Past exam of the general relativity and cosmology course of the University of Oxford / 2016 (A13484W1, B5)

Source paper: www.whatdotheyknow.com/request/physics_past_exam_papers_2015_16/response/973997/attach/3/Cowell%20Attachment.zip?cookie_passthrough=1. The dated cover identifies the Part B sitting of 18 June 2016. All four options are solved below.

Constants and radial motion. Write $m=GM/c^2$ and $f(r)=1-2m/r$. A Schwarzschild black hole has spherical symmetry, so its conserved angular momentum fixes an orbital plane through the centre. A rotation of the angular coordinates puts any nonradial orbit in the equatorial plane. A radial orbit can also be placed in such a plane. This is a choice of coordinates, not a restriction on the possible orbital inclinations.

The quadratic geodesic Lagrangian, with a dot denoting differentiation with respect to the satellite's proper time, isThe last equality is four-velocity normalization, imposed after deriving the Euler-Lagrange equations. The cyclic coordinates $t,\varphi$ give independent conserved specific Killing energy and angular momentum:Here $e$ is the conserved energy per unit rest energy and $\ell$ has dimensions of angular momentum per unit mass. Substituting them in the normalization condition givesThe two signs of $\dot{r}$ distinguish outgoing and ingoing branches; the expression must be nonnegative on the allowed radial interval.

For a nonradial orbit, put $u=1/r$. Since $\dot{r}=-\ell du/d\varphi$, the radial equation becomesDifferentiating and extending continuously through isolated radial turning points givesA circular solution can equivalently be obtained directly from the radial geodesic equation, avoiding division by $du/d\varphi =0$ everywhere on that special orbit.

Circular launch and the station's clock. For $u=1/R$ the orbit equation givesThe static astronaut measures circumferential proper length $Rd\varphi$ and station proper time $d{\tau }_s=\sqrt{f(R)}dt$. Thus the Schwarzschild circular-orbit local speed and return time areThis is a local speed measured with the astronaut's clock, not the satellite's proper velocity $\gamma v$. After one azimuthal revolution the circular satellite meets the fixed station again. For comparison, the satellite's own elapsed proper time would instead be $2\pi \sqrt{R^3/(GM)}\sqrt{1-3m/R}$.

A static station requires $R>2m$; a timelike circular satellite requires $R>3m$, since the local circular speed must be below $c$. The circular orbit is stable only for $R>6m$, marginal at $6m$, and unstable for $3m<R<6m$. Thus a request for a robust near-circular trajectory implicitly needs more than merely placing the station outside the horizon.

Slightly changing the launch speed. In the stable, bound regime, a small tangential increase makes the launch point a pericentre; a small decrease makes it an apocentre, provided the perturbed orbit remains outside its plunge separatrix. In either case the orbit is a precessing rosette, not an exactly closed Newtonian ellipse. For an explicit first-order description write $A=A_c+\delta A$ and $u=1/R+\delta u$. ThenThe tangential launch gives $\delta u(0)=\delta u^{\prime }(0)=0$. At the station, $\ell =R\gamma v$, so $\delta \ell /{\ell }_c={\gamma }_c^2\Delta v/v_{\mathrm{circ}}$ and $\delta A=-2A_c{\gamma }_c^2\Delta v/v_{\mathrm{circ}}$. ThereforeThis perturbative expression also requires the resulting radial excursion to be small compared with $R$; close to the innermost stable orbit, $|\Delta v|\ll v_{\mathrm{circ}}$ alone does not ensure that.

