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The time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\psi =E\psi$, with attractive Coulomb potential $V(r)=-\kappa /r$, where $\kappa =e^2/(4\pi {\epsilon }_0)$. Separating $\psi =R(r)Y_{\ell }^m(\theta ,\phi )$ uses the spherical harmonic angular eigenvalue $-\ell (\ell +1)$ of the unit-sphere Laplacian. The radial kinetic contribution is $-{\hslash }^2(R^{\prime \prime }+2R^{\prime }/r)/(2m)$, the angular kinetic contribution is the centrifugal term ${\hslash }^2\ell (\ell +1)R/(2m{r}^2)$, and the remaining term is the attractive potential energy. The right side is the energy eigenvalue times the radial wavefunction. Here the stated $m$ neglects nuclear recoil; including it replaces the electron mass by the reduced mass.

For $\ell =0$, substitution of the ground-state exponential givesIts $1/r$ coefficient must vanish and its constant coefficient determines the energy, soHere $a_0={\hslash }^2/(m\kappa )$ is the Bohr radius.

For $R_2=N_2(r+b)e^{-\beta r}$, direct differentiation yieldsCancel the nonzero exponential in the radial Schrödinger equation and compare coefficients of $r$, the constant term, and $1/r$. They give, respectively,ConsequentlyThe positive decay constants ensure normalizability; the second radial wavefunction has one positive-radius node at $2a_0$, as appropriate to the first radial excitation.

If the entire energy gap is carried by one photon, neglecting atomic recoil, conservation of energy gives the requested frequencyThere is a physical qualification: these are the $2s$ and $1s$ states, so the usual single-photon electric-dipole transition is forbidden by the electric-dipole selection rules for hydrogen. Explicitly, both angular factors are $Y_0^0$, and $\int |Y_0^0{|}^2\hat{\mathbf{r}}d\Omega =0$, so every component of $⟨1s|\mathbf{r}|2s⟩$ vanishes. The leading isolated-atom decay is instead two-photon emission, for which ${\nu }_1+{\nu }_2={\nu }_{\mathrm{gap}}$ and individual frequencies are not fixed. This metastability and the two-photon channel are documented in NIST's hydrogen spectral-data compilation. Thus the boxed answer is the energy-gap frequency under the printed single-photon assumption, rather than an assertion of an allowed electric-dipole line.

Restore the Fourier transform integral and the last-scattering distance:Insert the Rayleigh plane-wave expansion and use orthogonality of spherical harmonics with the complex conjugate on the projection harmonic. The angular integral selects the matching indices, givingThe complex conjugation is present in the original PDF. If $A_{\mathbf{k}}=A{k}^{n/2}$ denotes a mode coefficient with its random amplitude and phase included in $A$, substitution of ${\Phi }_{\mathbf{k}}=-6A_{\mathbf{k}}/k^2$ gives the equivalent factor $-8\pi i^{\ell }{A}_{\mathbf{k}}/k^2$ inside the Fourier integral. The monopole and local observer dipole removed in part (b) do not change this formula for $\ell \ge 2$.

Use the same negative-exponent Fourier transform as in the preceding part and let $f=d\log D_{+}/d\log a$. In linear theory, $\dot{\delta }=Hf\delta$, and the linearized cosmological continuity equation is $\dot{\delta }+a^{-1}{\nabla }_x\cdot \mathbf{v}=0$. Its irrotational solution isHere $\mathbf{r}$ labels the comoving spatial position for the Fourier expansion. The quantity being differentiated is also $-\Phi /(4\pi G{a}^2\bar{\rho })$, by the Poisson equation; thus this explicitly obtains the velocity from the Fourier components of the peculiar gravitational potential.

