Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 53 (Advanced Cosmology)

Write $h_{ij}={}^{(3)}{g}_{ij}$ for the positive spatial metric tensor and $h^{ij}$ for its inverse. Expanding the 3+1 decomposition gives $g_{00}=N^2-h_{ij}{N}^i{N}^j$, $g_{0i}=-h_{ij}{N}^j$ and $g_{ij}=-h_{ij}$. The inverse isFor example, $g^{00}{g}_{00}+g^{0i}{g}_{i0}=1$, and $g^{00}{g}_{0j}+g^{0i}{g}_{ij}=0$. The spatial block similarly gives $g^{i0}{g}_{0j}+g^{ik}{g}_{kj}={\delta }^i{}_j$. These checks determine all blocks without treating the spatial block alone as the inverse of the four-metric.

Raising the normal covector with this inverse metric tensor yieldsTake $N>0$ so it is future-pointing. Its norm is $n_{\mu }{n}^{\mu }=1$. The spatial projection tensor obeys $P^{\mu }{}_{\nu }{n}^{\nu }=0$, and multiplication givesThus it is an idempotent linear projection onto vectors tangent to the spatial hypersurface. The negative spatial components of the spacetime metric tensor do not change this idempotence.

With the $+---$ signature and $n^2=1$, direct expansion of the two spatial projection tensors givesThis restriction is negative definite on spatial vectors. The positive spatial induced metric is instead $h_{\alpha \beta }=n_{\alpha }{n}_{\beta }-g_{\alpha \beta }=-q_{\alpha \beta }$, whose pullback to a time slice is the $h_{ij}$ used in the line element. The spatial metric sign for a unit timelike normal is important here: the plus sign in the PDF's claimed equality is incompatible with its normal normalization: $g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ is not even transverse to $n^{\alpha }$.

The spatial covariant derivative of a spatial tensor projects every index, including its derivative index. Projection only on the derivative index is sufficient for a scalar but not for a general tensor. Using metric compatibility of the spacetime Levi-Civita connection, we obtainEach term contains a normal contracted with its spatial projection tensor. Consequently $\boxed{D_i{h}_{jk}=0}$, and the negative spatial restriction also satisfies $D_i{q}_{jk}=0$. This proves the requested metric compatibility of the spatial covariant derivative after correcting the source's metric sign. Contracting indices gives the particular expression written in the question.

Differentiate the spatial projection tensor before making any contractions:For the spatial projector derivative identity, its first term vanishes after projection on $\mu$. The definition of the extrinsic curvature of a spatial hypersurface then gives the stronger tensor identityHere $\lambda$ remains a free index. For a normal to a genuine foliation, the extrinsic curvature is symmetric: projecting ${\nabla }_{[\nu }{n}_{\mu ]}$ gives zero because locally $n_{\mu }$ is a scalar multiple of a time gradient. This is hypersurface orthogonality implies symmetric extrinsic curvature.

Contract $\rho$ with $\lambda$ in the stronger identity to obtain exactly the contraction displayed in the PDF:The last equality follows from the transversality of the extrinsic curvature. Thus the literal printed identity is valid, although both its sides vanish; the uncontracted identity explains its geometric origin.

For a centered Gaussian random field, Wick's theorem says that every odd moment vanishes and every even moment is the sum over all pairings of products of two-point functions. For six fields there are $15$ pairings. In the connected three-point function at nonzero external momenta, each of the three external fields must pair with a distinct field at the cubic interaction. There are $3!=6$ such Wick contractions. Choosing the undifferentiated field gives three possibilities; interchanging the two differentiated fields gives a further factor of two. The remaining nine pairings involve an external-external pair and an internal pair and belong to tadpole/disconnected contributions. Define the background so the one-point function vanishes, or equivalently subtract these contributions.

It is useful to keep a coefficient $\mathcal{C}$ multiplying the cubic Hamiltonian: its literal printed value is $1$, whereas the standard dimensionally normalized curvature interaction has $\mathcal{C}=M_{\mathrm{Pl}}^2$. This distinction will matter for the final amplitude. Withand $dt=ad\tau$, $\dot{\zeta }={\zeta }^{\prime }/a$, the interaction entering the time integral isIn the interaction picture, the unequal-time vacuum contraction required by the in-in formalism isFor a differentiated internal field replace $u_k^{\ast }$ by $u_k^{\ast \prime }$. The equal-time power spectrum fixes its magnitude; the displayed free De Sitter curvature mode functions and vacuum choice fix its unequal-time phase.

