Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 62 (Cosmology)

Use units with $c=1$. For the flat FRW metric, the Friedmann equation is $H^2=8\pi G\rho /3+\Lambda /3$, where $H=\dot{a}/a$ is the Hubble parameter. The cosmological continuity equation for a separately conserved component with constant equation of state $P=w\rho$ gives $\rho \propto a^{-3(1+w)}$. Consequently radiation in cosmology scales as $a^{-4}$, pressureless matter as $a^{-3}$, and cosmological constant energy is constant. With $a_0=1$, define the present critical density and cosmological density parameters byThe Hubble constant $H_0$ measures the expansion rate today. Here ${\Omega }_r$ includes photons and relativistic neutrinos, ${\Omega }_m$ includes baryons and cold dark matter, and ${\Omega }_{\Lambda }$ measures the vacuum contribution. Spatial flatness implies ${\Omega }_r+{\Omega }_m+{\Omega }_{\Lambda }=1$. ThusSince conformal time satisfies $d\tau =dt/a$, we have $a^{\prime }=a\dot{a}=a^{2}H$. Multiplication by $a^4$ provesFor an exam-level estimate, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ with $h\simeq 0.7$, ${\Omega }_m\simeq 0.3$, ${\Omega }_{\Lambda }\simeq 0.7$, and the specified radiation in cosmology gives ${\Omega }_r=4.2\times 10^{-5}{h}^{-2}\simeq 8.6\times 10^{-5}$. These are rounded illustrative flat-universe values; precision cosmological density parameters depend on the fitted model. For comparison, the base flat-model fit in Planck 2018 cosmological parameters gives $H_0\simeq 67.4\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and ${\Omega }_m\simeq 0.315$.

On the expanding branch of the Friedmann equation, dropping the vacuum term gives the conformal timeThe lower limit sets $\tau =0$ at the Big Bang. The approximation requires ${\Omega }_{\Lambda }{a}^4\ll {\Omega }_{m}a$, and remains safe earlier when radiation in cosmology dominates. At matter-radiation equality, ${\rho }_{m,0}{a}^{-3}={\rho }_{r,0}{a}^{-4}$, so $a_{\mathrm{eq}}={\Omega }_r/{\Omega }_m$. Substitution yieldsIn conformal coordinates the radial FRW metric is $d{s}^2=a^2(-d{\tau }^2+d{r}^2)$. A radial light ray therefore travels a comoving distance $dr=d\tau$. The maximum distance traversed since the Big Bang is ${\int }_0^{t_{\mathrm{eq}}}dt/a={\tau }_{\mathrm{eq}}$: this is the comoving particle horizon, rather than the instantaneous comoving Hubble radius or the acoustic horizon. Restoring $c$ in the numerical conversion givesFor ${\Omega }_m=0.3$ and $h=0.7$, the comoving particle horizon at matter-radiation equality is about $110\mathrm{Mpc}$. The corresponding scale factor is $a_{\mathrm{eq}}\simeq 2.9\times 10^{-4}$.

Dropping radiation in cosmology in the Friedmann equation givesTo evaluate this matter-vacuum conformal age integral, put $a=({\Omega }_m/{\Omega }_{\Lambda }{)}^{1/3}y$. The square root in the denominator becomes ${\Omega }_m^{2/3}{\Omega }_{\Lambda }^{-1/6}\sqrt{y(1+y^3)}$, whereas $da={\Omega }_m^{1/3}{\Omega }_{\Lambda }^{-1/3}dy$. ThereforeThe radiation-free approximation is not valid arbitrarily close to the lower limit. Its extension to zero nevertheless gives a good leading conformal time estimate: the early-time correction has size $O(H_0^{-1}\sqrt{{\Omega }_r}/{\Omega }_m)$, small compared with the total age when matter-radiation equality occurs sufficiently early. The apparent $y^{-1/2}$ endpoint singularity is integrable.

