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For a smooth ambient function near the unit sphere, the chain rule along the great circles ${\gamma }_i(t)=p\cos t+e_i\sin t$ gives $(F\circ {\gamma }_i{)}^{\prime \prime }(0)=D^{2}F(e_i,e_i)-DF(p)$. Sum over an orthonormal basis of the tangent space and supplement it by the radial vector $p$. The geodesic trace formula for the Laplace-Beltrami operator then gives the displayed formula, evaluated at radius one. In particular a degree-$\ell$ harmonic polynomial restricts to a spherical harmonic with positive eigenvalue $\ell (\ell +d-1)$.

The normalized angular average is over the unit sphere, with solid angle measure $d\Omega$. Rotational symmetry gives $⟨e^i{⟩}_{\Omega }=0$, $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$, and $⟨e^i{e}^j{e}^k{e}^{\ell }{⟩}_{\Omega }=({\delta }^{ij}{\delta }^{k\ell }+{\delta }^{ik}{\delta }^{j\ell }+{\delta }^{i\ell }{\delta }^{jk})/15$. The constants follow by contracting indices and using $e^i{e}_i=1$. These identities extract photon angular temperature moments. For normalized spherical harmonics, $\int Y_{\ell m}^{\ast }{Y}_{{\ell }^{\prime }{m}^{\prime }}d\Omega ={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}$; their angular average instead has an extra factor $1/(4\pi )$.

For a normalized spherical harmonic of degree $\ell \ge 1$, write ${\rho }_1=s(r)Y$ and ${\varphi }_1=F(r)Y$. Integration of the radial Poisson equation gives $A=-4\pi G\int r^2{F}^{\ast }sdr$, where $A=\int [r^2|F^{\prime }{|}^2+\ell (\ell +1)|F{|}^2]dr$. Cauchy-Schwarz inequality and $\int |F{|}^2\le A/[\ell (\ell +1)]$ give the displayed bound on $|E_g|=A/(4\pi G)$. For bounded density amplitudes with fixed radial support this decays as ${\ell }^{-2}$, giving a quantitative angular short-scale justification of the Cowling approximation.

The angular dipole is the degree-one spherical harmonic component. A real angular dipole is a function $\mathbf{d}\cdot \hat{\mathbf{n}}$ for a fixed vector $\mathbf{d}$. Direction-dependent observer velocity and gravitational endpoint terms can therefore be discarded from higher-multipole calculations when they have this form.

Associated Legendre functions solve $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+[\nu (\nu +1)-{\mu }^2/(1-x^2)]y=0$. For integers $\ell \ge 0$ and $0\le m\le \ell$, the regular solution is $P_{\ell }^m(x)=(-1{)}^m(1-x^2{)}^{m/2}(d/dx{)}^m{P}_{\ell }(x)$, using a Legendre polynomial. It enters normalized spherical harmonics. Despite the traditional term associated Legendre polynomial, the factor $(1-x^2{)}^{m/2}$ is not a polynomial when $m$ is odd. Negative integer orders satisfy $P_{\ell }^{-m}=(-1{)}^m(\ell -m)!P_{\ell }^m/(\ell +m)!$.

Each orbital angular momentum operator annihilates a radial factor and $r^2$. Direct differentiation gives $L^2{x}_3^2={\hslash }^2(6x_3^2-2r^2)$ and $L_3{x}_3^2=0$. Hence the displayed trace-free quadratic polynomial, times any admissible radial factor, has orbital labels $l=2,m=0$. Its angular dependence is proportional to the spherical harmonic $Y_{20}$; subtracting $r^2/3$ removes the rotational scalar part.

Writing $D_r=\mathbf{x}\cdot \nabla =r{\partial }_r$, the spherical-coordinate Laplacian gives $r^2\Delta =D_r^2+D_r+{\Delta }_{S^2}$. Thus the positive angular Laplace-Beltrami operator has the displayed Cartesian expression and eigenvalues $l(l+1)$ on spherical harmonics. Omitting $D_r$ is incorrect when the square means composition: on $x_1$, the truncated expression gives $x_1$, whereas the degree-one angular eigenvalue is two.

// Target: fluid-mechanics.bigb

Linear steady transport on a translating sphere gives $B=2A{C}_{0}a/(k{a}^2+2D_s)$. The components of its unit normal vector are degree-one spherical harmonics. The condition $2|A|Ua/(k{a}^2+2D_s)\ll 1$ ensures the concentration perturbation is small relative to $C_0$.

The integral of three spherical harmonics on the sphere. In the conventional complex basis it is real and equals a product of two Wigner 3j symbols and $\sqrt{(2{\ell }_1+1)(2{\ell }_2+1)(2{\ell }_3+1)/(4\pi )}$. It vanishes unless the multipoles satisfy a triangle condition, ${\ell }_1+{\ell }_2+{\ell }_3$ is even, and $m_1+m_2+m_3=0$. It carries the angular geometry of the reduced CMB bispectrum.

