Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 57 (Advanced Cosmology)

Use signature $(-,+,+,+)$ and write $h_{ij}$ for the spatial induced metric, $D_i$ for its spatial covariant derivative, and $q^2=h^{ij}{D}_i\varphi D_j\varphi$. The negative shift vector convention gives $n^{\mu }=N^{-1}(1,N^i)$ and $\Pi =n^{\mu }{\partial }_{\mu }\varphi$. Decomposing the scalar field derivative givesVarying the scalar field matter action with respect to the inverse metric tensor, including the variation of $\sqrt{-g}$, gives the stress-energy tensorRaising both indices gives the requested $T^{\mu \nu }$. Contracting with the unit normal twice, or once with the spatial projection tensor, givesProjecting both indices of the stress-energy tensor onto the spatial hypersurface givesTwo printed identifications require correction. The canonical momentum of the displayed density is ${\pi }_{\varphi }=\sqrt{h}\Pi$, not $\Pi$: explicitly ${\mathcal{L}}_m=N\sqrt{h}[{\Pi }^2/2-q^2/2-V]$ and differentiation with respect to $\dot{\varphi }$ supplies $\sqrt{h}$. Thus $\Pi$ is the normal scalar-field momentum. Also, the printed trace-free stress has an extra factor $1/2$ and uses ${\delta }_{ij}$ where the general spatial induced metric requires $h_{ij}$. In an orthonormal spatial frame the latter becomes ${\delta }_{ij}$, but the coefficient is still one. For a gradient $(q,0,0)$ in that frame, direct projection gives ${\tilde{S}}_{11}=2q^2/3$, rather than the printed $q^2/3$. The PDF's energy-density gradient is spatial ${\partial }_i\varphi {\partial }^i\varphi$; the TeX aid's spacetime index there is an OCR defect.

The unimodular spatial metric separates volume from shape: writing $h_{ij}=a^2{\tilde{h}}_{ij}$ and $a^6=h=\det h_{ij}$ gives $\det \tilde{h}=1$. Contract the extrinsic curvature with a negative shift with $h^{ij}$. Metric compatibility and the Jacobi determinant derivative formula giveConsequentlyFor zero shift vector, $\dot{h}/h=6\dot{a}/a$, soThe infinitesimal spatial volume is $\sqrt{h}{d}^{3}x=a^3{d}^{3}x$; its logarithmic change per unit normal proper time is $3H$. Thus this $H$ is the local volume Hubble parameter, reducing to the usual Hubble parameter in a homogeneous universe. The determinant-defined $a$ refers to fixed spatial coordinates, while this normal volume expansion has a geometric interpretation.

Retain zeroth and first spatial-gradient orders in the long-wavelength approximation in cosmology. Spatial Ricci curvature, $D_i{D}_{j}N$, $q^2$ and scalar anisotropic stress are of second gradient order. Put $A^i{}_j={\tilde{K}}^i{}_j$ and $A^2=A^i{}_j{A}^j{}_i$. The supplied Einstein field equations then reduce toThe scalar-field stress projections show that the momentum constraint is a first-gradient equation and must be retained. The square $A^2$ is not a spatial derivative and cannot yet be dropped merely by invoking the long-wavelength approximation in cosmology. The time-derivative dot on the trace-free evolution equation is visible in the PDF and missing from the TeX aid.

Using $NH=\dot{a}/a$, the last equation becomes ${\partial }_t(a^3{A}^i{}_j)\simeq 0$. HenceThe time-independent spatial tensor must also satisfy the momentum constraint and the symmetry inherited from $K_{ij}$. This is inflationary shear damping: expanding regions lose anisotropic expansion, without requiring the frozen spatial metric shape to vanish. The matter evolution, useful below, is $\dot{\varphi }=N\Pi$ and $\dot{\Pi }\simeq -N(3H\Pi +V_{,\varphi })$, the leading Klein-Gordon equation in each local region.