An object at matter-radiation equality lies at comoving radial distance $\chi ={\tau }_0-{\tau }_{\mathrm{eq}}$ in the flat FRW metric. Its angular diameter distance is $D_A=a_{\mathrm{eq}}\chi$, while one particle horizon radius has physical size $a_{\mathrm{eq}}{\tau }_{\mathrm{eq}}$. Its small angular radius is consequentlyThe last step uses ${\tau }_{\mathrm{eq}}\ll {\tau }_0$. If ${\Omega }_{\Lambda }/{\Omega }_m\gg 1$, the matter-vacuum conformal age integral tends to $I$: its integrand behaves as $y^{-1/2}$ at zero and $y^{-2}$ at infinity. Inserting the two conformal times givesThe negative matter exponent follows directly from ${\Omega }_m^{-1}/{\Omega }_m^{-1/3}$. This agrees with the original PDF; the TeX transcription drops the minus sign. If desired, $z=y^3$ evaluates the constant through the beta function and gamma function:The infinite-upper-limit approximation requires vacuum dominance much stronger than in the illustrative parameters of part (a), so it is an asymptotic formula rather than a precision angular prediction for those values.

Let $\mathcal{L}=-\frac{1}{2}{g}^{\alpha \beta }{\partial }_{\alpha }\varphi {\partial }_{\beta }\varphi -V$. In the metric variation specified here, the covariant metric tensor $g_{\mu \nu }$ is varied while the scalar field is held fixed. Differentiating the inverse and the determinant givesThus $\delta \mathcal{L}=\frac{1}{2}{\partial }^{\mu }\varphi {\partial }^{\nu }\varphi \delta g_{\mu \nu }$, and the action varies asThe functional derivative therefore gives $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$. Lowering both indices yields the stress-energy tensor of a canonical scalar field:The determinant variation has a positive sign because the variable is $g_{\mu \nu }$; varying the inverse metric tensor instead would reverse that intermediate sign.

For the flat FRW metric, $g^{00}=-1$, $g^{0i}=0$, and $g^{ij}=a^{-2}{\delta }^{ij}$. Consequently the kinetic contraction of the scalar field isInserting this in $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi +g^{\mu \nu }\mathcal{L}$, with ${\partial }^0\varphi =-\dot{\varphi }$ and ${\partial }^i\varphi =a^{-2}{\partial }_i\varphi$, givesThese are contravariant coordinate components of the stress-energy tensor; spatial components in an orthonormal frame carry different powers of $a$. In particular $T^{00}$ is the energy density seen by the comoving observer.

Vary the scalar field with compactly supported variation, leaving the metric tensor fixed. The action variation isIntegration by parts moves the derivative from $\delta \varphi$ and gives the Euler-Lagrange equationFor the FRW metric, $\sqrt{-g}=a^3$, so the covariant wave operator becomesRearranging gives the requested scalar field evolution:The damping term is Hubble friction; it comes from the expanding volume factor $a^3$, not from a microscopic dissipative force.

For a homogeneous scalar field, every spatial derivative vanishes. The stress-energy tensor consequently has no momentum density or directional stress:The only spatial tensor left is ${\delta }^{ij}$, so the stress is isotropic. This is the perfect fluid form in its comoving frame. Now expand the time component of stress-energy conservation using the covariant derivative:The first term is $\dot{\rho }$. In the second, only $\lambda =0$ contributes, and ${\sum }_i{\Gamma }^i{}_{i0}=3H$, giving $3H\rho$. The third isHere the quoted Christoffel symbols include their lower-index symmetry ${\Gamma }^i{}_{j0}={\Gamma }^i{}_{0j}$. Therefore the cosmological continuity equation isAs a check using the scalar field equation, $\dot{\rho }=\dot{\varphi }(\ddot{\varphi }+V_{,\varphi })=-3H{\dot{\varphi }}^2=-3H(\rho +P)$, with exactly the same sign.

With zero scalar potential and a homogeneous scalar field, the preceding perfect fluid identification givesThis equation of state is that of a stiff fluid. The field equation integrates to $a^3\dot{\varphi }=K$, since ${\partial }_t(a^3\dot{\varphi })=0$. Hence $\rho =K^2/(2a^6)$, agreeing with the cosmological continuity equation for $w=1$. If this nonzero component dominates a flat universe, the Friedmann equation giveson the expanding branch. Integrating once more, $a^3=3\sqrt{4\pi G/3}|K|(t-t_B)$. Choosing the Big Bang at $t_B=0$ yieldsThe homogeneous free scalar as a stiff fluid construction requires $K\ne 0$; a constant zero-potential field has zero energy density and cannot dominate this expansion.