Conjugating the Rayleigh plane-wave expansion, and using the symmetric addition formula for spherical harmonics, givesFor the radial component $U=\mathbf{v}\cdot \hat{r}$, differentiate at fixed angular position. Only the Spherical Bessel function depends on $r$, and $d{j}_{\ell }(kr)/dr=k{j}_{\ell }^{\prime }(kr)$. Therefore the radial peculiar velocity in spherical harmonics isThe coefficient is $4\pi /(2\pi {)}^3=1/(2{\pi }^2)$, and $(-i{)}^{\ell }=(i^{\ell }{)}^{\ast }$, so the phase and conjugation agree with the requested expression. At the present epoch, $a=1$ and $H=H_0$. Adopting the traditional approximation $f\simeq {\Omega }_m^{0.6}$ givesThe printed expression is therefore the present-epoch result in units with $H_0=1$, or with its velocity interpreted as $U/H_0$. For distances and velocities in ordinary physical units, it needs the factor $H_0$; at a general epoch it needs $aH$. The approximation $f\simeq {\Omega }_m^{0.6}$ is exact in the Einstein-de Sitter limit but is not an exact growth formula for arbitrary cosmologies.

Use units $c=1$, signature $(+---)$, and the weak-field metric $g_{ab}={\eta }_{ab}+h_{ab}$. Define the trace-reversed metric perturbation by ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$, with $h={\eta }^{ab}{h}_{ab}$. In the Lorenz gauge in linearized gravity, ${\partial }^a{\bar{h}}_{ab}=0$, the sign convention of Q2 givesFor completeness, the linearized Ricci tensor in this convention isSubstituting the Lorenz condition ${\partial }_c{h}^c{}_b={\partial }_{b}h/2$ and trace reversing gives the stated Einstein tensor. For a stationary source, $\square =-{\nabla }^2$, so ${\nabla }^2{\bar{h}}_{ab}=16\pi G{T}_{ab}$.

Keep the monopole terms of order $ϵ=GM/R$ and the leading term linear in the rotation rate. Since $\Omega R=O({ϵ}^{1/2})$, the latter metric term is of order $ϵ\Omega R=O({ϵ}^{3/2})$ on the shell. Thus it would disappear from a strictly order-$ϵ$ truncation; the requested approximation must retain the leading rotation correction as well. Terms $T_{ij}=\rho v_i{v}_j$ generate metric corrections of order $ϵ(\Omega R{)}^2=O({ϵ}^2)$ and are neglected. Corrections to four-velocity normalization and the supporting shell stresses affect the omitted higher orders. To retained order,The minus sign in $T_{0i}$ comes from lowering its spatial index; using $T^{0i}$ in its place would reverse the dragging direction.

The scalar Poisson solution regular at the centre and vanishing at spatial infinity is ${\bar{h}}_{00}=-4GM/r$ outside the shell and ${\bar{h}}_{00}=-4GM/R=-4ϵ$ inside. Continuity and the derivative jump ${\bar{h}}_{00}^{\prime }(R^{+})-{\bar{h}}_{00}^{\prime }(R^{-})=4GM/R^2$ verify the density $M\delta (r-R)/(4\pi R^2)$. Also ${\bar{h}}_{ij}=0$ at this order. Undoing trace reversal givesComparing with the scalar parts of the interior line element yields $A=-ϵ$ and $C=ϵ$.

For the mass-current contribution writeEach nonzero angular component is a degree-one spherical harmonic. The supplied Laplacian eigenvalue therefore givesAway from the shell, the two radial solutions are $r$ and $r^{-2}$. Regularity at $r=0$ and decay at infinity select $F_{\mathrm{in}}=ar$ and $F_{\mathrm{out}}=b/r^2$. Continuity at $R$ requires $b=a{R}^3$, and integrating the differential equation across the shell givesThus $a=4GM\Omega /(3R)=\omega$, with $F_{\mathrm{in}}=\omega r$. Trace reversal leaves mixed components unchanged, so $h_{0i}=\omega (\hat{\mathbf{z}}\times \mathbf{x}{)}_i=\omega (-y,x,0{)}_i$. The paper writes $g_{0i}=-B_i$, giving the frame dragging inside a slowly rotating spherical shellThis dipole matching is valid throughout $r<R$; it does not assume the observation point is very close to the centre. The stationary current is divergence-free, so this solution also satisfies the Lorenz gauge. The exterior result is $h_{0i}=2G(\mathbf{J}\times \mathbf{x}{)}_i/r^3$, where the thin shell has $\mathbf{J}=(2/3)M{R}^2\mathbf{\Omega }$, providing a check on the coefficient.

