Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 312 (Advanced Cosmology)

Write $h_{ij}$ for the spatial metric tensor and lower the shift vector with $h_{ij}$. For the negative-shift convention, the four-dimensional metric and its inverse have componentsThe normal has covariant components $n_{\mu }=(-N,0,0,0)$, so $n_{\mu }{n}^{\mu }=-1$ and it annihilates every tangent vector to the spatial slice. Since $n_i=0$, its covariant derivative gives $n_{i;j}=N{\Gamma }^0{}_{ij}$ and therefore $K_{ij}=-N{\Gamma }^0{}_{ij}$.

Compute the required Christoffel symbol using the symmetric connection formula. Its time-index term and spatial-index term combine to giveThe spatial expression in parentheses is ${\dot{h}}_{ij}+D_i{N}_j+D_j{N}_i$, where $D$ is the covariant derivative of $h$. Thus the extrinsic curvature with a negative shift isThis derivation fixes both signs from the actual 3+1 decomposition of spacetime, rather than transferring a formula using the opposite shift vector convention. The connection formula used here has $g_{\lambda \kappa ,\nu }$ as its second differentiated term; the printed repeated-index version is a typographical error.

Set $h=\det h_{ij}$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. The Jacobi determinant derivative formula and metric compatibility giveConsequentlyThe local volume element is $\sqrt{h}{d}^{3}x$, so for zero shift vector the proper-time derivative along the normal is $d/ds=N^{-1}{\partial }_t$ andHere $\theta$ is the expansion scalar. Writing $a_{\mathrm{loc}}=h^{1/6}$ defines a local linear scale from this volume, whence the local volume Hubble parameter isThe factor of three converts volume expansion into linear expansion, and $N$ converts coordinate time into proper time. This agrees with the usual Hubble parameter in a homogeneous FLRW metric.

For a volume-shape decomposition the unimodular spatial metric is ${\gamma }_{ij}=a_{\mathrm{loc}}^{-2}{h}_{ij}$, with determinant one. If instead the printed positive power $a_{\mathrm{loc}}^2{h}_{ij}$ is taken literally, its determinant is $a_{\mathrm{loc}}^{12}$; it is a conformal rescaling but not the unimodular shape metric. The trace calculation does not need that rescaling.

Decompose the mixed extrinsic curvature of a spatial hypersurface into $K^i{}_j=\frac{1}{3}K{\delta }^i{}_j+{\tilde{K}}^i{}_j$. In the long-wavelength approximation in cosmology, the inflationary shear damping law makes ${\tilde{K}}^i{}_j$ fall as the inverse local volume. During cosmic inflation, many e-folds therefore suppress the anisotropic expansion, while $K$ still measures the local isotropic expansion through the local volume Hubble parameter.

Neglecting spatial gradients at leading order gives approximately independent expanding patches. Restrict also to scalar cosmological perturbations, and neglect the independent frozen tensor cosmological perturbation. The remaining spatial scale can be written as a background scale times a local fluctuation:This is the nonlinear scalar metric in a cosmological gradient expansion. The exponential keeps the metric positive and permits finite $\zeta$; no expansion in its amplitude is needed. Scalar lapse and shift perturbations describe the remaining freedom in time slicing and spatial threading. Choose $N=1+\Psi$ and $N_i=-a_{\mathrm{bg}}^2{\partial }_{i}B$. The negative shift vector convention then produces the positive mixed term $2a_{\mathrm{bg}}^2{\partial }_{i}Bdtd{x}^i$.

The exact completed-square form hasThe quadratic shift expression must be understood as this norm. If the raised derivative in the displayed ansatz is intended as a flat derivative, the corresponding factors of the spatial scale are needed. In either convention the shift-square term is of second gradient order, and the leading long-wavelength approximation in cosmology is unaffected.

Damping of shear motivates locally isotropic scalar expansion, not the disappearance of every tensor perturbation. A time-independent anisotropic shape can survive; the conformally flat scalar ansatz additionally neglects that tensor sector.

