Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 55 (Advanced Cosmology)

Use signature $(+---)$, the positive spatial metric ${\gamma }_{ij}$, and a future-directed unit normal $n_{\mu }=(N,0)$, so $n^{\mu }=N^{-1}(1,-N^i)$. This fixes the sign of the normal to agree with the specified $\Pi$. The negative normal covector printed in the source would instead give $-\Pi$; its orientation and the displayed derivative cannot both be retained. Use the operational negative-expansion convention for the extrinsic curvature of a spatial hypersurface, $K_{ij}=-({\dot{\gamma }}_{ij}-D_i{N}_j-D_j{N}_i)/(2N)$, which is the convention giving the requested expanding-universe sign. With this future normal and signature it equals $n_{i;j}$; the source's $-n_{i;j}$ identity instead uses its past-pointing normal.

For the scalar-field matter projections, put $s_i=D_i\varphi$ and $s^2={\gamma }^{ij}{s}_i{s}_j$. Decomposing the scalar gradient into normal and tangent parts gives ${\partial }_{\mu }\varphi {\partial }^{\mu }\varphi ={\Pi }^2-s^2$. Contracting the Klein-Gordon scalar stress-energy tensor with the normal and the spatial projection tensor therefore yieldsIn particular, the positive spatial metric raises the $i,j$ indices in $s^2$; raising them with the four-metric would reverse that spatial sign. These quantities are the energy density, momentum density and spatial stress in the normal frame.

For the homogeneous Spatially flat FLRW metric, $N_i=0$, ${\gamma }_{ij}=a^2{\delta }_{ij}$ and $s_i=0$. Consequently $\Pi =\dot{\bar{\varphi }}/\bar{N}$, $J_i=0$, $\bar{\rho }={\Pi }^2/2+V$, $\bar{P}={\Pi }^2/2-V$, andThe intrinsic Ricci scalar is zero and $K^2-K_{ij}{K}^{ij}=9H^2-3H^2=6H^2$. The Hamiltonian constraint thus gives the scalar-field Friedmann equation:The momentum constraint is identically satisfied, since its homogeneous spatial derivatives and $J_i$ vanish. The scalar equation reduces to ${\bar{N}}^{-1}\dot{\Pi }+3H\Pi +V_{,\varphi }=0$. Substituting $\Pi$ and multiplying by ${\bar{N}}^2$ gives the background inflaton equation of motion:No derivation of the supplied gravitational constraints is needed.

Write $\mathcal{D}={\bar{N}}^{-1}{\partial }_t$, $\bar{\Pi }=\mathcal{D}\bar{\varphi }$ and $\Delta =a^{-2}{\nabla }^2$. Since the background scalar has no spatial gradient,Also $\delta P=\bar{\Pi }\mathcal{D}\delta \varphi -{\bar{\Pi }}^2\Psi -V_{,\varphi }\delta \varphi$. These are the linearized scalar-field matter projections; gradient energy first contributes quadratically.

There is a source convention defect in the shear formula. For the stated positive shift $N_i=a^2{B}_{,i}$ and spatial metric with $+2E_{,ij}$, direct substitution into the extrinsic curvature gives the positive-shift scalar shear conventionThere is no extra Laplacian in the definition of ${\chi }_{+}$. Equivalently use $\chi =-{\chi }_{+}$: then the trace-free term has the printed minus sign and $\kappa =3(\mathcal{D}\Phi +H\Psi )-\Delta \chi$. This latter convention retains the printed momentum-constraint form. The source mixes these two choices.

At first order $\delta (K^2)=-6H\kappa$ and $\delta (K^i{}_j{K}^j{}_i)=-2H\kappa$. Combining their difference with $\delta {}^{(3)}R=4\Delta \Phi$ givesFor the momentum constraint, differentiating the tensor above givesFor nonzero Fourier wavenumber, and absorbing the homogeneous integration mode into the background, the consistent momentum constraint is

Linearizing the scalar equation, the shift-advection term and lapse-gradient times scalar-gradient term vanish at this order because the background is spatially homogeneous. The remaining equation isInsert $\delta \Pi$, differentiate its lapse term, and use $\mathcal{D}\bar{\Pi }+3H\bar{\Pi }=-V_{,\varphi }$. The two background-acceleration terms combine into $2\Psi V_{,\varphi }+3H\bar{\Pi }\Psi$. Hence the perturbed Klein-Gordon equation with background lapse isThis scalar evolution equation and the Hamiltonian constraint agree with the displayed targets after the shear conventions are repaired.

