Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 64 (Advanced Cosmology)

The spatial coordinates are cyclic because the flat Friedmann-Lemaître-Robertson-Walker metric depends only on time. Differentiating the printed affine Lagrangian gives the canonical momentumThe Euler-Lagrange equations give $d{P}_i^c/d\tau =\partial L/\partial x^i=0$, soThis is the conserved quantity associated with spatial translation invariance. The overall minus sign in the stated action makes this canonical covector opposite to the usual future-directed mechanical momentum; it does not alter the geodesic trajectories or the conservation result. The canonical momentum sign in the affine particle action will be kept explicit below.

The mass-shell condition and $P_{\mu }^c=-m{g}_{\mu \nu }{\dot{x}}^{\nu }$ imply $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$, so the affine parameter is proper time up to orientation. Choose its future orientation. With conformal time $t$,In a local comoving orthonormal frame both time and spatial coordinate intervals acquire the same factor $a$, so the measured peculiar velocity is precisely $\mathbf{v}=d\mathbf{x}/dt$. Thus the physical momentum obeysThe fixed vector direction also gives $\mathbf{P}(t)=a(t_{\ast })\mathbf{P}(t_{\ast })/a(t)$. With the conventional opposite overall sign for the affine particle action, the familiar comoving momentum is $\mathbf{q}=a\mathbf{P}={\mathbf{P}}^c$ instead. The redshift law is identical under either sign convention.

For signature $(-+++)$, $\sqrt{-g}=a^4$ and $g^{\mu \nu }=a^{-2}{\eta }^{\mu \nu }$. The scalar action becomesVary $\varphi$ with compactly supported variations and integrate its derivative terms by parts:Setting this coefficient to zero gives the Klein-Gordon equation in the expanding background:The additional first-derivative term is the conformal-time expansion damping; the time coordinate here is not the proper cosmological time used in Question 2.

Put $\chi =a\varphi$. Direct differentiation of the previous equation yieldsIn radiation domination, $a\propto t$, and for a massless scalar field both extra terms vanish. Each nonzero Fourier mode therefore satisfies ${\chi }_{\mathbf{k}}^{\prime \prime }+k^2{\chi }_{\mathbf{k}}=0$. The general real radiative solution isSumming over both wavevector directions supplies the two traveling-wave solutions, with constant amplitudes. In infinite space the sum is replaced by the corresponding Fourier integral.

There is a qualification to calling the printed oscillatory expression completely general. At $k=0$, ${\chi }_0^{\prime \prime }=0$, so ${\chi }_0=C_1+C_{2}t$ rather than a single zero-frequency exponential. The zero mode of a radiation-era massless scalar isA constant nonzero field is an immediate counterexample to the completeness of the fixed-amplitude oscillatory formula at zero wavenumber. Include this homogeneous term, or explicitly restrict the displayed plane-wave expansion to the radiative modes with no homogeneous background.

The spatial momentum operator acts only on the plane wave:Thus the modes have fixed canonical/comoving wavevector momentum, not fixed physical momentum. In the usual positive-energy field-mode convention the local physical momentum and energy are $\hslash \mathbf{k}/a$ and $\hslash k/a$. This agrees with conserved spatial comoving momentum and cosmological redshift proportional to $a^{-1}$.

The overall particle-action sign noted in (i) reverses the literal worldline canonical label relative to this usual mechanical momentum: it is ${\mathbf{P}}^c=-a\mathbf{P}$ for a future-directed particle with that printed action. The conserved plane-wave label therefore has the opposite sign if it is identified with that literal worldline label. This sign convention does not change the wavenumber magnitude or either conservation/redshift conclusion.

This question uses proper cosmological time $t$. The future-directed unit normal isThe non-null hypersurface projection removes all normal components. The only relevant Christoffel symbols are ${\Gamma }_{ij}^0=a\dot{a}{\delta }_{ij}$. Since the components of $n_{\mu }$ are constant,ConsequentlyWith the printed positive extrinsic curvature convention, this is the expansion of the normal comoving congruence. A fixed coordinate cell has proper volume proportional to $a^3$, so $K=d\log (a^3)/dt={\nabla }_{\mu }{n}^{\mu }$: the fractional proper-volume expansion rate.