Use a dot for the coordinate-time derivative and let $\bar{N}$ be the homogeneous background lapse function. The passive linear transformation is $\delta {\tilde{g}}_{\mu \nu }=\delta g_{\mu \nu }-{\mathcal{L}}_{\xi }{\bar{g}}_{\mu \nu }$. Its time-time and time-space components yield the scalar gauge transformations with a background lapse:For example, ${\bar{g}}_{00}=-{\bar{N}}^2$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{00}=2\bar{N}\dot{\bar{N}}{\xi }^0+2{\bar{N}}^2{\dot{\xi }}^0$; ${\bar{g}}_{0i}=0$ gives $-({\mathcal{L}}_{\xi }\bar{g}{)}_{0i}={\bar{N}}^2{\partial }_i{\xi }^0-a^2{\partial }_i\dot{\lambda }$. These establish the signs directly.

To preserve synchronous gauge in cosmology, both transformed quantities must remain zero. HenceIntegrating gives the residual synchronous-gauge freedomThe integration lower limit can be absorbed into $D$. Spatially homogeneous additions to $\lambda$ generate no spatial displacement, so they are physically irrelevant to this scalar parametrization.

The spatial transformation also gives $\tilde{\Phi }=\Phi +(\dot{a}/a){\xi }^0$ and $\tilde{E}=E+\lambda$. Thus fixing the lapse and shift perturbations does not fix the entire coordinate system; the arbitrary functions $C,D$ can still change the remaining scalar cosmological perturbations.

Throughout this question dots denote conformal time derivatives, and $\mathcal{H}=\dot{a}/a$ is the conformal Hubble parameter. The monopole of the photon Boltzmann hierarchy gives ${\dot{\Theta }}_0+k{\Theta }_1/3=\dot{\varphi }$. With the specified velocity convention, the photon continuity equation is thereforeThe dipole equation is ${\dot{\Theta }}_1+k(2{\Theta }_2/5-{\Theta }_0)=\dot{\tau }({\Theta }_1+v_b)+k\psi$. Thus the photon Euler equation isIn particular, $\dot{\tau }<0$ makes the scattering term damp the photon-baryon velocity slip; it does not amplify it.

The linear momentum density of the photon-baryon fluid is weighted by enthalpy, not merely energy density:Elastic Thomson scattering exchanges momentum internally. Its photon force density is $(4{\bar{\rho }}_{\gamma }/3)\dot{\tau }(v_{\gamma }-v_b)$, so the baryon force density is its negative. Dividing by ${\bar{\rho }}_b$ gives the baryon Euler equation with Thomson dragThis $R$ is the baryon loading parameter, the baryon-to-photon inertia ratio. The two collision terms cancel exactly when the equations are weighted by their enthalpies. Background conservation gives ${\bar{\rho }}_b\propto a^{-3}$ and ${\bar{\rho }}_{\gamma }\propto a^{-4}$, so $\dot{R}=\mathcal{H}R$.

The tight-coupling approximation requires the positive scattering rate $\kappa =-\dot{\tau }$ to exceed both $k$ and the conformal Hubble parameter. The photon-baryon velocity slip and photon quadrupole are then small, and the leading fluid motion has $v_b=v_{\gamma }=v$. One must combine the equations before setting the slip to zero: a small slip times a large collision rate can exert a finite force.

Add the photon Euler equation to $R$ times the baryon Euler equation with Thomson drag. The collisions cancel and, neglecting ${\Theta }_2$ at this order, the result isSince $\dot{R}=\mathcal{H}R$, the common velocity obeysThe inertia of the baryons reduces the photon-baryon sound speed to $c_s^2=1/[3(1+R)]$.

Differentiate the photon continuity equation and substitute this velocity equation:Thus the photon-baryon acoustic oscillator isThe terms on the right drive the oscillation gravitationally; the first-derivative term comes from changing baryon inertia. Diffusion damping enters only beyond this leading tight-coupling approximation.

Expand the fractional CMB temperature fluctuation as $\Theta (\mathbf{n})={\sum }_{\ell m}{a}_{\ell m}{Y}_{\ell m}(\mathbf{n})$. Under statistical isotropy, its CMB angular power spectrum is defined byThe normalization is for dimensionless temperature, rather than temperature in kelvin. The spherical harmonics separate angular scales, with large $\ell$ corresponding to small angles.