For $W=\bar{\rho }+\bar{P}\ne 0$, the conserved uniform-density curvature in the spatial-metric convention $1-2\Phi$ isThe source's plus sign in front of both terms is not gauge invariant and cannot yield the claimed conservation. Choosing the opposite overall sign, $\Phi -\delta \rho /(3W)$, is equally valid, but changing only one term is necessary. A passive time displacement $t\mapsto t+T$ gives $\tilde{\Phi }=\Phi +\bar{N}HT$ and $\widetilde{\delta \rho }=\delta \rho +3\bar{N}HWT$, so their combination above is unchanged. This is the uniform-density curvature sign test.

The lapse perturbation in the perturbed energy-conservation source is $\Psi$, not the spatial-curvature perturbation $\Phi$. Indeed the perturbation of the proper-time density derivative is $\mathcal{D}\delta \rho -\Psi \mathcal{D}\bar{\rho }$, while the normal expansion changes by $-\kappa$. The consistent equation isFor a constant barotropic equation of state, $\delta P=w\delta \rho$ and $\mathcal{D}W=-3H(1+w)W$. Differentiate ${\zeta }_{\mathrm{ud}}$ and cancel the density terms:More generally, $\delta P_{\mathrm{nad}}=\delta P-(\mathcal{D}\bar{P}/\mathcal{D}\bar{\rho })\delta \rho$ givesRegular scalar modes have gradient terms suppressed by $(k/aH{)}^2$. Thus adiabatic uniform-density curvature is conserved to leading order on superhorizon scales, ${\dot{\zeta }}_{\mathrm{ud}}\simeq 0$. This is a long-wavelength limit, not exact finite-wavenumber conservation; it assumes a regular gradient expansion and excludes $W=0$, where this variable is undefined. In contrast the literal printed variable obeys $\mathcal{D}{\zeta }_{\mathrm{printed}}=2\mathcal{D}\Phi +O(k^2)$ even after repairing the energy equation, and generally evolves across changes of the background equation of state.

There are three practical benefits. First, uniform-density curvature is gauge invariant, so matching it does not depend on coordinate slicing. Second, superhorizon conservation of uniform-density curvature transports primordial amplitudes through changes in the expansion history without solving each short-scale dynamical equation. Third, when a mode re-enters the horizon it supplies initial conditions for density growth and Cosmic microwave background anisotropy; non-adiabatic pressure is also an explicit diagnostic of when that simple transport fails.

In the Spatially flat FLRW metric, neglect metric perturbations and write $\varphi =\bar{\varphi }+\delta \varphi$. The kinetic invariant is then exactly quadratic in field derivatives:Subtracting $\bar{X}={\dot{\bar{\varphi }}}^2/2$ gives the second-order expansionIn the derivative-dominated approximation, $\delta \varphi =-\dot{\bar{\varphi }}\zeta /H$ with $\dot{\bar{\varphi }}/H$ treated as slowly varying. Thus $\delta \dot{\varphi }\simeq -\dot{\bar{\varphi }}\dot{\zeta }/H$ andThis uses the linear flat-to-comoving slicing relation. Differentiating its slowly varying coefficient would add terms involving undifferentiated $\zeta$; it is not an exact nonlinear gauge transformation, and those neglected terms do not alter the leading pure ${\dot{\zeta }}^3$ coefficient computed here.

