Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 64 (Advanced Cosmology)

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the induced metric and lower the shift vector with it, $N_i=h_{ij}{N}^j$. In the stated negative-shift convention the four-dimensional components and inverse components areMultiplying the given future unit normal by this metric tensor gives $n_{\mu }=(-N,0,0,0)$. In particular the tangential normal components vanish, but their covariant derivatives do not:The relevant Christoffel symbol isFor the second equality, the last parenthesis is $2h_{k\ell }{}^{(3)}{\Gamma }^{\ell }{}_{ij}$, so its contraction subtracts the two spatial connection terms in the symmetrized derivative of $N_i$. ThereforeThis is extrinsic curvature with a negative shift. Both the future-normal orientation and the definition $K=-\nabla n$ matter: replacing the threading by $d{x}^i+N^{i}dt$ reverses the shift terms. The spatial induced metric is generally a function of time as well as position; restricting it to time-independent data would simply make the displayed time derivative vanish.

Retain the homogeneous lapse function $\bar{N}(t)$, and define the physical Hubble parameter $H=\dot{a}/(\bar{N}a)$. At first order the scalar metric perturbations give $\delta g_{00}=-2{\bar{N}}^2\Phi$, $\delta g_{0i}=a^2{B}_{,i}$, and $\delta g_{ij}=-2a^2(\Psi {\delta }_{ij}+E_{,ij})$. Applying the passive coordinate displacement component by component gives the scalar gauge transformations with a background lapseFor example the $0i$ transformation is $\delta {\tilde{g}}_{0i}=a^2{\partial }_{i}B-a^2{\partial }_i\dot{\lambda }+{\bar{N}}^2{\partial }_i{\xi }^0$. The spatial isotropic part acquires $-2a\dot{a}{\xi }^0{\delta }_{ij}$, explaining the positive sign in the transformation of $\Psi$.

Preserving synchronous gauge requires both transformed $\Phi$ and $B$ to remain zero. The first condition is ${\partial }_t(\bar{N}{\xi }^0)=0$, and the second is $\dot{\lambda }={\bar{N}}^2{\xi }^0/a^2$. Integrating them givesThus fixing the lapse function and shift vector perturbations does not exhaust the scalar coordinate freedom. The functions $C,D$ describe residual synchronous-gauge freedom; an integration-origin change in the integral is absorbed into $D$.

To move from synchronous gauge to Newtonian gauge, choose $\lambda =-E_S$ to set $\tilde{E}=0$. Requiring $\tilde{B}=0$ then givesA scalar energy density transforms as $\delta {\rho }_N=\delta {\rho }_S-\dot{\bar{\rho }}{\xi }^0$. Dividing by the background energy density, which changes the contrast only at second order if perturbed in that denominator, yields the synchronous-to-Newtonian density transformation with a lapseThe last form uses background stress-energy conservation, $\dot{\bar{\rho }}=-3\bar{N}H(\bar{\rho }+\bar{P})$. With cosmic time $\bar{N}=1$ this is ${\delta }_N={\delta }_S-3(1+w)a^{2}H{\dot{E}}_S$; with conformal time $\bar{N}=a$ it is ${\delta }_N={\delta }_S-3(1+w)\mathcal{H}{E}_S^{\prime }$, where $\mathcal{H}=a^{\prime }/a$. Keeping these clock conventions distinct is essential to the factors of $a$.

Use the spatial-potential convention of this paper and assume $\bar{\rho }+\bar{P}\ne 0$. A passive time displacement changes the spatial scalar by $\tilde{\Psi }-\Psi =\bar{N}H{\xi }^0$ and changes the energy density byThe two contributions to the uniform-density curvature perturbation therefore cancel:A spatial scalar displacement changes $E$ but not this combination. Hence $\zeta$ is gauge invariant at linear order. Pure cosmological-constant matter, for which $\bar{\rho }+\bar{P}=0$, does not supply a energy density clock and this formula is not defined for that component alone.

