Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 312 (Advanced Cosmology)

Using the normalization of the timelike unit normal, the spatial projection tensor satisfiesLikewise $P^{\mu }{}_{\alpha }{n}^{\alpha }=0$ and $P^{\mu }{}_{\alpha }{P}^{\alpha }{}_{\nu }=P^{\mu }{}_{\nu }$. The spacetime form of the induced metric is thereforeIts restriction to the slice is the given three-metric. Metric compatibility of the four-dimensional Levi-Civita connection givesApply the three projectors defining the spatial covariant derivative. In the first summand $P^{\beta }{}_j{n}_{\beta }=0$, and in the second $P^{\alpha }{}_i{n}_{\alpha }=0$. Both summands vanish, soThis proves the metric compatibility of the spatial covariant derivative without treating the spatial projection tensor itself as covariantly constant.

Differentiate the normalization of the timelike unit normal with the metric-compatible covariant derivative:Thus the derivative of the normal is already orthogonal to the normal in its second index. Expanding the second spatial projection tensor in the definition of the extrinsic curvature of a spatial hypersurface givesTo prove symmetry using Frobenius theorem, set $F_{\alpha \beta }={\nabla }_{\alpha }{n}_{\beta }-{\nabla }_{\beta }{n}_{\alpha }$. Hypersurface orthogonality is equivalent toContract this with $n^{\alpha }{P}^{\beta }{}_{\mu }{P}^{\lambda }{}_{\nu }$. The last two terms vanish because a projected normal vanishes, while $n^{\alpha }{n}_{\alpha }=-1$ in the first term. HenceThe antisymmetric part of the doubly projected covariant derivative of $n$ is therefore zero. This proves that hypersurface orthogonality implies symmetric extrinsic curvature:No geodesic assumption for the normal congruence is needed; its normal acceleration may be nonzero.

Write $h_{ij}={}^{(3)}{g}_{ij}$, and retain only first-order scalar cosmological perturbations. The inverse induced metric and inverse lapse function areThe flat background spatial connection vanishes, and $N_i=-a^2{B}_{,i}$ is already first order. ConsequentlyThe comma on $B_{,i}$ matters: this is the gradient of the scalar shift vector potential, not an unrelated vector field. Substituting the time derivative of $h_{ij}$ into the negative extrinsic curvature convention givesRaising the first index with the perturbed inverse induced metric is essential. Its $\Phi$ and $E$ terms cancel the corresponding terms multiplying the background Hubble parameter, leavingFor the scalar shear potential $\chi =a^2(\dot{E}-B)/\bar{N}$, raised spatial derivatives mean ${\partial }^i=a^{-2}{\delta }^{ij}{\partial }_j$ to the required order. Thus the final term is $-{\partial }^i{\partial }_j\chi$. Taking the trace identifies the scalar expansion perturbation:Separating the trace and trace-free parts now yieldsAll signs follow from the source's $d{x}^i-N^{i}dt$ shift vector convention and negative extrinsic curvature definition.

The result above gives $K=-3H+\kappa$. Since the shear is trace-free and already first order, it cannot contribute linearly to the quadratic tensor contraction $K_{ij}{K}^{ij}=\mathrm{tr}(K^2)$. Thereforeup to second-order scalar cosmological perturbations. Combining these with the supplied spatial Ricci scalar gives the Hamiltonian constraintIts background part is the flat Friedmann equation $6H^2=16\pi G\bar{\rho }$. Subtracting that equation proves the linearized cosmological Hamiltonian constraint:Here $\rho$ is the energy density measured by the hypersurface normal, as required by the normal-normal projection of the Einstein field equations.

Use a local orthonormal tetrad and units in which the speed of light is one. A particle has four-momentum $p^{\hat{a}}=E(1,\mathbf{e})$ and moves with velocity $\mathbf{e}$. Thus its energy density contributes $E$, its momentum density contributes $E{e}^i$, and its momentum flux contributes $E{e}^i{e}^j$. Integrating the phase-space distribution function gives the kinetic stress-energy tensor. The measure $d^{3}p/E$ is the future mass-shell Lorentz-invariant phase-space measure, so the same expression transforms as a spacetime stress-energy tensor; fixed state-counting or polarization factors are understood to be included in $f$.

Define $⟨F{⟩}_{\Omega }=(4\pi {)}^{-1}\int Fd\Omega$. The photon momentum measure is $d^{3}p=E^{2}dEd\Omega$, so every component of the kinetic stress-energy tensor has a radial factor $E^{3}dE=a^{-4}{ϵ}^{3}dϵ$. An isotropic background has $⟨e^i{⟩}_{\Omega }=0$ and $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$. HenceThis radiation pressure equation of state requires isotropy and masslessness, not a thermal spectrum.