Only $P_{,XX}\delta X_1\delta X_2$ and $P_{,XXX}\delta X_1^3/6$ contribute to the requested three-time-derivative term. Therefore the directly evaluated cubic coefficient isThis is the derivative-cubic coefficient of a noncanonical scalar. For a physical $P(X,\varphi )$ scalar field theory, differentiating $\rho =2X{P}_{,X}-P$ at fixed $\varphi$ gives ${\rho }_{,X}=P_{,X}+2X{P}_{,XX}$. Its sound speed is consequentlyThe minus sign printed in the denominator of $c_s^2$ reverses the asserted coefficient identity; it must be a plus. The printed background equation also lacks a power: it is $3H^2=\rho$ in the stated reduced-Planck units. A stable propagating scalar has positive time-kinetic coefficient $P_{,X}+2X{P}_{,XX}$ and positive gradient coefficient $P_{,X}$, as summarized by noncanonical scalar kinetic stability.

The unfactored expression for $C$ stays meaningful if $P_{,XX}=0$, although $\mathcal{A}$ separately then need not exist. For a canonical kinetic term with no higher kinetic derivatives, $C=0$. The cubic curvature action for a P(X, phi) scalar field can include other vertices and gravitational terms; they are outside this requested derivative coefficient.

For a centred Gaussian random field, Wick theorem expresses each even correlation function as a sum over all pairings of two-point functions; odd moments vanish. For example,For six fields there are $5\cdot 3\cdot 1=15$ pairings. In a tree-level connected three-point calculation with one cubic vertex, each of the three external fields must contract with a different vertex field. This gives $3!=6$ connected Wick contractions. Pairings connecting two external fields instead are disconnected tadpole contributions, omitted from the primordial bispectrum.

Put $\mathcal{F}=(1-c_s^2)\mathcal{A}/c_s^2$. Using $dt=ad\tau$ and $\dot{\zeta }={\zeta }^{\prime }/a$, the integrated interaction Hamiltonian becomesThe unequal-time Wick contraction required by the displayed operator order isThe conjugate is essential: the equal-time power spectrum alone does not specify the erroneous unstarred replacement printed in the target expression. The in-in bispectrum conjugation rule fixes the vacuum convergence and final sign.

Fourier transforming the vertex gives $(2\pi {)}^3{\delta }^{(3)}({\mathbf{p}}_1+{\mathbf{p}}_2+{\mathbf{p}}_3)$ with three measures $d^3{p}_j/(2\pi {)}^3$. The six pairings supply the product of three external-vertex delta functions, each carrying $(2\pi {)}^3$. Integrating all three internal momenta leaves one overall $(2\pi {)}^3{\delta }^{(3)}({\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3)$. Thus the normalized connected contraction formula isThis spells out what the printed permutations and delta functions must represent: all six permutations, full Fourier normalization, and conjugated vertex modes. Three cyclic permutations alone would miss a factor of two.

Use exactly the stipulated unit-sound-speed mode functions for this integral. Their late-time values are real and $u_k^{\prime }{}^{\ast }(\tau )=H{k}^2\tau e^{ik\tau }/\sqrt{4ϵk^3}$. ConsequentlyThe vacuum contour lies above the negative real time axis, so $e^{iK\tau }$ decays at its past endpoint. With $s=iK+0^{+}$, repeated differentiation of ${\int }_{-\infty }^0{e}^{s\tau }d\tau =1/s$ givesSubstitution gives the bispectrum with consistent operator ordering and the stated Hamiltonian:The sign is positive in front of $\mathcal{A}$, contrary to the target's negative sign. The printed negative result is obtained algebraically if one uses the unstarred derivatives, $e^{-iK\tau }$, and the conjugate past contour, which gives $-2i/K^3$, while retaining $-2i$ outside. That combination is not the stated vacuum contour or the stated external-before-vertex contraction. A consistently conjugated representation must also conjugate the external prefactor and has the positive result above. The unit-speed cubic time-derivative bispectrum records these conventions explicitly.

