Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 71 (Cosmology)

A photon loses energy by $1+z=a_0/a_e$ between emission and reception. The interval between successive photon arrivals is also longer by $1+z$, so the received energy per unit time is reduced by $(1+z{)}^2$. An isotropic wavefront at reception has area $4\pi (a_{0}r{)}^2$: the astronomical radial coordinate is an areal coordinate, even in curved space. HenceThis defines the luminosity distance. Photon absorption or conversion would require a further transmission factor; the relation assumes freely propagating conserved photons and bolometric luminosity. The radial cosmological proper distance is generally different from the areal distance $a_{0}r$.

Take zero cosmological constant and an open geometry, which means $k<0$ in the metric convention. Evaluating the Friedmann equations today gives $H_0^2=H_0^2{\Omega }_0-k/a_0^2$, soOn a constant-time slice the radial line element is $d\ell =a_{0}dr/\sqrt{1-k{r}^2}$. ThereforeThis is a cosmological proper distance on today's slice, not the distance traveled during the photon flight. The printed condition $k<1$ is insufficient to specify an open model; the derivation uses its intended $k<0$.

For radial null propagation, $dr/\sqrt{1-k{r}^2}=|dt|/a$. Since $dz/dt=-(1+z)H(z)$ and $a=a_0/(1+z)$, its present-distance integral isThus the printed extra $a_0$ before the redshift integral is inappropriate when $H$ is the physical Hubble parameter; it disappears only if $a_0$ has been set to one.

Dust conservation and the curvature relation giveTo evaluate it, put $s=\sqrt{1-{\Omega }_0}$ and $v=\sqrt{1+{\Omega }_{0}z}$. The dimensionless integral $I=H_0{d}_S$ isInverting the proper-distance expression in part b gives $r=\sinh (sI)/(a_0{H}_{0}s)$. Substituting the logarithm, writing the hyperbolic sine as half the difference of its exponential and inverse, and using $v^2-s^2={\Omega }_0(1+z)$ yieldsThis is the open dust distance-redshift relation. The printed square root lacks the factor $z$; its literal form would give a negative, nonzero coordinate at $z=0$. The corrected expression has $r\sim z/(a_0{H}_0)$ near zero. At fixed positive ${\Omega }_0$ and $z\gg {\Omega }_0^{-1}$, the term linear in $z$ dominates, givingThe zero-density limit is nonuniform: a pure Milne model has no such finite limiting radial coordinate.

Use radiation-era $H=C_{\ast }{T}^2/m_{\mathrm{pl}}$, with $C_{\ast }=1.66\sqrt{g_{\ast }}$. The stipulated interaction rate is ${\alpha }_Y^4{T}^5/m_Y^4$, not ${\alpha }_Y^2{T}^5/m_Y^4$. Decoupling therefore givesAt the threshold $T_D=m_{Y\mathrm{nr}}$, the assumed $g_{\ast }\simeq 100$ is self-consistent with the stated high-temperature condition. Smaller masses decouple nonrelativistically, with $x_D=m_Y/T_D=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}\gg 1$.

For an thermal equilibrium species with zero chemical potential, $n_Y=2(m_Y{T}_D/2\pi {)}^{3/2}{e}^{-x_D}$, while $s=2{\pi }^2{g}_{\ast s}{T}_D^3/45$. ThusThe numerical coefficient before rounding is about $2.9\times 10^{-3}$ for $g_Y=2$ and $g_{\ast s}=100$. This is mass-dependent fermion freeze-out. It assumes the rate maintains chemical equilibrium until sudden cosmological particle freeze-out and that entropy is conserved afterwards. Elastic scattering alone would not justify the thermal equilibrium number-density estimate. For decoupling below the high-temperature range, changes in effective degrees of freedom modify the numerical estimate; the requested formula treats them as roughly constant.