Write $q_{\alpha \beta }=g_{\alpha \beta }+n_{\alpha }{n}_{\beta }$ for the induced metric. By definition of the spatial covariant derivative,The spacetime Levi-Civita connection is metric compatible, so $\nabla g=0$. The remaining terms contain either $P_{\alpha }{}^{\sigma }{n}_{\sigma }$ or $P_{\beta }{}^{\lambda }{n}_{\lambda }$, each zero because $n^{\mu }{n}_{\mu }=-1$. ThereforeThis verifies metric compatibility of the spatial covariant derivative directly, without assuming that the unprojected derivative of $q$ vanishes.

For tangent vectors $X,Y$ on a slice, the extrinsic curvature is $K(X,Y)=g({\nabla }_{X}n,Y)$. Differentiating $g(n,Y)=0$ and using metric compatibility gives $K(X,Y)=-g(n,{\nabla }_{X}Y)$. Since the connection is torsion-free,The final equality holds because tangent fields on a hypersurface have a tangent Lie bracket. All normal components are projected out, so $K_{\mu \nu }$ is symmetric. This is precisely hypersurface orthogonality implies symmetric extrinsic curvature; without an integrable tangent distribution the corresponding antisymmetric vorticity need not vanish.

Double contraction of the projected Gauss equation, with the timelike-normal sign and curvature convention in the question, givesHere $q^{\mu \nu }=g^{\mu \nu }+n^{\mu }{n}^{\nu }$. Expanding the two projectors givesThe four-normal contraction vanishes by Riemann antisymmetry. Because $g_{\mu \nu }{n}^{\mu }{n}^{\nu }=-1$, the right side is $2G_{\mu \nu }^{(4)}{n}^{\mu }{n}^{\nu }$. Rearranging proves the Hamiltonian constraint identity:For flat FRW, $R^{(3)}=0$, $K^2=9H^2$ and $K_{\mu \nu }{K}^{\mu \nu }=3H^2$, so $G_{nn}=3H^2$. Einstein's equation is therefore the first Friedmann equation, $3H^2=8\pi G\rho$, or $3H^2=8\pi G\rho +\Lambda$ if a cosmological constant is placed separately in Einstein's equation.

The scalar $R^{(3)}$ measures intrinsic spatial curvature, $K^2$ measures squared volume expansion, and the norm of $K_{ij}$ measures directional expansion including shear. More generally $K_{ij}=K{q}_{ij}/3+s_{ij}$ with trace-free $s_{ij}$ gives $K_{ij}{K}^{ij}=K^2/3+s_{ij}{s}^{ij}$. The shear vanishes in FRW, leaving the positive $3H^2$ contribution. This is an instantaneous constraint on initial data, not the separate acceleration equation.

Use $\mathcal{H}=a^{\prime }/a$ and the Fourier convention $\nabla \to i\mathbf{k}$. A valid linearization requires all components of $h_{ij}$ to be small: the printed small-determinant condition alone is insufficient, as the matrix $\mathrm{diag}(1,1,0)$ has zero determinant without being small. This is small determinant does not imply a small metric perturbation.

Define the density contrast $\delta =(\rho -\bar{\rho })/\bar{\rho }$ and the scalar velocity potential by $v^i={\partial }^i\theta$, or $v_{\mathbf{k}}^i=i{k}^i{\theta }_{\mathbf{k}}$. Hence $\nabla \cdot \mathbf{v}=-k^2\theta$. To first order, $u^0=a^{-1}$, $u^i=a^{-1}{v}^i$ and $g^{ij}=a^{-2}({\delta }^{ij}-h^{ij})$. The product $u^i{u}^j$ is second order. Substitution into the perfect fluid stress-energy tensor givesHere the equation of state is used locally, so the pressure perturbation is $w\bar{\rho }\delta$.

