Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 311 3 c Solution Created 2026-10-03 Updated 2026-10-06
For the zero-angular-momentum curves, the fiber term in the induced metric tensor vanishes along the tangent, leavingfor a null geodesic. In the exterior, a future geodesic has positive conserved energy . Its equations areDefine the tortoise coordinate, separately on intervals bounded by horizon radii, byFor an outgoing curve ,for an ingoing curve , . Thus and are the respective constant labels. The derivative , with its sign, is required for continuation; taking would give the wrong labels between the horizons.
For a nonextremal horizon , the tortoise coordinate has leading behaviorAt the double root of the extremal black hole it instead has a pole. These singularities are coordinate effects at a regular Killing horizon; they are not the curvature singularity at . To cross the future horizon, the ingoing coordinatesgive the regular induced metric tensorThis also fixes the future branch used in the causal argument below.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 311 4 Solution Created 2026-10-03 Updated 2026-10-06
Work first in units . Black-hole thermodynamics starts with the laws of black-hole mechanics. For a connected stationary regular Killing horizon, the Zeroth law of black-hole mechanics makes its surface gravity constant when the Einstein field equations and dominant energy condition hold. This parallels uniform equilibrium temperature. For neighboring stationary asymptotically flat four-dimensional Einstein-Maxwell black holes, the First law of black-hole mechanics iswhere is the horizon angular velocity and its electric potential relative to infinity. The last terms are rotational and electromagnetic work, analogous to the work terms in . The second law of black-hole mechanics, expressed by Hawking's area theorem, makes the total future horizon area nondecreasing for classical matter obeying the null energy condition and global assumptions such as strong asymptotic predictability. The third law of black-hole mechanics is an unattainability statement: subject to the usual regularity assumptions, the weak energy condition, and a bounded stress-energy tensor, no finite physical process reduces the surface gravity of a regular horizon to zero. It does not assert that extremal black holes have zero area or zero entropy.
Classical geometry alone fixes an analogy, not a nonzero physical temperature. Quantum field theory supplies Hawking temperature . Comparing the area term of the first law with gives for nonextremal holes, and hence the standard Bekenstein-Hawking entropy. Restoring physical constants,Here restored is the physical acceleration surface gravity and is the squared Planck length. The first-law comparison determines the area coefficient, leaving an additive entropy constant unspecified; the displayed formula is the usual convention. Its area scaling, rather than ordinary volume scaling, signals that the available thermodynamic degrees of freedom of a gravitating system are constrained by the horizon geometry.
Particle production explains the quantum input. For a real scalar field obeying the massless covariant wave equation, the Klein-Gordon inner product on solutions isThe wave equation makes its current divergence-free, so the product is independent of a Cauchy hypersurface when the boundary flux vanishes. Use suitably normalized wave packets or the appropriate continuum distributions. Asymptotic past and future Minkowski spacetimes give preferred positive-frequency solution mode bases and , normalized by , and . Between those regions, a nonstationary geometry generally has no preferred positive-frequency splitting.
The mode bases are related by a Bogoliubov transformation,The last identities follow from conservation of the Klein-Gordon inner product and are the canonical identities for a bosonic Bogoliubov transformation. Expanding the field in either complete basis and extracting its positive-frequency coefficient givesThus the state annihilated by every hasThis particle number from Bogoliubov coefficients is nonzero when time evolution mixes positive and negative frequencies. It describes one state as vacuum in the early particle basis and populated in the late basis; it does not require an arbitrary choice of a vacuum at each intermediate time. For a finite implementable transformation the state remains a squeezed pure quantum state. In infinitely many modes, a common unitary bosonic Fock space implementation requires additional conditions, such as the beta map being a Hilbert-Schmidt operator; finite wave packet observables need not share every global divergence of an ideal continuum calculation.
For gravitational collapse, the future is not globally Minkowski: it contains a black hole. The relevant late out-basis includes modes reaching future null infinity together with modes entering the future event horizon. Modes on infinity alone are not a complete Cauchy basis. Nevertheless the same mode-mixing calculation determines the outgoing particle flux. Near a nonextremal horizon, the logarithm in the tortoise coordinate converts regular early null coordinates into the Hawking exponential ray mapwhere is the early affine null coordinate and is late retarded time at infinity. Backward propagation of a late outgoing mode gives a factor for . Its positive- and negative-frequency Fourier integrals differ by analytic continuation around the logarithmic branch. With a convergence regulator they reduce toTaking gives the thermal ratio of Hawking Bogoliubov coefficients, . Combining this with the canonical normalization yields the bosonic occupation . Late-time wave packets turn formal continuum coefficients into a finite number flux.