In the specified dust model $H_0=2/(3t_0)$, so ${\rho }_{\mathrm{crit}}=m_{\mathrm{pl}}^2/(6\pi t_0^2)$ and $n_B={\Omega }_{B0}{\rho }_{\mathrm{crit}}/m_B$. Photons and the six neutrino/antineutrino helicity states giveUsing the supplied GeV estimates literally givesThe rounded time/temperature conversions are unusually crude: converting $10$ Gyr to $4.8\times 10^{41}{\mathrm{GeV}}^{-1}$ and $3$ K to $2.6\times 10^{-13}$ GeV instead gives $Y_B\sim 10^{-10}$. Either set gives the same order-of-magnitude mass conclusion below, but the numerical abundance should identify which supplied units are used.

For a stable relic, the relevant comparison is energy density, not merely particle number. The matter density left over after the baryons is at most nine times the baryon density here. ConsequentlyPut $x=(m_{Y\mathrm{nr}}/m_Y{)}^{1/3}$ and use $Y_Y\simeq C_Y{x}^{3/2}{e}^{-x}$ with $C_Y\simeq 10^{-3}$. The condition becomesFor the literal supplied numbers the right side is about $30.5$, giving $x\gtrsim 25.6$ and $m_Y\lesssim 40$–$50$ GeV. With the physically consistent conversions the estimate is about $60$ GeV. The intended rough conclusion is therefore a sufficiently suppressed branch with $m_Y$ no larger than order $10^2$ GeV, within the toy cosmological particle freeze-out law. Comparing the two number yields alone gives a similar rough scale but misses the mass weighting. This is a thermal relic density bound for a tunable fermion.

For $m_Y\gtrsim m_{Y\mathrm{nr}}$, relativistic decoupling would instead leave $Y_Y\simeq 135\zeta (3)g_Y/(8{\pi }^4{g}_{\ast s})\simeq 4\times 10^{-3}$, incompatible with the matter-density bound for such heavy stable particles. These are conditional bounds on a populated, stable thermal relic. A short lifetime or an initially suppressed nonthermal population can change the conclusion; the supplied information alone does not constrain every possible production and decay history.

Above the transition,for $g_{\ast }\sim 100$. The ratio increases during cooling, rather than at higher temperature. A GUT transition at $T_c\sim 10^{15}$–$10^{16}$ GeV occurs well above this scale, so the interaction is too slow to establish or maintain thermal equilibrium throughout $T>T_c$. In fact its integrated interaction probability from an arbitrarily high temperature down to $T_c$ is at most of order $T_{\mathrm{eq}}/T_c\ll 1$, since $dt=-dT/(HT)$.

An initial Fermi-Dirac distribution with zero chemical potential therefore cannot be assumed from this high-temperature rate alone. If imposed initially and collisionlessly redshifted while the particles are massless, its thermal-looking form can persist, but this is not dynamically maintained thermal equilibrium. Other initial states can remain nonthermal. This is the thermalization temperature for a massless gauge mediator.

The below-transition rate stipulated in the model has $\Gamma /H=(T/T_D{)}^3$. If $T_c\gg T_D$ and it includes efficient number-changing reactions, it can restore chemical equilibrium after the transition and erase the initial-state memory, making the estimates in a and b applicable. If it only equilibrates momenta, or if reheating never populates the relevant range, the thermal relic calculation must be reconsidered.

Separate conservation of a fluid with fixed $w=P/\rho$ gives $\dot{\rho }+3H(1+w)\rho =0$, hence $\rho \propto a^{-3(1+w)}$. The $w=-1/3$ component therefore has ${\rho }_Q\propto a^{-2}$. Flatness today fixes its normalization to ${\rho }_{Q0}=3H_0^2(1-{\Omega }_{M0})/(8\pi G)$, soThis is a coasting fluid: its contribution to $\rho +3P$ is zero, so it does not itself drive accelerated expansion. Its background Friedmann contribution resembles curvature, although the actual spatial geometry remains flat.

Write $\Omega ={\Omega }_{M0}$, ${\beta }^2=H_0^2(1-\Omega )$ and $\mathcal{H}=a^{\prime }/a=\dot{a}$. The Friedmann equation isThe Raychaudhuri equation has only the matter component on its active-gravity side, soCombining these expressions provesThis is the conformal Riccati equation for matter and a coasting fluid. The prime denotes conformal time, whereas $H_0$ is the physical present Hubble parameter.

