/r)$, and the scalar disturbance must have the degree-two form $\mathbf{E}:\nabla \nabla (1/r)$. They are harmonic outside the bubble and have precisely the required rotational covariance. Dimensionless amplitudes may therefore be writtenFor a steady bubble, the no-penetration boundary condition gives $1+P-3Q=0$. Its tangential velocity gives $\alpha =1+2Q$. The tangential stress boundary condition then yieldsHence $\alpha =5/(5+2M)$, $Q=-M/(5+2M)$ and $P=-(5+5M)/(5+2M)$. The inviscid interior supplies only the constant pressure balancing $2{\gamma }_0/a$. Matching the remaining normal stress givesEliminating $P,Q,\alpha$ givesAs $M\to \infty$, $\alpha \to 0$: the Marangoni stress suppresses tangential motion and effectively immobilizes the interface. This limit must retain the small surface Péclet number and small-deformation assumptions; $M$ can grow through increasing ${\gamma }_1$ without invalidating the linear concentration approximation.

In a static spherical star, hydrostatic equilibrium and the Poisson equation giveor equivalently $r^{-2}d(r^{2}g)/dr=4\pi G\rho$.

Let $\mathbf{\xi }$ be the fluid displacement field, so the velocity perturbation is $\delta \mathbf{u}={\partial }_t\mathbf{\xi }$. Linearizing the ideal-fluid momentum equation about the static state and cancelling the background hydrostatic terms givesConservation of mass says that the Lagrangian density perturbation is ${\Delta }_L\rho =-\rho \nabla \cdot \mathbf{\xi }$. The relation between Eulerian and Lagrangian fluid perturbations and adiabatic compression gives, with $\Delta =\nabla \cdot \mathbf{\xi }$,while linearized self-gravity gives $\boxed{{\nabla }^2\delta \Phi =4\pi G\delta \rho }$.

For the stated spherical harmonic displacement, the radial divergence is $r^{-2}d(r^2{\tilde{\xi }}_r)/dr$, while $\nabla {\tilde{\xi }}_h$ is radial and orthogonal to the angular gradient of $Y_l^m$. HenceEquating the radial and horizontal coefficients of $Y_l^m$ and $\nabla Y_l^m$, and applying the separated Laplacian in spherical coordinates to $\widetilde{\delta \Phi }(r)Y_l^m$, gives

Define the stellar buoyancy frequency byEliminating $\tilde{\Delta }$ between the density and pressure perturbations givesSubstitution in the radial equation, followed by use of $dp/dr=-\rho g$, yieldsRegular spherical profiles near the center have $\rho ={\rho }_c+O(r^2)$ and $p=p_c+O(r^2)$, while $m(r)=4\pi {\rho }_c{r}^3/3+O(r^5)$ and hence $g=O(r)$. Both logarithmic gradients in the definition of $N^2$ are $O(r)$, so $N^2=A{r}^2+O(r^4)$. A Sun-like radiative central stratification is stable, making $A>0$.

Let $\mathbf{\xi }$ be the Lagrangian displacement. Linearized mass conservation and adiabaticity giveThe fixed potential has no Eulerian perturbation, so the linearized momentum equation isFor the stated spherical harmonic decomposition,Taking radial and horizontal components and using $g={\Omega }^{2}r$ gives

For an isolated solenoidal magnetic field, zero flux through every sphere removes the degree-zero spherical harmonic of $P=\mathbf{B}\cdot \mathbf{x}$. In a poloidal-toroidal decomposition, let ${\mathcal{L}}^2=-{\Delta }_{S^2}$. The spherical poloidal component obeysIf $P={\sum }_{l\ge 1,m}{p}_{lm}(r)Y_{lm}$, put $D_{lm}={\int }_0^{\infty }[r^2|p_{lm}^{\prime }{|}^2+l(l+1)|p_{lm}{|}^2]dr$. The two sides of the bound are respectively $\sum D_{lm}/[l(l+1)]$ and $\frac{1}{2}\sum D_{lm}$. Since $l(l+1)\ge 2$, the bound follows, with equality for degree-one radial scalars. Together with the radial-flow dynamo energy bound it gives $(q/\eta {)}^2\ge 2E_P/E_{\mathrm{total}}$ for a marginal or growing dynamo field.