The approximation treats $H$, $ϵ$ and the interaction coefficients as constant, uses the de Sitter approximation, retains only the stipulated vertex at tree level, and takes $k_i|{\tau }_{\mathrm{final}}|\to 0$ after doing the convergent early-time integral. The $i0$ prescription must be kept while imposing the past boundary condition; an arbitrary early-time cutoff leaves spurious oscillatory boundary terms. For a physical noncanonical quadratic action with $c_s\ne 1$, the mode functions also change to $H(1+i{c}_{s}k\tau )e^{-i{c}_{s}k\tau }/\sqrt{4ϵc_s{k}^3}$. They were not supplied here; the displayed calculation is conditional on the paper's specified modes, rather than a full self-consistent noncanonical power-spectrum calculation.

With the stipulated power spectrum $P_{\zeta }(k)=H^2/(4ϵk^3)$, an equilateral bispectrum configuration gives $B(k,k,k)/P(k{)}^2=2\mathcal{F}/9$. For the local-style normalization evaluated only at that triangle, $B(k,k,k)=(18/5)f_{\mathrm{NL}}^{\mathrm{effective}}P(k{)}^2$, this meansThis is a normalization convention for an equilateral amplitude, not a claim that the shape is local. In a squeezed bispectrum configuration, the ratio to the long-short power product is suppressed by $(k_L/k_S{)}^2$. Small sound speed or large higher-kinetic derivatives can make $|\mathcal{F}|$ large enough for potentially observable primordial non-Gaussianity, whereas $c_s\to 1$ with finite $\mathcal{A}$ suppresses this vertex. Detectability is conditional on amplitudes, the other cubic interactions, and perturbative control; it cannot be asserted from the given constants alone.

In Newtonian gauge, the coordinate-comoving observer has normalized four-velocity $u^{\mu }=a^{-1}(1-\psi ,0)$, hence $u_0=a(1+\psi )$. In the $(+---)$ convention the measured photon energy is $E=u_{\mu }{p}^{\mu }$. Thus to first order

Use the time component of the geodesic equation and divide by $p^0=d\eta /d\lambda$:Let $\mathcal{H}=a^{\prime }/a$ and take the photon path direction $\mathbf{e}$ at zeroth order. The left side divided by $ϵ/a^2$ is ${ϵ}^{\prime }/ϵ-2\mathcal{H}+2\mathcal{H}\psi -d\psi /d\eta$. The contraction on the right side, using the provided connection coefficients, isThe homogeneous expansion cancels. Therefore the scalar photon energy-redshift law isThe total derivative is $d/d\eta ={\partial }_{\eta }+\mathbf{e}\cdot \nabla$ on the unperturbed ray. Perturbing that path inside a first-order potential would contribute only at second order. The homogeneous $a^{-1}$ energy redshift is already removed by the comoving momentum variable $ϵ$.

Along the zeroth-order photon path, set $Y=\Theta +\psi$ and $\mathcal{S}=(3/4)\int d\hat{\mathbf{m}}\Theta (\hat{\mathbf{m}})[1+(\mathbf{e}\cdot \hat{\mathbf{m}}{)}^2]/(4\pi )$. The scalar photon energy redshift and the Photon Boltzmann equation with Thomson scattering giveThe integrating factor is $e^{-\tau }$, because ${\tau }^{\prime }<0$ in this convention. HenceTight coupling before decoupling makes the incident distribution nearly isotropic in the baryon frame. Neglecting its quadrupole, the angular kernel integrates to $\mathcal{S}={\Theta }_0$, since $(3/4)⟨1+(\mathbf{e}\cdot \hat{\mathbf{m}}{)}^2⟩=1$. The primordial term at the optically thick lower endpoint is erased. With instantaneous last scattering, no later scattering and $g=\delta (\eta -{\eta }_{\ast })$, the line-of-sight solution isThe approximation is linear, uses an unperturbed ray, neglects the scattering quadrupole and polarization feedback, and idealizes the cosmological visibility function as a delta function. It also excludes reionization or any other post-recombination opacity.