A Schwarzschild black hole has , so this spectrum has . The exterior curvature potential partly reflects the outgoing modes. Its transmission probabilities are the greybody factors, giving, for one massless scalar species in the stationary late-time approximation,Thus the horizon temperature is universal, while the spectrum received at infinity is not a perfect featureless blackbody radiation spectrum. The derivation assumes the near-horizon quantum state inherited from a regular collapse vacuum and ignores rapid backreaction over the timescale of the packets. Exponentially blueshifted precursor frequencies indicate the usual short-distance assumption in the semiclassical calculation, not an independently demonstrated quantum-gravity description of the endpoint.
The outgoing positive energy flux drives black-hole evaporation. At leading order for a large isolated Schwarzschild black hole, area scales as and temperature as , giving and a lifetime of order , with the coefficient depending on species and transmission factors. Its negative heat capacity of a Schwarzschild black hole, , means that it heats up as it loses mass and cannot be in stable canonical equilibrium with an unlimited thermal reservoir. The approximation fails when quantum-gravitational scales are reached and does not fix whether the endpoint is complete evaporation, a remnant, or something else.
Area loss does not contradict the classical area theorem, because the quantum stress-energy tensor need not obey the classical null energy condition; negative horizon energy accompanies the positive outgoing flux. The appropriate thermodynamic statement is the generalized second law, involving . Finally, Hawking radiation raises the black hole information paradox: if a pure quantum state completely evaporates into an exactly thermal mixed quantum state and its partners disappear, the result conflicts with ordinary unitary time evolution. Early outgoing radiation can be mixed simply because of its entanglement with interior modes while the complete state remains a pure quantum state; that fact alone is not information loss. Resolving the fate of all correlations through the evaporation endpoint requires more than the leading semiclassical flux calculation.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 312 2 iii Solution Created 2026-10-03 Updated 2026-10-06
In the local orthonormal tetrad, the components of the photon number current areThey are respectively the photon number density and the photon number flux. Thus is the photon number four-current, not an energy current or an electric charge current. Photons carry no electric charge. For the isotropic background,Integrating the covariant Boltzmann equation over the invariant mass-shell measure, with vanishing momentum-boundary flux, giveswhere is the collision operator in the covariant streaming equation. ThereforeCollisionless photon propagation satisfies this condition, as do Thomson scattering and ordinary Compton scattering when all participating photons are counted: each collision has one incoming and one outgoing photon. Photon emission, photon absorption, bremsstrahlung, double Compton scattering and electron-positron annihilation in cosmology or its inverse generally give a nonzero number source. A cancellation of net creation and destruction can nevertheless make the integrated source vanish even when individual number-changing reactions occur. Cosmological redshift itself changes photon energy, not photon number; in an expanding homogeneous universe a conserved photon number current gives .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 314 1 b Solution Created 2026-10-03 Updated 2026-10-06
Let the constant mass density be and put , the Alfvén velocity. Since both the velocity and the magnetic field have zero divergence, the magnetic tension and magnetic pressure decomposition of the Lorentz force givesThe Newtonian gravitational potential remains in ; uniform mass density does not justify dropping a prescribed gravitational acceleration. Adding and subtracting the two equations proves the Elsässer variable equationsHere ; each Elsässer variable is transported by the other.
Define . Taking the dot product with gives the Elsässer energy invariant balanceThe divergence theorem proves whenever the net boundary flux vanishes. In particular, makes , so both fluxes vanish pointwise. Periodic boundary conditions or decay at infinity are also sufficient. These boundary conditions matter: fixed volume alone gives no conservation.
The kinetic energy plus magnetic energy isExpanding and using the cross-helicity definition yieldsBoth energy and cross-helicity therefore follow from the two Elsässer energy invariants. The quantity here is exactly the requested kinetic energy plus magnetic energy, without an additional gravitational term.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 315 2 a i Solution Created 2026-10-03 Updated 2026-10-06
In the collisionless exosphere, an atom escapes if its outward trajectory has positive total mechanical energy. Neglect tides and stellar forces and use Newtonian gravity at exobase radius :With thermal speed , the Jeans escape parameter isA thermal distribution always has an escaping tail; Jeans escape is exponentially suppressed for , with Jeans escape flux proportional to . Efficient escape requires of order a few or smaller, with an order-unity energetic estimate .