The contraction in ${\nabla }_{\mu }{T}^{0\mu }=0$ isUsing the displayed connections gives, through first order,In particular ${\Gamma }_{\mu 0}^{\mu }=4\mathcal{H}+h^{\prime }/2$; the spatial connection contraction adds $3\mathcal{H}P+\bar{P}{h}^{\prime }/2$, while differentiating $a^{-2}\rho$ supplies $-2\mathcal{H}\rho$. Subtract the background relation ${\bar{\rho }}^{\prime }+3\mathcal{H}(1+w)\bar{\rho }=0$. The background terms multiplying $\delta$ cancel, leavingFor comoving cold dark matter, $w=0$ and $\theta =0$, so ${\delta }_c^{\prime }=-h^{\prime }/2$ and ${\delta }_c+h/2$ is constant at each spatial point. The proper-volume element of a comoving cell is $a^3\sqrt{\det (I+h)}{d}^{3}x=a^3(1+h/2)d^{3}x$. Conserved mass in the cell then gives the density change opposite to the perturbation of its volume. In comoving synchronous coordinates the linear CDM density change is a volume change, not transport across cell boundaries.

There is a conformal-time normalization error in the printed density parameter. For physical background density, $H=\mathcal{H}/a$, so the cosmological density parameter isThe printed definition lacks the numerator's $a^2$. The scale factor in a conformal-time density parameter is essential for ${\Omega }_c\simeq 1$ in the intended flat matter-dominated approximation.

During matter domination, pressureless matter supplies both the expansion and the gravitational source: $a\propto {\tau }^2$, $\mathcal{H}=2/\tau$ and ${\Omega }_c\simeq 1$ when CDM is taken to constitute that matter. ThusTry ${\delta }_c\propto {\tau }^p$. Its characteristic polynomial is $p(p-1)+2p-6=(p-2)(p+3)$, givingRadiation-era photon pressure and radiation perturbations affect the metric source, and matter alone does not dominate the expansion, so this matter-only closure and its power laws are inadequate before equality. At late acceleration the growth also ceases to follow $a\propto {\tau }^2$ and slows. The question's claim that the differential form itself is valid only in matter domination is too strong: for subhorizon pressureless matter with smooth dark energy it still applies with the correct time-dependent $\mathcal{H}$ and ${\Omega }_m$. Indeed, the pressureless growth equation in proper time is $\ddot{\delta }+2H\dot{\delta }-4\pi G{\bar{\rho }}_m\delta =0$ for smooth dark energy; substituting $d/d{t}_{\mathrm{proper}}=a^{-1}d/d\tau$ gives ${\delta }^{\prime \prime }+\mathcal{H}{\delta }^{\prime }-4\pi G{a}^2{\bar{\rho }}_m\delta =0$, the same conformal differential form. It is the matter-dominated coefficients and simple solutions that are restricted to the stated interval.

Outside the radiation-era horizon, the regular growing adiabatic mode has nearly constant curvature, while its comoving-synchronous density perturbation begins at gradient order $k^2{\tau }^2$. Equivalently its metric trace grows as ${\tau }^2$ and the volume relation in (i) gives ${\delta }_c\propto {\tau }^2$. This is a relativistic superhorizon solution, not instantaneous Newtonian communication between causally disconnected regions. Inside the horizon, pressure makes the dominant radiation oscillate and its gravitational potential decay, while CDM self-gravity is still subdominant. CDM then grows only weakly, logarithmically in the more accurate Mészáros effect; the stated freezing approximation replaces that slow growth by a constant.