// Target: fluid-mechanics.bigb

For radius $a$ and tangential velocity ${\mathbf{u}}_s={\mathbf{I}}_s\mathbf{Ax}$ with symmetric traceless $\mathbf{A}$, the surface divergence is $-3\mathbf{n}\cdot \mathbf{An}$. At small surface Péclet number $a^2|\mathbf{A}|/D_s$, neglecting advection of the concentration perturbation givesThe result follows because this quadratic is a degree-two spherical harmonic with surface Laplacian $-6/a^2$ times itself and zero spherical mean.

For a point-mass companion at separation $a$, subtract the constant and linear terms of its Newtonian gravitational potential in a frame accelerating with the target's center. The leading residual isIt is degree two in spherical harmonics. In a circular orbit with angular frequency ${\omega }_o$, its time-dependent $m=\pm 2$ components force at $2{\omega }_o$ in a frame with no spin; there is also a static $m=0$ component. This is the leading approximation in $r/a$, and the tidal amplitude may be absorbed into a harmonic-dependent constant.

A statistically isotropic Cosmic microwave background anisotropy has $⟨a_{{\ell }_1{m}_1}{a}_{{\ell }_2{m}_2}{a}_{{\ell }_3{m}_3}⟩={\mathcal{G}}_{m_1{m}_2{m}_3}^{{\ell }_1{\ell }_2{\ell }_3}{b}_{{\ell }_1{\ell }_2{\ell }_3}$, where $\mathcal{G}$ is the integral of the three spherical harmonics. The reduced CMB bispectrum removes this angular geometric factor and retains the primordial and transfer-function dependence. A Sachs-Wolfe projection of a constant bispectrum is an analytic large-angle example.

A regular solid harmonic is $r^l{Y}_l^m(\theta ,\varphi )$: it is homogeneous of degree $l$ and solves the Laplace equation. Thus $\mathbf{r}\cdot \nabla F=lF$ and ${\nabla }^{2}F=0$. These identities turn angular spherical harmonic expansions of displacement, density and Newtonian gravitational potential into radial equations.

Rotational Raman spectroscopy observes a frequency change of scattered light associated with molecular rotation. The coupling uses the induced electric dipole moment, ${\mathbf{d}}_{\mathrm{ind}}=\mathbf{\alpha }\mathbf{E}$, where $\mathbf{\alpha }$ is the electric polarizability tensor. For a linear molecule, the useful anisotropy is ${\alpha }_{\parallel }-{\alpha }_{\perp }$: the response differs for an electric field parallel or perpendicular to the molecular axis. A permanent electric dipole moment is not required.^[ref]

For the spin-free linear rigid rotor in one internal quantum state, the anisotropic tensor depends quadratically on the molecular direction. Its angular part has even parity and degree two in spherical harmonics. The triangle inequality for coupled angular momenta permits $|\Delta J|\le 2$, and even parity removes odd $\Delta J$, leaving $\Delta J=0,\pm 2$. The $\Delta J=0$ contribution is unshifted light; shifted rotational lines have $\Delta J=\pm 2$. Consequently the rigid-model Stokes shifts, in ordinary frequency units, areSuccessive allowed shifts are separated by $4B$, rather than the $2B$ absorption spacing of pure rotational electric-dipole selection rules. In spectroscopic wavenumber units replace $B$ by $\tilde{B}=B/c$ and $\Delta \nu$ by $\Delta \tilde{\nu }=\Delta \nu /c$, using the speed of light in $\mathrm{cm}{\mathrm{s}}^{-1}$ for $\mathrm{c}{\mathrm{m}}^{-1}$.

A homonuclear diatomic molecule such as molecular hydrogen has no permanent electric dipole moment in its symmetric internal state, so it lacks the ordinary electric-dipole microwave ladder. Its anisotropic electric polarizability can nevertheless give rotational Raman spectroscopy lines. Nuclear-spin symmetry can change the populations of even and odd $J$ levels, so the spin-free population example above must not be reused unchanged for molecular hydrogen. The method of coupling, as well as the energy eigenvalues, determines what the experiment can see.

For the spin-free linear rigid rotor in thermal equilibrium at temperature $T>0$, assume all integer $J\ge 0$ are allowed and ignore nuclear-spin restrictions. Each $J$ level has $2J+1$ quantum states sharing one energy eigenvalue, an instance of degenerate energy levels, so the Boltzmann distribution givesHere $Q_{\mathrm{rot}}$ is a rotational partition function, $h$ the Planck constant, $k_{\mathrm{B}}$ the Boltzmann constant, and $B$ a frequency in $\mathrm{Hz}$. Thus $hB/(k_{\mathrm{B}}T)$ is dimensionless. The $2J+1$ factor counts magnetic quantum numbers; it does not mean that a whole level is one quantum state. For the stipulated $B=5\mathrm{GHz}$, the most populated levels are $J=2$ at $3\mathrm{K}$ and $J=7$ at $30\mathrm{K}$.