Performing the three momentum integrations imposes ${\mathbf{p}}_a=-{\mathbf{k}}_a$ for each assignment and leaves one overall momentum delta function. Put $U=u_{k_1}(0)u_{k_2}(0)u_{k_3}(0)$, which is real here, and defineThe upper early-time contour ${\mathcal{C}}_{+}$ runs from $-\infty (1-i\delta )$ to $0$, with $\delta >0$. The Hamiltonian's minus sign and the six connected Wick contractions then giveThis is equivalently $\mathrm{Re}[-4i\mathcal{C}{ϵ}^{2}U\sum I_i]$ with unconjugated De Sitter curvature mode functions and the conjugate lower contour ${\mathcal{C}}_{-}$. Before momentum integration, the same result consists of the three cyclic delta assignments, each with the extra factor of two for the identical differentiated legs.

The PDF's intermediate formula writes only three cyclic assignments without that factor of two. Taken literally it undercounts the connected contractions. Its unconjugated modes must also use the conjugate contour, rather than the upper contour of the original in-in expression. Both points are required for a consistent in-in bispectrum conjugation rule.

Let $A_k=H/(2\sqrt{ϵ}{M}_{\mathrm{Pl}}{k}^{3/2})$ and $K=k_1+k_2+k_3>0$. For the unconjugated integrand on the lower contour, the De Sitter curvature mode functions giveHere the two derivative factors supply ${\tau }^2$, cancelling $a^2=1/(H^2{\tau }^2)$. The vacuum prescription for inflationary in-in integrals can be implemented by a factor $e^{\delta \tau }$ on the real negative axis, with $\delta >0$, followed by $\delta \downarrow 0$. In this notationThereforeAn undamped boundary evaluation on the real axis at $-\infty$ is not valid. On the upper contour the complex-conjugate integrand instead yields the conjugate value. Conjugating the modes without conjugating the contour would produce exponential growth.

Using all six connected Wick contractions from the preceding part and $U=A_{k_1}{A}_{k_2}{A}_{k_3}$ gives the literal-Hamiltonian resultThe connected correlator is $(2\pi {)}^3{\delta }^{(3)}(\sum {\mathbf{k}}_i)B_{\mathcal{C}}$. The three assignments constitute the curvature bispectrum from a zeta zeta-prime-squared interaction. The numerator contains six powers of $H$ from the six modes, two of which are cancelled by $a^2$; the six powers of $M_{\mathrm{Pl}}$ remain unless the vertex supplies two.

Consequently the printed final expression, with $M_{\mathrm{Pl}}^{-4}$, is obtained for $\boxed{\mathcal{C}=M_{\mathrm{Pl}}^2}$. With the literal printed Hamiltonian $\mathcal{C}=1$, the answer instead has $\boxed{M_{\mathrm{Pl}}^{-6}}$. The Planck normalization of a cubic curvature interaction requires a factor $M_{\mathrm{Pl}}^2$ in the Hamiltonian if its final bispectrum is intended. In addition, integrating the intermediate expression literally with only three assignments gives half the fully contracted amplitude. These are normalization defects, rather than changes in the momentum shape. For dimensionless comoving curvature perturbations, the usual interaction normalization is also required by the mass dimension of the action.

For the standard normalization, $P_{\zeta }(k)=H^2/(4ϵM_{\mathrm{Pl}}^2{k}^3)$, so $B/P_{\zeta }^2$ is of order $ϵ$. This slow-roll suppression means that the primordial non-Gaussianity from this interaction alone is not expected to be detectably large. For example, in the squeezed bispectrum configuration $k_1\ll k_2\simeq k_3$, matching this contribution to the local convention gives $f_{\mathrm{NL}}=5ϵ/24$. This is a single-vertex contribution, not the complete single-field slow-roll inflation prediction; other vertices and field redefinitions contribute at the same slow-roll order. A claim of detectability for enhanced interactions requires a model beyond this approximation.

The one-particle distribution function is a scalar function on the future mass shell. Choose its normalization so that, in a local orthonormal tetrad, $f{d}^{3}x{d}^{3}p$ is the expected photon number in a phase-space cell on a constant-time slice. Spin degeneracy and any phase-space normalization factors are included in $f$; with an occupation-number convention they would instead appear explicitly in the measure.

