Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 49 2 Solution Created 2026-10-03 Updated 2026-10-07
Use a selected translationally invariant pure thermodynamic phase and define the connected lattice correlation functionAway from criticality its long-distance tail has an exponential factor , possibly multiplied by a power. Equivalently when this limit exists; different directions can have different lengths on an anisotropic lattice. At a thermodynamic critical point the correlation length diverges and the tail becomes algebraic. Differentiate with respect to the physical field appearing in the energy Statistical Hamiltonian. With , the correlation-function susceptibility sum rule isIn physical continuum coordinates the sum becomes . Differentiating with respect to the dimensionless source instead removes the explicit inverse-temperature factor. Below the transition, take a one-sided pure-phase limit rather than a macroscopic symmetric mixture.
A normalized blocking kernel assigns a probability or a delta constraint to a coarse configuration for each microscopic one, with . Define the blocked Boltzmann weight bySumming over proves exact equality of the partition functions. Repeated real-space renormalization group steps compose the kernels and integrate out shorter-scale variables. To retain equality one must keep all generated interactions and the field-independent constant; a finite truncation is an approximation. Long-distance observables are mapped through the corresponding coarse observables, not identified blindly with microscopic spins. In physical units,A subsequent coordinate rescaling restores the reference lattice spacing while expressing couplings in cutoff units.
Let be free energy per microscopic site. Invariance of givesThe constant transforms as . Therefore, with , the identity-operator contribution to renormalization-group free energy givesThe second equality follows by substitution and induction. The inhomogeneous term records eliminated-mode entropy, determinants, and normalization factors, all multiplying the identity operator; it is not an interaction with the retained spins. Strictly, this can contain regular background terms. After an analytic subtraction of an inhomogeneous renormalization recursion, the genuinely singular part satisfies the homogeneous scaling equation, except for possible logarithmic resonances.
A renormalization-group fixed point satisfies . Linearize in scaling coordinates : . A relevant operator has , an irrelevant operator has , and a marginal operator has and needs nonlinear analysis. These are logarithmic scaling eigenvalues; the discrete Jacobian eigenvalues are , not . The critical surface is the stable manifold obtained by tuning all relevant coordinates to zero. A repulsive renormalization-group trajectory leaves the fixed point along a relevant direction as the coarse-graining length grows. Stable and unstable manifolds become curved away from linear order.
Linear renormalization-group flows: attraction along a critical surface and repulsion in two relevant directions
. The left section has , one irrelevant coordinate and a relevant thermal coordinate . Its critical surface is in that section. With two relevant variables, the full critical surface has codimension two: the second section shows repulsion in the plane at .
Assume an isolated fixed point with two relevant scaling fields, analytic changes from physical controls to , short-range isotropic scaling, no dangerously irrelevant variable in the thermodynamic scaling function, and no marginal or resonant logarithm. Let denote their scaling eigenvalues. The homogeneous singular free-energy density obeysChoose a stopping scale so the renormalized thermal field is order one. Irrelevant fields vanish at this scale. With and ,Continuous scale interpolation absorbs the immaterial integer stopping-scale convention. Nonuniversal thermal, field and free-energy metric factors can be included explicitly; the normalized functions depend only on the universality class. The and labels refer to and , which flow to different sides of the critical surface.
For clarity about the additive terms, let solve . If , its analytic coefficients areprovided the denominators do not vanish. Subtracting makes the recursion homogeneous. A vanishing denominator is a resonance and generically produces logarithms, which the simple power-law hypothesis excludes. In a convention where the regular field-dependent background has been subtracted, write and . This yields the requested formThe contain branch-dependent matching contributions; only is physically determined. The are nonsingular functions analytic through the thermodynamic critical point and represent the same regular Taylor background on the two sides, though either branch may be used to describe them. For the full free energy at nonzero field, the regular background is generally , not just . Adding an analytic identity-operator contribution proportional to provides a direct counterexample to an exact unrestricted formula; it changes no singular exponents. The printed expression is justified for the leading singular scaling form with that analytic field background removed. Confluent corrections from irrelevant fields are also omitted in this leading form.