Population alone is insufficient to predict intensity. In weak, unsaturated absorption, the integrated absorption coefficient includes the transition frequency, the squared electric dipole moment matrix element, and the difference of lower and upper substate populations because stimulated emission returns energy to the field. A pulsed FTMW spectroscopy signal additionally depends on excitation, quantum coherence, cavity tuning, and receiver response.^[ref]

For an isotropic sample with polarization of an electromagnetic wave along $z$, write $d_z=d\cos \theta$. Normalized spherical harmonics giveThe last equality follows from ${\sum }_{M=-J}^J{M}^2=J(J+1)(2J+1)/3$. Combining this sum with the per-substate Boltzmann distribution and subtracting stimulated emission gives relative integrated absorption areasThe common factor includes $d^2/3$ and cancels upon normalization. This formula describes integrated weak absorption in a thermal, spin-free linear rigid rotor, not arbitrary peak heights. The example's strongest integrated lines originate at $J=3$ and $J=13$ for $3$ and $30\mathrm{K}$, respectively: they differ from the most populated levels. A missing line can reflect population, coupling, coverage, or sensitivity; it does not prove that the corresponding quantum state is absent.

A scalar spheroidal harmonic is a regular angular solution of $(\sin \theta S^{\prime }{)}^{\prime }/\sin \theta +[c^2{\cos }^2\theta -m^2/{\sin }^2\theta +\Lambda ]S=0$; with $e^{im\varphi }$ it is smooth at both axes. For real $c$, regularity gives discrete eigenvalues ${\Lambda }_{\ell m}(c)$. At $c=0$ it reduces to the angular part of a spherical harmonic, with $\Lambda =\ell (\ell +1)$. The sign of the $c^2$ term distinguishes the convention used here.

A solid harmonic extends a spherical harmonic to a harmonic function in space with either the regular radial power $r^n$ or the irregular radial power $r^{-n-1}$. In three dimensions these are the two separated radial solutions of the Laplace equation at angular degree $n$.

Spectral geometry studies which properties of a Riemannian manifold can be recovered from the spectrum of geometric differential operators, especially the Laplace-Beltrami operator. Explicit spherical harmonics, Fourier series on a flat torus, and isospectral manifolds illustrate three different aspects of the problem.

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.

Here $\mu =\cos \theta$ and $a_{\ell ,m}=\sqrt{({\ell }^2-m^2)/[(2\ell +1)(2\ell -1)]}$. This follows by combining the associated Legendre function recurrence $(2\ell +1)\mu P_{\ell }^m=(\ell -m+1)P_{\ell +1}^m+(\ell +m)P_{\ell -1}^m$ with the normalized spherical harmonic constants. It gives the adjacent-multipole couplings in a photon Boltzmann hierarchy.

The angular basis ${}_s{Y}_{\ell m}$ for a spin-weight-$s$ field generalizes ordinary spherical harmonics. It has $\ell \ge |s|$. For scalar-source CMB polarization about a wavevector, only $m=0$ survives and ${}_{\pm 2}{Y}_{20}\propto {\sin }^2\theta$.

For a uniform-density sphere with incompressible flow of radius $R$ with one surface spherical harmonic displacement ${\xi }_r(R)=lA{R}^{l-1}{Y}_l^m$, the bulk density perturbation vanishes but the surface contributes $\delta \rho =\rho {\xi }_r(R)\delta (r-R)$. Continuity of the Newtonian gravitational potential and the Poisson equation give $[{\partial }_r\delta \Phi {]}_{\mathrm{out}-\mathrm{in}}=4\pi G\rho {\xi }_r(R)$. The regular interior and decaying exterior solutions areThe coefficient follows because the derivative jump is $-(2l+1)C{R}^{l-1}{Y}_l^m$. This surface source is why neglecting the bulk density perturbation does not justify neglecting the perturbed gravity.

The surface Laplacian is the Laplace-Beltrami operator of a surface with its induced Riemannian metric. It is ${\Delta }_{s}f={\nabla }_s\cdot {\nabla }_{s}f$, using the surface gradient and surface divergence. On a sphere of radius $a$, a degree-$l$ spherical harmonic has eigenvalue $-l(l+1)/a^2$.