Assume a Neptune-like mass and radius, , and atomic hydrogen. Using the supplied rounded constants givesHenceUsing the mean kinetic energy instead gives an order-unity coefficient and . An expanded exobase has weaker binding and lowers the estimate by . A comet-like tail can also be shaped by radiation pressure and stellar-wind interactions; it does not by itself measure or prove that a hydrostatic Jeans model is valid. At , and hydrostatic equilibrium fails as a global description: substantial mass loss must usually be treated as hydrodynamic atmospheric escape.
The exobase is defined by mean free path , not by a universal pressure. For a neutral hydrostatic gas with collision cross-section ,For example, explicitly assuming gives , or . These are representative extremely dilute neutral-exobase pressures, with orders of magnitude varying with composition, cross-sections and expansion. The supplied constants contain no collision information, so they cannot uniquely determine an exobase pressure; ionization or a non-hydrostatic density profile changes this estimate.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 315 3 a i Solution Created 2026-10-03 Updated 2026-10-06
For a quasi-static self-gravitating object of fixed mass, negligible surface pressure, and no accretion, nuclear source or additional deep heating, conservation of energy gives the intrinsic planetary luminosityHere is internal energy, and is negative gravitational potential energy. Gravitational contraction makes more negative and releases power . Some of that release raises the internal energy; it cannot all be radiated while the object maintains hydrostatic equilibrium.
To quantify the split assume thermal gas pressure with constant specific-heat ratio and negligible rotation or magnetic support. This excludes the dynamically unstable gas-supported regime. The stellar virial theorem givesIt follows thatFor a monatomic ideal gas, , so : half the released gravitational energy heats the gas and half escapes. If the density profile remains homologous, write , with a fixed structure constant. Thenup to the structure and virial coefficients, where is the Kelvin-Helmholtz cooling time.
For a diatomic gas with the idealized fraction is , already demonstrating that the half-factor is not universal. Partial electron degeneracy pressure, dissociation, changing structure, accretion and deep heating require the full energy equation. The general link between luminosity and Kelvin-Helmholtz contraction is , with any genuine source terms added explicitly.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 316 1 i Solution Created 2026-10-03 Updated 2026-10-06
Use the separation vector from the star to the comet, and let . The two-body problem separates into uniform centre of mass motion andTaking a cross product with proves conservation of angular momentum; taking the scalar product with proves conservation of energy. The relative constants areThe total relative angular momentum and energy are and , respectively. The conserved eccentricity vector is . Its scalar product with gives , with the true anomaly. The periapsis and apoapsis distances are therefore and . Their sum is twice the semi-major axis, soAt either apsis ; substituting an apsidal radius gives . Thus the specific orbital energy implies the vis-viva equationThese are exact relative-coordinate formulas for an elliptic orbit, . In the barycentric frame the comet's speed is times the relative speed; the printed identification with the comet's stellar orbital speed uses .
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 316 4 iii Solution Created 2026-10-03 Updated 2026-10-06
The incoming and outgoing velocity vectors have the same magnitude , by conservation of energy. The chord between them has lengthFor fixed put . Then , and proves . Equality occurs at .
A non-colliding comet must have , where in the point-comet approximation and for finite bodies. Thus the non-colliding gravitational velocity-kick bound isThe optimizing grazing orbit has , and . If contact itself is excluded, this is a supremum approached from larger . The finite collision radius is essential: an ideal point planet with unrestricted has no finite upper bound.
A constant Galilean boost by the planet's stellar orbital velocity leaves the vector difference unchanged. It changes the comet's incoming and outgoing heliocentric energies, which is the basis of a gravitational assist.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 316 4 i Solution Created 2026-10-03 Updated 2026-10-06
Let be the separation of the two centres at release. The two-body problem has relative specific orbital energy . Marginal escape has , soThis follows equally by equating the relative kinetic energy to the magnitude of the Newtonian gravitational potential energy , where .
For a pointlike comet released at the planet's surface, , giving the usual . If the finite comet touches that surface, its centre is at ; the same formula uses this larger separation. The assumption by itself does not imply . Subsequent formulas written in terms of the planet-only escape velocity use the additional small-comet approximation.
This is local escape from an isolated planet. In a stellar Newtonian gravitational field, reaching the edge of the Hill sphere and escaping the stellar system are different energy thresholds.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 316 4 iv Solution Created 2026-10-03 Updated 2026-10-06
Let , with . The Safronov number is . In the instantaneous planet frame let the asymptotic vectors be and ; their magnitudes agree. In the stellar frame the corresponding specific orbital energy change during a short encounter isThe quadratic relative-speed terms cancel. A comet with semi-major axis comparable to has stellar binding energy per unit mass of order . Consequently measures whether a single strong encounter can change a substantial fraction of that binding.