To identify the freely falling frames inside, let ${\mathcal{R}}_z(\theta )$ be the ordinary spatial rotation by angle $\theta$. Define the inertial coordinates inside a slowly rotating spherical shell byDifferentiating and dropping ${ϵ}^2$, products beyond the retained rotation order, and ${\omega }^2$ giveswhich is precisely the interior metric. Thus the retained interior curvature is zero, despite its nontrivial relation to the nonrotating coordinates anchored at infinity. Timelike geodesics in these local inertial coordinates are straight lines $\mathbf{X}={\mathbf{X}}_0+\mathbf{V}T$, with $|\mathbf{V}|<1$. In the original coordinates this isunderstood to the same perturbative order. Their coordinate equation is $d^2\mathbf{x}/d{t}^2=2\mathbf{\omega }\times d\mathbf{x}/dt$ to first rotation order. There is no interior Newtonian acceleration from the constant potential; the velocity-dependent term is the apparent Coriolis term relative to the locally dragged inertial axes.

A freely transported gyroscope at rest in the interior to the retained order has constant spatial direction in the $\mathbf{X}$ frame, hence its direction in the original axes obeys $d\mathbf{S}/dt=\mathbf{\omega }\times \mathbf{S}$. Local inertial axes precess in the same sense as the shell, at $\omega =(4ϵ/3)\Omega$ relative to distant nonrotating axes. This is the interior Lense-Thirring precession. For a weak shell the dragging is only a small fraction of its angular velocity, not perfect corotation. It is compatible with the equivalence principle: rotation relative to distant axes is a comparison between frames, and local flatness at the retained order does not make that comparison vanish. Quadratic rotation terms, which would include centrifugal effects and additional curvature, are outside the requested leading approximation.

There is a minor operator misprint to resolve first. Interpreting $(\mathbf{x}\cdot \nabla {)}^2$ as composition, the Cartesian formula for the angular Laplacian isThe last term is missing from the printed Cartesian expression. For example, its truncated expression maps $x_1$ to $x_1$, whereas a degree-one spherical harmonic must have angular eigenvalue two. In what follows, use the stated angular spectrum $l(l+1)$ and the supplied poloidal-energy identity; these are consistent with the corrected operator.

The divergence theorem and the solenoidal magnetic-field constraint imply zero magnetic flux through every sphere. Consequently $P=r{B}_r$ has zero angular mean and no $l=0$ component. Expand in orthonormal spherical harmonics:By angular orthogonality, withthe gradient norm and the supplied poloidal-toroidal decomposition identity becomeSince $l(l+1)\ge 2$, this proves the poloidal magnetic energy bound by the radial scalar gradient:Equality holds when the radial scalar contains only degree-one angular components.

Define the poloidal and total magnetic energies by $E_P=M_P/(2{\mu }_0)$ and $E_{\mathrm{total}}=M/(2{\mu }_0)$. Combining the radial-flow dynamo energy bound with $D\ge 2M_P$ gives the necessary conditionIt has the same temporal interpretation as the bound in the preceding part; statistically steady versions use the corresponding time-averaged energies. It quantifies why dynamo action with a substantial poloidal fraction requires radial motion, while a purely tangential flow has $Q=0$.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. The resistive induction equation, together with the solenoidal conditions on both fields, isSince $D_t\mathbf{x}=\mathbf{u}$, the product rule givesAlso $\Delta P=\mathbf{x}\cdot \Delta \mathbf{B}+2\nabla \cdot \mathbf{B}=\mathbf{x}\cdot \Delta \mathbf{B}$. Substitution proves the radial magnetic induction scalar equation:

In the insulating exterior the quasistatic magnetic field is current-free, so $\nabla \times \mathbf{B}=0$ and $\nabla \cdot \mathbf{B}=0$. Thus $\Delta \mathbf{B}=0$ and $\Delta P=0$ for $r>a$, with decay at infinity for an isolated field. This is an instantaneous exterior Laplace equation, rather than a diffusion equation with the interior value of $\eta$.

