Past exam of the general relativity and cosmology course of the University of Oxford / 2018 (A13484W1)

Source paper: www-thphys.physics.ox.ac.uk/people/AndreiStarinets/TUTORIALS_NOTES_GR/gr_cosmology_exam_paper_2018.pdf

The strong equivalence principle extends universality of free fall to bodies with appreciable gravitational self-energy, and states that the outcome of any local experiment, including a gravitational experiment, in a freely falling laboratory is independent of its location and velocity. In a sufficiently small freely falling laboratory the external gravitational influence can be removed to first order by choosing a local inertial frame.

This does not remove tidal forces or the Riemann curvature tensor. The laboratory must be small compared with the length scale over which the external gravitational field varies. These qualifications distinguish a local equivalence statement from the incorrect assertion that gravity can be transformed away throughout an extended curved region.

In general coordinates the basis changes along the world line. Differentiating only the components of a vector omits that basis change, and does not transform as a vector under a position-dependent coordinate transformation. The appropriate rate is the covariant derivative along a curve:For the metric-compatible torsion-free connection, the Christoffel symbols are determined by the metric tensor and its first derivatives:Thus the ship's four-acceleration isIn Cartesian coordinates of an inertial frame in flat spacetime, the connection coefficients vanish and the ordinary component derivative happens to suffice. In curved spacetime it suffices at a chosen event in a local inertial frame, but not generally along an extended trajectory.

Use metric signature $(-,+,+,+)$ and $x^0=ct$. At an event in the ship's instantaneous rest local inertial frame, $u^{\alpha }=(c,0,0,0)$. Since its squared norm is a scalar,in every coordinate system. Equivalently it follows directly from $d{s}^2=-c^{2}d{\tau }^2$ and $u^{\alpha }=d{x}^{\alpha }/d\tau$.

Metric compatibility lets us differentiate this identity covariantly:In the instantaneous rest local inertial frame, the last equation gives $a^0=0$. The spatial components are the acceleration recorded by an onboard accelerometer, ${\mathbf{a}}_{\mathrm{rest}}$, whose magnitude is $g$. ThereforeThis proper acceleration is zero in ideal free fall, even when the coordinate position has a nonzero second derivative. A coordinate acceleration and an accelerometer reading are different quantities.

Put $F(r)=1-r_s/r$ and $r_s=2GM/c^2$. For radial motion $u^{\theta }=u^{\varphi }=0$. The given Christoffel symbols implyandThe four-velocity normalization of part c isIts bracketed combination is therefore exactly $c^2$, independently of the instantaneous velocity or the thrust history. ConsequentlyThe derivative here is with respect to proper time. This expression does not say that the ship measures a gravitational acceleration $GM/r^2$ during free fall; setting $a^r=0$ gives its radial proper-time equation of motion in Schwarzschild spacetime.

For radial motion outside the Schwarzschild event horizon, the equations needed to determine the acceleration components areThey are respectively the four-velocity norm, velocity-acceleration orthogonality, and the maximum proper acceleration norm. Together with part d they give the stated thrust equation; no integration of the trajectory is required. The positive radial sign corresponds to outward thrust.

Let $r_I=\sqrt{GM/g}$. A first transition from inward to outward motion requires $u^r=0$ at a minimum radius and $d{u}^r/d\tau \ge 0$ there. The largest possible radial derivative at such an event isAt $r=r_I>r_s$, this isThus the ship cannot reverse its inward motion there. It cannot reverse at any smaller exterior radius either: $GM/r^2$ grows while $\sqrt{1-r_s/r}$ decreases. If it first reaches $r_I$ while falling inward, it cannot return home using the stipulated bounded radial thrust. If $r_I\le r_s$, it would instead have reached the black-hole horizon before that radius, which already prevents escape.

The minimum radius for radial turnaround with bounded proper acceleration lies strictly outside the proposed radius. Define $r_c>r_s$ byThe right side is the proper acceleration required to hover, decreases monotonically from infinity to zero, and exceeds $GM/r^2$. A genuine turnaround needs $r_{\mathrm{turn}}>r_c>r_I$; equality gives no positive initial outward acceleration at rest. A sufficiently early burn is also needed to remove the pre-existing inward velocity. The failure at $r_I$ follows from the turning-point condition, not from treating the moving ship as static throughout its fall.

