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For an outward sightline, Fourier velocity $v_b^i=i{k}^i{\theta }_b$, and sudden last scattering, intrinsic photon temperature projects with a Spherical Bessel function, while the Doppler term projects with its derivative. Density and velocity acoustic sources are approximately a quarter cycle out of phase. Their amplitude is suppressed by photon diffusion, giving a power damping envelope $e^{-2k^2/k_D^2}$. Projection maps the sound scale to peak spacing approximately $\pi D_{\ast }/r_s$ and the damping scale to angular multipoles of order $k_D{D}_{\ast }$.

For a spherical square well of radius $a$, depth $V_0$ and reduced mass $\mu$, the regular zero-orbital angular momentum solution has reduced radial wavefunction $U=A\sin kr$ inside and $U=B{e}^{-qr}$ outside. Wavefunction matching at a finite potential step gives $k\cot (ka)=-q$, where $k^2=2\mu (V_0+E)/{\hslash }^2$ and $q^2=-2\mu E/{\hslash }^2$. The $n$th radial branch first reaches $E=0$ at $ka=(n+1/2)\pi$, for $n=0,1,\dots$. A normalizable negative-energy state on that branch requires $V_0>V_{s,n}$. Thus exactly one such state exists for $V_{s,0}<V_0\le V_{s,1}$; the new zero-energy $s$-wave solution at the upper equality is not normalizable.

For the first $\ell =1$ branch, the zero-energy exterior solution has $U\propto r^{-1}$. The regular interior solution is $U\propto r{j}_1(kr)$, where $j_1$ is a Spherical Bessel function. Matching logarithmic derivatives gives $1+z{j}_1^{\prime }(z)/j_1(z)=-1$, or $j_0(z)=0$, with $z=a\sqrt{2\mu V_0}/\hslash$. Its first threshold is $z=\pi$, so $V_{p,0}={\pi }^2{\hslash }^2/(2\mu a^2)$. Higher values of orbital angular momentum increase the positive centrifugal potential, so none can bind below this first $p$-wave threshold. Spin-dependent nuclear forces are absent from this central-well model.

For $x^2{y}^{\prime \prime }+2x{y}^{\prime }+[x^2-\ell (\ell +1)]y=0$ with integer $\ell \ge 0$, the Frobenius method gives $\sigma =\ell$ or $-\ell -1$. The general coefficient equation is $[(j+\sigma )(j+\sigma +1)-\ell (\ell +1)]a_j=-a_{j-2}$, taking negative-index coefficients as zero. The regular branch has $a_1=0$ and $a_j=-a_{j-2}/[j(j+2\ell +1)]$. The singular branch has $a_j=-a_{j-2}/[j(j-2\ell -1)]$ except at $j=2\ell +1$. That resonant equation is compatible because its earlier odd coefficient vanishes; the newly free coefficient adds a multiple of the regular solution. Setting it to zero gives a pure even series times $x^{-\ell -1}$, independent of the regular branch, with no logarithmic term. For $\ell =0$ the resonance occurs already at $j=1$.

The normalization $a_0=2^{\ell }\ell !/(2\ell +1)!$ on the regular branch gives the Spherical Bessel functionIn particular $j_0(x)=\sin x/x$, with removable value one at zero.

For the Fourier transform convention ${\delta }_{\mathbf{k}}=\int d^{3}x\delta (\mathbf{x})e^{i\mathbf{k}\cdot \mathbf{x}}$ and covariance $⟨{\delta }_{\mathbf{k}}{\delta }_{{\mathbf{k}}^{\prime }}⟩=(2\pi {)}^{3}P(k){\delta }_D(\mathbf{k}+{\mathbf{k}}^{\prime })$, inversion gives $\xi (\mathbf{r})=\int d^{3}kP(k)e^{i\mathbf{k}\cdot \mathbf{r}}/(2\pi {)}^3$. Isotropy reduces the angular integral to $4\pi j_0(kr)$, where $j_0(x)=\sin x/x$ is the order-zero Spherical Bessel function. At zero separation this gives the variance if the spectrum integral converges. A nonnegative spectrum can produce negative correlations away from the origin.