Assume all modes being compared start outside the horizon at common ${\tau }_i$, and define ${\tau }_h=2\pi /k$, $k_{\mathrm{eq}}=2\pi /{\tau }_{\mathrm{eq}}$ and $G_0=({\tau }_0/{\tau }_i{)}^2$. Modes with $k\le k_{\mathrm{eq}}$ never undergo the radiation-era frozen interval, so their growth is $G_0$. For $k>k_{\mathrm{eq}}$, multiply growth before horizon entry, freezing until equality, and subsequent matter growth:Thus the frozen-radiation approximation to the matter transfer function isThis $T$ includes total growth because the question defines it directly between two density contrasts. The usual transfer normalized to one on large scales is $T/G_0$.

Power is the variance of a Fourier amplitude, so $P_0=|T{|}^2{P}_i$. The prescribed primordial density spectrum givesIt turns over at the comoving equality-horizon scale $2\pi /k_{\mathrm{eq}}={\tau }_{\mathrm{eq}}$. The turnover and reduced small-scale growth help organize the broad spectrum of galaxy clustering, clusters and superclusters; they do not select one unique object size. The dimensionless power $k^3{P}_0/(2{\pi }^2)$ is approximately flat above equality in this crude approximation, so a decreasing dimensional spectrum does not imply no small-scale structures. Restoring radiation-era logarithmic growth adds logarithmic corrections to the small-scale tail. Baryon acoustic features, a smooth equality transition, late growth suppression and nonlinear halo formation are absent from this simple linear broken-power approximation.

For an adiabatic mode the perturbation of baryon number per photon is zero. Since $n_{\gamma }\propto T^3$, ${\rho }_{\gamma }\propto T^4$ and nonrelativistic ${\rho }_b\propto n_b$, the common temperature/entropy perturbation gives ${\delta }_b=3\delta T/T$ and ${\delta }_{\gamma }=4\delta T/T$. Hence ${\delta }_{\gamma }=4{\delta }_b/3$.

Let $q=a{n}_e{\sigma }_T$, $\mathcal{H}=a^{\prime }/a$ and $R=4{\bar{\rho }}_{\gamma }/(3{\bar{\rho }}_b)$. In tight coupling, the relative velocity is small but its drag force need not vanish, so it must be eliminated before setting the two velocities equal. Multiply the photon Euler equation by $R$ and add the baryon Euler equation. Their opposite drag terms cancel, even with the common drag-prefactor typo in the printed equations. The result at common velocity potential $\theta$ isTogether with continuity this givesThe same photon-baryon sound speed follows from $\delta P=\delta {\rho }_{\gamma }/3$ and $\delta \rho =\delta {\rho }_{\gamma }+\delta {\rho }_b$: using the adiabatic density relation gives $\delta P/\delta \rho =c_s^2$.

The Euler bracket in the original PDF prints $a/a^{\prime }$; the derivation requires $a^{\prime }/a=\mathcal{H}$. Its baryon continuity equation also needs a prime on ${\delta }_b$. Finally, for the defined velocity potential $\mathbf{v}=i\mathbf{k}\theta$, physical Thomson drag is $-q({\theta }_{\gamma }-{\theta }_b)$, without an extra $k^{-2}$: taking a velocity divergence introduces $-k^2$ on both sides, which cancels on returning to the potential equation. This is photon opacity normalization for a velocity potential. These errors should not be concealed by reproducing a dimensionally inconsistent equation.

The inequalities are compatible: CDM can dominate the total density while photons dominate baryons. With $R\gg 1$, $c_s^2\simeq 1/3$ and the baryon-inertia expansion term is negligible. Differentiate photon continuity and substitute the combined Euler equation:Thus ${\delta }_{\gamma }^{\prime \prime }+k^2{\delta }_{\gamma }/3-4k^2{\sigma }_{\gamma }/3=0$. The quasistatic quadrupole equation givesSubstitution yields the shear-only photon diffusion damping equationTo solve it, put $\Gamma =4k^2{\tau }_c/15$ and ${\delta }_{\gamma }=u\exp [-\frac{1}{2}{\int }^{\tau }\Gamma (s)ds]$. Direct differentiation, including both derivatives of the prefactor, givesNeglect the requested ${\tau }_c^{\prime }$ and ${\tau }_c^2$ terms. The remaining equation is the free acoustic oscillator, so its general solution isA different lower limit is absorbed into the amplitudes. The damping integral is positive and increases with time: shorter wavelengths are smoothed more strongly by photon diffusion. A subhorizon but still tightly coupled mode satisfies $k\tau \gg 1$ and $k{\tau }_c\ll 1$, with slowly varying scattering time; the formula is not valid for arbitrarily short waves after tight coupling fails. The truncated shear coefficient is the one implied by the supplied equations, not the more complete damping coefficient including photon polarization and photon-baryon slip. For constant ${\tau }_c$, the exact oscillator frequency differs from $k/\sqrt{3}$ only at order ${\tau }_c^2$, consistent with this expansion.