Choose the Big Bang at $\tau =t=0$. For $0<\Omega <1$, the expanding singular solution of the Riccati equation is $\mathcal{H}=\beta \coth (\beta \tau /2)$. Integrating $a^{\prime }/a=\mathcal{H}$ gives $a=C{\sinh }^2(\beta \tau /2)$. The Friedmann constraint fixes $C=\Omega /(1-\Omega )$, rather than leaving an independent amplitude. ThusNow $dt=ad\tau$. Integrating from the singular origin givesThis derives the flat matter-coasting-fluid Friedmann solution. For small $\tau$, $a\propto {\tau }^2$ and $t\propto {\tau }^3$, hence $a\propto t^{2/3}$. At late time $a\propto t$, as expected when the coasting fluid dominates.

Set $a=1$ in the parametric solution. With $b=1-\Omega$,ConsequentlyThe endpoint values are obtained by limits. A particularly direct proof of the age bound avoids subtracting nearly equal terms: from the Friedmann equation,For nonnegative component densities, $0\le \Omega \le 1$ and $a\le \Omega +(1-\Omega )a\le 1$. The integrand lies between $\sqrt{a}$ and $1$. Integrating therefore givesEquality holds for pure matter and pure coasting fluid, respectively. These are the age of a flat matter-coasting-fluid universe and the coasting-fluid age bound; the positivity range is necessary for the stated universal inequality.

For a homogeneous canonical scalar, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. The acceleration equation requires $V>{\dot{\varphi }}^2$ for cosmic inflation, equivalently ${ϵ}_H=-\dot{H}/H^2<1$ in the flat scalar-dominated limit. Inflation need not be exactly de Sitter.

A standard sufficient regime for a long period is positive potential energy dominating the kinetic energy, physical gradient energy and curvature. Homogeneity over an expanding approximately Hubble-sized patch allows the gradient term to redshift away. In comoving coordinates the gradient energy is $|{\nabla }_{\mathrm{com}}\varphi {|}^2/(2a^2)$; the gradient in the displayed energy formula must be interpreted physically if no $a^{-2}$ is written there. RequireThe slow-roll equations are then $3H\dot{\varphi }\simeq -V^{\prime }$ and $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$. Sufficient potential conditions areThe potential must remain sufficiently flat over a large enough field interval to produce the required expansion. These slow-roll conditions are stronger than the bare necessary condition for acceleration.

Write $V=V_0{e}^{-A\varphi /m_{\mathrm{pl}}}$, where the stated model has $V_0=m_{\mathrm{pl}}^4$, and take $A>0$. Slow roll givesThus $d[e^{A\varphi /(2m_{\mathrm{pl}})}]/dt=A^2\sqrt{V_0}/(\sqrt{96\pi }{m}_{\mathrm{pl}})$. Absorb the integration constant into $b$ to obtainIt follows that $H=16\pi /[A^2(t+b)]$, soThe exponential potential has constant ${ϵ}_V=A^2/(16\pi )$ and ${\eta }_V=A^2/(8\pi )$. Inflation requires $p>1$, whereas reliable slow-roll approximation requires $A^2/(8\pi )\ll 1$. This is the slow-roll normalization for exponential inflation.

The stated field normalization is a slow-roll result, not an exact solution of the original full equations. Keeping the acceleration and kinetic terms yields the same $p$ but $C^2=A^4{V}_0/[2m_{\mathrm{pl}}^2(48\pi -A^2)]$, as follows by substituting $\dot{\varphi }=2m_{\mathrm{pl}}/[A(t+b)]$ into both full equations. It agrees to leading order for small slope and is the exponential-potential power-law inflation solution.

There is no minimum about which this scalar can undergo the damped oscillations normally used for reheating. Its slow-roll parameters are constant, so an inflating exponential solution has no automatic exit. An independent stopping and energy-transfer mechanism is needed.

At the assumed exit, $V_R=m_{\mathrm{pl}}^4{e}^{-30}$. Equating this to the thermal radiation density givesfor $g_{\ast }=10^3$. Corrections from the small kinetic fraction do not change this rough estimate.