Now take the definite constant-$w$ barotropic equation of state and use physical-time differentiation $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$. Put $A=\bar{\rho }+\bar{P}$. Background stress-energy conservation gives $\mathcal{D}A=-3H(1+w)A$. Differentiating $\zeta =-\Psi +\delta \rho /(3A)$, then substituting the two supplied perturbation equations, givesThus the lapse and expansion perturbations cancel without imposing a gauge. In terms of the non-adiabatic pressure perturbation $\delta P_{\mathrm{nad}}=\delta P-w\delta \rho$,For the stipulated barotropic equation of state, $\delta P=w\delta \rho$, so the first term vanishes. For a Fourier mode, $\Delta =-k^2/a^2$. For regular long-wavelength solutions the remaining gradients are suppressed when $k\ll aH$. Consequently $\dot{\zeta }\simeq 0$ on superhorizon scales, with exact conservation in the leading zero-gradient limit. It is not an exact finite-$k$ identity. Entropy perturbations or a singular gradient expansion can invalidate that conclusion; the supplied adiabatic assumption is what removes the unsuppressed pressure source.

This superhorizon conservation of uniform-density curvature is useful because inflationary fluctuations leave the horizon during cosmic inflation and may re-enter only much later. In an adiabatic single-clock evolution, the conserved uniform-density curvature perturbation transports their initial amplitude through reheating and the later radiation in cosmology and matter eras without solving every short-distance process. It then determines the initial conditions for large-scale structure and Cosmic microwave background anisotropy. Extra fluctuating fields or fluids can introduce isocurvature perturbations and convert them into curvature, so conservation must be checked rather than assumed during such transitions.

Let $\mathcal{H}=a^{\prime }/a$ and ${\theta }_N=i\mathbf{k}\cdot {\mathbf{v}}_N$. Choose the synchronous gauge rest frame of cold dark matter, so ${\theta }_C=0$. Its continuity equation then gives $h^{\prime }=-2{\delta }_C^{\prime }$. Substitution in the trace Einstein equation givesFor radiation in cosmology the continuity and Euler equations are ${\delta }_R^{\prime }+4{\theta }_R/3+2h^{\prime }/3=0$ and ${\theta }_R^{\prime }=k^2{\delta }_R/4$. Differentiate the first, eliminate ${\theta }_R^{\prime }$ and $h^{\prime \prime }$, and obtainradiation in cosmology pressure supplies the acoustic restoring term while its energy density contributes twice as strongly to the trace gravitational source through $1+3w_R=2$.

Deep in radiation domination, $a\propto \tau$, $\mathcal{H}=1/\tau$, ${\Omega }_R\simeq 1$ and ${\Omega }_C\simeq 0$. At long wavelengths $k\tau \ll 1$, the adiabatic initial conditions require ${\delta }_C=3{\delta }_R/4$. The cold-matter equation reduces toA power ${\tau }^p$ gives $p^2-4=0$. The regular growing solution therefore hasThe other adiabatic solution diverges as ${\tau }^{-2}$ and is discarded for the regular primordial growing mode. The relation between the density contrasts fixes their relative entropy to zero, not their subsequent equality inside the acoustic horizon.

One can also solve the leading radiation-era equations explicitly, which proves that the same regular mode develops logarithmic cold-matter growth after horizon entry. Put $x=k\tau /\sqrt{3}$. The first equation gives ${\delta }_R=(x^2{\delta }_{C,xx}+x{\delta }_{C,x})/3$. Substituting in the second givesDefineIts useful derivative is $F^{\prime }=1/x-2\sin x/x^2+2(1-\cos x)/x^3$. Direct differentiation verifies the fourth-order equation and gives the associated radiation in cosmology solution:At zero, $F=x^2/8+O(x^4)$ and ${\delta }_R=K{x}^2/6+O(x^4)$, so this is precisely the regular adiabatic radiation-era cold-dark-matter transfer solution with $K=24A/k^2$. The other indicial behaviors of the fourth-order equation are $1,x,x^{-2}$; they are either nonadiabatic at leading order or singular and do not add another regular growing adiabatic mode.