For approximately constant $R,\varphi ,\psi$, a useful solution of the photon-baryon acoustic oscillator isHere $r_s$ is the comoving sound horizon. With slowly varying $R$, the oscillatory phase is still $k{r}_s$ and its homogeneous amplitude varies approximately as $(1+R{)}^{-1/4}$ under the short-period approximation. Evolving potentials add gravitational driving rather than erase this phase coherence.

The regular growing adiabatic initial conditions produce an initial density displacement and negligible initial acoustic velocity outside the horizon. They therefore select the cosine phase, with $B$ approximately zero, for all modes. At last scattering, extrema of compression and rarefaction occur at $k{r}_s({\eta }_{\ast })\simeq n\pi$. The observed monopole-plus-potential Sachs-Wolfe combination is ${\Theta }_0+\psi ={\delta }_{\gamma }/4+\psi$, so baryon loading shifts its equilibrium and makes alternating peaks unequal. The velocity produces a phase-shifted Doppler CMB anisotropy.

Projection from the Cosmic microwave background last-scattering surface places a mode near $\ell \simeq k{\chi }_{\ast }$, where ${\chi }_{\ast }={\eta }_0-{\eta }_{\ast }$ is its comoving radial distance in a flat universe. The resulting adiabatic acoustic-peak spacing isCoherent acoustic phases turn successive compressions and rarefactions into the CMB peak series. Finite-width last scattering, Doppler projection, evolving potentials and diffusion shift or broaden actual peaks, so this is the requested approximate spacing, not an exact prediction for every peak.

Let $q^i=e^i-\frac{1}{2}{h}^i{}_j{e}^j$, so $p^0=ϵ/a^2$ and $p^i=(ϵ/a^2)q^i$. The orthonormal tetrad gives ${\delta }_{ij}{e}^i{e}^j=1$. To first order,Hence the time-index Christoffel symbols contract toThe $\mathcal{H}{h}_{ij}$ term cancels the perturbation in the coordinate direction norm. This cancellation is why the local tetrad direction must not be mistaken for the coordinate direction.

Along the photon null geodesic, $d/d\lambda =p^{0}d/d\eta$, andThe time component of the geodesic equation now gives the photon energy redshift from a tensor metric perturbation:The derivative on the left follows the ray; the dot on $h$ is a partial conformal time derivative. The background redshift has already been removed by using comoving energy. The calculation only needs ${\Gamma }^0{}_{00}$ and ${\Gamma }^0{}_{ij}$; the unrelated printed ${\Gamma }^i{}_{j0}$ should contain $\frac{1}{2}{\dot{h}}^i{}_j$, rather than the full derivative.

To first order, propagate along the unperturbed ray $\mathbf{x}(\eta )={\mathbf{x}}_0-({\eta }_0-\eta )\mathbf{e}$ and keep its orthonormal tetrad direction fixed. Corrections to direction or trajectory multiply an already first-order temperature source and are second order. Define the angular scattering sourceUsing the photon energy redshift from a tensor metric perturbation, the Photon Boltzmann equation with Thomson scattering along this ray becomesMultiply by the survival probability $e^{-\tau }$. The cosmological visibility function is $g(\eta )=-\dot{\tau }{e}^{-\tau }$, so integration givesEvery source here is evaluated on the same ray, and $\tau ({\eta }_0)=0$.

The instantaneous-visibility approximation makes $e^{-\tau }$ a step from zero to one at ${\eta }_{\ast }$. It suppresses the early boundary term and localizes the scattering source to $S_{\ast }$. To obtain only the requested tensor CMB line-of-sight source, additionally neglect this last-scattering angular source: in leading tight coupling, the tensor-induced incident quadrupole is small, and a tensor mode has no scalar monopole or dipole. Ignore later reionization and its scattering as well. ThenInstantaneous visibility alone would leave the term $S_{\ast }$; neglect of that term is an additional approximation. The time derivative on $h$ is essential and is present in the PDF, although it was lost in the extracted TeX.