Work in natural units $\hslash =c=k_B=1$, at temperatures far below the $W$-boson mass but of order or above the electron mass. The Fermi interaction then supplies a local four-fermion amplitude. The Fermi constant has dimension energy${}^{-2}$. In scattering of a thermal lepton from a nonrelativistic nucleon, the nucleon current matrix element has size $m_N$, while the lepton current has size $T$. Thus $\mathcal{M}\sim G_F{m}_{N}T$. The two-body relativistic scattering cross-section contains the kinematic factor $s^{-1}\sim m_N^{-2}$; when the final and initial lepton energies are comparable, their momentum ratio is of order one. Consequently $\sigma \sim |\mathcal{M}{|}^2/m_N^2\sim G_F^2{T}^2$. This estimates temperatures above the electron mass and mass-splitting thresholds; those scales only affect the dimensionless rate function near freeze-out.

The light leptons in the thermal bath have number density $n\sim T^3$ and speed of order unity. Their conversion rate per nucleon is thereforeFor example, $n+{\nu }_e\leftrightarrow p+e^{-}$ and $n+e^{+}\leftrightarrow p+{\bar{\nu }}_e$ change the neutron-to-proton ratio. The bath density here is the lepton density, not the much smaller baryon density. Dimensionless weak couplings, phase-space integrals and the several reaction channels set a numerical coefficient. Electron masses and the neutron-proton splitting also become important for a precision rate near cosmological weak freeze-out.

During radiation domination, the effective number of relativistic energy degrees of freedom, denoted $g_{\ast }$, gives the energy densityThe paper's ${\mathcal{N}}_{\mathrm{eff}}$ is this $g_{\ast }$; it is not the conventionally normalized effective number of neutrino species. With the specified reduced Planck mass, the Friedmann equation reads $H^2={\rho }_R/(3M_{\mathrm{Pl}}^2)$, whenceWrite the weak interaction conversion rate as $\Gamma =A{G}_F^2{T}^5$. The ratio $\Gamma /H$ scales as $T^3$, so it falls during cooling: conversion eventually becomes too slow to maintain neutron-proton chemical equilibrium. Equating the conversion and expansion rates givesOn suppressing the numerical coefficient, the cosmological weak freeze-out temperature is consequentlyThe one-sixth power reflects a balance between a conversion rate proportional to $T^5$ and a Hubble parameter proportional to $\sqrt{g_{\ast }}{T}^2$.

Before electron-positron annihilation in cosmology, photons contribute two helicity states; electrons and positrons contribute two spin states each. Each of the three light neutrino flavours contributes one active neutrino helicity and one antineutrino helicity. At a common temperature the effective number of relativistic energy degrees of freedom is thereforeUsing $G_F=1.17\times 10^{-5}{\mathrm{GeV}}^{-2}$ and $M_{\mathrm{Pl}}\simeq 2.435\times 10^{18}\mathrm{GeV}$ in the scaling formula alone gives $T_F\sim 2\mathrm{MeV}$. Including the explicit expansion coefficient from (ii), but setting the unknown rate coefficient $A$ to one, gives $T_F\simeq 1.5\mathrm{MeV}$. Thus the warranted conclusion is a freeze-out temperature of order $1\mathrm{MeV}$, above the $0.511\mathrm{MeV}$ electron mass. The omitted weak-rate coefficient prevents a more precise prediction from this estimate.

Neutrino decoupling occurs before most electron-positron pairs annihilate. Afterwards neutrinos redshift freely, $T_{\nu }a=\mathrm{constant}$, while the photons receive the pair entropy. For the still-coupled photon-pair gas, cosmological entropy conservation gives $g_{\ast s}{T}_{\gamma }^3{a}^3=\mathrm{constant}$. Its entropy degrees of freedom drop from $2+(7/8)4=11/2$ to $2$. Comparing this with the freely cooling neutrinos givesAfter annihilation, photon and neutrino temperatures differ. To write the total energy density as $({\pi }^2/30)g_{\ast }{T}_{\gamma }^4$, the neutrino contribution must include that temperature ratio:This is the paper's post-annihilation ${\mathcal{N}}_{\mathrm{eff}}$. Merely deleting the electron term while keeping all species at the same temperature would be incorrect. For comparison the entropy count, normalized to $T_{\gamma }^3$, is $g_{\ast s}=2+(7/8)6(4/11)=43/11\simeq 3.91$. These results assume instantaneous neutrino decoupling and relativistic neutrinos; the two counts represent different thermal moments.