For fields projected along the line of sight, the Limber approximation replaces the oscillatory radial Spherical Bessel function coupling by a local radial contribution with $k\chi \simeq \ell +1/2$. The displayed cross-power spectrum formula applies to $u(\hat{\mathbf{n}})=\int d\chi q_u(\chi )U(\chi \hat{\mathbf{n}},z)$ and similarly for $v$, in a spatially flat background. It requires radial kernels and the three-dimensional cross-power spectrum to vary slowly across the radial coupling width. It suppresses correlations of disjoint radial kernels but can fail at low multipoles or for very narrow selections; exact unequal-time correlations can remain nonzero there.

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 $16/15$.

With the specified Fourier transform convention, the inverse is $\delta (\mathbf{x})=\int d^{3}k{\delta }_{\mathbf{k}}{e}^{-i\mathbf{k}\cdot \mathbf{x}}/(2\pi {)}^3$. Inserting both inverse transforms into the cosmological two-point correlation function and applying the given Dirac delta covariance yieldsFor an isotropic matter power spectrum, align the polar axis with $\mathbf{r}$. The angular integral is $2\pi {\int }_{-1}^1{e}^{ikr\mu }d\mu =4\pi \sin (kr)/(kr)$. Hence the isotropic cosmological correlation-power-spectrum relation isThe angular kernel is the order-zero Spherical Bessel function $j_0(kr)$. At $r=0$, replace it by its limit one, giving the unsmoothed variance when the integral converges. The argument assumes an ordinary integrable spectrum, or an explicitly specified distributional Fourier transform interpretation otherwise.

Let $q=\mathbf{k}\cdot \hat{\mathbf{n}}$ and retain the negative-spatial-perturbation convention of the preceding part. Since $h=h_S$, the isotropic pieces of the supplied Fourier mode contraction cancel:For $h=A(\mathbf{k}){\tau }^2{k}^2$, the factor $1/2$ in the line-of-sight integral cancels the derivative's factor two. ThusAn antiderivative convenient even at $q=0$ isIt follows directly, or by two integrations by parts, thatThe source sign and both endpoint coefficients are fixed by the metric convention; no division by $q$ is needed.

To interpret the large-angle Cosmic microwave background anisotropy, restore the observer-centered propagation phase $e^{-iq{\tau }_0}$, which can equivalently be absorbed into the definition of each Fourier mode amplitude. Put $D={\tau }_0-{\tau }_{\mathrm{dec}}$. One mode contributesThe first term is a local angular monopole plus a dipole, so it is removed when discussing observed angular multipoles $\ell \ge 2$. For modes outside the horizon at decoupling, $k{\tau }_{\mathrm{dec}}\ll 1$, the higher-multipole emission signal is dominated byIts transfer amplitude has no leading positive power of $k$: the two explicit powers in $h$ have canceled in the integral. Projection onto the sky supplies Spherical Bessel functions $j_{\ell }(kD)$, with modes of order $kD\sim \ell$ contributing to each angular scale. This is the large-angle Sachs-Wolfe effect; the emission velocity-like correction is relatively of order $k{\tau }_{\mathrm{dec}}$. For wavelengths much larger than the entire observed region, $kD\ll 1$, removing the monopole and dipole leaves a signal starting at order $A(kD{)}^2$, rather than a finite observable anisotropy from a spatially uniform mode.

The stochastic scale dependence of $A(\mathbf{k})$ is not specified by the deterministic metric solution. If one additionally assumes a scale-invariant primordial spectrum, $k^3{P}_A(k)=\text{constant}$, thenfor $\ell \ge 1$, giving the familiar large-angle Sachs-Wolfe plateau, $\ell (\ell +1)C_{\ell }\simeq \text{constant}$, for the measured $\ell \ge 2$ multipoles. A tilted primordial spectrum changes that trend. The full numerical temperature transfer also includes intrinsic last-scattering and Doppler terms, omitted from the stipulated metric-only integral.