In coordinates $(ct,r,\varphi ,z)$ the diagonal metric tensor is $\mathrm{diag}(-1,1,f^2,1)$, and only $g_{\varphi \varphi }$ depends on position. Substitution into the connection formula giveswith all other Christoffel symbols zero away from singular axes.

Use the explicit Ricci tensor formula and the Einstein equation sign printed on the source formula sheet. They giveFor example, in $R_{rr}$ the two terms proportional to $(f^{\prime }/f{)}^2$ cancel. For $R_{\varphi \varphi }$ the derivative of $-f{f}^{\prime }$ contributes $(f^{\prime }{)}^2+f{f}^{\prime \prime }$, the product terms contribute $-2(f^{\prime }{)}^2$, and the determinant term contributes $(f^{\prime }{)}^2$.

Formula-sheet qualification. Directly contracting the separately printed Riemann curvature tensor as instructed on that sheet instead gives $R_{rr}=-f^{\prime \prime }/f$, the opposite sign. For the concrete example $f(r)=\ell \sin (r/\ell )$, that contraction gives $+1/{\ell }^2$, whereas the explicit Ricci tensor formula gives $-1/{\ell }^2$. The two source formulas cannot both hold with the stated contraction. The calculation here uses the explicit Ricci tensor formula, whose pairing with the negative Einstein equation coupling yields the requested stress-energy tensor.

The desired matter tensor assumes $\Lambda =0$; otherwise the cosmological-constant contribution must also be included. With that assumption,and raising both indices givesThe static observer measures energy density $T^{00}=-c^4{f}^{\prime \prime }/(8\pi Gf)$ and equal-magnitude negative axial pressure when that energy is positive. The axial stress is a tension. Keeping the source's curvature sign and its negative Einstein-equation coupling together is essential to this result.

Assume $\varphi$ has period $2\pi$, $f>0$ in the regular bulk, and consider a unit proper length in $z$. The proper transverse area element is $dA=fdrd\varphi$, while the mass density is $T^{00}/c^2$. The axial boundary term in the mass of a cylindrical spacetime is thereforeChoose the usual deficit branch $0\le \lambda <1$, so $f_{\mathrm{out}}=(1-\lambda )r$. Continuity of the induced metric requires $f(R^{-})=(1-\lambda )R$. If there is no distributional thin shell at $R$, continuity of the extrinsic curvature also requires $f^{\prime }(R^{-})=1-\lambda$. A jump in $f^{\prime }$ would add a surface-layer stress and cannot simply be discarded.

Using the printed inner condition $f^{\prime }(0^{+})=0$, these assumptions yieldThis is the literal smooth-bulk answer. It is negative for the ordinary deficit range, so it must not be silently replaced by the usual positive cosmic-string mass formula.

For comparison, a regular axis with proper radial coordinate and angular period $2\pi$ requires $f(0)=0$ and $f^{\prime }(0)=1$, since the circumference-to-radius ratio must approach $2\pi$. Under that different regular-axis condition the same integral givesThe original PDF really prints the derivative limit zero. It describes either a nonregular axis or an inner geometry different from an ordinary smooth string core; any additional singular-axis matter is not included in the bulk integral above and would need its own stress specification. Standard regular-core matching can also require an offset in the exterior radial origin. Thus the regular-axis formula is a qualified physical interpretation, not a derivation from the printed zero. For a general exterior branch replace $1-\lambda$ by $|1-\lambda |$ in the boundary derivative.

Outside the core, $f^{\prime \prime }=0$ and the metric is locally flat: set $\tilde{\varphi }=(1-\lambda )\varphi$. Static test particles have no Newtonian-like attraction because $g_{00}$ is constant and ${\Gamma }^r{}_{00}=0$. Local trajectories are straight lines in locally unwrapped coordinates, so a very thin dark core is hard to identify through ordinary orbital accelerations or local tidal forces.

The global angular range is nevertheless $2\pi (1-\lambda )$ rather than $2\pi$. The missing angle $\delta =2\pi \lambda$ can give gravitational lensing with two nearly identical, essentially undistorted images of a background galaxy, and multiple such pairs aligned along the projected object. For a small deficit, a straight string and favourable geometry, their angular separation is of order $\delta D_{ls}/D_s$, with an additional inclination factor for a tilted string. The matching spectra and common line of image pairs distinguish this from an ordinary compact lens.