The nucleons are nonrelativistic, so the Maxwell-Boltzmann distribution givesBoth spin degeneracies are two. Taking the ratio and using the specified negligible chemical potentials yields the neutron-to-proton ratioThe mass prefactor differs from unity only at the per-mille level. The dependence on $T_F$ is much stronger: $d\ln r_F/d\ln T_F=Q/T_F>0$. Thus earlier, hotter weak freeze-out leaves more neutrons relative to protons. For illustration, $T_F=1\mathrm{MeV}$ gives $r_F\simeq 0.28$; $T_F=0.8\mathrm{MeV}$ gives $r_F\simeq 0.20$. Free-neutron decay after freeze-out lowers the ratio further before most Big Bang nucleosynthesis takes place.

The small baryon-to-photon ratio suppresses nuclear equilibrium abundances through the factor ${\eta }^{A-1}$. Cooling eventually makes the binding-energy exponential favourable, but the deuterium bottleneck delays substantial nuclear burning until well after cosmological weak freeze-out. In ordinary Big Bang nucleosynthesis, nearly every neutron surviving that delay ends up in helium-4. This makes the final primordial helium mass fraction chiefly a neutron budget rather than an unrestricted equilibrium abundance.

Let $r_F$ be the neutron-to-proton ratio at freeze-out, let $\Delta t$ be the delay until effective nuclear burning, and let ${\tau }_n$ be the neutron mean lifetime. The initial fraction of baryons which are neutrons is $x_{n,F}=r_F/(1+r_F)$. Because neutron decay converts a neutron into a proton, total baryon number remains constant, while the neutron fraction becomes $x_{n,B}=x_{n,F}{e}^{-\Delta t/{\tau }_n}$. Each helium nucleus uses two neutrons and four baryons. Hence neutron-limited helium synthesis givesEquivalently, $Y_4=2r_B/(1+r_B)$ with $r_B=x_{n,B}/(1-x_{n,B})$. This accounts for the extra protons produced by neutron decay; multiplying $r_F$ alone by the survival factor would miss their contribution. The following responses refer to ordinary neutron-limited burning, without exotic particle injection or a change to the nuclear network.

Extra relativistic neutrino species increase the effective number of relativistic energy degrees of freedom. At fixed temperature the Hubble parameter rises as $g_{\ast }^{1/2}$, while the standard conversion rate is unchanged. Consequently cosmological weak freeze-out moves to higher temperature, $T_F\propto g_{\ast }^{1/6}$, and the neutron-to-proton ratio $r_F=e^{-Q/T_F}$ increases. The initial neutron fraction $r_F/(1+r_F)$ therefore increases as well.

The quicker expansion also reduces the cosmic time required to cool to the deuterium bottleneck temperature: during radiation domination, $t\propto g_{\ast }^{-1/2}{T}^{-2}$ while the thermal count is roughly constant. Fewer neutrons decay before burning. Both factors in the neutron budget increase, giving a larger final helium-4 mass fraction. This is the primordial helium response to extra relativistic species, under the assumption that the additional species contribute mainly to expansion rather than independently changing the conversion rates.

A longer neutron half-life means a larger mean lifetime ${\tau }_n={\tau }_{1/2}/\ln 2$. Holding the initial neutron-to-proton ratio and the nuclear-burning delay fixed, the derivative of the neutron survival factor is positive:Thus fewer neutrons disappear during the deuterium bottleneck, and the final helium-4 mass fraction increases.