The distribution perturbation is $\delta f=-ϵ{\bar{f}}^{\prime }(ϵ)\Theta$. Assume finite energy density and endpoint behavior ${ϵ}^4\bar{f}(ϵ)\to 0$ at zero and infinity, as for the Planck photon distribution. Integration by parts then givesConsequently the perturbed energy density and momentum density areSince $q^i=(\bar{\rho }+\bar{P})v^i=(4\bar{\rho }/3)v^i$, the photon angular temperature moments give $\delta =4⟨\Theta {⟩}_{\Omega }$ and $v^i=3⟨\Theta e^i{⟩}_{\Omega }$.

The sign of anisotropic stress must be specified. Direct kinetic integration gives the conventional trace-free spatial stress ${\pi }^{ij}=T^{ij}-P{\delta }^{ij}=+4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3){⟩}_{\Omega }$. The PDF uses the opposite sign, ${\Pi }^{ij}=-{\pi }^{ij}$, equivalently $T^{ij}=P{\delta }^{ij}-{\Pi }^{ij}$. With that convention, all requested moments areTaking the spatial trace also givesThe angular temperature description assumes $\Theta$ is independent of photon energy; independent spectral distortions would require additional energy-dependent moments.

The collisionless Boltzmann equation is conservation of $f$ along the photon trajectory. To linear order, use the unperturbed trajectory $d\mathbf{x}/d\eta =\mathbf{e}$ when differentiating $\Theta$. Deflection of $\mathbf{e}$ is first order and multiplies the first-order angular dependence of $\delta f$, while the background has no angular dependence; it therefore contributes only at second order. Similarly, the energy derivative of $\delta f$ times the perturbed energy change is second order. ThusFor a nontrivial spectrum this identifies the linear Free-streaming photon Boltzmann equation:In Newtonian gauge in cosmology, substitute the specified scalar gravitational-redshift source to writeMultiply its angular average by four. Since $⟨e^j{⟩}_{\Omega }=0$, the photon angular temperature moments imply the photon continuity equationFor the photon Euler equation, multiply by $3e^i$ and average. The source is $-3⟨e^i{e}^j{⟩}_{\Omega }{\partial }_j\psi =-{\partial }^i\psi$, while the second angular moment isIt follows thatThe background energy density is spatially homogeneous, so no spatial derivative acts on $\bar{\rho }$. The minus sign of its anisotropic stress term follows from the PDF's $\Pi =-\pi$ convention. Switching to the conventional trace-free stress $\pi$ changes that term to a plus sign.

In the local orthonormal tetrad, the components of the photon number current areThey are respectively the photon number density and the photon number flux. Thus $J^{\mu }$ is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, giveswhere $C[f]$ is the collision operator in the covariant streaming equation. ThereforeCollisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives ${\bar{n}}_{\gamma }\propto a^{-3}$.

Evaluate the metric inner products of the proposed orthonormal tetrad to first order in the vector cosmological perturbation:The orthonormal tetrad therefore has metric signature $(-,+,+,+)$ to the required order. Put $q=ϵ/a^2$ and $b=\mathbf{e}\cdot \mathbf{B}$. The photon momentum components areThey satisfy the null vector condition to first order, with $|\mathbf{e}|=1$.

Use $\eta$ as parameter in the time component of the geodesic equation:The derivative of the direction is first order, since the background photon moves on a straight comoving line; $B_{i}d{e}^i/d\eta$ is consequently second order. With $d\mathbf{x}/d\eta =\mathbf{e}+O(B)$,The supplied Levi-Civita connection coefficients giveThe spatial gradients and all background expansion terms cancel. Since ${ϵ}^{\prime }/ϵ$ vanishes in the background, its product with $b$ is also second order. The photon energy redshift from a vector metric perturbation is thereforeAn independent check uses the covariant component $p_0=g_{0\mu }{p}^{\mu }=-ϵ+O(B^2)$. The covariant geodesic equation gives $d{p}_0/d\eta =(2p^0{)}^{-1}{\partial }_{\eta }{g}_{\alpha \beta }{p}^{\alpha }{p}^{\beta }=ϵe^i{\dot{B}}_i$; the conformal expansion term vanishes by the null vector condition and reproduces the same result.

For a nonzero Fourier mode $\mathbf{k}=k\hat{z}$, the divergence-free condition implies $B_z=0$. The complex vectors $m^{(\pm )}=(\hat{x}\pm i\hat{y})/\sqrt{2}$ form a Hermitian orthonormal basis of this transverse plane. Their opposite rotation phases give the helicity decomposition of a transverse Fourier mode; the factor $i/\sqrt{2}$ in the amplitudes is a convention. The baryon velocity in this vector sector has the analogous decomposition.