Choose the arbitrary normalization of the complex plane wave polarization so that $h_{ij}=u_i{u}_{j}h(\eta )e^{ikz}$ with $\mathbf{u}=(1,i,0)$. For $\mathbf{e}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$, contraction givesAt the observer origin, write $\mu =\cos \theta$. The tensor CMB line-of-sight source isThe rest of the expression is independent of $\phi$. Projection onto $Y_{\ell m}^{\ast }\propto e^{-im\phi }$ therefore gives zero unless $m=2$. This is the helicity selection for tensor CMB multipoles. It applies to the specified complex helicity mode; taking a physical real field also introduces the conjugate mode and its conjugate azimuthal dependence.

For the very long wavelength solution during matter domination, differentiation gives $\dot{h}=-h_{\mathrm{prim}}{k}^2\eta /5+O(k^4{\eta }^3)$. Expand the plane wave factor once to obtainwhereLet ${\mathcal{N}}_{22}=\frac{1}{4}\sqrt{15/(2\pi )}$. Then $Y_{22}={\mathcal{N}}_{22}{\sin }^2\theta e^{2i\phi }$ and $Y_{32}=\sqrt{7}{\mathcal{N}}_{22}\mu {\sin }^2\theta e^{2i\phi }$. Reading off the long-wavelength tensor CMB quadrupole and octupole yieldsIn the limit ${\eta }_{\ast }/{\eta }_0\to 0$,A perfectly frozen tensor produces no redshift; the first contribution comes from its order-$k^2{\eta }^2$ evolution. The extra spatial phase makes the octupole one power of $k{\eta }_0$ smaller than the quadrupole.

For a centered Gaussian random field, Wick theorem takes the classical form of the Isserlis theorem: every even correlator is a sum over pairings of two-point correlators, and every odd correlator vanishes. For example, with $G_i={\zeta }_G({\mathbf{x}}_i)$,A $2n$-point correlator has $(2n-1)!!$ pairings. Equivalently, all connected correlation functions beyond order two vanish. The covariance completely determines the centered Gaussian statistics; higher ordinary correlators can be nonzero, but contain no independent connected information. For a field with nonzero mean, apply these statements to the centered field and then restore its mean.

Use the Fourier transform convention $\zeta (\mathbf{x})=\int d^{3}k{e}^{i\mathbf{k}\cdot \mathbf{x}}\zeta (\mathbf{k})/(2\pi {)}^3$. Statistical homogeneity and isotropy define the primordial bispectrum byThe Dirac delta function enforces momentum conservation; the three magnitudes describe a closed triangle. Set $\alpha =3f_{\mathrm{NL}}/5$. For the centered quadratic Gaussian transformation, its quadratic Fourier term isAt first order in $\alpha$, choose the quadratic field at one of the three external positions. For the first position, Wick contractions pair the two internal fields with the two remaining external fields in two ways. The self-pairing is precisely removed by the subtracted mean. The surviving contribution is $2\alpha P(k_2)P(k_3)$ times the momentum delta. Adding the other positions gives the tree-level local-type primordial non-Gaussianity resultThis expression is at leading order in $f_{\mathrm{NL}}$. The exact quadratic map also has a loop correction to the local primordial bispectrum, from three quadratic vertices:There is no quadratic-in-$\alpha$ bispectrum term, because it would involve five centered Gaussian fields. A regulator may be needed for this higher-order integral. The displayed leading formula uses the Gaussian power $P$, consistently neglecting its order-$f_{\mathrm{NL}}^2$ correction.

Distinguish the slow-roll approximation parameter $ϵ$ from the infinitesimal positive contour regulator $\epsilon$. In the interaction picture, the ordered unequal-time Wick contraction isA differentiated vertex field instead supplies $u_k^{\ast \prime }(\tau )$. These conjugates are fixed by the annihilation operators and creation operators and cannot be dropped while retaining the same contour.