There can be a second effect if the increased lifetime represents a uniform reduction of the relevant weak interaction strength. With fixed masses and other couplings, neutron decay and the conversion reactions both scale as $G_F^2$. Thus ${\tau }_n\propto G_F^{-2}$, whereas $T_F\propto G_F^{-2/3}\propto {\tau }_n^{1/3}$. Weaker conversion then freezes out earlier and increases $r_F$, reinforcing the survival effect. The survival argument alone already establishes the requested neutron half-life effect on primordial helium; linking lifetime to all conversion rates additionally requires this microscopic assumption.

A larger baryon-to-photon ratio favours bound nuclei at a fixed temperature through the equilibrium factor ${\eta }^{A-1}$. In particular it helps deuterium survive photodissociation sooner: there are fewer energetic photons per baryon, and nuclear reaction rates per baryon also rise. A schematic deuterium bottleneck condition is $\eta e^{B_D/T_B}\sim \mathrm{constant}$, suppressing slowly varying powers of temperature. It givesThus higher $\eta$ permits burning at a higher temperature. During radiation domination that means earlier cosmic time, $t_B\propto T_B^{-2}$, leaving less time for free-neutron decay after freeze-out. More neutrons enter helium, so the final helium-4 mass fraction increases, ordinarily rather weakly. This is the baryon-density effect on primordial helium.

The equilibrium helium factor ${\eta }^3$ indicates the direction of the change but is not a prediction that the final helium abundance grows without bound as ${\eta }^3$. Once almost all surviving neutrons are captured, the final primordial helium mass fraction is limited by their supply, as the shared neutron budget shows.

Write $\delta ={\delta }_{C,\mathbf{k}}$ and $\mathcal{H}=a^{\prime }/a$, the conformal Hubble parameter. The flat Friedmann equation givesOutside the horizon, the stated adiabatic cosmological perturbation relation is ${\delta }_R=(4/3)\delta$. It expresses equality of ${\delta }_i/(1+w_i)$ for dust and radiation in the density convention of the supplied equations. During radiation domination, $a\propto \tau$, so $\mathcal{H}=1/\tau$ and ${\rho }_C$ is negligible. Consequently $8\pi G{a}^2{\rho }_R=3/{\tau }^2$, and the linear cosmological density perturbation equation becomesThe trial solution $\delta \propto {\tau }^q$ gives $q(q-1)+q=4$, hence $q=2,-2$.

During matter domination, $a\propto {\tau }^2$, so $\mathcal{H}=2/\tau$. Now the radiation source is negligible and $4\pi G{a}^2{\rho }_C=(3/2){\mathcal{H}}^2=6/{\tau }^2$. The equation becomeswhose exponents satisfy $q(q-1)+2q=6$, or $(q-2)(q+3)=0$. Thus the growing solutions in both eras areThe other modes are ${\tau }^{-2}$ in the radiation limit and ${\tau }^{-3}$ in the matter limit. These are statements in the density and gauge convention of the supplied evolution equation; a density contrast outside the horizon is not a gauge-independent observable by itself.

After cosmological horizon crossing, the rapidly oscillating radiation source can be averaged away as specified. Well inside radiation domination, the remaining cold dark matter self-gravity is also subleading because ${\rho }_C/{\rho }_R\ll 1$. To leading order the linear cosmological density perturbation equation is thereforeTwo integrations giveThe reference time makes the logarithm dimensionless. A growing superhorizon mode reaches entry with a nonzero derivative, so matching generally excites $B\ne 0$. For example, an abrupt leading-order match from $\delta \propto {\tau }^2$ gives $B={\tau }_H{\delta }_H^{\prime }\simeq 2{\delta }_H$, whereas its exact numerical coefficient depends on the transition and the radiation forcing.

This logarithmic growth is the Mészáros effect. It is an asymptotic radiation-era result: after dropping the oscillating radiation perturbation, the original equation still contains $4\pi G{a}^2{\rho }_C\delta$. Retaining that term and the changing background leads to the Mészáros equation, which provides the interpolation near matter-radiation equality. It would be incorrect to claim that a pure logarithm solves the complete mixed-background equation exactly.