Let $q_{\ast }=k{\tau }_{\mathrm{dec}}/\sqrt{3}$ and $E_D=\exp (-k^2/k_D^2)$ at last scattering. For random oscillator amplitudes, define spectra $P_A,P_B$ and cross-spectrum $P_{AB}$ using the same Fourier normalization as the density spectrum. Averaging the solution's squared modulus givesConsequently the intrinsic-temperature source has variance spectrum $P_{{\delta }_{\gamma }}/16$. A common acoustic phase is necessary for pronounced peaks in the variance. Equal independent sine/cosine amplitudes would erase them. A regular growing adiabatic mode begins with negligible velocity outside the sound horizon and selects approximately the cosine phase, $B\simeq 0$, preserving a ${\cos }^2{q}_{\ast }$ sequence. This illustrates coherent initial phases are needed for acoustic variance peaks.

From continuity, ${\theta }_{\gamma }=3{\delta }_{\gamma }^{\prime }/(4k^2)$. Ignoring the small damping correction in this velocity relation, the coherent cosine solution givesThe Doppler CMB anisotropy source therefore has directional variance $3{\mu }^2{P}_A{E}_D^2{\sin }^2{q}_{\ast }/16$, where $\mu =\hat{\mathbf{n}}\cdot \hat{\mathbf{k}}$. Its velocity maxima lie between density maxima: compression and bulk motion are a quarter acoustic period apart.

Angular projection must also be included. Let ${\chi }_{\ast }={\tau }_0-{\tau }_{\mathrm{dec}}$, assume a thin last-scattering surface, and choose the emission phase $e^{ik{\chi }_{\ast }\mu }$. The coherent intrinsic-plus-Doppler radial response is proportional toThe derivative follows from $i\mu e^{ix\mu }={\partial }_x{e}^{ix\mu }$ and the velocity sign above. A reversed line-of-sight convention reverses the corresponding Doppler sign consistently. The CMB angular power spectrum is $C_{\ell }=(2/\pi )\int k^{2}dk{P}_A(k)|{\Theta }_{\ell }(k){|}^2$ in this simplified model. Squaring the combined response keeps its correlated density–velocity cross term; adding two independent angular powers would discard that correlation. The derivative Bessel projection also spreads the Doppler contribution differently from the density contribution.

The sound horizon here is $r_s\simeq {\tau }_{\mathrm{dec}}/\sqrt{3}$. Intrinsic acoustic extrema occur near $k{r}_s=n\pi$, mapping roughly to ${\ell }_n\simeq n\pi {\chi }_{\ast }/r_s$, with interleaved velocity extrema and projection broadening. Cosmic microwave background diffusion damping suppresses both contributions exponentially at $k\gtrsim k_D$, or multipoles $\ell \gtrsim k_D{\chi }_{\ast }$. Modes outside the sound horizon have no completed acoustic oscillation.