Large escape-to-orbital-speed ratios favour ejection; small ratios favour collision or accretion during repeated encounters. For , a suitably oriented gravitational assist can eject the comet before it strikes the planet. For , most individual kicks are too weak; many close passages may be needed, providing repeated chances of collision.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 316 4 viii Solution Created 2026-10-03 Updated 2026-10-06
The two-boundary map omits several effects that can alter planetary scattering.
The initial semi-major axis, orbital eccentricity, orbital inclination and orbital phase determine whether encounters occur and their relative velocities. A strongly bound comet needs more energy to escape; a nearly parabolic one needs less. The planetary radius and mass density, the finite comet radius, and gravitational focusing determine collision probabilities. Tidal disruption, atmospheric gas drag, sublimation and physical fragmentation can destroy a body before a nominal point-particle scattering sequence is completed.
Other planets can hand a comet from one scatterer to another, eject it, or lift its periapsis clear of the original scatterer's orbit. Mean-motion resonances and secular perturbations can protect objects from encounters or correlate kicks, contradicting the independent random walk assumption. Planetary migration changes the encounter geometry over time.
Stellar flybys and a galactic tide can change distant comet periapses, allowing new encounters or detaching an object from the planetary region. Stellar mass evolution changes both binding and planetary orbits. Finally, the age and the supply rate of new comets determine whether the observed population is a residual one or continuously replenished. Thus the mass-radius map is a useful conditional classification, not a complete survival law.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 317 2 Solution Created 2026-10-03 Updated 2026-10-06
For a quasistatic spherical star with fixed total mass, the stellar structure equations in radius areHere is the specific entropy of a moving mass shell, and the entropy term accounts for release or storage of thermal and gravitational energy. In thermal equilibrium it vanishes. An equation of state, opacity, energy-generation law and composition evolution close the equations. The absence of mass loss fixes the outer mass coordinate; it does not require the star to be in thermal equilibrium.
The radiative temperature gradient needed to carry the full luminosity isHere the actual temperature gradient is . For uniform composition the Schwarzschild criterion says that a radiative layer is stable if , with marginal stability at equality, where the adiabatic temperature gradient is . Use there. If , convection carries some of the flux. Efficient convection gives ; inefficient surface convection needs a transport prescription and can be superadiabatic. For composition gradients, use the Ledoux criterion, with threshold instead. Here , and , using the mean molecular weight .
Above the thin burning shell, take constant luminosity , constant mean molecular weight , ideal-gas pressure with , and the Kramers opacity law . Then the temperature equation isTo implement the printed assumption that and are all power laws while retaining mass conservation, write , , and . The mass equation gives ; the ideal gas law gives ; hydrostatic equilibrium gives . The diffusion equation gives . Solving this linear system gives the self-gravitating Kramers power-law envelope:The coefficients can also be matched consistently. If is the core boundary and its enclosed mass,Radiative transfer fixes the constant luminosity throughwhose right-hand side is independent of radius for these exponents. Thus this is a solution of all four envelope equations, not just a dimensional estimate.
Matching the boundary temperature to the isothermal core and extrapolating the idealized power law to the specified photosphere givesFor reference , so the envelope mass is not negligible in this particular solution. The given fixes the core-radius normalization ; a numerical absolute radius needs the unspecified molecular weight. The exponent ratio also gives , so the fully ionized monatomic version of this idealization is convectively stable.
A different common approximation neglects the envelope's self-gravity and sets in the force equation. It cannot obey the exact mass equation with a nonzero density and exactly constant . Under that additional approximation, the same calculation instead givesThis is the constant-core-mass power-law opacity radiative envelope limit, not the full power-law solution with changing enclosed mass. Stating which approximation is used resolves the otherwise different numerical radius ratios.
A compact nearly isothermal core, a thin luminosity-producing shell and a much larger cool envelope describe a shell-burning red giant. The full power-law profile is a deliberately simple model. The Kramers opacity law and an ionized ideal gas are not reliable at a photosphere; partial ionization, other opacity sources and a convective envelope usually modify real cool giants. The quoted radius ratios are extrapolations within the stipulated model.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 317 4 Solution Created 2026-10-03 Updated 2026-10-06
Let be particle number per unit volume in a momentum element, including the spin-state count in . This is the local momentum part of a phase-space distribution function, normalized by the number density . For an isotropic distribution with particle energy , the kinetic pressure of an isotropic gas is the average momentum flux:The kinetic energy density, explicitly excluding rest mass, isFor , and , giving . For , and , giving . ThusNeither relation assumes a Maxwellian distribution. At intermediate momenta neither constant ratio is exact, and anisotropic distributions require a pressure tensor rather than this scalar pressure.