If ${\mathcal{P}}_{\ell }$ is the space of degree-$\ell$ homogeneous polynomials and ${\mathcal{H}}_{\ell }$ its harmonic polynomials, then ${\mathcal{P}}_{\ell }={\mathcal{H}}_{\ell }\oplus |x{|}^2{\mathcal{P}}_{\ell -2}$. Under the Fischer inner product, the second summand is orthogonal to the kernel of the Laplacian. Iteration expresses every polynomial restricted to the unit sphere as a sum of spherical harmonics.

For $\mathbf{k}=k\hat{z}$, these complex vectors form an orthonormal basis of the transverse plane using the Hermitian inner product. They have opposite phases under rotations about $\hat{z}$, so they describe the two helicity components. Since $\mathbf{e}\cdot m^{(\pm )}=\sin \theta e^{\pm i\phi }/\sqrt{2}$, transverse-vector sources carry only the angular sectors $m=\pm 1$ of spherical harmonics. Streaming by $ik\cos \theta$ preserves $m$ and couples neighboring angular degrees.

A complex helicity-$+2$ plane wave along the $z$ axis has polarization contraction ${\sin }^2\theta e^{2i\phi }$. Its remaining tensor CMB line-of-sight source depends only on $\cos \theta$. Orthogonality of the azimuthal factors of spherical harmonics therefore restricts its temperature coefficients to $m=2$. A real perturbation also includes the conjugate mode.

Helioseismology infers the Sun's structure and dynamics from its stellar oscillations. Spatial projection of resolved surface observations onto spherical harmonics identifies angular degree and azimuthal order; a temporal Fourier transform measures mode frequencies. Stellar acoustic modes constrain the adiabatic sound speed, while rotational splitting constrains stellar rotation. Their finite wavelength, surface reflection phases and finite observational coverage limit inversion resolution.

At a spherical interface with an insulating exterior, equal magnetic permeability on both sides and no surface current, all components of the magnetic field are continuous. The solenoidal magnetic-field constraint then makes $P=\mathbf{B}\cdot \mathbf{x}$ and ${\partial }_{r}P$ continuous. A current-free exterior has $\Delta P=0$. For an isolated field decaying at infinity, its degree-$l$ spherical harmonic satisfies ${\partial }_r{P}_{lm}=-(l+1)P_{lm}/a$ at a sphere of radius $a$. This boundary condition incorporates the exterior field without setting the radial field to zero.

// Target: fluid-mechanics.bigb

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

A hydrogen atom state with principal quantum number $n=l+1$ has no radial nodes and radial factor $R(r)=A{r}^l{e}^{-r/[a(l+1)]}$, where $a$ is the Bohr radius. Its energy is $-{\hslash }^2/[2m{a}^2(l+1{)}^2]$. With unit-normalized spherical harmonics, the radial normalization gives $|A{|}^2=2^{2l+3}/[a^{2l+3}(l+1{)}^{2l+3}(2l+2)!]$.

Projecting free-streaming neutrinos onto the tensor spherical harmonic givesHere $j_2$ is a Spherical Bessel function. The continuous value at zero follows by integrating $(1-{\mu }^2{)}^2$, whose integral is $16/15$.

Every orbital angular momentum operator annihilates a radial function: substituting ${\partial }_{i}f(r)=f^{\prime }(r)x_i/r$ makes the two terms cancel. It also annihilates $r^2$. Thus the radial factor can be carried through the angular operators. Since $x_3^2$ is independent of $x_1,x_2$, $L_3\psi =0$ for every value of $a$.

Direct differentiation givesConsequently $L^2{x}_3^2={\hslash }^2(6x_3^2-2r^2)$ and $L^2{r}^2=0$. For a nonzero admissible radial factor, comparison with $\lambda (x_3^2+a{r}^2)f(r)$ forces $\lambda =6{\hslash }^2$ and $6a=-2$. Therefore the simultaneous eigenfunction occurs atIt is the axially symmetric quadratic orbital eigenfunction, with quantum numbers $l=2$, $m=0$ and angular dependence proportional to the spherical harmonic $Y_{20}$. The subtraction of $r^2/3$ removes its rotationally invariant scalar component.

A spherical harmonic of angular degree $\ell$ has horizontal wavenumber $k_h=L/r$, with $L=\sqrt{\ell (\ell +1)}$. The local acoustic dispersion relation therefore gives the radial wavenumberIn the high-order regime a travelling stellar acoustic mode accumulates phase $\int k_{r}dr$ across its radial cavity. Reflection at the inner turning point and at the near-surface boundary fixes additional phases. A round trip must return the same complex amplitude, so WKB quantization givesHere $n$ counts radial oscillations, up to the choice of origin for radial order, and $\alpha$ incorporates both reflection phases. The outer acoustic reflection need not coincide exactly with the geometric stellar radius; replacing it by $R$ and absorbing its correction in $\alpha$ is the leading asymptotic approximation. The model neglects buoyancy, perturbations of self-gravity and rapid background variations on a wavelength.

