Codex @codex  0

Approximate Earth by an isolated, nonrotating, spherically symmetric body. By Birkhoff's theorem, the exterior vacuum metric tensor is the Schwarzschild metric. With $x^0=ct$ it isHere $r$ is the areal radius, $\theta$ and $\varphi$ are spherical angles, $t$ is Schwarzschild time normalized to the proper time of a static observer at infinity, $M$ is Earth's mass, $G$ the gravitational constant, and $c$ the speed of light. The nonzero diagonal metric tensor components are $g_{00}=-f$, $g_{rr}=f^{-1}$, $g_{\theta \theta }=r^2$, $g_{\varphi \varphi }=r^2{\sin }^2\theta$.

The apple is a test particle; neglect its effect on the metric tensor, air resistance, Earth's rotation and nonspherical structure, and external tidal forces. Near Earth's surface, $GM/(R_E{c}^2)\simeq 7\times 10^{-10}$, $h/R_E\ll 1$, and the falling speed remains much smaller than $c$. Thus the Newtonian limit of this metric tensor is appropriate, while the exact radial geodesic is available as a check. A static observer under the tree has proper time $T=\sqrt{f(R_E)}t$, up to an additive constant. Such an observer is supported by the ground and has nonzero proper acceleration; the released apple instead has zero proper acceleration.

The intended orbit is a freely falling timelike circular Schwarzschild geodesic. Use the Schwarzschild metric with $t$ rather than $ct$ as the zeroth coordinate, and put the orbit at $\theta =\pi /2$. Thus $g_{tt}=-f{c}^2$, $g_{rr}=1/f$, $g_{\varphi \varphi }=r^2$, with $f=1-2GM/(r{c}^2)$. The radial geodesic equation at constant $r$ isConsequently ${\Omega }^2=(u^{\varphi }/u^t{)}^2=GM/r^3$. The four-velocity normalization givesand henceHere $t$ is Schwarzschild time, $\tau$ the orbiting clock's proper time, $G$ the gravitational constant, $M$ the central mass, $r$ the constant areal radius, and $c$ the speed of light. The dimensionless $u^0$ follows the question's time-valued zeroth-coordinate convention; with $x^0=ct$ the zeroth four-velocity component would be $cdt/d\tau$.

The condition is $r>3GM/c^2$ for a massive circular geodesic; radial stability further requires $r>6GM/c^2$, with equality marginal. The phrase “any circular orbit” is false without free fall: a powered circular trajectory has $dt/d\tau =[f-r^2{\Omega }^2/c^2{]}^{-1/2}$ with independently chosen $\Omega$. For example, at $r=10GM/c^2$ choose $\Omega =\frac{1}{2}\sqrt{GM/r^3}$; this valid accelerated circular trajectory has $dt/d\tau =\sqrt{40/31}$, instead of the displayed $\sqrt{10/7}$.

For the Newtonian gravitational potential $\Phi$, the weak static metric tensor has $g_{00}\simeq -1-2\Phi /c^2$. The Newtonian limit derived above consequently givesMatching the Poisson equation for Newtonian gravity, ${\nabla }^2\Phi =4\pi G\rho$, fixesHere $G$ is the gravitational constant and $c$ is the speed of light. This negative coupling agrees with the source formula sheet. The familiar positive coupling uses the opposite curvature sign convention for the Ricci tensor; the physical Poisson equation for Newtonian gravity is the same.

Source: released examination PDF, bundle pages 92–95.

We use metric signature $(-,+,+,+)$ and set the speed of light to one. In question 2, $M$ and $Q$ have the geometric normalization of the Reissner-Nordstrom metric, with $M>0$ and $r>0$. Every numbered part is treated, including all three cases in question 2(b).

Complete solutions to the three questions of the Part C General relativity II course of the University of Oxford paper, sat on 1 June 2021. Original released PDF, bundle pages 63–65; paper pages 1–3. The bundle filename says 2020, but the paper's cover explicitly says Trinity Term 2021. The metric signature is $(-,+,+,+)$ and the speed of light is set to one. All questions and both Roman-numbered parts of question 1(d) are covered.

Apply the product-to-sum formula $2\cos u\cos v=\cos (u+v)+\cos (u-v)$ to the amplitude modulation:The amplitude modulation sidebands have angular frequencies $\omega +{\omega }_m$ and $\omega -{\omega }_m$, each with amplitude $b/2$ in this real cosine convention. The carrier wave has angular frequency $\omega$ and amplitude $a$.