For $x\gg 1$, the integral is $\gamma +\ln x-\mathrm{Ci}(x)$, where $\gamma$ is the Euler's constant and $\mathrm{Ci}$ the cosine integral. Its oscillatory tail cancels the explicit leading $\sin x/x$ term, givingThus the requested late sound-horizon behavior isAn additive constant in ${\delta }_C$ is absorbed into the reference ${\tau }_{\ast }$; the regular adiabatic solution fixes the acoustic phase, with $D=0$ in the explicit time-origin convention above. A general acoustic mixture has both terms. The physical expansion parameter is $k\tau /\sqrt{3}$; comparisons with $2\pi /k$ describe the same long- and short-wavelength regimes up to fixed numerical factors.

The contrast grows much more slowly after entry during radiation domination: radiation in cosmology undergoes pressure-supported acoustic oscillations and cold matter only the logarithmic Mészáros effect. Modes entering well before matter-radiation equality consequently have suppressed late amplitudes relative to modes that stay outside the horizon until near equality. After matter domination begins, pressureless growing modes can grow as $a$. This scale-dependent history produces the turnover and small-scale suppression of the matter transfer function and affects the initial conditions for large-scale structure of the universe. The early subhorizon limit requires both $k\tau \gg 1$ and $\tau \ll {\tau }_{\mathrm{eq}}$; it is not applicable to a mode that enters only in the matter domination.

Let $w=c_s^2$ be constant and small, and write $\theta =i\mathbf{k}\cdot \mathbf{v}$. The supplied synchronous gauge equations implyDifferentiating the first and eliminating $\theta$ givesFor the dominant single component, the Einstein equation gives $h^{\prime \prime }+\mathcal{H}{h}^{\prime }=-3{\mathcal{H}}^2(1+3w)\delta$. Therefore the exact algebraic intermediate relation isThe requested nonrelativistic density equation in synchronous gauge follows to leading nonrelativistic order: discard pressure corrections to the background, Hubble damping, and metric tensor terms while retaining $c_s^2{k}^2\delta$, which can be comparable to gravity at the Jeans wavenumber. Since $w\ll 1$, that scale lies well inside the Hubble radius, so keeping its enhanced gradient term is consistent. The resulting equation iswhere the Friedmann equation gave $3{\mathcal{H}}^2/2=4\pi G{a}^2{\bar{\rho }}_m$. Now ${\delta }^{\prime }=a\dot{\delta }$ and ${\delta }^{\prime \prime }=a^2(\ddot{\delta }+H\dot{\delta })$, soThis is the specified approximation, not an exact finite-$w$ relativistic identity. The pressure-gradient term stabilizes short wavelengths, while gravity drives the long-wavelength Jeans instability.

Now apply the stipulated equation to $P=K{\rho }^{4/3}$ with $K>0$. In the pressure-negligible matter background,Thus $c_s^2{k}^2/a^2$ is proportional to $t^{-2}$, exactly like the gravitational term. DefineThe comoving Jeans wavenumber and $q$ are constant in this era. The equation becomes the Euler-Cauchy equationSetting $\delta \propto t^p$ gives $p^2+p/3+2(q-1)/3=0$. For distinct real roots the full pair of modes isThese are the Jeans modes of a four-thirds polytropic cosmological fluid. The physical Jeans length isHere ${\lambda }_J\propto a$, so its comoving value is constant. For $\lambda >{\lambda }_J$, $q<1$ and $p_{+}>0$, giving genuine growth; the other solution decays. In the long-wavelength pressure-free limit, $p_{+}=2/3$ and $p_{-}=-1$, the familiar matter-era modes. At $\lambda ={\lambda }_J$, the two powers are $0$ and $-1/3$, so the leading mode is constant rather than growing.

For $1<q<25/24$ both powers are negative: pressure prevents growth, although the modes still decay without oscillation. At $q=25/24$ they coincide and the complete solution isFor $q>25/24$ they become a complex-conjugate pair; a real basis isThese short-wavelength acoustic oscillations have a decaying envelope. The onset of instability is $q=1$, while the threshold for oscillatory time dependence is slightly larger, $q=25/24$; expansion damping explains why these two thresholds are not identical.