More covariantly, the photon number crossing a spacelike element is $f{p}^{\mu }d{\Sigma }_{\mu }{d}^{3}p/E$. The factor $p^{\mu }d{\Sigma }_{\mu }$ reduces to $E{d}^{3}x$ on a local constant-time slice, recovering the cell interpretation. The measure $d^{3}p/E$ is Lorentz invariant on the mass shell.

A photon carries energy $E$ and momentum $p^i$, while its velocity is $p^i/E$. Energy density, momentum density and momentum flux are thereforeThey assemble into the stress-energy tensorThe tetrad momenta are related to coordinate components by the tetrad basis. For an isotropic distribution, angular averaging gives $T^{\hat{i}\hat{j}}={\rho }_{\gamma }{\delta }^{ij}/3$ and $T^{\hat{0}\hat{i}}=0$. Thus the photon gas has pressure ${\rho }_{\gamma }/3$, consistent with its massless equation of state.

A small direction-dependent temperature change $T=\bar{T}(1+\Theta )$ shifts a thermal spectrum according to $\bar{f}(ϵ/(1+\Theta ))=\bar{f}(ϵ)-\Theta ϵ{\bar{f}}^{\prime }(ϵ)$ at first order. Hence $\boxed{\Theta =\delta T/\bar{T}}$ is the angular photon temperature perturbation. For a general spectrum the same ansatz describes an energy-independent brightness dilation; a literal thermodynamic temperature interpretation additionally assumes a thermal shape without spectral distortions.

Use $E=ϵ/a$ in the tetrad stress-energy tensor. The background photon energy density isIntegration by parts gives $-\int dϵ{ϵ}^4{\bar{f}}^{\prime }=4\int dϵ{ϵ}^3\bar{f}$, assuming the boundary term vanishes. Thus each directional energy perturbation is four times its temperature perturbation. Writing $⟨\cdot {⟩}_{\Omega }=(4\pi {)}^{-1}\int d\Omega$, we haveIn the paper's convention for photon angular temperature moments, there is no $(2\ell +1)$ factor multiplying ${\Theta }_{\ell }$. Legendre polynomial orthogonality gives $⟨\Theta {⟩}_{\Omega }={\Theta }_0$ and $⟨e^i\Theta {⟩}_{\Omega }=-i{\hat{k}}^i{\Theta }_1/3$. Comparing with the stated density and velocity conventions yieldsFor completeness, $⟨e^i{e}^j{P}_2(\hat{\mathbf{k}}\cdot \mathbf{e}){⟩}_{\Omega }=({\hat{k}}^i{\hat{k}}^j-{\delta }^{ij}/3)/5$. The $(-i{)}^2$ quadrupole phase then gives $\boxed{{\Pi }_{\gamma }=-3{\Theta }_2/5}$. These moment relations are photon angular temperature moments and keep the anisotropic stress convention consistent with the subsequent hierarchy.

All dots in this solution denote differentiation with respect to conformal time $\eta$. This is required by comoving $k$, the streaming term $k{\Theta }_{\ell }$, and $\dot{\tau }=-a{\bar{n}}_e{\sigma }_T$. The PDF's reference to proper-time derivatives is inconsistent with these coefficients. Proper-time equations follow by dividing the conformal-time equations by $a$, replacing streaming $k$ by $k/a$ and the collision rate by $-{\bar{n}}_e{\sigma }_T$.

The monopole equation of the photon Boltzmann hierarchy is ${\dot{\Theta }}_0+k{\Theta }_1/3=\dot{\varphi }$; its collision term vanishes. Consequently the photon continuity equation isFor the dipole the hierarchy givesUsing the relations from the preceding part, the photon Euler equation becomesBecause $\dot{\tau }<0$, the collision term drives the photon velocity towards the baryon velocity, as Thomson scattering should.

For pressureless baryons, stress-energy conservation without energy exchange gives $\boxed{{\dot{\delta }}_b-k{v}_b-3\dot{\varphi }=0}$. At this order Thomson scattering changes momentum but not the local photon energy: energy transfer by the relative bulk motion is second order, and recoil/thermal energy exchange is neglected. Total momentum conservation fixes the baryon force to be the negative of the photon force. The momentum densities are proportional to their enthalpy densities, $4{\bar{\rho }}_{\gamma }/3$ and ${\bar{\rho }}_b$. ThusIncluding the pressureless expansion and gravitational terms yields the baryon Euler equation with Thomson dragThe expansion coefficient for photons cancels against the changing background pressure, while pressureless baryons retain the Hubble drag term. The opposite drag signs and the enthalpy ratio ensure that the combined photon-baryon fluid conserves momentum.