Convert the interaction time integral to conformal time. Since $dt=ad\tau$ and each cosmic-time derivative is $a^{-1}{\partial }_{\tau }$, the integrated cubic vertex isThe in-in formalism supplies $-2i$ multiplying the expectation of the Hamiltonian, so its negative sign gives $+2i$ multiplying this vertex. At nonzero external momenta, the connected contractions join each of the three external fields to one vertex field. There are six bijections, not three. Choosing which external field meets the undifferentiated vertex gives three cyclic choices; swapping the two differentiated fields gives a further factor of two. This is the Wick-pairing multiplicity for a zeta zeta-prime-squared vertex.

For example, before the internal momenta are integrated, one cyclic contribution includesThe factors of $2\pi$ cancel to leave one overall $(2\pi {)}^3$ and the external momentum delta. Thus the properly normalized reduced expression isThe vacuum prescription for inflationary in-in integrals damps the early-time oscillations and selects the Bunch-Davies vacuum. Internal self-contractions correspond to disconnected zero-momentum tadpole terms and are excluded from this connected three-point function.

The printed intermediate expression has unconjugated modes and only three literal cyclic terms with coefficient $-2i$. The conjugate form of the result above would use $-4i$, together with the conjugated contour. A literal $-2i$ with only three terms gives half the final printed answer. The six-contraction expression above is the consistent reduction of the supplied Hamiltonian and leads to that final answer.

Write $A_k=H/\sqrt{4ϵM_{\mathrm{Pl}}^2{k}^3}$, so $u_k(0)=A_k$. For the cyclic term with the first field undifferentiated, insertion into the in-in formalism giveswhere $K=k_1+k_2+k_3>0$. The two differentiated modes contribute ${\tau }^2$, cancelling the ${\tau }^{-2}$ in $a^2$.

Along the vacuum-selected contour, the lower boundary vanishes. The elementary integrals areand thereforeMultiplying by $4i{M}_{\mathrm{Pl}}^2{ϵ}^2$, taking the real part and summing the three cyclic choices gives the curvature bispectrum from a zeta zeta-prime-squared interaction:This matches the final normalization. The coefficient requires all six connected Wick contractions.

The calculation assumes the Gaussian Bunch-Davies vacuum, tree order in the specified cubic interaction, effectively constant $H$ and $ϵ$ during the integral, the approximate de Sitter scale factor, and observation after the modes freeze as $\tau \to 0^{-}$. The contour tilt is kept until the early boundary has been eliminated; an undamped real-axis integral cannot simply discard that boundary. The reduced Planck mass convention is the one in the supplied De Sitter curvature mode function. Other cubic interactions and nonlinear field redefinitions are outside the specified model, so this is not by itself the complete bispectrum of a general single-field action.

At late time the Gaussian curvature power spectrum is $P(k)=H^2/(4ϵM_{\mathrm{Pl}}^2{k}^3)$. Consider the squeezed bispectrum configuration $k_1=q\ll k_2\simeq k_3=k$. The local-type primordial non-Gaussianity result is dominated byIn the curvature bispectrum from a zeta zeta-prime-squared interaction, the leading cyclic term has the long mode in the undifferentiated slot. Its momentum numerator tends to $k^4/(2k)=k^3/2$, while the other two cyclic terms contain $q^2$. HenceThis is the slow-roll amplitude of the squeezed zeta zeta-prime-squared bispectrum. It retains the local-like $q^{-3}$ enhancement: assigning the soft momentum only to differentiated legs would incorrectly miss the dominant permutation. Comparing squeezed amplitudes givesFor comparison, in the equilateral configuration $B^{\mathrm{sf}}(k,k,k)=(4ϵ/3)P(k{)}^2$, so the signal is slow-roll suppressed away from the squeezed limit as well.

Use the quoted observational bound as the sensitivity benchmark supplied by this problem, rather than as a new current measurement. A local-template sensitivity at the order-ten level is far above this order-$ϵ$ signal; detection of this interaction with comparable CMB measurements is consequently very unlikely. The entire shape is not identical to the local template, so its bound is not an exact constraint on every single-field shape, but the squeezed amplitude already exposes the large suppression. The obstacle is the slow-roll amplitude, even though this specified interaction has a squeezed enhancement.