The comoving observer measures $E=a{p}^0$, so $q=aE=a^2{p}^0$. Write $\mathcal{H}=a^{\prime }/a$. The null geodesic condition impliesThe time component of the geodesic equation, using $d\tau /d\lambda =p^0$, is thereforeDifferentiate $q=a^2{p}^0$. The homogeneous expansion terms cancel:At zeroth order $p^i/p^0=n^i$, with ${\delta }_{ij}{n}^i{n}^j=1$. The difference between coordinate velocity and physical unit direction is already first order and multiplies $h^{\prime }$, so it can be discarded in this expression. We obtain the synchronous photon momentum redshiftIn the unperturbed universe $q$ is constant and physical photon energy redshifts as $a^{-1}$; the displayed term is the additional anisotropic first-order redshift.

To check the direction, let $v^i=p^i/p^0$. Combining the spatial and temporal geodesic equations cancels their homogeneous Hubble terms and givesEvery remaining connection or explicit metric-derivative term is first order. A local orthonormal spatial frame defines the physical unit vector by $n^i=v^i+\frac{1}{2}{h}^i{}_j{v}^j+O(h^2)$. Its derivative adds only first-order terms, soConsequently direction deflection multiplied by an already first-order temperature anisotropy is second order. The first-order brightness calculation may use a straight unperturbed ray without neglecting the first-order gravitational frequency change.

Before decoupling, rapid Thomson scattering makes the photon distribution nearly isotropic in the local photon-fluid rest frame. A local blackbody temperature perturbation changes its energy density by ${\delta }_{\gamma }=4\delta T/T$. A first-order boost by the fluid velocity adds a temperature photon dipole $\mathbf{n}\cdot \mathbf{v}$; it does not generate a photon quadrupole until second order in that boost. Thus the photon brightness perturbation at instantaneous decoupling isHigher angular moments are suppressed by the photon mean-free-path to perturbation-length ratio in the tight-coupling approximation. The idealized equilibrium assumption neglects those finite-mean-free-path corrections, polarization effects and finite thickness of the last-scattering surface. It is not an assertion that collisionless photons remain isotropic after decoupling. The distinction between a fluid treatment and the subsequent hierarchy is discussed in the primary synchronous and Newtonian perturbation treatment.

Let $\mu =\hat{\mathbf{k}}\cdot \mathbf{n}$ and $W(\tau )=e^{ik\mu (\tau -{\tau }_0)}$. Multiplication of the synchronous photon brightness equation by its integrating factor givesIntegration from ${\tau }_{\ast }={\tau }_{\mathrm{dec}}$ to ${\tau }_0$ and division by four yieldFor the real-space observation point ${\mathbf{x}}_0$, the corresponding background ray is $\mathbf{x}(\tau )={\mathbf{x}}_0-\mathbf{n}({\tau }_0-\tau )$. Fourier inversion gives the synchronous Sachs-Wolfe line-of-sight formulaEvery source is evaluated at its appropriate point on the ray, rather than at the observation position. Here $\mathbf{n}$ is photon propagation direction, opposite to the outward observer-to-source sky direction; changing that convention changes the written sign of the Doppler projection consistently.

The first term is the intrinsic photon temperature at last scattering. On wavelengths larger than the sound horizon, it retains primordial adiabatic information; inside that horizon the photon-baryon fluid has acoustic energy density oscillations, producing the acoustic temperature peaks. The second term is the Doppler effect from the fluid's bulk motion at emission. A regular long-wavelength velocity is gradient-suppressed; the Doppler contribution is important near and below the acoustic horizon, with its acoustic phase displaced from that of the energy density oscillation.