Now impose the stipulated negligible shear and work where $\Pi \ne 0$. The Hamiltonian constraint, momentum constraint and local expansion evolution giveTime differentiation of the first equation, followed by $\dot{\varphi }=N\Pi$, gives $\dot{\Pi }=-N(3H\Pi +V_{,\varphi })$. Spatial differentiation of the same equation and use of the momentum constraint givesThese are the temporal and spatial identities needed for the nonlinear curvature covector. Differentiate its definition, keeping the spatially varying lapse function:The terms proportional to ${\partial }_{i}N$ cancel. The terms $-N{\partial }_{i}H$ and $-4\pi GN\Pi {\partial }_i\varphi$ cancel by the momentum constraint, and the remaining two terms cancel by the spatial momentum identity. Thereforeto the retained gradient order. No homogeneous-lapse assumption was needed.

For linear adiabatic cosmological perturbations, expansion of the local scale and scalar field gives ${\zeta }_i={\partial }_i[\Psi +\bar{H}\delta \varphi /\bar{\Pi }]$. For nonzero spatial modes, linearizing the Hamiltonian constraint and the momentum constraint givesThusThis uses the paper's overall sign for the uniform-density curvature perturbation; the opposite overall sign is also common. For an adiabatic single-field attractor, superhorizon conservation of single-field comoving curvature allows the primordial amplitude to be transported through later epochs without following every short-scale process. This proof specifically discarded the shear contribution to the momentum constraint and assumed a usable field clock. It is not a claim that every single-field background conserves every curvature mode: growing curvature perturbation in ultra-slow-roll inflation is a non-attractor counterexample, and $\Pi =0$ makes this particular covector undefined.

Take $w$ constant and $w\ne -1$, with a barotropic perturbation $\delta P=w\delta \rho$. Define the density contrast $\delta =\delta \rho /\bar{\rho }$ and use the scalar velocity potential convention $v_i(\mathbf{k})=i{k}_i\theta (\mathbf{k})$, so ${\partial }_i{v}^i=-k^2\theta$. This $\theta$ is a potential, not the sometimes-used velocity-divergence variable. In synchronous gauge in cosmology, the inverse metric is $g^{00}=-a^{-2}$ and $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$ to first order. Discarding products of perturbations in the perfect fluid in general relativity givesFor the linearization, each $h_{ij}$ must be small: the PDF's determinant condition alone is insufficient, since $h=\mathrm{diag}(L,0,0)$ has zero determinant even for large $L$.

Write $\mathcal{H}=a^{\prime }/a$. The time component of covariant conservation of stress-energy isHere ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}+h^{\prime }/2$. To first order, ${\Gamma }_{ij}^0{T}^{ij}=a^{-2}w\bar{\rho }[3\mathcal{H}(1+\delta )+h^{\prime }/2]$: the two metric-$h$ contributions cancel. Multiplying the conservation equation by $a^2$, including ${\Gamma }_{00}^0{T}^{00}$, givesThe background cosmological perfect-fluid continuity equation, ${\bar{\rho }}^{\prime }=-3\mathcal{H}(1+w)\bar{\rho }$, cancels the first and third terms. ThusThe $h^{\prime }$ term accounts for perturbation of the spatial volume expansion in synchronous gauge in cosmology. The PDF's correct connection ${\Gamma }_{0j}^i=\mathcal{H}{\delta }_j^i+h^{\prime i}{}_j/2$ is garbled in the TeX aid.

In the decoupled perfect fluid in general relativity limit, photons have $w=1/3$ and no anisotropic stress, while cold baryons have $w\simeq 0$ with a small pressure correction $c_b^2{\delta }_b$. The density equations express particle and energy conservation; Thomson scattering transfers momentum rather than leading-order rest-frame energy. The photon Euler equation consequently gains a drag toward the baryon velocity potential, and the baryon Euler equation gains the equal-and-opposite drag weighted by their inertias.

Let ${\Gamma }_T=a{n}_e{\sigma }_T>0$. With the velocity potential used in part (i), these forces are $-{\Gamma }_T({\theta }_{\gamma }-{\theta }_b)$ and $-R{\Gamma }_T({\theta }_b-{\theta }_{\gamma })$, with $R=4{\bar{\rho }}_{\gamma }/(3{\bar{\rho }}_b)$. Multiplication by the photon and baryon enthalpies, $4{\bar{\rho }}_{\gamma }/3$ and ${\bar{\rho }}_b$, makes their sum zero. Note that this paper's $R$ is the reciprocal of the usual baryon loading parameter. Thomson scattering also relaxes the photon quadrupole, giving the $-{\Gamma }_T{\sigma }_{\gamma }$ term in the stipulated simplified hierarchy. This quadrupole is an anisotropic stress, omitted by a perfect-fluid model; finite relaxation creates Silk damping.