The photon monopole ${\Theta }_0$ is the intrinsic emission-temperature perturbation. The emission potential ${\psi }_{\ast }$ gives gravitational redshift on leaving the potential well; together they form the Sachs-Wolfe combination. The baryon peculiar velocity supplies the Doppler CMB anisotropy. The integral is the Integrated Sachs-Wolfe effect, from changes in gravitational potentials during travel. The observer potential is a direction-independent endpoint shift absorbed into the observed monopole. An observer peculiar velocity, if included, contributes an additional kinematic dipole; the stated observer is coordinate-comoving.

For the growing adiabatic mode in an epoch of constant $w$, the superhorizon potential is constant. The Friedmann equation gives $4\pi G{a}^2(\bar{\rho }+\bar{P})=\frac{3}{2}(1+w){\mathcal{H}}^2$, so the given relation impliesThus ${\varphi }_{\mathrm{rad}}=-2\mathcal{R}/3$ and ${\varphi }_{\mathrm{mat}}=-3\mathcal{R}/5$. Photon continuity, neglecting its gradient term on superhorizon scales, gives $({\Theta }_0-\varphi {)}^{\prime }=0$. The radiation-era initial condition fixes ${\Theta }_0-\varphi =\mathcal{R}$. Through the matter-radiation transition it follows that ${\Theta }_{0\ast }=\mathcal{R}+{\varphi }_{\mathrm{mat}}=2\mathcal{R}/5$. Since $\psi =\varphi$, the matched large-scale emission perturbation isKeeping the radiation value of ${\Theta }_0$ unchanged across the transition would give the wrong coefficient. This is the Sachs-Wolfe radiation-to-matter matching for the paper's sign convention for comoving curvature perturbation.

Neglect the Integrated Sachs-Wolfe effect, the superhorizon-suppressed Doppler term, observer monopole, and any observer kinematic dipole. Put ${\mathbf{x}}_0=0$ by statistical homogeneity, and defineThe plane-wave expansion on ${\mathbf{x}}_{\ast }=-{\chi }_{\ast }\mathbf{e}$ gives angular multipolesStatistical isotropy and orthonormal spherical harmonics then imply $⟨a_{\ell m}{a}_{{\ell }^{\prime }{m}^{\prime }}^{\ast }⟩={\delta }_{\ell {\ell }^{\prime }}{\delta }_{m{m}^{\prime }}{C}_{\ell }$. Radial integration supplies $(4\pi {)}^2(2{\pi }^2)/(2\pi {)}^3=4\pi$, giving the large-angle angular power spectrumThis assumes adiabatic growing modes, matter domination at emission, negligible anisotropic stress, and wavenumbers that are outside the Hubble radius then. It treats recombination as instantaneous, omits late potential evolution and rescattering, and omits lensing at this linear order. The formal primordial contribution is defined for $\ell >0$; the observed dipole has a large additional kinematic contribution, so the clean large-angle primordial comparison is usually $\ell \ge 2$. A scale-invariant spectrum also yields the Sachs-Wolfe plateau $\ell (\ell +1)C_{\ell }=2\pi {\mathcal{P}}_{\mathcal{R}}/25$ for $\ell \ge 1$.

A scalar cosmological perturbation at fixed wavevector carries no helicity or distinguished transverse direction. Its metric and scalar sources are unchanged by rotations around $\mathbf{k}$, and the scalar baryon peculiar velocity is longitudinal. With scalar initial conditions, rotational invariance of the linear evolution therefore makes $\Theta$ a function only of $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$. Expand it using the stated normalization, without an extra $(2\ell +1)$ factor:In the Thomson kernel, $1+v^2=4/3+(2/3)P_2(v)$. The spherical harmonic addition theorem and Orthogonality of Legendre polynomials giveOnly the monopole and quadrupole survive; the quadrupole's $(-i{)}^2=-1$ gives the angular scattering integralThis also explains why an isotropic distribution has zero net collision term.