Curvature obeys $|\Omega -1|\propto (aH{)}^{-2}$. For constant ${ϵ}_H=A^2/(16\pi )$ and $N=\log (a_R/a_i)$, its suppression during cosmic inflation is $e^{-2(1-{ϵ}_H)N}$. Starting from order-one curvature and requiring present curvature no larger than order one givesAfter instantaneous reheating, entropy conservation implies $a_R/a_0=(T_0/T_R)(g_{\ast s,0}/g_{\ast s,R}{)}^{1/3}$, with $H_R\simeq 1.66\sqrt{g_{\ast }}{T}_R^2/m_{\mathrm{pl}}$. Present scales make the numerator roughly $60$; using the literal approximate units from question2 gives about $63$. A tighter present curvature tolerance adds its corresponding half-logarithm. In slow-roll approximation this is the conventional requirement of roughly sixty e-folds.

Because the field increases toward ${\varphi }_R$, enough expansion requiresThis is the flatness bound on the initial field in exponential inflation. It is an upper bound on the initial field, not a nonzero minimum: starting earlier at a smaller field produces more expansion. If one additionally restricts the initial potential below $m_{\mathrm{pl}}^4$, then ${\varphi }_i\ge 0$ supplies a lower bound, but that is a separate restriction. Without initial-curvature and final-tolerance choices there is no unique numerical initial-field value.

Put $\mathcal{H}=a^{\prime }/a$ and choose the residual synchronous frame comoving with cold dark matter, so ${\mathbf{v}}_C=0$. Its Euler equation preserves this choice. The continuity equation then gives ${\delta }_C^{\prime }=h^{\prime }/2$. Neglect background baryon pressure and all unenhanced corrections of order $c_s^2$, but retain $c_s^2{k}^2$ because a short wavelength can make it important. The trace equation becomesUsing $h^{\prime }=2{\delta }_C^{\prime }$ provesFor baryons let ${\theta }_B=i\mathbf{k}\cdot {\mathbf{v}}_B$. The retained continuity and Euler terms are ${\delta }_B^{\prime }=-{\theta }_B+h^{\prime }/2$ and ${\theta }_B^{\prime }+\mathcal{H}{\theta }_B-c_s^2{k}^2{\delta }_B=0$. Differentiate the first and substitute the second:HenceThe pressure sign is stabilizing. The drag terms contain first derivatives of the density perturbations, as printed in the PDF; those primes are lost in the converted TeX. The approximation assumes slowly varying nonrelativistic background baryon pressure and the growing physical mode, not a residual synchronous gauge mode.

For small baryon fraction the CDM growing mode is ${\delta }_C\propto a$, and its gravity forcesFor an adiabatic cosmological perturbation to follow the unsuppressed growing mode, pressure must be below the gravitational forcing frequency: $c_s^2{k}^2\lesssim 4\pi G{a}^2{\bar{\rho }}_C$. With physical wavelength $\lambda =2\pi a/k$, this becomesThis is Jeans support versus forced baryon response. The wording “only grow” must refer to unsuppressed gravitational structure growth; it is not a theorem that all shorter-wavelength baryon perturbations are absent or constant. In the pressure-dominated regime a forced response ${\delta }_B\simeq (k_J/k{)}^2{\delta }_C$ remains. In particular, a falling sound speed can make this small response grow even before it catches up to CDM.

Before photon decoupling, the photon-baryon fluid has sound speed of order $c/\sqrt{3}$. Its Jeans scale is roughly a horizon scale, so smaller perturbations oscillate acoustically instead of freely following the growing CDM structures. After decoupling the baryon sound speed drops sharply. Above the new Jeans scale, pressure is negligible and subtraction of the CDM equation gives $({\delta }_B-{\delta }_C{)}^{\prime \prime }+\mathcal{H}({\delta }_B-{\delta }_C{)}^{\prime }=0$. For $a\propto {\tau }^2$, the difference is a constant plus a decaying $1/\tau$ term; its ratio to the growing CDM contrast tends to zero. Baryons therefore catch up. Under the stipulated post-decoupling $c_s\propto T\propto a^{-1}$, the physical Jeans length scales as $a^{1/2}$, the comoving scale as $a^{-1/2}$ and the corresponding Jeans mass as $a^{-3/2}$.