Differentiate the singular scaling form, with one-sided derivatives in the ordered phase. It givesThe length transforms as , so . Thus the Rushbrooke scaling relation and hyperscaling relation follow by direct substitution:The same reasoning also gives . Hyperscaling relation relies on the assumptions above and can fail above an upper critical dimension because of a stabilizing dangerously irrelevant coupling.
For the universal specific-heat amplitude ratio, write at . Two temperature derivatives multiply both branches by the same dimensional and thermal metric factors. Away from logarithmic resonances and exceptional zero amplitudes, with the same heat-capacity sign convention on both sides. HenceThe functions and their matched branch constants are tied to the fixed-point theory; the constants are not independent arbitrary experimental parameters. If , this means the amplitudes of the subleading singular contribution after subtracting the analytic specific-heat background. A finite background ratio itself need not be universal.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 62 1 Solution Created 2026-10-03 Updated 2026-10-07
Write and . In this geometry the Lorentz force density is purely longitudinal and equals . The ideal magnetohydrodynamic induction equation becomes . Together with the continuity equation and adiabatic pressure evolution, the four primitive-variable equations areThey have the required quasilinear form withHere the notation in derivatives means , whereas denotes the longitudinal velocity. Spatial dependence is one-dimensional; time dependence is allowed in the wave problem.
Put , and assume and . Expansion of the characteristic polynomial givesTwo independent eigenvectors for can be chosen as and . For the two propagating branches, useful eigenvectors areThe repeated branch advects an entropy disturbance and a disturbance with unchanged total gas-plus-magnetic pressure. The branches are perpendicular fast magnetosonic waves, with speed relative to the fluid.
On a uniform background, a Fourier amplitude obeys . Thus the eigenvalues are the phase velocities for , and the four dispersion branches are twice and . The reduced geometry retains neither an independent transverse velocity nor an additional propagating Alfvén wave.
To construct each finite-amplitude simple wave in magnetohydrodynamics, trace a state-space curve with tangent , using as its parameter. Integrating the pressure and magnetic components givesThe constants and are the entropy and transverse magnetic flux freezing invariants along this state curve. HenceSince , substituting this curve into the original system reduces it to . By the chain rule,Thus the characteristic speed itself satisfies the Inviscid Burgers equation. For initial speed , its smooth solution is along , until the characteristic curves intersect. This derivation supplies genuine state curves for simple waves, not merely a linear dispersion relation.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 7 4 Solution Created 2026-10-03 Updated 2026-10-07
A kinetic model of a dilute gas. In the Boltzmann equation, is the particle number density in phase space, so counts particles near position and velocity . For unit particle mass in three dimensions, without external forces, the nonlinear equation isThe left side is the free transport equation. The quadratic right side describes binary elastic collisions occurring locally in space. A prescribed acceleration would add ; all conclusions below about energy or entropy must then include the appropriate force terms. Assume enough smoothness, velocity decay and finite moments to justify the calculations, and either periodic space or decay that removes spatial boundary fluxes. These are assumptions on a classical solution, not a general existence theorem.
The dilute-gas approximation neglects simultaneous many-body collisions and treats the interaction region as small relative to macroscopic spatial scales. Molecular chaos for the Boltzmann equation factorizes the distribution of two incoming particles as . It applies to incoming collision pairs; it does not assert that an interacting many-particle system stays exactly independent. This approximation closes the equation at the one-particle level and explains its nonlinearity.
Collision geometry and the gain-loss operator. Let and . A useful parametrization of an equal-mass elastic collision isThe center velocity is unchanged and the relative velocity is reflected in the plane normal to . ConsequentlyThe fixed- pre/post-collision transformation is an involutive orthogonal map of , so it preserves . Take a nonnegative collision kernel invariant under interchange of particles and reversal of the collision; for example . The nonlinear Boltzmann collision operator isAny convention about counting and is absorbed in . For a hard-sphere gas, this reflection parametrization has , with a positive constant determined by the sphere size and counting convention. Maxwell molecule collision operators instead have a collision frequency independent of after angular integration. A Grad angular cutoff requires a finite angular integral; singular kernels need additional cancellation analysis and cannot be handled by every elementary gain-loss argument below.