For helicity $\sigma =\pm 1$, the angular source isThe same formula holds for ${\mathbf{v}}_b^{(\sigma )}$. Thus these sources have only angular order $m=\sigma$. Streaming multiplies by $ik\cos \theta$ and the stated Thomson scattering operator is diagonal in $m$, so neither mixes this sector with other angular orders. Assuming the initial anisotropy is in the same vector-helicity sector, or considering the anisotropy generated from zero initial data, the solution therefore contains only $Y_{\ell \sigma }$ with $\ell \ge 1$. Arbitrary independently imposed anisotropies of other $m$ would instead evolve in their own homogeneous sectors.

There is a genuine normalization inconsistency in the PDF. To retain its printed temperature expansion, writewhere $q_{\ell }$ denotes exactly the multipoles called ${\Theta }_{\ell }$ in that expansion. The spherical-harmonic streaming recurrence gives the coefficient of $Y_{\ell \sigma }$ in $ik\cos \theta \Theta$, divided by $C_{\ell }$, asHere $C_{\ell +1}/C_{\ell }=-i\sqrt{(2\ell +3)/(2\ell +1)}$ and $C_{\ell -1}/C_{\ell }=i\sqrt{(2\ell -1)/(2\ell +1)}$. Both streaming denominators are therefore $2\ell +1$, rather than the PDF's $2\ell +3$ and $2\ell -1$. The dipole source also changes: $C_1=-i\sqrt{3/(8\pi )}$, so $-(\mathbf{e}\cdot \dot{\mathbf{B}}+\dot{\tau }\mathbf{e}\cdot {\mathbf{v}}_b)/C_1$ is $-\sigma (4\pi /3)({\dot{B}}^{(\sigma )}+\dot{\tau }{v}_b^{(\sigma )})Y_{1\sigma }$. The quadrupole collision correction is still $-\dot{\tau }{q}_2{\delta }_{\ell 2}/10$.

Thus the vector photon Boltzmann hierarchy consistent with the literal printed expansion isA concrete counterexample to the printed hierarchy with this expansion is a collisionless instant with $q_2=1$ and every other multipole zero. Direct streaming gives ${\dot{q}}_1=-k\sqrt{3}/3$, whereas the printed hierarchy would give $-k\sqrt{3}/5$.

Alternatively, preserve the PDF's desired hierarchy by making the photon multipole normalization changeThis is the reciprocal angular normalization, not the printed expansion. Multiplying the corrected hierarchy by $(2\ell +1)/(4\pi )$ gives preciselyThis supplies both consistent conventions explicitly. The lower coupling vanishes at $\ell =1$, so no vector monopole is introduced. Cosmological optical depth to the observer has $\dot{\tau }<0$, giving damping of the dipole and the factor $9/10$ quadrupole damping in this temperature-only collision model; no polarization collision term has been added to the supplied equation.

Use the comoving curvature perturbation mode normalization. Take $H,ϵ,c_s$ to be effectively constant at leading slow-roll approximation order, with $H,ϵ,c_s>0$, and use the Bunch-Davies vacuum selected by the early-time $i\epsilon$ contour. The regulator $\epsilon$ is distinct from the Hubble slow-roll parameter $ϵ$.

Let $C=ϵ+1-c_s^2$ and use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k\zeta (\mathbf{k})e^{i\mathbf{k}\cdot \mathbf{x}}/(2\pi {)}^3$. Conversion to conformal time givesThe last sign follows from substituting $a=-1/(H\tau )$, which is positive for $\tau <0$; dropping the explicit minus sign in that substitution would reverse the final primordial bispectrum.

At the order needed by the in-in formalism, the unequal-time Wick contraction isIt uses the Gaussian quantum vacuum two-point function with the displayed operator ordering; an arbitrary classical unequal-time covariance cannot replace it. For the connected three-point function, each external field contracts with one of the three fields at the vertex. There are $3!=6$ such Wick contractions, all equal. The spatial integral conserves total momentum and leaves one factor $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. The other nine pairings are disconnected tadpole terms supported on a zero external momentum; they are excluded from the connected primordial bispectrum, equivalently by subtracting the one-point function or normal ordering the vertex.

The free Gaussian odd correlator is zero. Applying the stated first-order in-in formalism therefore givesBy definition, the full connected correlator is $(2\pi {)}^3{\delta }^{(3)}({\sum }_r{\mathbf{k}}_r)B$. The assumptions are a Gaussian Bunch-Davies vacuum, the given leading cubic vertex, tree order, the late-time limit $\tau \to 0^{-}$, effectively constant background parameters, and connected nonzero external modes. No symmetry factor $1/3!$ is inserted into the given Hamiltonian.