The Schwarzschild launch-orbit recurrence needs care when deciding whether a collision is inevitable. The satellite returns to radius $R$ once per radial cycle, after an azimuthal advance approximately ${\Phi }_r=2\pi /{\kappa }_{\varphi }$. It returns to the station's exact point only if an integer number of radial cycles also gives an integer number of azimuthal revolutions. Rational ${\Phi }_r/(2\pi )$ gives an exact return; an irrational ratio gives arbitrarily close approaches but no exact post-launch return for ideal point objects. At $R=12m$, for example, the linearized ratio is $\sqrt{2}$, illustrating why precession alone does not prove an exact collision. For generic bound orbits the dense recurrence eventually threatens any station and satellite of nonzero size; a resonant closed rosette returns exactly. A small speed adjustment is therefore not a robust permanent collision-avoidance manoeuvre for a bound satellite.

There is no unconditional collision conclusion without specifying the regime. An unstable circular orbit cannot support the small bounded oscillations used above. In particular, near $R=3.5m$ the circular specific energy is already $e_c>1$; a small positive speed increment can send the satellite outward on an unbound escape orbit, with no subsequent return to $R$. Such an escape is a different way to avoid a future encounter. Radiation reaction and manoeuvres have been excluded throughout the ideal test-particle calculation.

Curvature and conservation. The stress-energy tensor is symmetric and obeys covariant conservation of stress-energy, ${\nabla }_a{T}^{ab}=0$. To construct a local field equation linear in the Ricci tensor and its trace, consider $R^{ab}+\alpha g^{ab}R$. The contracted Bianchi identity and metric compatibility giveThe divergence must vanish identically, rather than constrain every matter source to have constant scalar curvature. Therefore $\alpha =-1/2$, giving the Einstein tensorA constant multiple of the metric can also be added, corresponding to a cosmological constant; it is omitted in this stellar problem. Proportionality to matter by itself does not exclude arbitrary higher-curvature theories. The stated linear combination assumes the usual second-order metric field equation, linear in curvature, with its coupling fixed by the Newtonian limit.

The source's displayed curvature components use the convention $G_{ab}=-8\pi G{T}_{ab}/c^4$, opposite to another common curvature convention. Retain both that sign and those components together: replacing only the sign would give an incorrect negative stellar mass.

Fluid frame and mass function. Because the metric and the Einstein tensor are diagonal, the mixed time-space components of the stress-energy tensor vanish. For the perfect fluid in general relativity, $T^{0i}=(\rho +p/c^2)u^0{u}^i$. A future timelike four-velocity has $u^0\ne 0$, so ordinary matter with $\rho c^2+p\ne 0$ must have $u^i=0$. Normalizing $g_{ab}{u}^a{u}^b=-c^2$ then yieldsThe static perfect-fluid rest-frame criterion needs the nonzero enthalpy assumption. A vacuum-like component with $p=-\rho c^2$ has $T^{ab}=p{g}^{ab}$ for every normalized timelike $u^a$, so diagonality cannot pick its rest frame. That is a concrete exception to the unqualified inference about velocity; it does not describe the ordinary stellar fluid considered next.

Lowering the indices of the static-fluid tensor gives $T_{00}=\rho c^2{e}^{2\Phi }$ and $T_{11}=p{e}^{2\Lambda }$. The time component of the field equation therefore impliesRegularity at the centre fixes the integration constant to zero, givingA central point mass would introduce an additional constant, but is excluded by the regular-star boundary condition. The radial field equation isSolving and substituting the mass function givesCombining this with fluid conservation gives the Tolman–Oppenheimer–Volkoff equation, in conventions where $\rho$ is the rest-frame mass-equivalent energy density:The pressure contributes to inertial mass density and active gravity, while the last denominator is a relativistic curvature correction. The weak-field, low-pressure limit recovers $p^{\prime }\simeq -G\rho M/r^2$.

Constant-density integration. Set $w=p/(\rho c^2)$ and $\alpha =8\pi G\rho /(3c^2)$. Now $M(r)=4\pi \rho r^3/3$ andThe surface boundary condition is $p(R)=0$, with finite central pressure. Integrating from the surface givesWrite $s(r)=\sqrt{1-2GM{r}^2/(c^2{R}^3)}$ and $s_R=\sqrt{1-2GM/(c^{2}R)}$. Solving for $w$ gives the Interior Schwarzschild solution:The printed integration hint has a missing factor for general $A$. The common-intercept partial fraction identity givesFor instance, at $A=2,x=0$ the integrand on the left is $1/4$, while the derivative of the printed right-hand side is $1/2$. The normalized $w$ calculation above uses $A=1$, where that missing factor happens not to matter.