Writing , wheremakes the physical meaning clear: gain into velocity competes with loss out of . Along a free characteristic curve, put . Then . The integrating factor givesFor finite collision integrals, this representation shows preservation of nonnegativity. It is implicit, because both the gain and damping depend on ; it does not alone prove existence.
The weak collision identity and conserved quantities. The weak Boltzmann collision identity follows from particle interchange and the measure-preserving pre/post-collision substitution: it gives, for a velocity test function ,The factor comes from averaging the two particle labels. Equivalently the right side isA collision invariant has zero square-bracketed change in every collision. The functions , the three velocity coordinates, and are collision invariants by the geometric identities above. Hence the collision term conserves particle number, momentum and kinetic energy locally in position. Multiplying the full Boltzmann equation by these functions and integrating in velocity gives local conservation laws; integrating in space yields constant total mass, momentum and energy when boundary fluxes vanish.
The smooth collision invariants are exactly their linear span, provided the kernel allows all scattering directions. Here is a direct proof. If is unchanged by every collision, then for fixed center and relative radius , the sum is independent of the unit vector . Taylor expansion at forces to be independent of , so . Mixed derivatives vanish. Differentiating in a direction and commuting derivatives gives . Thus is constant andThis proof can be read for invariants; weaker regularity is handled distributionally. The multidimensional scattering hypothesis matters: in one dimension equal-mass elastic collisions only exchange the two velocities, and every function is then a collision invariant.
Entropy dissipation and the H theorem. Define the Boltzmann H functional , with . The physical kinetic entropy has the opposite sign, up to constants and the particle-number normalization. Particle-number conservation removes the in the derivative of ; spatial transport contributes only a boundary flux. With and , the symmetrized weak collision identity givesThe sign follows from monotonicity of the logarithm: . Zeros are treated by the usual extended nonnegative limit. This is the Boltzmann H theorem for the present kernel normalization, and is the Boltzmann entropy dissipation.
For a positive smooth and a kernel positive on all collision directions, equality requires in every collision, so is a collision invariant. Finite velocity mass then forces a negative quadratic coefficient and gives a local Maxwellian:Here is number density, is mean velocity, and is the temperature expressed in velocity-squared units, . Completing the square proves the formula; conservation of momentum and kinetic energy makes , so . Vacuum is another collision equilibrium.
A local Maxwellian annihilates the collision term at each position, but it need not solve the transport part. A spatially uniform time-independent Maxwellian is a full equilibrium on periodic space. On all of it has infinite total mass unless it is vacuum. Spatially varying local Maxwellians must satisfy their additional transport equations; collision equilibrium alone is insufficient.
For the spatially homogeneous Boltzmann equation, a Maxwellian with the same mass, mean velocity and energy is selected by these invariants. The relative entropy is nonnegative. Indeed, writing and using equal mass givesMoreover is a collision invariant, so the derivative of this relative entropy is . Thus minimizes at fixed conserved moments. Entropy decrease and the characterization of its equality case are structural information; convergence to equilibrium or a decay rate additionally needs compactness or a quantitative coercive estimate.
Moment equations and fluid closure. LetThe tensor is the kinetic pressure tensor and the kinetic heat flux. Put . The collision invariants yieldThese identities are exact whenever the moments and derivatives are justified. They are not a closed fluid system: and contain information beyond . For a local Maxwellian, Gaussian integration gives , and . Substitution gives the compressible Euler equations for a monatomic ideal gas, with pressure and ratio of specific heats .
This connection is naturally organized by the Knudsen number , the ratio of the mean free path to a macroscopic length. In an appropriate fluid scaling the equation reads . Formally, as , the dominant collision term imposes , hence local Maxwellian form, and the conserved moments evolve by the Euler equations. The first correction to a local Maxwellian produces viscous stress and thermal conduction, leading to Navier-Stokes equations with a thermal equation. This is a formal hydrodynamic limit of the Boltzmann equation; a rigorous limit requires estimates, preparation of data and treatment of boundaries, and is not justified merely by inserting .