Use the usual nonmagnetic-interface assumptions: equal magnetic permeability on both sides and no surface current. The electromagnetic boundary conditions then make $B_r$ and the tangential magnetic field continuous. Hence $P=r{B}_r$ is continuous. In spherical coordinates, the solenoidal magnetic-field constraint readswhere the last term is the angular divergence on the unit sphere. Its right-hand side is also continuous, giving the insulating boundary condition for the radial magnetic scalar:Equivalently, the decaying exterior degree-$l$ spherical harmonic is proportional to $r^{-(l+1)}$, so its boundary amplitude satisfies ${\partial }_r{P}_{lm}(a)=-(l+1)P_{lm}(a)/a$.

The material derivative follows surfactant carried by the moving interface. The term $-C{\nabla }_s\cdot u_s$ is dilution or concentration by tangential area expansion or contraction. The term $-C(u\cdot n){\nabla }_s\cdot n$ is the additional area change produced by normal motion of a curved interface. The term $D_s{\nabla }_s^{2}C$ represents surface diffusion, while $-k(C-C_0)$ relaxes the concentration towards a reservoir value through exchange. All these terms refer to concentration per surface area, as in surfactant transport with exchange relaxation.

In the steady bubble frame, the spherical interface has $u\cdot n=0$. Write $C=C_0+C^{\prime }$ with uniform $C_0$ and regard both $u$ and $C^{\prime }$ as first order in the rising speed. Advection of $C^{\prime }$ and its product with surface dilation are second order. Hence the linear steady equation isSpherical symmetry and linearity leave only the tangential projection of the imposed translation vector: $u_s=A{I}_{s}U$. An azimuthal term $n\times U$ would change sign under reflection and is excluded for this non-chiral rising bubble. The degree-one forcing does not excite other spherical-harmonic degrees in the linear problem.

On $r=a$, differentiating $n=x/r$ givesEach component of $n$ is a degree-one spherical harmonic, soOne may also obtain the coefficient from $n\cdot n=1$: its surface Laplacian gives $n\cdot {\nabla }_s^{2}n=-|{\nabla }_{s}n{|}^2=-2/a^2$, while rotational symmetry makes the vector Laplacian parallel to $n$. Expanding $I_{s}U=U-n(U\cdot n)$ then yieldsThe candidate $C^{\prime }=BU\cdot n$ therefore solves the linear transport equation ifIts spherical mean is zero. This is the dipolar surfactant distribution. Assuming $k{a}^2+2D_s>0$, its consistency condition isDiffusion or exchange must restore uniform concentration faster than the interfacial flow redistributes it: this holds with a small surface Péclet number, fast exchange, or sufficiently immobilized surface velocity. These same conditions justify neglecting the perturbation's advection. Small capillary number and sufficiently small buoyancy-induced shape distortion are separate requirements for a spherical interface.

Take $+$ to mean the outer liquid, $-$ the bubble interior and $n$ outward from the bubble. The general interfacial stress balance with variable surface tension isFor $\gamma ={\gamma }_0-{\gamma }_1{C}^{\prime }$, its tangential part is $-{\nabla }_s\gamma ={\gamma }_{1}B{I}_{s}U/a$. Thus the requested form isThis positive parameter measures the strength of the restoring Marangoni stress relative to viscous stress, including the relaxation of concentration by diffusion and exchange.