This identity is exact for the supplied space-time expression. A physical qualification is needed if every term is meant to be a freely propagating electromagnetic wave with speed of light $c$: the dispersion relation requires $k_{\pm }=(\omega \pm {\omega }_m)/c$, whereas the prescribed modulation at every position leaves all three wavenumbers equal to $k=\omega /c$. A freely propagating modulated wave instead has envelope $a+b\cos ({\omega }_m(t-x/c))$; its analogous expansion has $k_{\pm }$. For example, at $x=t=0$ the prescribed expression has $A_{tt}-c^2{A}_{xx}=-b{\omega }_m^2$, which is nonzero for nontrivial modulation. The given formula can describe modulation imposed throughout space or the temporal signal at one position.

The vertical hydrostatic equilibrium equation balances the gradient of the total pressure against gravity. In a thin disk with Keplerian rotation, expanding the central gravitational field about the midplane gives vertical acceleration $-{\Omega }^{2}z$, so $p_z=-\rho {\Omega }^{2}z$. Disk self-gravity is neglected in this approximation.

The flux-gradient equation is local thermal balance: the upward radiative flux increases with height because differential rotation dissipates energy. For a Newtonian effective dynamic viscosity $\mu$, the volumetric heating rate is $\mu (r{\Omega }^{\prime }{)}^2$, which becomes $9\mu {\Omega }^2/4$ for Keplerian rotation. The symbol $\mu$ here is dynamic viscosity, not kinematic viscosity; the latter is $\nu =\mu /\rho$.

The radiative diffusion equation applies to an optically thick medium in local thermodynamic equilibrium. Its diffusion coefficient is $c/(3\kappa \rho )$, and the equilibrium radiation energy density is $u_{\mathrm{rad}}=4\sigma T^4/c$. HenceThe minus sign means energy flows down the temperature gradient. Here $\sigma$ is the Stefan-Boltzmann constant, $\kappa$ the opacity per unit mass, and $c$ the speed of light. Constant Thomson scattering opacity is the adopted transport approximation; local thermodynamic equilibrium is an additional assumption behind the radiation equation of state.

Finally the equation of state adds ideal gas gas pressure $p_g=\rho kT/({\mu }_m{m}_H)$ and isotropic radiation pressure $p_r=u_{\mathrm{rad}}/3=4\sigma T^4/(3c)$. The Boltzmann constant is $k$, and ${\mu }_m{m}_H$ is the mean mass per gas particle. Both contributions supply the total pressure entering vertical hydrostatic equilibrium.

During radiation domination, conservation of energy gives $\rho \propto a^{-4}$: dilution contributes $a^{-3}$ and the cosmological redshift of each photon contributes another $a^{-1}$. Thus $\alpha =4>2$ and $a=A\sqrt{t}$ in the flat radiation-dominated approximation. The particle horizon measured as proper radial distance on a constant-time slice isEvery local comoving observer measures the speed of light to be $c$. However, the spatial separations traversed earlier have expanded by the time the final distance is measured; the present separation from the earliest accessible comoving worldline is not the integral of local path lengths measured at their different traversal times. Indeed ${\dot{s}}_{\mathrm{hor}}=H{s}_{\mathrm{hor}}+c=2c$ in this era, explicitly separating cosmic expansion from local propagation. Curvature or a later transition to matter domination changes the exact formula; $2ct$ is the radiation-dominated result.

Use the inverse Lorentz transformationAlong the particle's worldline $d{x}^{\prime }=u^{\prime }d{t}^{\prime }$, so the relativistic velocity-addition formula isFor subluminal collinear velocities, put $u/c=\tanh {\phi }_u$, $u^{\prime }/c=\tanh {\phi }_{u^{\prime }}$ and $v/c=\tanh {\phi }_v$. The hyperbolic tangent addition formula, obtained by dividing the supplied sine/cosine addition formulas, givesBecause hyperbolic tangent is injective on the real line,Thus rapidity is additive even though velocity is not.

Each positive increment in the instantaneous rest frame adds the same rapidity $\alpha =\mathrm{artanh}(1/2)=\frac{1}{2}\log 3$. Starting from zero rapidity, iterated collinear boosts with equal rapidity giveEvery finite number of increments leaves $u_n<c$; the speed of light is approached as $n\to \infty$.