In tight coupling, the scattering time $|\dot{\tau }{|}^{-1}$ is short compared with an acoustic period and an expansion time: $k/|\dot{\tau }|\ll 1$ and $\mathcal{H}/|\dot{\tau }|\ll 1$. Photons and baryons have nearly the same velocity; higher photon multipoles and the photon-baryon velocity slip are suppressed by powers of this small ratio. Here retain the collision operator as printed, which ignores CMB polarization. Its diffusion coefficient differs from the polarized result.

Neglect gravity and expansion, so $R$ is constant on the timescale under consideration. The quadrupole equation of the photon Boltzmann hierarchy isTo first order in $k/|\dot{\tau }|$, ${\dot{\Theta }}_2$ and $k{\Theta }_3$ are subleading relative to the dipole source. HenceThis is the temperature-only tight-coupling quadrupole. The negative collision rate is essential to its sign.

Put $\Delta =v_{\gamma }-v_b$. Subtracting the baryon Euler equation with Thomson drag from the photon Euler equation givesThe zeroth-order common velocity obeys $(1+R)\dot{v}=-k{\delta }_{\gamma }/4$. Solving the slip equation to its first nonzero order therefore givesThe second expression assumes that $R$ and $\dot{\tau }$ vary only on the neglected background timescale. Terms from their variation would need retaining if cosmic expansion were restored. Multiplying the baryon Euler equation with Thomson drag by $R$ and adding it to the photon equation eliminates drag. Since $v_b=v_{\gamma }-\Delta$,Using $v_{\gamma }=3{\dot{\delta }}_{\gamma }/(4k)$ from the photon continuity equation, the quadrupole term contributes $(2/9)k{\dot{\delta }}_{\gamma }/\dot{\tau }$, and the slip term contributes $R^{2}k{\dot{\delta }}_{\gamma }/[4(1+R)\dot{\tau }]$. HenceDifferentiating the photon continuity equation gives the photon-baryon diffusion damping equationThe sound speed reflects baryon inertia; the $R^2$ term is heat conduction through velocity slip, and the $8/9$ term is photon shear viscosity for the unpolarized hierarchy.

For constant coefficients the characteristic roots are $-D/2\pm \sqrt{D^2/4-c_s^2{k}^2}$. On the acoustic branch where $D<2c_{s}k$,Thus the solutions are damped acoustic oscillations, with positive diffusion damping growing as $k^2$. This is Silk damping. Slowly varying coefficients give an approximate envelope $\exp [-\frac{1}{2}\int Dd\eta ]$ and acoustic phase $k\int c_{s}d\eta$. The mathematical critically damped and overdamped cases follow from the roots, but extrapolation beyond the tight-coupling range is not reliable. For very large $R$, the acoustic and damping scales must be compared explicitly rather than inferring underdamping from $k/|\dot{\tau }|\ll 1$ alone.

For a statistically homogeneous comoving curvature perturbation, define the connected primordial bispectrum byStatistical isotropy makes $B$ depend only on the three magnitudes, which must form a triangle. A Gaussian random field has zero connected bispectrum. Canonical attractor single-field slow-roll inflation with the usual vacuum produces only slow-roll-sized primordial non-Gaussianity. A measurable signal can arise from additional light fields and nonlinear conversion of isocurvature perturbations, or from enhanced interactions such as a small sound speed, departures from an attractor, or suitable features/excited initial states. The resulting model must still reproduce the observed nearly scale-invariant power spectrum and remain under perturbative control, with acceptable backreaction and late-time conversion to the observed perturbations. Large non-Gaussianity is therefore possible but is not automatic in a viable model.

Put $c=3f_{\mathrm{NL}}/5$ and use the consistent Fourier transform convention with ${\zeta }_G(\mathbf{x})$, rather than the repeated momentum argument printed in the integrand. The quadratic part has transformThe subtraction sets the mean to zero and removes the internal contraction of a single quadratic factor. At first order in $f_{\mathrm{NL}}$, choose one of the three external factors to be quadratic. For example, the two connected Wick contractions at the third leg pair its two fields with the first two legs, givingSumming the three placements provesThis is the leading local primordial bispectrum. The factor $6/5$ is the quadratic coefficient $3/5$ multiplied by the two cross-pairings. Terms of order $f_{\mathrm{NL}}^2$ have an odd Gaussian moment and vanish.