The printed baryon density sign and collision normalization are inconsistent with part (i). The density equation must be ${\delta }_b^{\prime }-k^2{\theta }_b=-h^{\prime }/2$. As a direct check, with $h^{\prime }=0$, nonzero common velocity, and initially ${\delta }_{\gamma }=4{\delta }_b/3$, the printed plus sign gives $({\delta }_{\gamma }-4{\delta }_b/3{)}^{\prime }=8k^2\theta /3$, violating the tight-coupling adiabatic condition immediately. Likewise division of the collision forces by $k^2$ belongs to neither this potential convention nor the usual common velocity-divergence convention; the physical relaxation rate is ${\Gamma }_T$, independent of $k$. Both printed drag terms still cancel in the weighted sum used below, so that particular algebraic combination is unaffected by this normalization error.

For a nonrelativistic distribution, $T^{ij}=\int d^{3}p{p}^i{p}^{j}f/E$ is down by $v^2$ relative to the rest-mass energy density, and the corresponding anisotropic stress is also suppressed. The streaming terms coupling successive angular moments are proportional to $kv$, and stresses from the velocity distribution disappear in the cold limit. Thus, on scales where velocity dispersion is negligible, density and bulk velocity suffice; the photon Boltzmann hierarchy has no analogous small-speed suppression. This argument needs a cold distribution or a controlled small-dispersion approximation. Collisionless nonrelativistic matter with appreciable dispersion or multistreaming need not be an exact perfect fluid, and near photon decoupling the higher photon moments cannot all remain negligible.

Use the corrected baryon density equation from part (ii). Subtracting $4{\delta }_b^{\prime }/3$ from the photon equation gives $({\delta }_{\gamma }-4{\delta }_b/3{)}^{\prime }=(4/3)k^2({\theta }_{\gamma }-{\theta }_b)$. Tight coupling therefore preserves adiabatic initial conditions to leading order: ${\theta }_b={\theta }_{\gamma }=\theta$ and ${\delta }_b=3{\delta }_{\gamma }/4$.

Multiply the photon Euler equation by the paper's inertia ratio $R$ and add the baryon Euler equation before setting the velocities equal. The Thomson scattering terms cancel exactly. With $\mathcal{H}=a^{\prime }/a$ and $c_b^2$ denoting the baryon sound speed, this yieldsNo derivative of $R$ occurs in this instantaneous weighted addition; writing a derivative of total momentum instead would also differentiate the background inertia. Define the photon-baryon sound speedThen the coupled equations areThe sound speed is radiation pressure divided by the total photon-baryon inertia. Baryons add inertia with little pressure, reducing it below $1/\sqrt{3}$. The combined formula is meaningful despite the printed collision-factor error because the same erroneous factor cancels; maintenance of the assumed adiabatic relation does require correcting the baryon continuity sign.

For $R\gg 1$, $3{\tilde{c}}_s^2\simeq 1$, and the baryon-inertia corrections to the common Euler equation vanish. Differentiating the photon continuity equation givesSet aside the gravitational driving terms and balance the quadrupole streaming source against Thomson scattering. With ${\tau }_c={\Gamma }_T^{-1}$ and ${\sigma }_{\gamma }^{\prime }\simeq 0$,The positive friction is Silk damping; this particular truncation gives shear-only photon diffusion damping. Its coefficient follows the supplied simplified quadrupole equation; adding polarization and the full velocity-slip correction changes that coefficient, as in the general photon-baryon diffusion damping equation.