Use a local orthonormal tetrad and units in which the speed of light is one. A particle has four-momentum $p^{\hat{a}}=E(1,\mathbf{e})$ and moves with velocity $\mathbf{e}$. Thus its energy density contributes $E$, its momentum density contributes $E{e}^i$, and its momentum flux contributes $E{e}^i{e}^j$. Integrating the phase-space distribution function gives the kinetic stress-energy tensor. The measure $d^{3}p/E$ is the future mass-shell Lorentz-invariant phase-space measure, so the same expression transforms as a spacetime stress-energy tensor; fixed state-counting or polarization factors are understood to be included in $f$.

Define $⟨F{⟩}_{\Omega }=(4\pi {)}^{-1}\int Fd\Omega$. The photon momentum measure is $d^{3}p=E^{2}dEd\Omega$, so every component of the kinetic stress-energy tensor has a radial factor $E^{3}dE=a^{-4}{ϵ}^{3}dϵ$. An isotropic background has $⟨e^i{⟩}_{\Omega }=0$ and $⟨e^i{e}^j{⟩}_{\Omega }={\delta }^{ij}/3$. HenceThis radiation pressure equation of state requires isotropy and masslessness, not a thermal spectrum.

The distribution perturbation is $\delta f=-ϵ{\bar{f}}^{\prime }(ϵ)\Theta$. Assume finite energy density and endpoint behavior ${ϵ}^4\bar{f}(ϵ)\to 0$ at zero and infinity, as for the Planck photon distribution. Integration by parts then givesConsequently the perturbed energy density and momentum density areSince $q^i=(\bar{\rho }+\bar{P})v^i=(4\bar{\rho }/3)v^i$, the photon angular temperature moments give $\delta =4⟨\Theta {⟩}_{\Omega }$ and $v^i=3⟨\Theta e^i{⟩}_{\Omega }$.

The sign of anisotropic stress must be specified. Direct kinetic integration gives the conventional trace-free spatial stress ${\pi }^{ij}=T^{ij}-P{\delta }^{ij}=+4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3){⟩}_{\Omega }$. The PDF uses the opposite sign, ${\Pi }^{ij}=-{\pi }^{ij}$, equivalently $T^{ij}=P{\delta }^{ij}-{\Pi }^{ij}$. With that convention, all requested moments areTaking the spatial trace also givesThe angular temperature description assumes $\Theta$ is independent of photon energy; independent spectral distortions would require additional energy-dependent moments.

Use the Minkowski metric ${\eta }_{mn}=\mathrm{diag}(-1,1,\dots ,1)$ and units with speed of light one. The contractions are $P^2={\eta }^{mn}{P}_m{P}_n$ and $(T{X}^{\prime }{)}^2=T^2{\eta }_{mn}(X^m{)}^{\prime }(X^n{)}^{\prime }$; $(X^m{)}^{\prime }{P}_m$ and ${\dot{X}}^m{P}_m$ pair a vector with a covector without another metric. The Lagrange multipliers impose the two first-class constraintsThey generate worldsheet diffeomorphisms, so the phase space contains both constrained directions and gauge redundancy. Two first-class constraints remove two canonical pairs, leaving $D-2$ physical degrees of freedom per point.

In Monge gauge, $X^0=t$ and $X^1=\sigma$. Write the transverse canonical variables as $\mathbf{X}=(X^2,\dots ,X^{D-1})$ and $\mathbf{P}=(P_2,\dots ,P_{D-1})$. Solving the first-class constraints givesThe negative root selects positive energy. Substitution into the phase-space action gives the Hamiltonian reductionFor a static segment, $\mathbf{P}=0$, its proper length element is $d\ell =\sqrt{1+|{\mathbf{X}}^{\prime }{|}^2}d\sigma$, and $\mathcal{E}d\sigma =Td\ell$. Thus the string tension is the rest energy per unit proper length. In particular a straight resting segment has $\mathcal{E}=T$. Monge gauge is a local choice on a string embedding map for which $X^1$ is a valid coordinate; it need not cover folded strings or all endpoint configurations.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

Write $E=|\mathbf{E}|$ and $\hat{\mathbf{E}}=\mathbf{E}/E$. The constant electric force gives momentumCombining $\mathbf{p}=\gamma m\dot{\mathbf{x}}$ with the relativistic energy-momentum relation givesThe dimensional analysis from part (a) confirms that $qEt$ has the dimensions of momentum, so $qEt/(mc)$ is dimensionless and the expression has the dimensions of velocity.