A moving cosmic string can also induce a step in the cosmic microwave background temperature across its path. That signature needs motion in addition to the static geometry supplied here. These are consequences of the exterior conical geometry; interpreting the inferred deficit as a positive mass per length requires the regular-axis/core assumptions distinguished in part b.

The cosmic distance ladder starts by calibrating a physical baseline. Radar echoes from planets or spacecraft give a round-trip light time and hence a distance $d=c\Delta t/2$, after accounting for orbital motion and propagation delays. Combining this with orbital geometry fixes the Earth-Sun orbital scale; directly timing a reflected pulse from the solar surface is not required.

That baseline supplies annual stellar parallax: for small parallax angle $p$, $d\simeq b/p$, or $d$ in parsecs equals $1/p$ when $p$ is in arcseconds and $b$ is one astronomical unit. Nearby parallax distances calibrate stellar absolute magnitudes. Main-sequence fitting extends the scale to star clusters, and the calibrated Cepheid period-luminosity relation gives luminosities for Cepheids visible in nearby galaxies.

Cepheid-host galaxies then calibrate Type Ia supernovae. After light-curve and colour standardization, their high luminosities permit distances to much more remote galaxies. For a source of calibrated luminosity $L$ and observed bolometric flux $F$,with consistent bandpass definitions and corrections. Interstellar extinction, composition, stellar-population differences and calibration uncertainty must be propagated through the steps. Redshift can subsequently provide a distance within a fitted cosmological model; using it alone cannot independently calibrate the Hubble constant.

Normalize the scale factor to $a(t)=R(t)/R(t_0)$, with $a(t_0)=1$. Along a radial null geodesic, $d\chi =cdt/a(t)$. Two successive wave crests sent by the same comoving emitter and received by the same comoving observer traverse the same comoving distance, soTo first order, $\delta t_0/a(t_0)=\delta t_e/a(t_e)$. Frequencies are inverse proper periods for the comoving observers, giving the cosmological redshiftExpansion stretches wavelengths, so a spectral line of known rest wavelength gives $z={\lambda }_{\mathrm{obs}}/{\lambda }_{\mathrm{rest}}-1$.

At small $z$, the light travel time is short compared with the expansion time. Expanding $R(t_e)$ gives $z\simeq H_0(t_0-t_e)$ and $d\simeq c(t_0-t_e)$, henceCombine independently calibrated luminosity distances with galaxy spectral redshifts and fit the slope. Nearby peculiar velocities must be allowed for, while high-redshift data require the full distance-redshift relation rather than the linear approximation.

Use standardized Type Ia supernovae as distance indicators and measure their redshifts. The curvature of the luminosity-distance relation constrains the deceleration parameter:Fainter supernovae at a given redshift imply larger distances, and comparison with the predicted curve tests accelerated expansion. In practice one fits the full expansion history over the observed redshift range and controls luminosity evolution, extinction and calibration, rather than relying only on a short low-$z$ series.

The cosmological fate from local expansion data is conditional on a matter model. For the spatially flat model of part b, if the only relevant components are pressureless matter and a constant $\Lambda$, put $A_0=\ddot{R}(t_0)/R(t_0)$. The two Friedmann equations giveHere $\Lambda$ follows the sign of the supplied Friedmann equations, where positive $\Lambda$ contributes positive acceleration. These fix the parameters governing subsequent evolution. Positive $\Lambda$ yields eternal late-time accelerated expansion; zero $\Lambda$ gives the flat matter solution; negative $\Lambda$ leads to eventual recollapse in this dust model. If instead one assumes dust and zero $\Lambda$ while allowing curvature, $q_0={\Omega }_{m0}/2$ determines whether the matter universe is closed and recollapsing or expands indefinitely.

Without a fixed equation of state and its future evolution, two present derivatives do not uniquely fix the future. For a concrete counterexample in units with $t_0=0$, the positive scale factors $a_{+}(t)=1+t+t^2/2$ and $a_{-}(t)=1+t+t^2/2-t^3$ have the same $a,\dot{a},\ddot{a}$ now. The former expands forever; the latter turns around and later reaches zero. Different time-dependent matter stresses can support their different flat expansion histories. Thus the printed sufficiency statement needs the cosmological-model assumption.