For completeness, put ${\omega }_0=k/\sqrt{3}$, $b=4k^2{\tau }_c/15$ and $D=\frac{1}{2}{\int }_{{\tau }_0}^{\tau }b(s)ds$. Substitution ${\delta }_{\gamma }=e^{-D}y$ gives the exact transformed equationFor constant ${\tau }_c$ the two independent solutions are $e^{-b(\tau -{\tau }_0)/2}\cos [\sqrt{{\omega }_0^2-b^2/4}(\tau -{\tau }_0)]$ and the corresponding sine. Their frequency correction begins at second order in ${\tau }_c$. For variable ${\tau }_c$, retaining first order alone does not automatically remove $b^{\prime }/2$: a general first-order representation isThe intended cosmological approximation additionally uses slow variation on an acoustic period, $\mathcal{H}/k\ll 1$, with tight coupling $k{\tau }_c\ll 1$. Then the $b^{\prime }$ correction to the frequency is subleading and the two independent photon-baryon acoustic oscillator solutions take the resummed damping formA shift of the time origin is absorbed in $A,B$. If the ionization fraction is constant, $n_e\propto a^{-3}$ gives ${\tau }_c\propto a^2\propto {\tau }^4$ during matter domination, and an initial origin at zero gives $k_D^{-2}=2\tau {\tau }_c/75$. Across recombination the ionization fraction varies, so the integral is the general answer. The PDF prints $O({\tau }_c^{-2})$, while the TeX aid prints $O({\tau }_c^2)$; the small-mean-free-time expansion requires the latter interpretation.

The intrinsic photon temperature fluctuation is $\delta T/T={\delta }_{\gamma }/4$. The cosine and sine produce the Cosmic microwave background acoustic peaks; the envelope suppresses their small-scale amplitudes. The corresponding contribution to the Cosmic microwave background power spectrum contains $e^{-2k^2/k_D^2}$, before angular projection and the other temperature sources are included. Thus Silk damping erases the high-multipole acoustic structure progressively rather than shifting every peak to a new frequency.

Let $X={\partial }^{-2}\zeta$, with an inverse spatial Laplacian defined by a boundary or zero-mode prescription. It depends on $\zeta$, not on its time derivative. Differentiation gives the canonical momentumCount $\zeta ,\pi ,X$ as first order. Inverting perturbatively givesFor the Legendre transform in mechanics, write $v=\dot{\zeta }=\pi +\Delta$ with $\Delta$ second order. The quadratic kinetic contribution is $\pi v-v^2/2={\pi }^2/2-{\Delta }^2/2$, whose correction is fourth order; in the cubic terms one may consequently replace $v$ by $\pi$. The Hamiltonian density is thereforeThe cubic interaction Hamiltonian density isIn the free interaction picture, ${\pi }_I={\dot{\zeta }}_I$. Thus here ${\mathcal{H}}_{\mathrm{int}}=-{\mathcal{L}}_3$ with the time derivatives interpreted as free fields. This equality follows from the cubic Legendre transform for scalar derivative interactions; at higher orders additional terms can appear.

For a statistically homogeneous isotropic dimensionless field, write its two-point autocorrelation function of a random field as $C(r)$ andThe formal real-space scale invariance condition is $C(\lambda \mathbf{r})=C(\mathbf{r})$ for all $\lambda >0$. Changing variable $\mathbf{q}=\lambda \mathbf{k}$ in the first expression and equating the nonzero-mode Fourier transforms gives ${\lambda }^{-3}P(q/\lambda )=P(q)$, equivalently $P(\lambda k)={\lambda }^{-3}P(k)$. Taking $\lambda =k/k_0$ yieldsThus a scale-invariant inflationary power spectrum has equal variance per logarithmic wavenumber interval, rather than equal power per Fourier volume.

There is an important mathematical qualification to the literal covariance condition. A nonzero $k^{-3}$ power spectrum over all scales has $C(0)=A\int dk/k/(2{\pi }^2)$, divergent at both endpoints; it is not the covariance of a finite-variance ordinary field. Moreover, exact dilation invariance of a continuous isotropic $C(r)$ makes it constant for $r>0$ and, by continuity, at zero: its spectrum can then consist only of a zero-mode delta measure. The nontrivial cosmological statement consequently concerns nonzero modes, with cutoffs or subtraction of an unobservable constant. For example the finite subtracted covarianceis invariant under simultaneous rescaling of $r,r_0$. This is the precise infrared qualification of a scale-invariant covariance; a regulated unsubtracted $C(r)$ generally shifts by an additive constant under dilation. The formal derivation above gives the intended nonzero-mode scaling, with this qualification rather than an impossible finite-variance premise.