The integral is the gravitational energy shift produced by the changing spatial metric tensor along the photon path. In synchronous gauge it includes endpoint gravitational redshifts as well as evolving-potential effects. It must not be identified solely with the Integrated Sachs-Wolfe effect: in Newtonian gauge the same total separates into the ordinary last-scattering Sachs-Wolfe effect, local observer terms, and an integral of time-varying potentials. The ordinary contribution is important on scales outside the sound horizon at decoupling. A late integrated contribution is strongest on large angular scales when dark energy or curvature changes the potentials; evolution around matter-radiation equality also contributes near the first acoustic scales.

The relevant comoving sound scale is $r_s({\tau }_{\ast })={\int }^{{\tau }_{\ast }}{c}_s(\tau )d\tau$, while the projection distance is $D_{\ast }\simeq {\tau }_0-{\tau }_{\ast }$. Modes project roughly to photon temperature multipole $\ell \sim k{D}_{\ast }$, and the acoustic scale is $\ell \sim D_{\ast }/r_s$, with successive peak spacing of order $\pi D_{\ast }/r_s$. At much shorter wavelengths Silk damping and the finite last-scattering thickness suppress energy density and velocity anisotropies; the instantaneous-decoupling approximation alone does not model that cutoff. These scale distinctions explain why the three terms cannot be assigned one common range of important wavelengths.

Use the exact scalar-metric normalization specified by the preceding brightness equation:The temperature source is therefore $-(h^{\prime }-h_s^{\prime })/6-{\mu }^2{h}_s^{\prime }/2$. Let $W=e^{ik\mu (\tau -{\tau }_0)}$, so $W^{\prime }=ik\mu W$ and $W^{\prime \prime }=-k^2{\mu }^{2}W$. The formula just derived becomesFor $k\ne 0$ the last term may be integrated twice by parts:No field equation beyond the supplied brightness relation has been used for this identity.

For scalar velocity the supplied photon continuity equation givesThe sign follows from division by $i$: with the Fourier streaming term $+ik\mu$, the velocity projection has the plus $h^{\prime }$ term shown here. Combining it with the lower endpoint from the integrations by parts gives the complete endpoint terms in the synchronous Sachs-Wolfe formula:The local observer bracket is an angular photon monopole plus photon dipole. Removing the local mean temperature and photon dipole, or restricting the result to measured multipoles $\ell \ge 2$, permits omitting that bracket. It must be retained for the literal full brightness amplitude. The $k=0$ limit should be taken in the original line-of-sight integral, rather than treating the separate inverse-$k$ endpoint terms as independently defined.

The final target formula printed in the PDF does not follow from its preceding equations. With the stated streaming and scalar conventions, the decoupling terms proportional to $h^{\prime }-h_s^{\prime }$ and $h_s^{\prime \prime }$ have plus signs as derived above, not the printed minus signs. Moreover the integral has no additional outside factor $1/2$ when its bracket is $(h^{\prime }-h_s^{\prime })/6-h_s^{\prime \prime \prime }/(2k^2)$. Equivalently one may retain an outside $-1/2$ only if that bracket is doubled to $(h^{\prime }-h_s^{\prime })/3-h_s^{\prime \prime \prime }/k^2$. Omitting the observer photon monopole and photon dipole does not fix these discrepancies.

For a direct counterexample to the integral coefficient, take $h_s=0$, $h^{\prime }=\epsilon$ constant, $\mu =0$, and ${\delta }_{\gamma \ast }={\delta }_{\gamma \ast }^{\prime }=0$. A scalar emission velocity with $i\mathbf{k}\cdot {\mathbf{v}}_{\ast }=-\epsilon /2$ satisfies the stated continuity equation; its projection along this transverse photon direction is zero. Keep $\epsilon ({\tau }_0-{\tau }_{\ast })\ll 1$ so perturbation theory is valid. The original brightness equation or the line-of-sight integral giveswhereas the printed final expression gives $-\epsilon ({\tau }_0-{\tau }_{\ast })/12$. All $h_s$ observer terms vanish in this example, so an observer-term convention cannot remove the contradiction. The boxed corrected identity supplies the requested integration-by-parts result with every boundary contribution and its observable higher-multipole version specified.