The streaming term $ik\mu \Theta$ and the Legendre polynomial recurrence relation yield, after division by $(-i{)}^{\ell }$, $k[(\ell +1){\Theta }_{\ell +1}/(2\ell +3)-\ell {\Theta }_{\ell -1}/(2\ell -1)]$. With ${\mathbf{v}}_b=i\hat{\mathbf{k}}{v}_b$, the velocity collision source is $+{\tau }^{\prime }{v}_b$ in the dipole equation; the gravitational gradient source is $+k\psi$. Therefore the photon Boltzmann hierarchy isSet the absent ${\Theta }_{-1}$ contribution to zero. This is algebraically the printed hierarchy. In particular,Since ${\tau }^{\prime }<0$, these collision terms damp the nonzero multipoles and drive the dipole to ${\Theta }_1=-v_b$. The latter sign follows from the paper's velocity and $(-i{)}^{\ell }$ conventions, not a reversed physical photon velocity.

In the tight-coupling approximation, the Thomson-scattering rate ${\Gamma }_T=-{\tau }^{\prime }$ is much faster than both spatial streaming and cosmological evolution: $k/{\Gamma }_T\ll 1$ and $\mathcal{H}/{\Gamma }_T\ll 1$. Frequent scattering locks the photon and baryon velocities and suppresses higher angular multipoles. The photon dipole is leading order, while the photon quadrupole is first order in the mean-free-time expansion.

Use the quadrupole equation. Its time derivative is smaller than its collision term by an additional expansion parameter, and ${\Theta }_3$ starts one order beyond ${\Theta }_2$. The leading balance is thereforeThus for the temperature-only collision operator supplied here,The temperature-only tight-coupling quadrupole is not numerically interchangeable with a hierarchy that includes polarization feedback. The latter changes the leading collision coefficient and gives the existing photon quadrupole in tight coupling with polarization instead. Here the stated temperature-only approximation is the one being expanded.

For $\mathbf{k}=k\hat{\mathbf{z}}$, scalar axisymmetry leaves only $m=0$ in the photon quadrupole: ${\Theta }_{20}=-\sqrt{4\pi /5}{\Theta }_2$ with the normalization used above. The two spin-weighted spherical harmonics satisfy ${}_{+2}{Y}_{20}={}_{-2}{Y}_{20}\propto {\sin }^2\theta$. Translation from last scattering supplies $e^{-ik{\chi }_{\ast }\mu }$. Thus the two helicity combinations have the same scalar source:where $\mathcal{C}$ is a common normalization constant independent of $\mu$. In this meridian polarization basis the scalar contribution has $U=0$; this does not mean that every rotated polarization basis has zero $U$.

Apply the spin-raising and lowering operators to an axisymmetric scalar potential $f$. The first action is $-{\partial }_{\theta }f$. In the second action, the intermediate spin is respectively $+1$ or $-1$; both giveIt follows thatSubtracting the two equations gives $P_B^{\prime \prime }=0$. The possible affine $P_B=A+B\mu$ consists solely of the unobservable $\ell =0,1$ kernel of the twice-applied spin operators. Defining polarization potentials to have no such components sets the physical scalar B mode to zero, $\boxed{P_B=0}$. This is the low-multipole kernel of polarization potentials, not a physical dipole polarization.

For $x\ne 0$ the remaining equation integrates to $P_E=-\mathcal{C}{\Theta }_2{e}^{-ix\mu }/x^2+A_E+B_E\mu$. Remove the same monopole/dipole kernel and use the plane-wave expansion. Thus the scalar E-mode radial projection isThe overall minus sign is absorbed in the stated proportionality constant. Although the unprojected antiderivative has $x^{-2}$, its divergent monopole and dipole pieces are discarded; the physical $\ell \ge 2$ limit is finite because $j_{\ell }(x)=O(x^{\ell })$.

The temperature-only tight-coupling quadrupole is suppressed by $k/{\Gamma }_T$. For regular superhorizon scalar modes, the photon dipole is itself proportional to $k$, so the recombination quadrupole and the resulting polarization are especially small at large angular scales. Last-scattering polarization is generated by the small departure from perfect tight coupling, not by an isotropic monopole. Reionization can generate a separate large-angle signal after recombination and is absent from the instantaneous-last-scattering approximation here. The no-B-mode conclusion applies to linear scalar sources; it does not include tensor sources or conversion by lensing.