In an Einstein-de Sitter universe, take $a(t)=(t/t_0{)}^{2/3}$ and $t_0=2/(3H_0)$. The galaxy seen today at $z=1$ emitted the light now reaching us when $a(t_e)=1/2$, so $t_e/t_0=2^{-3/2}$. Its fixed comoving radial distance isPut $b=1-1/\sqrt{2}$. A new message sent now reaches the galaxy at $t_1$, whereAssume an immediate reply and negligible local handling delays. The return consumes the same additional conformal time, giving the Einstein-de Sitter round-trip message timeFor the numerical value $H_0=72\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ supplied later in the paper, $H_0^{-1}=13.58\mathrm{Gyr}$ and the waiting time is about $27.1\mathrm{Gyr}$. The age $t_0=9.05\mathrm{Gyr}$ here belongs to the stipulated matter-only model, not a claim about the actual measured cosmic age.

After $n$ one-way legs, $t_n=t_0(1+nb{)}^3$. Since ${\int }_{t_0}^{\infty }dt/a(t)=\infty$, every finite number of messages is possible and there is no finite fundamental upper count. The exchange takes arbitrarily long, with successively longer intervals; infinitely many messages are not completed in finite cosmic time.

Spatial flatness constrains the total density, not matter alone. A flat model can havewith a substantial dark energy component. A positive cosmological constant has pressure $p_{\Lambda }=-{\rho }_{\Lambda }{c}^2$ and does not dilute like matter; it can produce the observed accelerated expansion.

If positive $\Lambda$ persists, $a(t)$ becomes exponential and the future conformal time intervalis finite. For two comoving civilisations separated by $\chi$, each one-way signal consumes $\chi /c$. Thus a message completing $n$ legs at finite time requiresOnly finitely many legs, and hence finitely many full exchanges, can then occur. A reply to even the first new message requires $2\chi /c<\Delta {\eta }_{\mathrm{future}}$, the cosmological round-trip communication horizon condition. Depending on the density parameters, the initial outward message might arrive while its reply can never reach us, as in the right-hand panel of part d.

The source gives no matter and dark-energy fractions, so it does not specify an exact count for the $z=1$ pair in the real Universe. Flatness and lack of matter dominance alone do not prove a finite horizon: a radiation-dominated flat universe, for example, has divergent future conformal time if that power-law expansion persists. The finite-message conclusion uses persistent accelerated expansion such as positive constant $\Lambda$.

For a photon gas in thermal and chemical equilibrium, photon number is not a conserved thermodynamic charge. Number-changing processes can create and absorb photons without a corresponding conserved-number constraint, so the equilibrium photon chemical potential is zero.

At $T=5\times 10^9\mathrm{K}$, $k_{\mathrm{B}}T\simeq 0.431\mathrm{MeV}$. The other light species are electrons, Positrons, three neutrino species and their antiparticles. Neutrinos are ultrarelativistic; electrons and Positrons are only moderately relativistic at this particular temperature because $m_e{c}^2=0.511\mathrm{MeV}$. The later simple counting formulas use the intended ultrarelativistic approximation for them. Baryons, muons and heavier particles are not an abundant relativistic bath at this temperature.

Pair equilibrium implies ${\mu }_{e^{-}}+{\mu }_{e^{+}}=2{\mu }_{\gamma }=0$. Approximate charge symmetry, with the net electron excess tied to the much smaller baryon asymmetry, then makes both electron and Positron chemical potentials small compared with $k_{\mathrm{B}}T$. For neutrinos one additionally assumes negligible lepton asymmetry. Their near-zero chemical potentials are not implied merely by their statistics. After neutrino decoupling, a distribution initially having negligible chemical potential retains a redshifted Fermi-Dirac distribution, with its own temperature.

Let $V$ be proper volume, let $g_i$ count the populated internal states, and take ${\mu }_i=0$ and $ϵ=pc$. Spherical integration in momentum space givesUsing the fermion-to-boson thermal integral ratio, the minus-denominator boson integral is $2\zeta (3)$ and the plus-denominator fermion integral is $3\zeta (3)/2$. HenceThe energy integral has one extra factor $pc=k_{\mathrm{B}}Tx$:The supplied distribution identity gives the fermion integral as $(1-2^{-3}){\pi }^4/15=7{\pi }^4/120$. ThereforeFor an isotropic ultrarelativistic gas, momentum flux gives $P_i=\frac{1}{3}\int pv{f}_i{d}^{3}p$. Since $v=c$ and $pc=ϵ$,Photons have $g_{\gamma }=2$, electrons and Positrons each have $g=2$, and each neutrino or antineutrino species has one populated helicity in the standard thermal bath.