During matter domination, the radiation source can be neglected irrespective of whether the mode is inside or outside the horizon in the supplied linear cosmological density perturbation equation. The Friedmann equation gives $\mathcal{H}=2/\tau$ and $4\pi G{a}^2{\rho }_C=6/{\tau }^2$, just as in part (a). ThereforeSubstituting ${\tau }^q$ gives $(q-2)(q+3)=0$, so for every Fourier modeThus the matter-era growing and decaying density modes have growing behaviour ${\delta }_{C,\mathbf{k}}\propto {\tau }^2\propto a$ for all $k$. There is no pressure-gradient term introducing a $k$-dependent growth rate for cold dark matter. The coefficients retain their different earlier histories; common growth does not erase the spectral shape. A specially chosen pure decaying solution is of course possible, so the growing-mode statement does not assert that every initial condition grows.

Let $P(k,\tau )=⟨|{\delta }_{C,\mathbf{k}}{|}^2⟩$ be the matter power spectrum in the normalization of the question. The Harrison-Peebles-Zeldovich spectrum gives $P\sim \mathcal{C}{\tau }^{4}k$ before entry. Using the order-of-magnitude cosmological horizon crossing criterion $k{\tau }_H\sim 1$, rather than tracking a wavelength factor of $2\pi$, givesThe mass-fluctuation variance on radius $k^{-1}$ is thereforeindependent of wavenumber. Equal logarithmic ranges of scales enter with comparable dimensionless fluctuation strength. This is the scale invariance encoded by the HPZ spectrum; it is not the claim that the dimensional matter power spectrum is independent of $k$ at a fixed time.

First consider $k{\tau }_{\mathrm{eq}}<1$. Such a mode remains outside the horizon through matter-radiation equality, grows as ${\tau }^2$ on both sides, and keeps the same matter-era growth after entry. Squaring this growth preserves the shape of the HPZ spectrum:up to a scale-independent matching coefficient.

For $k{\tau }_{\mathrm{eq}}\gg 1$, entry occurs in radiation domination at ${\tau }_H\sim k^{-1}$, with $P_H\sim \mathcal{C}{k}^{-3}$. The density amplitude then grows only logarithmically until equality. ThusDuring matter domination, the matter-era linear growth factor of the density is $(\tau /{\tau }_{\mathrm{eq}}{)}^2$, so its square multiplies the matter power spectrum:The turnover at $k\sim {\tau }_{\mathrm{eq}}^{-1}$ records whether a mode entered during radiation domination. Earlier-entering small-scale modes miss the rapid ${\tau }^2$ growth and receive only the logarithmic Mészáros effect. This is the origin of the logarithmic small-scale matter transfer. The formula is a large-$k$ asymptote: the additive constant in the logarithmic solution and smooth equality matching must be retained near $k{\tau }_{\mathrm{eq}}=1$, where the spectrum does not vanish.

Multiplying by $k^3$ gives the small-scale mass-fluctuation varianceThis gives the requested estimate for modes entering before equality, with the logarithm replaced by a smooth order-one factor in the transition region. On comoving scale $L\sim k^{-1}$, unit variance today requiresFor a numerical illustration faithful to this question's flat, matter-plus-radiation background, take $h=0.7$ and ${\Omega }_r=4.2\times 10^{-5}{h}^{-2}$, hence ${\Omega }_C=1-{\Omega }_r\simeq 1$ rather than the $0.3$ appropriate to a vacuum-dominated model. Part 1(b), now with ${\Omega }_m={\Omega }_C$, gives ${\tau }_{\mathrm{eq}}\simeq 33\mathrm{Mpc}$. In this two-component model the same integral is valid all the way to today, givingWith $L=3\mathrm{Mpc}$, ${\tau }_0/{\tau }_{\mathrm{eq}}\simeq 258$ and $\ln ({\tau }_{\mathrm{eq}}/L)\simeq 2.39$. Therefore $\mathcal{C}\sim [258^4(2.39{)}^2{]}^{-1}\simeq 4\times 10^{-11}$, corresponding to a horizon-entry root-mean-square fluctuation of roughly $6\times 10^{-6}$. A very small primordial amplitude, of order $10^{-10}$ in this schematic normalization, can reach unit small-scale variance by today. Factors from the Fourier convention, smoothing window and equality matching preclude a precise normalization here. The calculation marks the onset of nonlinear structure formation, where extrapolating the linear solution ceases to be valid. For a universe with late vacuum domination, one must replace the matter-only growth law by its actual growth factor before making a numerical inference.