For the requested instantaneous mass estimates, define the baryonic Jeans mass as the baryon content of a sphere of radius ${\lambda }_J/2$:Using the supplied matter-era age gives $t_{\mathrm{dec}}=t_0(1+z_{\mathrm{dec}}{)}^{-3/2}\simeq 9.5\times 10^{13}$ s and ${\bar{\rho }}_C\simeq 1/(6\pi G{t}_{\mathrm{dec}}^2)\simeq 6\times 10^{-23}\mathrm{g}\mathrm{c}{\mathrm{m}}^{-3}$. ThenThe drop is by $(\sqrt{3}\times 10^{-5}{)}^3$. A definition using the total gravitating mass instead of baryon content gives values about ten times larger. The pre-decoupling number is an acoustic Jeans order estimate, not an accurate application of the nonrelativistic equations from part a when $c_s\sim c$.

Holding the background metric fixed, the linearized homogeneous-field equation isConsequently each comoving Fourier coefficient obeysThe unscaled Laplacian in the starting equation must be interpreted as the physical spatial Laplacian; for the explicitly comoving wavevector it gives $k^2/a^2$, not $k^2$. Metric fluctuations have been discarded as permitted. The mode written later is a massless, minimally coupled approximation, requiring $|m_{\mathrm{eff}}^2|\ll H^2$; it is not a solution for an arbitrary potential curvature.

Take a periodic comoving box of volume $L^3$, with $\mathbf{k}=2\pi \mathbf{n}/L$. A real field has the expansionwhere $[{\hat{a}}_{\mathbf{k}},{\hat{a}}_{\mathbf{q}}^{†}]={\delta }_{\mathbf{kq}}$ and the other ladder-operator commutators vanish. The coefficient of $e^{i\mathbf{k}\cdot \mathbf{x}}$ is therefore $w_k{\hat{a}}_{\mathbf{k}}+w_k^{\ast }{\hat{a}}_{-\mathbf{k}}^{†}$. This opposite momentum is required by reality; the single-oscillator shorthand with the same label on both operators must not be read literally as every traveling-wave Fourier coefficient.

The quadratic action has canonical momentum $\hat{\pi }=a^3\delta \dot{\hat{\varphi }}$. Its equal-time field commutator requiresChoose the vacuum annihilated by all $a_{\mathbf{k}}$ and the subhorizon positive-frequency condition. For constant $H$ and negligible effective mass, put $x=k/(aH)$, $C_k=L^{-3/2}H/\sqrt{2k^3}$. The candidate is $w_k=C_k(i+x)e^{ix}$. Since $\dot{x}=-Hx$,Direct substitution makes ${\ddot{w}}_k+3H{\dot{w}}_k+(k^2/a^2)w_k$ identically zero. Its Wronskian is $2iH{C}_k^2{x}^3=i/(L^3{a}^3)$, which has the required normalization. ThusThis is the existing Bunch-Davies mode normalization in a finite comoving volume. It is exact for a massless minimally coupled test field in de Sitter and the leading approximation for a light slowly evolving field. The homogeneous zero mode needs separate treatment.

For $k\gg aH$, the dominant term isThis is the positive-frequency harmonic-oscillator normalization with the physical box volume $a^3{L}^3$. Over a period much shorter than $H^{-1}$, the scale factor and physical frequency are effectively constant. Equivalently, the rescaled field $a{L}^{3/2}{w}_k$ approaches $e^{-ik\eta }/\sqrt{2k}$ exactly in the short-wavelength limit.