Let $⟨{\Phi }_{\mathbf{k}}{\Phi }_{{\mathbf{k}}^{\prime }}^{\ast }⟩=(2\pi {)}^3{\delta }^3(\mathbf{k}-{\mathbf{k}}^{\prime })P_{\Phi }(k)$. Statistical isotropy and $\int |Y_{\ell m}{|}^{2}d\Omega =1$ applied to the preceding expression give the CMB angular power spectrumFor $n=1$, define the constant dimensionless potential power ${\mathcal{A}}_{\Phi }=k^3{P}_{\Phi }(k)/(2{\pi }^2)$. ThenThe needed logarithmic spherical Bessel square integral can be proved directly. Put $I_{\ell }={\int }_0^{\infty }{j}_{\ell }^{2}dx/x$ and $J_{\ell }={\int }_0^{\infty }{j}_{\ell }{j}_{\ell -1}dx$. From the Spherical Bessel function recurrences,integration of the squares' derivatives gives $J_{\ell }=(\ell +1)I_{\ell }$, and for $\ell >1$ also $J_{\ell }=(\ell -1)I_{\ell -1}$. At $\ell =1$, $j_0(0)=1$ and $j_0(\infty )=0$ give $J_1=1/2$, hence $I_1=1/4$. Induction now proves $I_{\ell }=1/[2\ell (\ell +1)]$. ConsequentlyThis is the flat large-angle Sachs-Wolfe plateau. The useful integral printed in the PDF is missing the square on $j_{\ell }$: without that square the formula is false, since at $\ell =1$ its left-hand side is $\pi /4$, whereas its asserted value would be $1/4$. The squared identity above is the one required by the variance calculation. The plateau describes the matter-era large-angle approximation, not acoustic peaks or late integrated-potential corrections.

Vary the action with compactly supported variations and integrate by parts. Its overall normalization does not affect the equation of motion:Here $\sqrt{-g}=e^{3t}$, so the massless scalar field obeysSeparate a spatial Fourier mode as $\varphi (t,\mathbf{x})=F_k(t)e^{i\mathbf{k}\cdot \mathbf{x}}$. Its mode equation is $F_{k,tt}+3F_{k,t}+k^2{e}^{-2t}{F}_k=0$. With conformal time $\tau =-e^{-t}$, $a=-1/\tau$ and $\mathcal{H}=-1/\tau$, it becomesFor $k>0$, put $z=-k\tau =k{e}^{-t}$ and $F_k=z^{2}G(z)$. Direct differentiation reduces the equation tothe Spherical Bessel function equation with order one. Hence the complete classical mode basis isSince $z^2{j}_1(z)=\sin z-z\cos z$ and $z^2{y}_1(z)=-\cos z-z\sin z$, an equivalent complex basis isThe classical action does not select one frequency branch or a quantum vacuum. A real field has conjugate-related coefficients at opposite wavevectors. The zero mode, which is not obtained by treating these two $k>0$ basis functions as independent at $k=0$, isAt late times one independent solution freezes and the other decays as $e^{-3t}$. These are minimally coupled massless modes; the supplied action has no curvature-coupling term, so it should not be replaced by the conformally coupled wave equation.

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.

To illustrate Tikhonov regularization precisely, take the unknown to be the trace $g$ on a chosen enclosing sphere of radius $R$. Use Hilbert spaces $X=Y=L^2(S^2)$ and the forward linear operatorAt real positive $kR$, $d_n\ne 0$, because the real Spherical Bessel functions $j_n,y_n$ have a nonzero Wronskian and cannot vanish simultaneously. Moreover $|d_n|\to \infty$. Thus $A$ is an injective compact operator: truncating to finitely many harmonic orders approximates it in operator norm, since the remaining diagonal entries tend uniformly to zero. Its inverse is unbounded. This is a linear field reconstruction; estimating the obstacle from the reconstructed field remains a further nonlinear step.

Given coefficients $b_n^m$ of measured data $y^{\delta }$, minimize the quadratic functionalVarying $g$ in every direction gives $(A^{\ast }A+\alpha I)g_{\alpha }^{\delta }=A^{\ast }{y}^{\delta }$. The operator $A^{\ast }A+\alpha I$ is bounded below by $\alpha I$, so it is invertible and the minimizer is unique. In coefficients the minimization is separable: $A^{\ast }y$ has coefficient $b_n^m/\overline{d_n}$ and $A^{\ast }A$ has multiplier $|d_n{|}^{-2}$. Hence the Tikhonov regularization of spherical far-field continuation isThe implied regularized far-field coefficients are $b_n^m/(1+\alpha |d_n{|}^2)$. This suppresses precisely the high-order coefficients that make backward continuation unstable.

