Past exam of the general relativity course of the University of Oxford / 2022 / 4

For the hypothetical Hubble flow, Hubble's law gives $v=H_{0}d$, where $d=4\times 10^5\mathrm{km}$ and $H_0$ is the Hubble constant. The accompanying constants sheet is not present in the supplied PDF, so retain its dependence explicitly. Taking $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ as an illustrative input, and using one million parsecs equal to $3.08568\times 10^{19}\mathrm{km}$,A year of $365.25$ days contains $3.15576\times 10^7\mathrm{s}$. ThereforeEquivalently, $H_0=100h\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ gives $v=4.09h\mathrm{cm}\mathrm{y}{\mathrm{r}}^{-1}$. This is a counterfactual recession rate for independently comoving bodies, rather than a prediction for the actual Earth-Moon orbit.

The Earth-Moon pair is a gravitationally bound system, so its separation does not obey $\dot{r}=H_{0}r$. The Hubble flow describes the motion of the approximately homogeneous cosmological matter distribution; it is not a compulsory extra velocity assigned to the members of every local orbit. Thus the numerical proximity of the counterfactual rate to $3.8\mathrm{cm}\mathrm{y}{\mathrm{r}}^{-1}$ is not evidence for that explanation.

The observed recession arises principally from tidal dissipation. The Earth rotates faster than the Moon orbits, so the displaced tidal response transfers angular momentum from the Earth's spin into the lunar orbit. For a nearly circular orbit, increasing orbital angular momentum increases its size. A model of this effect must include the appropriate tidal response, rather than add a universal Hubble flow recession.

A residual local cosmological tidal acceleration can exist, but its background scale is $|(\ddot{a}/a)r|\sim H_0^{2}r$, not an imposed first-order secular drift $H_{0}r$. At $r=4\times 10^8\mathrm{m}$,using $G(M_E+M_M)\simeq 4\times 10^{14}{\mathrm{m}}^3{\mathrm{s}}^{-2}$. The cosmological scale is about $10^{-24}$ of the mutual gravitational acceleration. Its detailed coefficient depends on how the local system is embedded in the cosmological background. Cosmological expansion can be neglected for the lunar recession calculation at the stated precision; this conclusion follows from binding and the tiny residual local cosmological tidal acceleration, rather than from comparing two recession speeds.

Let $t_0$ be the initial epoch, let $r(t)$ be the physical radius of the outer matter shell, and put $r(t_0)=r_0$. The background Einstein-de Sitter universe has critical density ${\rho }_b=3H_i^2/(8\pi G)$ at that epoch. Its mass inside $r_0$ is $M_b=M-\delta M$, where $M$ is the actual conserved mass. Hence the background Friedmann equation impliesFor spherical collapse of pressureless matter, with no shell crossing, the shell theorem gives $\ddot{r}=-GM/r^2$. Its conservation of energy first integral isConsequentlyFor $\delta M>0$, the initially expanding branch reaches cosmological turnaround at $r_{\mathrm{ta}}=A/B=M{r}_0/\delta M$. Since $dt=dr/\dot{r}$ on that branch, the elapsed cosmic time isThis is spherical turnaround from an unperturbed Hubble velocity. The integral measures the comoving matter observers' time interval in the pressureless model. The integration variable is the physical radius, not the radius divided by the background scale factor. If an absolute cosmic time is wanted, add $t_0$ to the displayed interval.

Set $u=Br/A$, retaining $r_0$ in $B$. Replacing only the lower integration limit by zero givesSubstitution of the constants yieldsTo quantify the approximation, write $ϵ=\delta M/M\ll 1$. The true lower limit is $u_0=B{r}_0/A=ϵ$, and an elementary integration by substitution, $u={\sin }^2\chi$, givesSince $\arcsin \sqrt{ϵ}-\sqrt{ϵ(1-ϵ)}=\frac{2}{3}{ϵ}^{3/2}+O({ϵ}^{5/2})$, the approximate time is too large by a relative fraction $\frac{4}{3\pi }{ϵ}^{3/2}+O({ϵ}^{5/2})$. This explains why changing the lower limit introduces little error for a small excess mass. The cosmological turnaround time diverges as $\delta M\to 0^{+}$, consistently with the absence of finite cosmological turnaround in an exactly critical Einstein-de Sitter universe.

There are two first-order effects: an intrinsic temperature change from adiabatic compression, and the gravitational redshift of the emerging photons. Evaluate the excess Newtonian gravitational potential at the boundary of the spherical fluctuation, with the exterior reference potential subtracted:In the weak-field approximation, gravitational time dilation gives $d{\tau }_{\mathrm{local}}/dt\simeq 1+\delta \Phi /c^2$. A local emitter's successive wave crests therefore arrive farther apart at a reference observer outside the potential well. Their frequency changes by $\delta \nu /\nu =\delta \Phi /c^2<0$. Since a redshifted black-body radiation spectrum has its temperature scaled by the same factor as its frequency, the gravitational contribution is

For the intrinsic effect, compare the local scale factor $R+\Delta R$ with the unperturbed value $R$ at the reference epoch. An equilibrium photon gas has energy density $u=a_{\mathrm{rad}}{T}^4$ and pressure $p=u/3$. For adiabatic compression, the first law of thermodynamics applied to a comoving volume $V\propto R^3$ givesThus $TR$ is constant along this adiabatic process, and the initial intrinsic contribution is $\Delta T/T=-\Delta R/R$. The smaller local expansion associated with the fluctuation leaves its radiation hotter; this is the compression or local expansion-history contribution, distinct from the clock-rate effect on the escaping radiation.

Multiplying the two first-order factors, and cancelling the common subsequent background redshift, givesFor the overdense region produced by adiabatic compression, $\Delta R<0$, so the final term is positive. It competes with the negative gravitational redshift term. Their sum is the intrinsic-plus-endpoint part of the Sachs-Wolfe combination; computing a full observed Cosmic microwave background anisotropy can also require Doppler CMB anisotropy and the Integrated Sachs-Wolfe effect.^[ref]