For a classical Maxwell-Boltzmann distribution, . Three Gaussian component integrals give , so the ideal gas hasIn a fully ionized mixture, sum over independent species to obtain , where the mean molecular weight counts ions and free Electrons. This excludes partial-ionization and interaction corrections. A dilute classical ultrarelativistic gas still has but ; its momentum distribution is proportional to instead of the nonrelativistic Gaussian.
For a thermal photon gas, the two polarization states and zero photon chemical potential give the Planck photon distributionUse and . The radiation constant and photon equation of state are thereforeLikewise . Unlike a gas with a fixed particle number, photons do not have pressure proportional to baryonic mass density.
For fully degenerate Electrons, the Fermi-Dirac distribution becomes a filled momentum sphere with two spin states. Counting them gives the Fermi momentumHere is the mean molecular weight per electron. The equation of state of a cold electron gas follows by integrating momentum flux up to :Its two limits areThe corresponding kinetic energy densities are and , respectively. Thus in the high-density relativistic limit the Electron pressure scales as and is nearly independent of temperature. This is the ideal noninteracting, fixed-composition Electron result, not a universal equation of state at nuclear densities where captures, interactions and the composition change.
The boundaries on a stellar equation-of-state regime diagram concern the Electron component. Define the electron relativistic density threshold and thermal electron relativistic threshold byA degenerate Electron gas is nonrelativistic well below and ultrarelativistic well above it: this is an approximately vertical division on a log-density plot. A nondegenerate Electron gas instead becomes thermally relativistic near : this is an approximately horizontal division. The regimes have broad crossovers rather than a discontinuity at either line.
The exact electron Fermi temperature, subtracting Electron rest energy, isFor Electrons with negligible thermal pairs, strong degeneracy requires , whereas gives a nondegenerate gas. In the pair-rich regime the actual Electron and Positron distributions must instead be treated with their chemical potentials, as discussed below. The two limiting degeneracy boundaries areThese slopes and explain the bent degeneracy boundary on the logarithmic graph. A thermal-wavelength test gives the same nonrelativistic density/temperature scaling, with an order-one definition of the crossover.
The radiation-to-gas pressure boundary in the nondegenerate fully ionized regime isRadiation dominates above this line. Once Electrons are strongly degenerate, compare radiation pressure with instead of continuing the ideal-gas comparison into that regime. The radiation-to-degeneracy pressure boundary is , with logarithmic slopes and in the nonrelativistic and ultrarelativistic limits. Degeneracy of the Electrons and dominance of their pressure are distinct criteria.
For the pair curve, distinguish baryonic net Electrons from thermally created Electrons and Positrons. Chemical equilibrium with photons requires opposite Electron/Positron chemical potentials when the one-particle energy includes rest energy, as in the distributions below. At low temperature and low degeneracy the zero-chemical-potential density per charge species isCharge neutrality gives , while the nondegenerate equilibrium product is . HencePairs become important when is comparable with or exceeds , approximately the electron-positron thermal pair abundance curve . Below its density at a fixed temperature, pairs dominate over the net charge Electrons. The exponential makes the low-temperature portion steep in a log-log plot. For , the full zero-potential integral gives approximate pair-marker densities at and at . At relativistic temperature use the full zero-chemical-potential Fermi-Dirac distribution instead; it gives , so the high-temperature pair curve approaches slope three in log-density versus log-temperature. Strong net-electron degeneracy suppresses Positrons and requires the full chemical-potential-dependent distribution.
Approximate stellar equation-of-state regimes and thermal pair boundary
. The original diagram uses fully ionized helium, and , only to set numerical locations. It plots the full curve and the zero-chemical-potential pair integral, rather than extending their asymptotes into the crossover. The pair boundary is deliberately an approximate abundance marker, not a phase transition; at high temperature the nonrelativistic ion approximation and fixed-composition model also have limits. For with abundant nondegenerate pairs, the combined Electron/Positron energy density is , so photons plus pairs have and . Real stellar matter adds Coulomb effects, partial ionization, nuclear reactions and, at sufficiently high density, nuclear-matter physics beyond the ideal regime map.