Let $g^{†}$ be the exact trace and $y=A{g}^{†}$, with $\Vert y^{\delta }-y\Vert \le \delta$. The noise-bias decomposition for linear regularization isFor a singular value $s=|d_n{|}^{-1}$, the noise multiplier is $s/(s^2+\alpha )$. Since $s^2+\alpha \ge 2s\sqrt{\alpha }$, its supremum is at most $1/(2\sqrt{\alpha })$. ConsequentlyThis Tikhonov stability bound measures the trace norm, not a distance between obstacle boundaries. The bias coefficient is $\alpha /(|d_n{|}^{-2}+\alpha )$ times the exact coefficient. It tends to zero for each fixed $n,m$, is bounded by one, and $\sum |g_n^{†,m}{|}^2<\infty$. Splitting off a finite sum and bounding its tail therefore proves $B(\alpha )\to 0$ as $\alpha \to 0$.

A convergent regularization parameter choice thus satisfiesIn nondimensionalized variables, $\alpha ={\delta }^p$ with $0<p<2$ is one such rule. Without a smoothness assumption on the exact trace there is no universal bias rate and no universal optimal value of $\alpha$. For a concrete normal-operator source condition, suppose $g^{†}=A^{\ast }Aw$ and $\Vert w\Vert \le C$. Each bias multiplier on $w$ is $\alpha s^2/(s^2+\alpha )\le \alpha$, so $B(\alpha )\le C\alpha$. Minimizing the resulting bound gives, for $C>0$,This optimizes the displayed worst-case bound under that additional source condition, rather than guaranteeing an exact optimum for every datum.

If only the effect of measurement error is minimized, the multiplier bound decreases when $\alpha$ increases. Indeed $\Vert (A^{\ast }A+\alpha I{)}^{-1}{A}^{\ast }\Vert \le \Vert A\Vert /\alpha \to 0$ as $\alpha \to \infty$. But the reconstruction then approaches zero and its bias approaches $\Vert g^{†}\Vert$. Suppressing noise alone does not optimize reconstruction: $\alpha$ must balance noise amplification against regularization bias. A geometrically accurate obstacle estimate additionally needs justified shape constraints and a stable procedure for extracting a boundary from the regularized field.

A blackbody photon density perturbation gives an intrinsic temperature fluctuation ${\Theta }_0={\delta }_{\gamma }/4$. The baryon/electron bulk velocity produces a Doppler effect at last scattering. In the common-potential convention of part (ii),Ignoring the slowly varying damping derivative at leading acoustic order, differentiating the density cosine produces a velocity sine. Thus density and velocity acoustic sources are approximately a quarter cycle out of phase. For regular adiabatic initial conditions the density oscillation has approximately a common cosine phase across modes; this phase coherence is what makes a sequence of peaks instead of a smooth incoherent superposition.

Let $D_{\ast }={\tau }_0-{\tau }_{\ast }$ be the comoving distance to last scattering and $r_s({\tau }_{\ast })={\int }^{{\tau }_{\ast }}{c}_{s}d\tau$ its sound horizon. In the present limit $r_s\simeq {\tau }_{\ast }/\sqrt{3}$. Density extrema have $k{r}_s\simeq m\pi$, whereas velocity extrema have $k{r}_s\simeq (m+1/2)\pi$. The projection from spatial wavenumber to high angular multipoles is roughly $\ell \simeq k{D}_{\ast }$, giving an acoustic spacingThe Doppler effect fills some density troughs and modifies the detailed peak shapes; its different projection prevents treating it as an identical second series of narrow peaks.