Positive finite central pressure requires $3s_R>1$. ConsequentlyEquality is a singular limiting solution with divergent pressure. The constant-density model is incompressible and is not a realistic causal neutron-star equation of state. Nevertheless, Buchdahl's theorem extends this compactness bound to regular static spherical isotropic fluids with nonincreasing density, under its usual assumptions.

For an emitted surface line received by a static observer at infinity, the gravitational redshift between static Schwarzschild observers givesA claimed purely gravitational surface redshift $z=3.1$ would require compactness $1-1/(4.1{)}^2\simeq 0.9405$, exceeding $8/9$. It is inconsistent with the regular isotropic static neutron-star interpretation, even though the surface would formally remain outside its Schwarzschild radius. Misidentification, additional kinematic shifts or departures from the model would need examination before treating that line as a neutron-star surface measurement.

Spatial geometry. An embedding of the three-sphere isTaking differentials and collecting terms gives the induced metric tensorPut $\bar{r}=A\sin \chi$. Then $d\bar{r}=A\cos \chi d\chi$, soThe areal radius $\bar{r}$ is not a global radial coordinate on the entire three-sphere: its maximum occurs at the equator and two hemispheres share each smaller value. Use separate charts there; the coordinate singularity at $\bar{r}=A$ is not a physical singularity.

For Euclidean space and hyperbolic space of curvature $-1/A^2$, respectively,For the latter, $\bar{r}=A\sinh \chi$ gives $d{\ell }_{-}^2=A^2[d{\chi }^2+{\sinh }^2\chi d{\Omega }^2]$. In all three cases the induced two-metric at fixed $\bar{r}=\bar{R}$ is ${\bar{R}}^{2}d{\Omega }^2$. Its surface area is thereforeThis equality compares the same areal radius, not the same proper radial distance. For a proper radial distance $\ell$, the corresponding areas are $4\pi A^2{\sin }^2(\ell /A)$, $4\pi {\ell }^2$, and $4\pi A^2{\sinh }^2(\ell /A)$.

Homogeneity, isotropy and redshift. The FRW metric has constant-time spatial slices of constant curvature $k/a^2(t)$, with $k=+1,0,-1$ selecting the three geometries above. Their isometries act transitively on points and isotropically about every point, implementing the cosmological principle. Multiplication by one common time-dependent scale factor preserves these properties. With the source's $k=\pm 1$ normalization, $r$ is dimensionless and $a$ has units of length; using a dimensionless scale factor instead rescales the spatial curvature parameter. The origin of the spherical coordinates is arbitrary; it does not identify a physical centre. Cosmic time is the proper time of the comoving observers, and the absence of time-space cross terms makes their worldlines orthogonal to the homogeneous slices.

For a radial null geodesic between comoving endpoints,The next wave crest travels between the same endpoints. Subtracting the two path integrals, for wave periods short compared with the expansion time, gives $\delta t_0/a_0=\delta t_E/a_E$. Thus the cosmological redshift isBoth observers' clocks are cosmic proper-time clocks. Peculiar velocities and local gravitational-potential shifts have been left out of this homogeneous calculation.