Write $K=k_1+k_2+k_3>0$ and $C=ϵ+1-c_s^2$. The inflationary scalar Fourier mode has real late-time value $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{c}_s{k}^3}$ and $u_k^{\prime \ast }(\tau )=A_k{c}_s^2{k}^2\tau e^{ik{c}_s\tau }$. Thus the time integrand, before the factor $-2i$, isThe Bunch-Davies vacuum prescription supplies damping in the past. Equivalently, use a positive regulator $d$ and then take the limit:The real part of $-2i$ times this integral is $4/(c_s^3{K}^3)$. Including the six Wick contractions yields the cubic time-derivative curvature bispectrumThe resulting primordial bispectrum is a homogeneous function of degree $-6$ in the wavenumbers, as expected for scale-independent background parameters.

The printed limit $c_s\ll 0$ is not a physical sound-speed limit for these modes. The normalization and positive-frequency choice assume $c_s>0$; a negative value is outside the positive sound-speed convention used by these modes. The likely intended limit is $0<c_s\ll 1$, which is analyzed explicitly here rather than silently substituted for the PDF.

To compare primordial non-Gaussianity, normalize by the power spectrum, not only by the raw three-point amplitude. HereAt an equilateral bispectrum configuration, the exact normalized amplitude isThus $c_s\to 1$ gives an amplitude of order $ϵ$, consistent with weak non-Gaussianity in single-field slow-roll inflation. For $0<c_s\ll 1$, this vertex gives an amplitude of order one, enhanced relative to the slow-roll limit but not diverging as $c_s^{-2}$ after power-spectrum normalization. At fixed $H,ϵ,M_{\mathrm{Pl}}$ the raw bispectrum does scale as $c_s^{-2}$, but $P_{\zeta }^2$ has the same scaling. The often-quoted larger small-sound-speed amplitudes of more general inflationary actions require their actual interaction coefficients and other vertices; they do not follow from the lone supplied vertex. These conclusions concern its tree contribution within the stated weak-perturbation approximation.

The bispectrum shape function isAt fixed perimeter, the arithmetic-geometric mean inequality bounds this by $1/27$, attained only at an equilateral bispectrum configuration. In a squeezed bispectrum configuration with $k_1=q\ll k_2\simeq k_3=k$, it behaves as $q/(8k)\to 0$. At a flattened bispectrum configuration $(k,k,2k)$ it is finite, $1/32$, with no boundary singularity. Thus the shape prefers equilateral triangles and suppresses squeezed triangles, with a broad finite flattened contribution. This is a shape preference, not a complete signal-to-noise calculation: actual observational weighting also depends on cosmological transfer functions, covariance and accessible triangles. If $C=0$, this vertex produces no bispectrum at any configuration; if $C<0$, the signed shape reverses while its absolute triangle preference is unchanged.

Let $R=\Delta \tau >0$, $L={\ell }_1+{\ell }_2+{\ell }_3$ and $D={\prod }_{r=1}^3(2{\ell }_r+1)$. Use exactly the given constant primordial bispectrum normalization, without an additional local-template convention factor. Combining it with the large-angle Sachs-Wolfe effect cosmological transfer function gives the reduced CMB bispectrumwhere $I_{\ell }(x,R)={\int }_0^{\infty }{j}_{\ell }(kR)j_{\ell }(kx)dk$. Each $1/5$ transfer factor has been retained; their product is $1/125$.

Apply the supplied spherical Bessel product integral after rescaling $q=kR$:The two branches agree at $x=R$. In the radial integral put $r=x/R$. The factor $R^3$ from $x^{2}dx$ cancels the $R^{-3}$ from the three kernels. The remaining integral isprovided $L>0$. This evaluation uses the momentum integrals at fixed radial parameter in the projection, then the radial integral; it does not require an unjustified global exchange of all oscillatory integrals.

The factors $(2/\pi {)}^3$ and ${\pi }^3/8$ cancel exactly. Hence the Sachs-Wolfe projection of a constant bispectrum isThe all-monopole case $L=0$ has a logarithmically divergent outer radial integral and is not covered by the printed finite formula. Observable CMB analyses remove the monopole and dipole, normally using ${\ell }_i\ge 2$, so this issue is absent there. Angular triangle and parity selection are carried by the triple-spherical harmonic geometric factor multiplying the reduced bispectrum.

There is no dependence on the last-scattering distance $R$, and no extra physical scale appears when ${\Delta }_{\zeta }^2$ is held constant. At fixed triangle shape and in the range $1\ll {\ell }_i\ll 200$, common rescaling of all multipoles givesThis is angular scale invariance in the usual weighted sense; a constant primordial shape does not make the unweighted reduced CMB bispectrum independent of angular scale. The exact finite-multipole expression retains the $+1$ and $+3$ terms shown above.