For clarity, choose the sightline to point outward from the observer and let its angle to $\mathbf{k}$ have cosine $\mu$. The intrinsic-plus-emitter-Doppler source is then $[{\delta }_{\gamma }/4-ik\mu {\theta }_b]e^{ik\mu D_{\ast }}$. Expanding its angular dependence gives the angular projection of density and Doppler acoustic sourcesThe second sign follows from the chosen outward sightline and $v_b^i=i{k}^i{\theta }_b$; it changes with the sightline convention together with the velocity projection. The $j_{\ell }$ are Spherical Bessel functions. Both sources are proportional to the primordial curvature. If ${\mathcal{T}}_{\ell }={\Theta }_{\ell }/{\zeta }_{\ast }$, their angular spectrum in the usual continuum Fourier normalization isThis formula includes the density-velocity cross term, not just a sum of independent source powers.

Photon diffusion multiplies both leading source amplitudes by $e^{-k^2/k_D^2}$, so their power carries $e^{-2k^2/k_D^2}$. The approximate angular damping scale isThis produces the damping tail in the Cosmic microwave background power spectrum. Finite last-scattering thickness adds further small-scale averaging. Gravity, baryon loading and later propagation alter detailed peak locations and heights, but the oscillatory peak sequence, phase-shifted velocity contribution and diffusion damping follow directly from the simplified solution obtained here.

For a single barotropic perfect fluid in general relativity, the perturbations obey $\delta P=w\delta \rho =w\bar{\rho }\delta$. This adiabatic closure is needed: constant background $P/\rho$ alone would not eliminate an independent entropy perturbation. The absence of scalar anisotropic stress allows the common potential used in Newtonian gauge in cosmology.

Substituting the density constraint into the pressure equation gives the gravitational potential evolution of a barotropic fluidSince $a\propto {\tau }^{2/(1+3w)}$, the conformal Hubble parameter is $\mathcal{H}=2/[(1+3w)\tau ]$. The bracket cancels identically, leavingFor a Fourier transform mode, ${\nabla }^2$ becomes $-k^2$.

During radiation domination, $w=1/3$ and ${\Phi }_k^{\prime \prime }+4{\Phi }_k^{\prime }/\tau +(k^2/3){\Phi }_k=0$. Set $x=k\tau /\sqrt{3}$ and write $\Phi =u/x$. The resulting equation is $u_{xx}+2u_x/x+(1-2/x^2)u=0$, so the two Spherical Bessel functions in the hint giveAt $x\ll 1$, the two solutions approach a constant and a mode proportional to ${\tau }^{-3}$. The regular adiabatic mode, normalized to its primordial potential, isAfter entry into the sound horizon, $x\gg 1$, the potential oscillates at cosmological sound speed $1/\sqrt{3}$ with envelope $x^{-2}\propto a^{-2}$. The Hubble radius and sound horizon differ by the sound-speed factor; outside the Hubble radius the regular potential is constant, while well inside it radiation supports acoustic oscillations.

During matter domination, $w=0$ gives ${\Phi }_k^{\prime \prime }+6{\Phi }_k^{\prime }/\tau =0$ at every wavenumber. HenceThe growing density mode has a constant potential both outside and inside the Hubble radius; the other potential mode decays. Pressureless matter has zero cosmological sound speed, so horizon entry does not produce the radiation acoustic decay. These formulas cover both independent solutions, while the subsequent sketches select the regular adiabatic growing mode.

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.

Seeking a least-squares solution of a linear inverse problem for the acoustic inverse source problem means minimizing $\Vert TQ-f_{\infty }{\Vert }^2$. Its normal equation for a linear inverse problem isFor compatible data, the formal minimum-norm solution can be written with the Moore–Penrose inverse of an operator:The inverse in this expression is defined on the range of the restricted operator, not as a bounded everywhere-defined inverse. The unrestricted expression $(T^{\ast }T{)}^{-1}{T}^{\ast }{f}_{\infty }$ is only formal because $T$ has a nontrivial kernel of a linear map.

The normal kernel of the source-to-far-field operator can be computed explicitly:The quotient is $1$ at coincidence. This smooth kernel already exhibits the loss of information.