The far velocity $-U$ is produced by the constant harmonic vector potential $\Phi =U$. The two independent decaying degree-one modes are the force-carrying vector potential $\alpha aU/r$ and the scalar dipole $\chi =\beta a^{3}U\cdot \nabla (1/r)$. They are harmonic outside the bubble and have precisely the symmetry of uniform translation. The Unscaled Papkovich–Neuber representation gives, at arbitrary radius,The velocity at $r=a$ agrees with the printed expression. However, the original PDF's traction formula has a factor-of-two error. Direct differentiation with $\sigma =-pI+\mu (\nabla u+\nabla u^T)$ givesnot the printed prefactor $12\mu /a$. This is traction of translating-sphere Papkovich–Neuber potentials. The $\alpha$ term follows immediately from the Stokeslet stress, and the scalar-potential part gives $6\mu \beta [U-3(U\cdot n)n]/a$; neither contribution contains the extra factor two.

With this consistent traction, no penetration requires $\beta -\alpha =1/2$, tangential velocity requires $A=-(1+\alpha +\beta )$, and tangential stress balance requires $\beta =A\lambda$. Solving,For comparison, taking the printed $12\mu /a$ literally would instead giveThose solve the supplied algebraic pair but not the Newtonian stress calculated from the supplied potentials. The factor-six result is the physically consistent translating surfactant-coated bubble solution.

As $\lambda \to 0$, $A\to -1/2$ and $\beta \to 0$: the clean bubble has a mobile interface and zero tangential traction. As $\lambda \to \infty$, $A\to 0$, $\alpha \to -3/4$ and $\beta \to -1/4$: the interface becomes immobilized and the exterior flow approaches that past a no-slip rigid sphere. The limiting tangential traction is $-3\mu I_{s}U/(2a)$, a useful check on the correct normalization.

The normal dynamic traction isapart from an arbitrary constant pressure. In the normal stress balance on a translating bubble, its degree-one part is balanced by the buoyant hydrostatic pressure difference and by the variation $-2{\gamma }_1{C}^{\prime }/a$ of normal capillary stress. A uniform internal gas pressure balances the constant Laplace-pressure term but cannot by itself balance this angular variation. If $U$ is upwards and $\Delta \rho$ is liquid minus bubble mass density, the degree-one balance can be writtenIt fixes $U=\Delta \rho g{a}^2(1+2\lambda )/[3\mu (1+3\lambda )]$, tending to the clean-bubble and rigid-sphere rise speeds. A degree-one radial displacement translates a sphere rather than changing its curvature, so a supposed dipolar shape distortion cannot replace this force balance.

Use the $L^2(S^2)$ data norm. The coefficients of the measurement error are $e_n^m=b_n^m-a_n^m$, so Parseval's identity givesAn unregularized continuation of these measurements to $r=R$ replaces $a_n^m$ by $b_n^m$ in the outgoing spherical harmonic expansion. Whenever this continuation has a finite trace, its error isChoose a measurement error supported on a single normalized spherical harmonic, with $e_N^0=\delta$ and all other coefficients zero. The data error has norm exactly $\delta$, butThis perturbation theory is a finite expansion, so it defines an actual outgoing solution of the Helmholtz equation outside any sphere of positive radius. It is not merely an example of nonconvergent formal coefficients. Taking instead $e_N^0=|d_N{|}^{-1}$ gives data errors tending to zero and trace errors of norm one, which directly proves discontinuity of inversion. General square-summable measurement noise can even violate the spherical far-field range condition, leaving no $L^2$ continuation to the chosen sphere.

For obstacle reconstruction one would continue the field toward the unknown boundary and look for a surface on which $u_i+u_s=0$, the Dirichlet boundary condition. The unbounded spherical far-to-near-field continuation shows why small far-field errors can cause arbitrarily large errors in the field used to estimate that surface. Locally, a differentiable normal displacement $\eta$ of a candidate boundary must obey the linearized condition $\Delta u_s+\eta {\partial }_n(u_i+u_s)=0$. Thus a large continued-field error, or a small normal derivative, defeats an uncontrolled boundary estimate.