Angular size and brightness. For a galaxy small compared with its distance, the angular diameter distance is $D_A=a_{E}R$ and its angular diameter isHere the emitted proper diameter is measured transverse to the line of sight; radial proper distance is a different integral when $k\ne 0$. Ignoring absorption, lensing caustics and intrinsic evolution, photon-number conservation spreads an isotropic-equivalent luminosity across the sphere $4\pi a_0^2{R}^2$. Each photon loses energy by $1+z$, and its arrival interval is stretched by the same factor. The received bolometric flux isThe disc's proper projected area is ${\mathcal{A}}_d=\pi d^2/4$, and its observed solid angle is ${\Omega }_d\simeq {\mathcal{A}}_d/D_A^2$. Therefore its surface brightness isThis is cosmological surface-brightness dimming. The normalization uses $L$ as isotropic-equivalent luminosity along the viewing direction; another disc emission law changes that normalization but not the redshift power. This is a bolometric result. In a specified observed frequency band the emitted spectrum is sampled at a shifted frequency and requires a spectral correction.

The distinction in telescope aperture and resolved surface brightness matters here. A larger telescope collects more photons and improves angular resolution, but cannot restore the incoming radiance lost by cosmological dimming. For aperture diameter $D$ and a diffraction resolution element of solid angle ${\Omega }_{\mathrm{res}}\sim (\lambda /D{)}^2$, the collected power from uniform resolved brightness scales asindependent of $D$. This is consistent with conservation of etendue. It explains why increasing aperture while keeping every gain in resolution does not increase counts per native resolution element from a smooth extended source. It does not make larger telescopes useless: at a fixed angular bin the counts grow as $D^2$, so a photon-noise-limited or uniform-sky-noise-limited signal-to-noise ratio grows roughly as $D\sqrt{t_{\exp }}$. They also resolve smaller structures and improve point-source sensitivity. Aperture, exposure and angular binning help measure a dim galaxy, without changing its intrinsic $(1+z{)}^{-4}$ brightness penalty.

Two stages of the thermal history. Big Bang nucleosynthesis is nuclear assembly in an almost fully ionized, radiation-dominated plasma. Around $t\sim 1\mathrm{s}$, at temperatures of order $1\mathrm{MeV}/k_B$, the weak neutron–proton conversion rate falls below the expansion rate. The neutron-to-proton abundance freezes out at order $1/5$ and then decreases by neutron decay. The deuterium bottleneck delays effective nuclear assembly until $k_{B}T\sim 0.07$–$0.1\mathrm{MeV}$, at ages of a few minutes: the high photon-to-baryon ratio permits photodissociation despite the deuteron binding energy exceeding the mean photon energy. Once deuterium survives, reactions rapidly put most remaining neutrons into helium-4. A neutron-to-proton ratio near $1/7$ gives a helium mass fraction $Y_p\simeq 2(n/p)/(1+n/p)\simeq 1/4$, with hydrogen dominant by number and small residual deuterium, helium-3 and lithium-7 abundances. Expansion and low density prevent substantial heavy-element production. Nuclear binding is involved; neutral atoms do not yet survive.

Photon decoupling occurs much later, near a few thousand kelvin and a few hundred thousand years, after matter has overtaken radiation in the expansion dynamics. Electrons and nuclei form neutral atoms during cosmological recombination. The resulting decline of free-electron density reduces Thomson opacity, and photons begin to free stream over cosmological distances. Those photons form the observed cosmic microwave background. Atomic binding and scattering opacity, rather than new nuclear reactions, control this stage. Helium forms neutral atoms earlier than hydrogen. Recombination, photon last scattering, baryon drag release and the later thermal decoupling of the gas are related but not identical events.

The temperature falls through expansion: after the relevant entropy releases have ended, $T_{\gamma }\propto a^{-1}$ and number densities scale as $a^{-3}$. Big Bang nucleosynthesis lies in the radiation-dominated universe; photon decoupling lies after matter–radiation equality, when a matter-dominated approximation is useful but radiation is not exactly negligible.

Photon density and baryon abundance. The cosmic microwave background has a black-body radiation spectrum obeying Planck's law:where $u_{\nu }$ is spectral energy density, not a photon-number density. The measured temperature is about $2.72548\mathrm{K}$.^[ref] The mean photon energy in a black body isUsing the supplied rounded energy density givesThe exact black-body number integral at $2.725\mathrm{K}$ gives $4.1\times 10^8{\mathrm{m}}^{-3}$; the difference reflects rounding of the stated energy density. Either estimate is adequate.