Put $k_1=q\ll k$ and $k_2,k_3\to k$ in the squeezed bispectrum configuration. Then $K\to 2k$ and the three symmetric combinations in the supplied expression have limitsThe first two terms inside the $ϵ$ bracket cancel, leaving $4k^3$. The full bracket is therefore $(\eta +2ϵ)k^3/4$, andThe $(2\pi {)}^3$ is kept in the normalization of $F$ used by the paper; it should not be introduced a second time by changing the primordial bispectrum convention.

Momentum conservation makes the three vectors close into a triangle. As $q/k\to 0$, the two short-wavelength vectors approach equal magnitude and opposite direction, producing a thin, nearly collapsed triangle. In real space the $q$ mode varies slowly over a region containing many short wavelengths, so it modulates their local statistics. For a single adiabatic inflationary clock this is a local spatial dilation, the basis of the single-field inflation squeezed-limit consistency relation. If $\eta$ is the Hubble-flow convention $\dot{ϵ}/(Hϵ)$, the scalar spectral index in Hubble slow-roll parameters gives $\eta +2ϵ=1-n_s$ to first order. This identification must not be made with the potential slow-roll $\eta$ without converting conventions.

Let $G={\Phi }_G$ and $f=f_{NL}^{\mathrm{local}}$. The Gaussian random field covariance in Fourier space isThe Fourier transform of the centered quadratic term isThe Gaussian three-point function vanishes. At first order in $f$, the only terms are $f⟨Q_1{G}_2{G}_3⟩$ and its two cyclic placements. For the first, Wick theorem pairs four Gaussian factors in three ways. Pairing the two factors inside $Q_1$ together is canceled by the mean subtraction. Pairing one of those factors with $G_2$ and the other with $G_3$ gives $P(k_2)P(k_3)$ and the overall momentum delta; exchanging the two factors gives the same contribution. ThusAdding the three placements givesThere is no second-order term because it would contain five centered Gaussian factors. The two-point function receives its first correction at $f^2$, whereas the displayed local-type primordial non-Gaussianity gives a three-point correlation already at $f$. For $P\propto k^{-3}$ the squeezed limit is $B_{\Phi }(q,k,k)\simeq 4fP(q)P(k)$; this explains the strong correlation between a long potential and short-scale power. A finite variance or regulator is needed to define the subtracted $⟨G^2⟩$, as noted in part (b).

Use the paper's mode normalization, effectively $M_p=1$, and set $C=ϵ+1-c_s^2$, $K=k_1+k_2+k_3$, $s_2=k_1{k}_2+k_1{k}_3+k_2{k}_3$, $s_3=k_1{k}_2{k}_3$. Write $⟨{\zeta }_1{\zeta }_2{\zeta }_3{⟩}_c=(2\pi {)}^3{\delta }^3(\sum {\mathbf{k}}_i)B$. There are two source consistency issues: the supplied interaction picture expansion makes an external-before-vertex Wick contraction contain $u_k({\tau }^{\prime })$, with positive exponential $e^{ik{c}_s{\tau }^{\prime }}$; it converges at $-\infty (1-i0)$, whereas the PDF prints $-\infty (1+i\epsilon )$. Also the stated positive interaction has the opposite overall bispectrum sign from the printed target. The following calculation identifies both issues directly.

For the stated mode expansion, the ordered contraction isA spatial derivative at the vertex supplies $i{p}_j$; the two differentiated legs consequently give $-{\mathbf{k}}_i\cdot {\mathbf{k}}_j$. There are two Wick contractions for each choice of the undifferentiated external leg. The extra $a({\tau }^{\prime })$ in the stipulated in-in formalism integral makes the vertex coefficient $a^2ϵC/c_s^2$, using $a^2=1/(H^2{\tau }^{\prime 2})$. This is consistent if the named $H_{\mathrm{int}}$ is the cosmic-time generator expressed as a function of $\tau$, since $dt=ad\tau$; if it were already the conformal-time generator, that extra $a$ would have to be removed.