For $k\ll aH$, $w_k\to i{L}^{-3/2}H/\sqrt{2k^3}$ and ${\dot{w}}_k=O(x^2)$, so the growing field mode freezes. A single Fourier coefficient has variance proportional to $k^{-3}$, which by itself is not scale-independent. The number of modes in a logarithmic shell is proportional to $L^3{k}^3$. Their product givesThus equal logarithmic intervals have equal variance in exact de Sitter. Slow evolution of $H$ or nonnegligible effective mass produces a tilt. Integrating an exactly flat spectrum over arbitrarily many logarithmic intervals still requires infrared and ultraviolet cutoffs; the statement concerns power per interval, not a finite all-scale total variance.

Use the convention consistent with the supplied fluid transformations: $\delta g_{ij}=a^2{h}_{ij}$, so the synchronous spatial metric is $g_{ij}=-a^2{\delta }_{ij}+a^2{h}_{ij}$. For a passive coordinate displacement ${\xi }^0=T{e}^{i\mathbf{k}\cdot \mathbf{x}}$, ${\xi }^i=i{\hat{k}}^{i}L{e}^{i\mathbf{k}\cdot \mathbf{x}}$, the metric perturbation changes by ${\widetilde{\delta g}}_{\mu \nu }=\delta g_{\mu \nu }-({\mathcal{L}}_{\xi }\bar{g}{)}_{\mu \nu }$. Computing its components givesThe Newtonian gauge has no mixed component and no traceless spatial part. ThusMatching ${\widetilde{\delta g}}_{00}=2a^2\Phi$ and ${\widetilde{\delta g}}_{ij}=2a^2\Psi {\delta }_{ij}$ givesThis is the synchronous-to-Newtonian transformation with positive spatial perturbations. The original PDF uses $i\hat{\mathbf{k}}L$ in the spatial displacement; the converted TeX loses the hat. Using $i\mathbf{k}L$ would require $L=h_S/(2k^2)$ instead. The nonzero modes $k>0$ are understood.

There is a sign error in the printed synchronous line-of-sight integrand. Direct contraction of the stated decomposition gives $h_{ij}^{\prime }{n}^i{n}^j=(h^{\prime }-h_S^{\prime })/3+(\hat{\mathbf{k}}\cdot \mathbf{n}{)}^2{h}_S^{\prime }$, with a plus sign in its last term. In the positive spatial-perturbation convention of part a, the redshift term is half this contraction. The printed minus sign cannot yield the requested Newtonian expression with the supplied transformations.

The consistent identity can be derived explicitly. Work mode by mode with $q=\mathbf{k}\cdot \mathbf{n}$, $E(\tau )=e^{iq(\tau -{\tau }_0)}$ along the unperturbed ray; the shift of the phase origin is immaterial. Put $A=(h-h_S)/6$. Part a gives $A=\Psi -\mathcal{H}T$, while $\Phi =-\mathcal{H}T-T^{\prime }$. Therefore $A^{\prime }={\Phi }^{\prime }+{\Psi }^{\prime }+T^{\prime \prime }$. The corrected synchronous integral issince $[E(T^{\prime }-iqT){]}^{\prime }=E(T^{\prime \prime }+q^{2}T)$. This is the photon redshift identity with positive spatial perturbations.

The supplied fluid shifts are ${\delta }_{\gamma }^N={\delta }_{\gamma }+4\mathcal{H}T$ and ${\mathbf{v}}_{\gamma }^N={\mathbf{v}}_{\gamma }+i\mathbf{k}T$. At emission, the synchronous intrinsic and Doppler terms are consequently $E_d[{\delta }_{\gamma }^N/4+{\mathbf{v}}_{\gamma }^N\cdot \mathbf{n}-\mathcal{H}T-iqT{]}_d$. Adding the lower integral endpoint cancels the velocity shift and leaves $-\mathcal{H}T-T^{\prime }=\Phi$. The upper endpoint is the local observer monopole/dipole $[T^{\prime }-iqT{]}_0$, removed by temperature calibration and the observer-rest-frame convention. With zero anisotropic stress, $\Phi =\Psi$, the observable higher-multipole result isAll fields are evaluated along the ray; the integral contains the time derivative at fixed spatial position. Retaining the printed negative angular term would leave the additional $-2q^2\int ETd\tau$, generally a gauge-dependent, nonzero term. It is therefore not repaired merely by discarding observer monopole and dipole.