There is a necessary qualification to the wording about errors in the scatterer itself. The calculation proves an unbounded field-continuation error. An arbitrary perturbed outgoing field need not correspond to any Dirichlet obstacle, and no admissible class or metric on obstacle shapes is specified. It therefore does not by itself prove an arbitrarily large geometric error in every constrained reconstruction method. Such a claim requires a shape model and an analysis of the boundary-extraction step. The instability displayed here is exactly the mechanism that an unregularized reconstruction must control.

Write $r=|\mathbf{x}|$, $\theta =\mathbf{x}/r$, and choose $R$ so that the sphere $r=R$ strictly encloses the scatterer. Normalize the spherical harmonics to be an orthonormal basis of $L^2(S^2)$, with surface measure on the unit sphere. The outgoing spherical Hankel function is denoted here by $h_n^{(1)}$, corresponding to the printed $H_n^{(1)}$. For fixed $n$, its large-argument behavior isConsequently the coefficient of $Y_n^m$ in the outgoing expansion is asymptotic to $a_n^m{e}^{ikr}/r$: the factors satisfy $i^{n+1}(-i{)}^{n+1}=1$. With the far-field pattern defined by $u_s(r\theta )=e^{ikr}{u}_{\infty }(\theta )/r+O(r^{-2})$, we obtainOne can first project onto each spherical harmonic and then take the large-$r$ limit. This avoids assuming that the fixed-order asymptotic is uniform in $n$. The Sommerfeld radiation condition selects the outgoing spherical Hankel functions rather than their incoming counterparts.

Let $g(\theta )=u_s(R\theta )$. Its coefficients are $g_n^m=d_n{a}_n^m$, where $d_n=k{i}^{n+1}{h}_n^{(1)}(kR)$. Since the field is smooth on a sphere outside the scatterer, $g\in L^2(S^2)$; Parseval's identity givesThe required coefficient condition uses fixed-argument growth of spherical Hankel functions, a different limit from the far-field limit. For fixed $t>0$,For completeness, the series of the ordinary Bessel function gives $J_{-n-1/2}(t)\sim (t/2{)}^{-n-1/2}/\Gamma (1/2-n)$. The identities $y_n(t)=(-1{)}^{n+1}\sqrt{\pi /(2t)}{J}_{-n-1/2}(t)$ and $\Gamma (1/2-n)=(-4{)}^{n}n!\sqrt{\pi }/(2n)!$ give the first formula; the corresponding positive-order series gives the estimate for $j_n$. The successive series corrections at fixed $t$ are $O(1/n)$. Applying Stirling's approximation to the factorial ratio yieldsIn particular there are positive constants giving upper and lower bounds by the last weight for all sufficiently large $n$. The finite trace norm is therefore equivalent to the spherical far-field range conditionwhere the $n=0$ weight is defined to be one. The same conclusion holds at every larger sphere on which the outgoing field is defined.

The printed comparison with $\infty$ must mean a finite sum: a literal non-strict inequality allows a divergent sum. Also, the supplied one-sided big-$O$ bound alone does not imply this necessary condition from a finite trace norm; a lower bound is needed in that direction. The two-sided asymptotic above supplies it. Conversely, its upper bound makes the weighted condition sufficient for an $L^2$ trace at the chosen sphere. This is a condition on exterior-field continuation, not a sufficient condition that the coefficients come from a particular Dirichlet obstacle.

The Time-independent Schrödinger equation is $[-{\hslash }^2{\nabla }^2/(2m)+V]\Psi =E\Psi$. For the hydrogen atom, separate variables as $\Psi =g(r)Y_{\ell m}(\theta ,\phi )$. The spherical-coordinate Laplacian has radial part $r^{-2}{\partial }_r(r^2{\partial }_r)$ and angular part $r^{-2}{\Delta }_{S^2}$. A spherical harmonic obeys ${\Delta }_{S^2}{Y}_{\ell m}=-\ell (\ell +1)Y_{\ell m}$. The first displayed term is therefore radial kinetic energy; the second is the attractive Coulomb's law potential energy $-e^2/(4\pi {\epsilon }_{0}r)$ of electron and proton; the third is the angular kinetic energy, also called the centrifugal potential. Regular single-valued angular states have $\boxed{\ell =0,1,2,\dots }$, with orbital angular momentum squared ${\hslash }^2\ell (\ell +1)$.