For the spatial-gradient cubic curvature bispectrum, define the late-time scalar gradient vertex integralWith $\omega =c_{s}K$, the first two terms of the product obey $({\tau }^{\prime -2}-i\omega {\tau }^{\prime -1})e^{i\omega {\tau }^{\prime }}=-{\partial }_{{\tau }^{\prime }}(e^{i\omega {\tau }^{\prime }}/{\tau }^{\prime })$. The other two use ${\int }_{-\infty }^0{e}^{i\omega {\tau }^{\prime }}d{\tau }^{\prime }=-i/\omega$ and ${\int }_{-\infty }^0{\tau }^{\prime }{e}^{i\omega {\tau }^{\prime }}d{\tau }^{\prime }=1/{\omega }^2$, with the same vacuum prescription. The endpoint divergence $-1/{\tau }_f$ is real, whileIn particular the finite $-K$ term comes from the endpoint of the total derivative and must not be discarded along with the real divergence.

Let $S={\mathbf{k}}_1\cdot {\mathbf{k}}_2+{\mathbf{k}}_1\cdot {\mathbf{k}}_3+{\mathbf{k}}_2\cdot {\mathbf{k}}_3$. The ordered vertex expectation, after stripping the momentum delta, is $Z=-2S{H}^{4}CI/(64{ϵ}^2{c}_s^5{k}_1^3{k}_2^3{k}_3^3)$. Since $\mathrm{Re}(-2iZ)=2\mathrm{Im}Z$, the literal positive Hamiltonian givesThe printed positive-sign target is obtained by replacing the displayed interaction Hamiltonian by its negative, with the convergent contour above. This is also the expected sign if a positive cubic $\zeta (\partial \zeta {)}^2$ coefficient was intended as a Lagrangian term, by the interaction Hamiltonian relation proved in question 3(a). Under that explicit repair,Restoring the overall $(2\pi {)}^3{\delta }^3(\sum {\mathbf{k}}_i)$ reproduces all three cyclic terms in the requested expression. An equilateral bispectrum configuration is a concrete sign check: for $k_i=k$, $S=-3k^2/2$, $Q=-17k/9$, so $SQ=17k^3/6>0$. The literal interaction gives $B=-17H^{4}C/(96{ϵ}^2{c}_s^4{k}^6)$, while the printed target is positive. This cannot be repaired by merely changing the overall definition of $\zeta$, because its field expansion and interaction were specified together. The contour typo and Hamiltonian sign are therefore documented source repairs, rather than silently altered contractions.

Factor the momentum dependence intoWrite $s=+1$ for the intended target, obtained with the Hamiltonian sign repair in part (i), and $s=-1$ for the literal stated Hamiltonian. The same sign choice applies to both limits and cancels in their amplitude ratio. The primordial bispectrum contribution is $s{H}^4(ϵ+1-c_s^2)\mathcal{G}/(16{ϵ}^2{c}_s^4)$. HenceAt fixed $H,ϵ$ and momentum triangle, the ratio to the unit-sound-speed amplitude isKeeping only the leading slow-roll order makes this $100/c_s^4$. To compare the two pieces of this same interaction at a fixed $c_s$, their ratio is instead $(1-c_s^2)/ϵ\simeq 100(1-c_s^2)$. Thus the noncanonical piece overtakes the slow-roll piece once $1-c_s^2\gg 0.01$, while both have the common raw $c_s^{-4}$ enhancement from the interaction and its external mode normalization. For instance $c_s^2=0.1$ gives an exact ratio $9100$ to the canonical limit with these fixed parameters.

A raw primordial bispectrum amplitude is not the same as normalized primordial non-Gaussianity: here $P_{\zeta }(k)=H^2/(4ϵc_s{k}^3)$, so the ratio $B/P_{\zeta }^2$ scales as $(ϵ+1-c_s^2)/c_s^2$ at fixed shape. The small-sound-speed enhancement in normalized non-Gaussianity is therefore $c_s^{-2}$. These are tree-level weak-interaction predictions; approaching arbitrarily small $c_s$ eventually requires checking the perturbative regime.