There are two distinct causes of ill-posedness. First, a fixed-frequency nonradiating source gives exact nonuniqueness. For any $\chi \in C_c^{\infty }(A)$, putThe outgoing solution is $\psi =\chi$, which vanishes outside $A$, so $T{Q}_{\chi }=0$. Equivalently, integration by parts in the source-to-far-field operator at fixed frequency givessince the exponential solves the homogeneous Helmholtz equation and the boundary terms vanish. This provides an infinite-dimensional family of invisible sources. Single-frequency data determine only the spherical restriction of a three-dimensional Fourier transform.

Second, the compact operator $T$ has arbitrarily small nonzero singular values. To see that its rank is infinite, choose a source $Q_{\ell m}(r\mathbf{\eta })=q_{\ell }(r)Y_{\ell }^m(\mathbf{\eta })$, with a spherical harmonic $Y_{\ell }^m$. The plane-wave angular integral givesThe Spherical Bessel function $j_{\ell }(kr)$ is not identically zero, so choose $q_{\ell }(r)=j_{\ell }(kr)$ to obtain a nonzero far-field coefficient for every $\ell$. The mutually orthogonal angular modes prove infinite rank. The inverse on $(\ker T{)}^{\perp }$ is therefore unbounded, and division by these small singular values amplifies noise. Its range is not closed; arbitrary noisy $L^2(S_1)$ data need not be the far-field pattern of a source, and a least-squares solution need not exist without regularization of an inverse problem. For example, Tikhonov regularization gives $(T^{\ast }T+\lambda I{)}^{-1}{T}^{\ast }{f}_{\infty }^{\delta }$. It stabilizes the recovered visible component but cannot identify the missing fixed-frequency nonradiating source component.

Write $S_{\ast }(\mathbf{x})=({\Theta }_0+\psi )({\eta }_{\ast },\mathbf{x})$ and $E=e^{-{\tau }_{\mathrm{re}}}$. Under the stated matter-era growing-mode approximation, $\dot{\varphi }+\dot{\psi }=0$, and ignoring velocities removes the Doppler CMB anisotropy. The two peaks of the cosmological visibility function give the Cosmic microwave background line-of-sight solutionApply the same solution with observation time ${\eta }_{\mathrm{re}}$, just before the thin reionization screen. Between recombination and that screen there is no further scattering, so free propagation givesA direction-independent potential commutes with the angular average. Thus the incoming photon monopole plus the potential at the screen isSubstitution givesThis derivation inherits the neglected scattering quadrupole and polarization assumptions of part (ii); the screen is an idealized isotropizing approximation.

For a plane Fourier mode $S_{\ast }(\mathbf{x})=S_{\mathbf{k}}{e}^{i\mathbf{k}\cdot \mathbf{x}}$, the angular average is explicit:where $j_0$ is the zeroth Spherical Bessel function. The mode's observed transfer isIf $k\Delta \chi \gg 1$, the rescattered monopole is suppressed by at least $1/(k\Delta \chi )$, because the incoming directions sample incoherent phases. If $k\Delta \chi \ll 1$, $j_0=1+O((k\Delta \chi {)}^2)$ and the two propagation phases agree to leading order. The two weights then sum to one. ConsequentlyThese are the small-scale damping and large-scale coherence limits of reionization damping of cosmic microwave background temperature anisotropy.

Each small-scale primary temperature amplitude is multiplied by $e^{-{\tau }_{\mathrm{re}}}$, so its Cosmic microwave background power spectrum is multiplied by $e^{-2{\tau }_{\mathrm{re}}}$. Holding transfer parameters fixed,Many measured high-$\ell$ modes determine this combination accurately. Their derivatives $\partial \log C_{\ell }/\partial \log A_s=1$ and $\partial \log C_{\ell }/\partial {\tau }_{\mathrm{re}}=-2$ are degenerate: a change $d\log A_s=2d{\tau }_{\mathrm{re}}$ leaves the leading temperature spectrum unchanged. The undamped large-scale modes can distinguish the parameters, but there are few of them and cosmic variance gives fractional full-sky uncertainty $\sqrt{2/(2\ell +1)}$ per multipole. This explains the primordial-amplitude–optical-depth degeneracy. Polarization from reionization, lensing, and more complete late-time effects partly break it. The TeX's final ${\tau }_e$ is a transcription mismatch; the PDF uses ${\tau }_{\mathrm{re}}$ consistently.

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 $K(0)=1/15$.