Resolved observations of the Sun measure line-of-sight surface velocity through the Doppler effect, or brightness fluctuations. Projection on spherical harmonics identifies $\ell$ and azimuthal order $m$; a temporal Fourier transform identifies the mode angular frequencies $\omega$. The sequences of peaks of successive radial order form ridges in the frequency-degree diagram. Their ordering and comparison with a reference stellar oscillation model identify $n$; the surface image alone does not count inaccessible radial nodes. Fit the residual phase of the radial quantization relation across the observed mode set to estimate $\alpha$. A simultaneous fit with the sound-speed profile, or a calibrated reference model, is needed: $\alpha$ is not determined independently by one mode, and an integer shift of the numbering of $n$ shifts $\alpha$ oppositely. In a more accurate fit the surface phase depends on frequency. These steps constitute helioseismology.

Set $a(r)=c(r)/r$ and $w=\omega /L$. Assume $a$ decreases outwards in the region to be inverted. With the surface approximation above, the Duvall law isThus a suitably scaled mode set samples one function of $w$. Smooth the measured values before differentiating; inversion amplifies observational noise. At a finite outer reference point $r_s$, the already known contribution exterior to $r_s$ must be subtracted. For the remaining cavity, let $w_s=a(r_s)$ and $h(b)=-d\ln r/db$. Changing integration variable from radius to $b=a(r)$ givesThe boundary term vanishes because the square root vanishes at $b=w$. Now multiply the derivative by $w/\sqrt{a^2-w^2}$ and interchange the integrals. The kernel identityfollows by setting $w^2=b^2+(a^2-b^2){\sin }^2\theta$ and then integrating $1/[b^2{\cos }^2\theta +a^2{\sin }^2\theta ]$. It yields the Abel inversion of stellar acoustic travel times:Knowing the monotone relation $r(a)$ then gives $c=ar$. Observed turning points delimit the region recoverable without extrapolation; an unknown surface contribution or nonmonotone $a$ prevents this simple unique inversion.

There are genuine inconsistencies in the printed numerical example. Since $w$ and $w_0$ have dimensions of angular frequency, $F$ must have dimensions of time, but the displayed empirical right-hand side is dimensionless. Moreover its derivative supplies a factor $1/w_0$, whereas the requested inversion supplies $1/w_0^2$. The dimensionally repaired relation that produces the requested inversion integral isSubstitution in the boxed Abel transform inversion gives precisely the exponent coefficient $2a^2/(\pi w_0^2)$, with the square-root denominator inside the integral. This repairs the first normalization but does not repair the later algebra.

Indeed, put $x_s=w_s/a$ and $b=\sqrt{1-x_s^2}$. The required elementary integrals evaluate toThe finite-reference-point answer following from $F_c$ is thereforeIt still depends on $w_s$. The later printed profile suppresses that dependence, which requires the additional surface limit $w_s\to 0$, $r_s\to R$. Even in that limit it is not the result of this integral. To see this exactly, write $\mathcal{L}=\ln (R/r)$ and $y=a^2/(\pi w_0^2)$. The integral givesThe minus branch is selected by $a\to 0$ at the surface. Its decreasing-radius branch extends only to $\ln (R/r)=1/6$, where $a^2=\pi w_0^2/6$; it cannot describe $r=R/e$. The stated lower bound on the largest sampled $w$ does not fix this algebraic obstruction. In contrast, squaring and rearranging the printed closed profile would require $\mathcal{L}=18y-81y^2$. This differs identically from $2y-6y^2$, so the two printed steps cannot both follow from the same data.

The surface polytrope test can nevertheless be performed exactly, and clarifies the final discrepancy. For a plane-parallel polytropic atmosphere of index $m=3$, take $g=GM/R^2$ constant and $z=R-r$. The stellar hydrostatic equation and $p=K{\rho }^{4/3}$ give $4p/\rho =gz$, hence the surface sound speed of a polytropic star isThe exponent in the equilibrium polytropic equation of state need not equal the perturbation adiabatic exponent $\gamma$. For the corrected inversion above, expansion at $z/R\ll 1$ gives $c^2\simeq \pi w_0^{2}Rz/2$, so matching the surface polytrope fixes $w_0^2=\gamma GM/(2\pi R^3)$.

Alternatively, if the later printed closed profile is adopted as a separate imposed model, its surface expansion gives $c^2\simeq \pi w_0^{2}Rz/18$. It also has the correct square-root depth dependence, but requiresAt $r=R/e$ that imposed profile then givesThe claimed coefficient $\pi /3$ is inconsistent with this substitution; $1/e\ne \pi /3$. Thus the surface scaling is demonstrable, but the empirical normalization, the requested closed inversion and the final numerical coefficient cannot all be asserted as printed. These discrepancies are visible in the original PDF, not introduced by its TeX conversion.