The speed increases monotonically from zero and approaches the speed of light asymptotically:Integrating in the direction of the field givesAt small times, the Taylor expansion of the square root gives the nonrelativistic limitwhich is motion from rest with the constant acceleration $q\mathbf{E}/m$.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a linear system with constant coefficients, so

PutThe initial four-velocity is $(u^0,u^1,u^2,u^3)={\gamma }_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,Taking the initial spacetime point to be the origin and integrating $u^{\mu }=d{x}^{\mu }/d\tau$ gives the relativistic motion in parallel electric and magnetic fieldsThe expressions have their continuous $a\to 0$ or $b\to 0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.

Eliminating proper time between the first two equations yieldsAt early coordinate time,which is uniformly accelerated motion with the transverse relativistic inertia factor ${\gamma }_0$. At late positive time,so the longitudinal speed approaches the speed of light.

The $y$-$z$ projection satisfiesIt is therefore a circle of radius and proper-time period

For the pitch magnitude assume $v_0>0$ and put $\alpha =|q|E/(mc)$, $\Omega =|q|B/m$, andNumber the $n$th proper-time period as $[(n-1)\Delta \tau ,n\Delta \tau ]$. Its longitudinal displacement isConsequently, as $n\to \infty$,where

Put $u=t-z$. Wherever $|u|>U$, the profile $A(u)$ vanishes and the displayed metric is exactly the Minkowski metric. Thus the two flat regions in the $(t,z)$-plane are the half-planes $t-z>U$ and $t-z<-U$, separated by the stripAt fixed $t$ this strip is $t-U\le z\le t+U$, so it has longitudinal width $2U$. A surface $u=\text{constant}$ obeys $z=t-u$ and travels in the positive $z$ direction at the speed of light. The curvature is confined to that moving strip, identifying the solution as a gravitational-wave pulse represented by a plane-fronted gravitational wave.

Take the total differential of the logarithm of the relativistic energy, with the speed of light fixed:Using the chain rule,Every term is a dimensionless fractional change. The differential relation is exact; applying it to finite small changes is a first-order Taylor expansion.

By Fermat principle, a light ray makes optical travel time stationary. Multiplication by the constant vacuum speed of light does not change the stationary paths, so the functional can be taken to be the refractive index integrated over arc length. The constant denoted $c$ below is an integration constant, not that vacuum speed.

The emission mechanism is Cherenkov radiation. An electron moving faster than the phase velocity $c/n$ of light in the material excites electric polarization disturbances whose radiation adds coherently on a cone. This does not require the electron to exceed the speed of light in vacuum. Taking $n=1.4$ as constant over the detected optical band, the Cherenkov cone angle satisfiesEmission requires $\beta >1/n$. The detector accepts ${\theta }_C<20^{\circ }$, soNumerically,The lower endpoint has vanishing Cherenkov radiation yield, and the upper endpoint is excluded by the stated strict angular acceptance. Faster electrons emit at larger cone angles: at $v\simeq c$, ${\theta }_C\simeq {\cos }^{-1}(1/1.4)=44.4^{\circ }$, outside this detector's acceptance. The angular condition gives the accepted speed interval conditional on rays reaching the detector; no collection geometry or total signal yield is specified.

Cosmological variables and conventions. The scale factor $R(t)$ describes the relative separation of comoving observers; dots denote derivatives with respect to cosmic time $t$. The Hubble parameter is $H(t)=\dot{R}/R$, usually time dependent; “Hubble constant” conventionally means its present value. Here $\rho$ is mass-equivalent density, so $ϵ=\rho c^2$ is energy density, while $P$ is isotropic pressure. $G$ is the gravitational constant and $c$ the speed of light. The constant $K$ specifies cosmological spatial curvature: with $R$ a length, $K=k{c}^2$ and $k=+1,0,-1$ describe closed, flat and open spatial geometries. Alternative normalizations of $R$ move a constant factor into $K$. In the printed equations $\Lambda$ has units ${\mathrm{s}}^{-2}$: it is $c^2$ times a cosmological constant defined in inverse square metres. This time-unit convention for the cosmological constant prevents a missing factor of $c^2$.

Continuity and the first law. Differentiating the first Friedmann equation with constant $K$ and $\Lambda$, and then eliminating $\ddot{R}/R$ with the second, givesThus, on an expanding or contracting branch,The relation extends continuously to a regular turning point. For a physical volume $V(t)=V_{\ast }{R}^3(t)$ bounded by fixed comoving coordinates, $\dot{V}=3HV$. Multiplication by $c^{2}V$ givesThis adiabatic work in a comoving volume is the first law of thermodynamics for adiabatic evolution with fixed entropy and no additional chemical work. It expresses local energy conservation, including pressure work. It does not assert that $U$ is constant: radiation loses internal energy as it does work; negative pressure makes $U$ grow during expansion. A generic expanding spacetime has no global time-translation symmetry from which a conserved total energy follows.