For $g=r^{\alpha }{e}^{-\beta r}$, direct differentiation givesSubstitution into the Radial Schrodinger equation for the hydrogen atom and equality of the coefficients of $r^{-2}$, $r^{-1}$ and the constant term giveThe regular radial solution selects $\alpha =\ell$, rather than the singular branch $-\ell -1$. At $\ell =0$, $r^{-1}$ is locally square integrable but is still inadmissible as a regular Coulomb eigenfunction: its Laplacian has a point-source contribution at the origin. Bound-state decay requires $\beta >0$. Thus, with the Bohr radius $a_0=4\pi {\epsilon }_0{\hslash }^2/(m{e}^2)$,These are the nodeless states with principal quantum number $n=\ell +1$. Increasing $\ell$ raises the energy towards zero, so emission is from the state labelled $\ell +1$ to the one labelled $\ell$. Using the Planck constant $h=2\pi \hslash$ and photon energy $h\nu$, the frequency is

Use the nonnegative Laplacian convention $\tilde{\Delta }=-{\sum }_{j=1}^{d+1}{\partial }_j^2$ and $\Delta =-\mathrm{div}\mathrm{grad}$ on the unit sphere. In polar coordinates the Euclidean Riemannian metric is $d{r}^2+r^2{g}_{S^d}$, its volume form is $r^{d}drd{V}_{S^d}$, and therefore the Laplacian in polar coordinates isConsequently the required restriction identity isFor the opposite sign convention all eigenvalues below change sign, and the restriction identity changes accordingly.

Let ${\mathcal{P}}_n$ be the vector space of degree-$n$ homogeneous polynomials in $d+1$ variables, and let ${\mathcal{H}}_n$ be its subspace of harmonic polynomials. Write $H(r\omega )=r^{n}Y(\omega )$ for $H\in {\mathcal{H}}_n$. Substitution in the Laplacian identity givesThus each restricted harmonic polynomial is a spherical harmonic with eigenvalue ${\lambda }_n=n(n+d-1)$.

To count these spherical harmonics, give polynomials the Fischer inner product $⟨x^{\alpha },x^{\beta }{⟩}_F=\alpha !{\delta }_{\alpha \beta }$. Multiplication by $x_j$ is adjoint to ${\partial }_j$, so multiplication by $r^2$ is adjoint to the ordinary signed Laplacian ${\sum }_j{\partial }_j^2$. In these finite-dimensional inner product spaces, the orthogonal complement of $r^2{\mathcal{P}}_{n-2}$ is exactly ${\mathcal{H}}_n$. Hence the Fischer decomposition isMultiplication by $r^2$ is injective, while restriction of a homogeneous polynomial to the sphere is injective: homogeneity recovers its value at every nonzero point. Counting monomials now gives the eigenvalue multiplicitywhere the second term is zero for $n<2$.

It remains to establish that these are all the eigenvalues and their full multiplicities. Iterating the Fischer decomposition expresses every restricted polynomial as a sum of the restricted harmonic polynomials. Restricted polynomials separate points and contain the constants, so the Stone-Weierstrass theorem makes their span dense in continuous functions on the sphere, hence dense in its L2 space. The Laplace-Beltrami operator is self-adjoint, and the distinct numbers ${\lambda }_n$ strictly increase for $d\ge 1$. Its different eigenspaces are therefore orthogonal. Any additional eigenfunction, or any additional vector in one of these eigenspaces orthogonal to its known spherical harmonics, would be orthogonal to a dense subspace and hence zero. The standard compact elliptic spectral theorem then shows that the spectrum of the round sphere is precisely ${\lambda }_n=n(n+d-1)$ with multiplicity $m_n$. For $S^1$ this gives multiplicity one at zero and two at $n^2$ for $n\ge 1$. If the degenerate case $d=0$ is included, $S^0$ has two points and its Laplacian is zero on a two-dimensional function space.