The metric integral represents gravitational redshift, including contributions that in another gauge are separated into endpoint Sachs-Wolfe effects and Integrated Sachs-Wolfe effects. It is not correct to call the entire synchronous metric integral only a late-time integrated effect. Large-scale metric/curvature fluctuations remain important where photon pressure cannot redistribute matter causally and produce the broad large-angle signal rather than just the pressure-oscillation sequence. Potential evolution during radiation-to-matter transition also drives and shifts acoustic amplitudes near horizon entry; late accelerating-era evolution adds a large-angle integrated signal. These effects are excluded by the oscillator's instruction to ignore metric perturbations. Even if the Newtonian-gauge potential is constant in matter domination, synchronous $h_{ij}^{\prime }$ need not vanish, and the complete observable requires the consistent density, velocity and metric combination.

Write $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. Vary the covariant metric while holding the scalar fixed. The supplied determinant and inverse-metric variations giveBy the definition $\delta S=\frac{1}{2}\int \sqrt{-g}{T}^{\mu \nu }\delta g_{\mu \nu }$, lowering its indices yields the stress-energy tensor of a canonical scalar field:The sign in the determinant variation is positive for varying $g_{\mu \nu }$ and would be negative for varying the inverse metric; matching that convention is essential.

For a homogeneous scalar in conformal time, $(\partial \bar{\varphi }{)}^2=-{\bar{\varphi }}^{\prime 2}/a^2$. The time component of the stress-energy tensor is $T_{00}={\bar{\varphi }}^{\prime 2}/2+a^{2}V$, while $T_{ij}={\delta }_{ij}[{\bar{\varphi }}^{\prime 2}/2-a^{2}V]$. Raising one index and applying the prescribed definitions givesThese are the scalar kinetic energy plus/minus potential energy. In proper cosmological time $d{t}_{\mathrm{proper}}=adt$, the same kinetic term is ${\dot{\bar{\varphi }}}^2/2$. Potential domination gives negative pressure, while kinetic domination gives $P\simeq \rho$.

The mixed stress tensor has the convenient exact formSince the background scalar has no spatial gradient, retaining first-order terms gives ${\partial }^0\varphi =-{\bar{\varphi }}^{\prime }/a^2$ and ${\partial }_i\varphi ={\partial }_i\delta \varphi$ in the product. ThereforeIt is a gradient momentum density, hence a scalar velocity perturbation. With a rolling background, uniform-field slices also have vanishing scalar momentum density and provide the field's comoving slicing.

Apply the stated passive coordinate change to the background metric. Its spatial variation isCompare this with the spatial scalar decomposition $\delta g_{ij}=-2a^2\Psi {\delta }_{ij}+2a^2{\partial }_i{\partial }_j\chi$. It gives $\tilde{\Psi }=\Psi +\mathcal{H}T$ and $\tilde{\chi }=\chi -L$, where $\mathcal{H}=a^{\prime }/a$. The scalar's background depends only on time, soThe spatial displacement $L$ changes the shear potential but not these two scalar amplitudes in this decomposition. Direct cancellation of $T$ gives either of the useful invariant combinations$Q$ is the field fluctuation on flat gauge in cosmology slices: choosing $T=-\Psi /\mathcal{H}$ sets the transformed curvature potential to zero and leaves $\widetilde{\delta \varphi }=Q$. The comoving curvature perturbation $\mathcal{R}$ is the spatial curvature potential on uniform-field slices: choose $T=\delta \varphi /{\bar{\varphi }}^{\prime }$ and use (iii) to see that these slices also have zero scalar momentum density. For this spatial sign convention the intrinsic curvature perturbation is $\delta R^{(3)}=4a^{-2}{\nabla }^2\mathcal{R}$ in that gauge; some conventions define the curvature perturbation with the opposite sign. The canonically useful Mukhanov-Sasaki variable is $v=aQ$.

The interpretation of $\mathcal{R}$ requires ${\bar{\varphi }}^{\prime }\ne 0$, and the flat-slice expression requires $\mathcal{H}\ne 0$. The polynomial combination ${\bar{\varphi }}^{\prime }\Psi +\mathcal{H}\delta \varphi$ remains invariant without division; at a stationary field background $\delta \varphi$ itself is invariant under the time shift, but that field cannot serve as a comoving clock.