Constant scalar-field energy density. Setting $\dot{\rho }=0$ in the continuity equation, with $H\ne 0$, givesFor a homogeneous canonical scalar field, in units $c=1$, $ϵ={\dot{\varphi }}^2/2+V(\varphi )$ and $P={\dot{\varphi }}^2/2-V(\varphi )$. A stationary potential-dominated field has precisely this equation of state. A general evolving scalar field need not have $w=-1$; the constant-density assumption is essential.

Expansion at the stipulated Higgs energy density. With zero explicit $\Lambda$ and zero $K$, constant ${ϵ}_H$ givesThe plus sign selects the expanding branch; there is also a contracting solution. The scalar-field energy is already included in $\rho$, so adding it again as an explicit cosmological constant would double count it. Using $1\mathrm{GeV}=1.602176634\times 10^{-10}\mathrm{J}$ gives ${ϵ}_H=2.4033\times 10^{45}\mathrm{J}{\mathrm{m}}^{-3}$, ${\rho }_H=2.6740\times 10^{28}\mathrm{kg}{\mathrm{m}}^{-3}$ and $H_H\simeq 3.87\times 10^9{\mathrm{s}}^{-1}$. ThereforeThis vacuum-dominated doubling time describes accelerated expansion in de Sitter spacetime, with $\ddot{R}/R=H_H^2>0$. The stipulated energy density exceeds the critical density in energy units used in Question 2 by $3\times 10^{54}$, so it cannot persist as the dominant component of the present universe. The calculation exhibits the cosmological constant problem: the absolute gravitational vacuum energy must be enormously smaller than such a particle-physics scale today. It does not establish that the Standard Model Higgs field actually remains constant or supplies a viable inflationary epoch.

The acoustic ruler and accelerated expansion. Before baryon drag becomes ineffective, pressure in the photon-baryon fluid drives acoustic waves around initial linear cosmological density perturbations. Their comoving sound horizon at the baryon drag epoch isAfter decoupling, this preferred separation survives in galaxy clustering as baryon acoustic oscillations: a peak in the two-point correlation function and oscillations in the matter power spectrum. Its value is calibrated by early-universe physics, commonly constrained by the cosmic microwave background. The SDSS luminous-red-galaxy measurement established a peak near $100h^{-1}\mathrm{Mpc}$; its use as a distance indicator depends on the ruler calibration rather than the length of an individual galaxy.^[ref]

For a ruler perpendicular to the line of sight and one parallel to it,where $D_M=(1+z)D_A$ is the transverse comoving distance and $D_A$ the angular diameter distance. These measure $D_M/r_d$ and $H{r}_d$, respectively. Comparing multiple redshifts tests the distance–redshift relation and expansion history. In particular,so sufficiently precise measurements of the shape of $H(z)$ can test $q<0$. In general relativity, accelerated expansion requires a sufficiently negative total $\rho +3P/c^2$; a positive cosmological constant supplies it.

A single acoustic angular scale does not independently determine acceleration, cosmological spatial curvature, the absolute sound horizon and the dark-energy equation of state. One combines redshift coverage with early-universe calibration and independent matter-density or distance information, comparing accelerating and nonaccelerating models. The galaxy selection function and redshift completeness must be modeled with random catalogues, survey covariance estimated with mock catalogues, and redshift-space distortions, galaxy bias and nonlinear broadening controlled. These tests distinguish a cosmological feature from a survey artifact. The evidence is the measured expansion history within an explicit cosmological framework, rather than the existence of an acoustic peak alone.

Cherenkov radiation is coherent electromagnetic emission by a charged particle moving through a dielectric faster than the phase velocity of light in that medium. The particle polarizes the material along its path. Wavelets interfere constructively on a cone whenA simple wavefront construction compares a particle displacement $vt$ with the light-wave displacement $ct/n$, giving the cone relation. The particle still travels slower than the vacuum speed of light; no violation of special relativity occurs. Dispersion makes the angle and threshold wavelength-dependent. The radiation has a continuous spectrum rather than the discrete line spectrum of an atomic transition.