Hawking radiation is a prediction of quantum field theory in curved spacetime, not of classical collapse alone. Consider a free massless Klein-Gordon field in the collapse vacuum for Hawking radiation: initially no incoming particles from past null infinity, with the short-distance state regular for freely falling observers through the forming event horizon. At late times use the approximately fixed exterior Schwarzschild metric; neglect backreaction during the observation interval. First set $G=c=\hslash =k_B=1$.

The exterior metric has $f(r)=1-2M/r$. Introduce the tortoise coordinate and outgoing and ingoing null coordinates,The surface gravity is $\kappa =1/(4M)$. The logarithm in $r_{\ast }$ makes a late escaping ray spend a long exterior coordinate time close to the event horizon. One of the regular outgoing Kruskal–Szekeres coordinates is proportional to $-e^{-\kappa u}$. Smooth propagation backwards through the collapsing body relates this regular coordinate to the incoming coordinate on past null infinity. If $v_H$ labels the last ray able to escape, the late-time Hawking exponential ray map is consequentlyHere the symbol $u(v)$ is the outgoing retarded time evaluated on the ray; the first equation is the leading asymptotic relation, with subleading corrections that vanish at late times. The constant $C$ depends on collapse history, whereas the exponential rate is determined by the final surface gravity. The relation follows from a nondegenerate Killing horizon; a linear ray map would not have the same effect.

An outgoing positive-frequency mode at future null infinity behaves as $p_{\omega }\sim e^{-i\omega u}/\sqrt{4\pi \omega }$, where $\omega >0$ is measured with respect to the asymptotic stationary time. In the near-horizon geometric optics approximation its backward-propagated form on past null infinity isThis is not purely positive frequency in $v$. Its decomposition into incoming positive-frequency modes $f_{{\omega }^{\prime }}\sim e^{-i{\omega }^{\prime }v}/\sqrt{4\pi {\omega }^{\prime }}$ and their complex conjugates is a Bogoliubov transformation:The coefficients follow from the Klein-Gordon inner product, or equivalently from the Fourier transform with the indicated flux normalization. Writing $y=v_H-v$ and $a=\omega /\kappa$, the two Fourier integrals, after a convergence regulator, areThe equality is the defining integral of the Gamma function, continued to these arguments from the right half-plane. Up to phases of unit modulus, $\alpha$ is $\sqrt{{\omega }^{\prime }/\omega }{I}_{+}/(2\pi )$ and $\beta$ is $\sqrt{{\omega }^{\prime }/\omega }{I}_{-}/(2\pi )$. Taking $ϵ\downarrow 0$ on the principal branches, the arguments of $ϵ\pm i{\omega }^{\prime }$ tend to $\pm \pi /2$. Thus $|I_{+}|\propto e^{\pi a/2}$ and $|I_{-}|\propto e^{-\pi a/2}$, giving the thermal ratio of Hawking Bogoliubov coefficientsIn particular, using $|\Gamma (1+ia){|}^2=\pi a/\sinh (\pi a)$,The collapse-dependent phases and $C$ disappear from the ratio. The canonical identities for a bosonic Bogoliubov transformation state that the positive-frequency norm minus the negative-frequency norm is one for a normalized outgoing mode. Combining that identity with the thermal ratio yields the late-time mean occupationThese are the Bose-Einstein distribution and the Hawking temperature. Strictly, monochromatic modes have delta-function normalization, and $\int d{\omega }^{\prime }|{\beta }_{\omega {\omega }^{\prime }}{|}^2$ contains the infinite-duration divergence $\int d{\omega }^{\prime }/{\omega }^{\prime }$. Normalized wave packets with a finite frequency band and a finite retarded-time window remove this artifact. For sufficiently late packets the occupation is the frequency average of the displayed thermal factor, tending to it for a narrow band. Fermionic fields instead give the corresponding denominator $e^{2\pi \omega /\kappa }+1$ because their canonical normalization uses anticommutators.