A numerical photon-to-baryon ratio additionally needs a baryon density, which is not printed in this question. Make that observational input explicit rather than infer baryons from radiation alone. For illustration take the standard rounded physical density ${\Omega }_b{h}_H^2\simeq 0.022$, with $H_0=100h_H\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$. This is consistent with the cosmological parameter fits available around the paper's date.^[ref] It givesCounting baryons, including those bound in helium, rather than counting nuclei then givesBoth populations dilute as $a^{-3}$ after electron–positron annihilation, so this ratio also applies at hydrogen recombination in the standard adiabatic history.

Estimating decoupling. The large photon excess explains why setting $k_{B}T$ equal to the Hydrogen binding energy ${\chi }_H=13.6\mathrm{eV}$ is wrong. Even a small ionizing tail can oppose neutral-atom formation. The crude exponential balance $e^{-{\chi }_H/(k_{B}T)}\sim \eta$ gives ${\chi }_H/(k_{B}T)\sim \ln (1/\eta )\sim 21$, or $T\sim 7\times 10^3\mathrm{K}$. This only establishes the scale and the large suppression below ${\chi }_H/k_B$; it neglects phase-space factors and the opacity criterion.

A more controlled elementary Saha-Thomson decoupling estimate uses the hydrogen-only Saha equation. With $X_e=n_e/n_H$ and ground-state statistical weights combined in the usual way,The electron translational phase-space factor is significant inside the logarithm. This equilibrium relation places the sharp neutralization interval around $T\sim (3\text{–}4)\times 10^3\mathrm{K}$. Combine it with the approximate Thomson decoupling rate criterionFor $Y_p\simeq 0.25$, $n_{b0}\simeq 0.25{\mathrm{m}}^{-3}$, $H_0\simeq 68\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$, and ${\Omega }_m\simeq 0.31$, solving these two approximate relations gives $T\simeq 3.1\times 10^3\mathrm{K}$, with equilibrium $X_e$ of order $10^{-2}$. Thus the appropriate estimate isThis is not a precision recombination calculation. Hydrogen-only equilibrium omits the kinetic bottleneck in escaping Lyman photons, two-photon processes, the detailed helium history and freeze-out; $\Gamma /H\sim 1$ is not identical to the maximum of the last-scattering visibility function. These approximations must be improved to predict an accurate free-electron fraction or a percent-level last-scattering redshift. The high photon-to-baryon ratio alone does not fix that detailed history.

The electrons are highly nonrelativistic: $k_{B}T/(m_e{c}^2)\sim 5\times 10^{-7}$, and their thermal root-mean-square speed is only about $10^{-3}c$. Photon coupling does not make their kinetic energies relativistic.

Age–temperature relation. During a nearly flat pressureless matter-dominated phase, with radiation, curvature and vacuum energy neglected in the expansion equation,Here the Big Bang time is chosen as $t=0$. Conserved photon entropy gives $T\propto a^{-1}$, soTo normalize this relation, use the matter-era expansion rate expressed in present-day measured parameters, rather than assume the late universe is still matter dominated:The same illustrative parameters give $t_{\mathrm{dec}}\sim 4.5\text{–}5\times 10^5\mathrm{yr}$ for $T_{\mathrm{dec}}\sim 3.0\text{–}3.1\times 10^3\mathrm{K}$. Radiation was still a meaningful correction; retaining it reduces the age, and a more complete recombination calculation gives the familiar scale of roughly $4\times 10^5\mathrm{yr}$.^[ref] A numerical age cannot follow from a proportionality $t\propto T^{-3/2}$ alone: an expansion or density normalization is essential. The small amount of dark energy at decoupling can be neglected even though it affects the present age.