For zero chemical potential, each species contributes $S_i=(U_i+P_{i}V)/T=4U_i/(3T)$. ThusEntropy density scales as $T^3$ at fixed relativistic particle content; total entropy also includes the changing proper volume. An isolated adiabatically expanding comoving gas has conserved total entropy and $aT$ constant, rather than its total entropy falling as $T^3$.

At equal temperature, single electrons and photons each have two spin/polarization states. In the ultrarelativistic approximation,Positrons are a separate species with the same contribution, so $(S_{e^{-}}+S_{e^{+}})/S_{\gamma }=7/4$. This distinction is needed for the heating count in part d. At the exact supplied temperature the electron rest mass gives corrections to the ideal $7/8$ ratio; it is the limiting relativistic value requested by the problem.

There is also a sign convention issue in the introductory entropy formula. With the chemical-potential convention used in the supplied distribution, the Euler relation is $U=TS-PV+{\sum }_i{\mu }_i{N}_i$, so $S=(U+PV-{\sum }_i{\mu }_i{N}_i)/T$. Its chemical-potential term has the opposite sign to the printed one. For a concrete counterexample, a dilute monatomic ideal gas with one internal state and $n{\lambda }_T^3=e^{-1}$, where ${\lambda }_T$ is the thermal de Broglie wavelength, has $\mu =-k_{\mathrm{B}}T$, $U=3N{k}_{\mathrm{B}}T/2$ and $PV=N{k}_{\mathrm{B}}T$. Its entropy is $7N{k}_{\mathrm{B}}/2$; the printed plus sign would instead give $3N{k}_{\mathrm{B}}/2$. All chemical potentials have been neglected here, so this source error does not change any of the above results.

As the plasma cools through electron-positron annihilation, the pairs transfer their entropy to the coupled photons. Neutrinos have largely decoupled and do not share that heating in the instantaneous-decoupling approximation. Before annihilation the electromagnetic entropy count iswhile after it $g_{\ast s,\mathrm{after}}=2$. Conservation of electromagnetic comoving entropy, $g_{\ast s}{T}_{\gamma }^3{a}^3=\text{constant}$, gives the relative photon heating factorEquivalently $T_{\gamma }/T_{\nu }=(11/4{)}^{1/3}$ afterward. The exact $11/4$ count compares an asymptotic relativistic pair bath with the pair-free bath; at the supplied temperature some finite-electron-mass entropy loss has already begun.

The word “rises” requires care. The photons become hotter than an otherwise freely redshifting bath at the same scale factor; their absolute temperature need not rise during the expansion. In the smooth equilibrium process,since the pair contribution grows with temperature. This monotonic photon cooling during pair annihilation is slower than $a^{-1}$ while $a{T}_{\gamma }$ increases. The factor $1.401$ compares the photon temperature with the freely redshifting neutrino bath, not two consecutive absolute temperatures along the expanding history.

Use the observed present photon temperature $T_{\gamma 0}\simeq 2.725\mathrm{K}$. This is an additional observational input, consistent with ^[ref]. Standard neutrino decoupling and electron-positron entropy transfer giveFor nonrelativistic relic neutrinos this is the temperature parameter of the redshifted momentum distribution, not the temperature of a newly thermalized gas of massive particles.

The photon number density from part b isOne neutrino mass eigenstate and its antineutrino have two populated helicity states. ThusThe total over three species is about $336\mathrm{c}{\mathrm{m}}^{-3}$, including antineutrinos. Comoving number conservation preserves these abundances after the particles become nonrelativistic.

For the supplied $H_0=72\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}=2.33\times 10^{-18}{\mathrm{s}}^{-1}$,Assuming nonnegative other energy components, relic neutrino rest-mass density cannot exceed the near-critical total density. With the common per-species abundance $n_{\ast }=n_{{\nu }_i}+n_{{\bar{\nu }}_i}$,Using the total $336\mathrm{c}{\mathrm{m}}^{-3}$ would instead bound the mean mass, not the sum. This critical-density bound on the sum of relic neutrino masses is deliberately conservative: baryons, dark matter and dark energy leave less density available. It assumes the standard thermal abundance, negligible chemical potentials, no later dilution or extra populated states, and the idealized decoupling count. Small non-instantaneous-decoupling corrections do not matter for this estimate.