The literal quadratic local model also has a connected cubic-in-$f_{\mathrm{NL}}$ contribution, from one quadratic factor at every leg:This is the loop correction to the local primordial bispectrum; regulators may be needed for idealized spectra. Thus the PDF's displayed formula is the tree/leading-order result, not an exact identity for arbitrary $f_{\mathrm{NL}}$. The weak-non-Gaussian expansion assumes these loop terms are small. The local shape is enhanced in the squeezed bispectrum configuration when a long-wavelength perturbation modulates small-scale power.

Insert the primordial bispectrum into the product of the three linear transfer integrals. Write $L={\ell }_1+{\ell }_2+{\ell }_3$ and abbreviate $g_{{\ell }_i}({\eta }_0,k_i)$ by $g_i$. The observer-position phase is one because the momentum delta function imposes $\sum {\mathbf{k}}_i=0$. Represent that delta function byThe Rayleigh plane-wave expansion and angular orthogonality give, for each momentum,The three $i^{\ell }$ factors cancel the $(-i{)}^L$ in the temperature multipoles. Angular integration over $\hat{x}$ leaves the complex conjugate Gaunt integral. In the conventional complex spherical harmonics, this integral is real, and it vanishes unless the angular momentum triangle, even-parity and $m_1+m_2+m_3=0$ selection rules hold. Therefore its conjugate equals itself.

The radial measure is $x^{2}dx$, and the momentum radial measures are $k_i^{2}d{k}_i$. The combined numerical prefactor is $(4\pi {)}^6/(2\pi {)}^9=8/{\pi }^3=64/(2\pi {)}^3$. Hence the reduced CMB bispectrum isand the angular three-point function factorizes asThis primordial-to-angular bispectrum projection separates dynamics and radial transfer from purely angular geometry. The spatial integration variable is auxiliary, not the observer position. Linear transfer is justified at leading order in the primordial signal; it does not require a large amplitude mathematically, although a signal must exceed measurement uncertainty to be detectable.

Let $b_{{\ell }_1{\ell }_2{\ell }_3}^1$ denote the template at unit $f_{\mathrm{NL}}$, and define $A_{123}=b_{{\ell }_1{\ell }_2{\ell }_3}^1{\mathcal{G}}_{m_1{m}_2{m}_3}^{{\ell }_1{\ell }_2{\ell }_3}$. All sums below are over ordered triples of the retained multipoles, with the monopole removed and $C_{\ell }>0$. A finite maximum multipole makes these expressions ordinary finite sums. Taking the expectation of the cubic numerator and using the template relation givesThus the full-sky cubic bispectrum estimator is unbiased forThe assumed template must have nonzero support, so $F>0$. Using the Gaunt sum rule, writeThe factor $1/6$ belongs to the ordered sum; an unordered triangle sum would instead use multiplicity factors for equal multipoles. This normalization is also the Gaussian Fisher information for the template amplitude.

For the cosmic variance, use the reality condition ${\Theta }_{\ell m}^{\ast }=(-1{)}^m{\Theta }_{\ell ,-m}$ and Gaussian covarianceThere are $15$ pairings of the six multipoles in the squared cubic numerator. Six pairings connect every factor in the first triple to a factor in the second. Each gives $\sum A_{123}^2/(C_{{\ell }_1}{C}_{{\ell }_2}{C}_{{\ell }_3})=6F$. The covariance phases cancel: nonzero Gaunt integrals have $m_1+m_2+m_3=0$, and simultaneous reversal of the three $m$ indices multiplies the Gaunt integral by $(-1{)}^{{\ell }_1+{\ell }_2+{\ell }_3}=1$.

The other nine pairings contain one contraction within each triple. To see why their sums vanish, contract two legs of a Gaunt integral and use the spherical harmonic addition theorem:The template and inverse-covariance weights are independent of $m$, so they do not spoil this cancellation. The only possible surviving unpaired mode is the monopole, and ${\Theta }_{00}=0$ removes it. This is monopole cancellation of internal cubic-estimator contractions. It explains why no linear correction is needed in this ideal full-sky isotropic problem; masks or anisotropic noise would spoil the argument.

The cubic numerator therefore has Gaussian variance $6(6F)=36F$. Dividing by $(6F{)}^2$ givesThe result concerns the Gaussian-limit covariance. Non-Gaussian connected four- and six-point terms can change the variance at finite amplitude. Unbiasedness uses the assumed linear template relation for the observed three-point function.