Linearization and its dissipative structure. Around a fixed positive Maxwellian , write . The linearized collision operator on the relative perturbation isThis follows by expanding the two quadratic products to first order and using . In the weighted Hilbert space with inner product , collision symmetrization givesThis dissipation form of the linearized Boltzmann operator and its corresponding polarized identity prove symmetry on a common domain where the integrals converge. Its nullspace is spanned by under the same nondegenerate scattering assumption, exactly the conserved modes. For unbounded collision frequencies, the domain and closed realization must be specified before calling the operator self-adjoint. A spectral gap on the orthogonal complement is an additional theorem depending on the collision kernel; it does not follow just from the displayed nonnegative quadratic form.
Analytic scope and irreversibility. With an angular cutoff and bounded collision kernel, the spatially homogeneous equation has a simple local well-posedness argument in velocity . The pre/post-collision substitution gives , so the quadratic map is locally Lipschitz. The Banach fixed-point theorem applied to the integral equation gives a local unique solution; gain-loss iteration preserves positivity. For nonnegative solutions the conserved mass keeps the norm bounded, allowing continuation for all time. This bounded-kernel homogeneous argument is a precise special case, not a proof of global classical well-posedness for arbitrary inhomogeneous hard-sphere data.
More difficult collision kernels and spatial dependence require different solution frameworks. One should distinguish classical solutions, weak solutions and renormalized solutions rather than infer smoothness from conservation and entropy alone. For instance, a renormalized Boltzmann solution uses nonlinear bounded transforms to make sense of collision and transport terms when their raw integrability is insufficient. No general smooth global existence conclusion for unrestricted large inhomogeneous data follows from the elementary calculations above.
Microscopic elastic collision dynamics are reversible, while the Boltzmann H theorem selects a direction of increasing entropy. The step that introduces this direction is the incoming-pair factorization: time reversal creates correlations in incoming pairs that the same factorization discards. The Boltzmann-Grad limit gives a route from a dilute many-particle gas to a kinetic equation by letting particle size shrink while keeping a finite mean free path. Propagation of an appropriate chaos hypothesis and control of recollisions are substantive parts of that limit; the collision geometry alone is not a derivation.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 6 5 d Solution Created 2026-10-03 Updated 2026-10-07
Strict positivity throughout phase space is incompatible with compact support. The following calculation uses positive smooth rapidly decaying solutions with all displayed integrals justified; nonnegative cases use the corresponding entropy limits where justified.
Write and use integration over . The two symmetrized weak collision identities areandThey follow from particle interchange and the angular exchange for elastic collisions. They are the weak identities needed here; no invalid fixed-angle Jacobian is used.
The collision invariants , and obey . Momentum follows from , and energy from . HenceFor the Boltzmann H functional , the spatial transport contribution is a boundary divergence and the term is zero. Set , . The second weak identity with gives the Boltzmann H theoremIndeed this integrand with its negative sign is for , which is nonpositive. The kinetic H functional decreases; the physical entropy with opposite sign increases.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 42 1 ii Solution Created 2026-10-03 Updated 2026-10-06
A first-order phase transition has a discontinuity in a first derivative of the equilibrium free energy, such as the entropy or order parameter. It can have latent heat when the entropy jumps. A continuous phase transition has a continuously vanishing order parameter and no latent heat, with singular higher derivatives and a diverging correlation length. The LG theory compares global minima, not merely the points where a local minimum loses stability.
For the uniform quartic Landau free energythe equation of state is . At zero field the stable minimum is for , and for . Thus tuning through zero gives a continuous transition. For a fixed , varying through zero instead switches between the two ordered minima and makes jump, giving a first-order phase transition in the conjugate field.
A temperature-like first-order transition can occur at zero field when and a positive sextic term stabilizes the potential. Write . Stationarity of a nonzero phase gives , while equality with gives . Solving these two conditions yieldsThe order parameter jumps from zero to . At this pointso the competing stationary points are genuinely global minima. This phase coexistence condition differs from the spinodal points and , which mark loss or creation of local stability, not equilibrium coexistence.
If the symmetry permits a cubic term , a positive quartic coefficient does not preclude a first-order transition. For with , stationarity and coexistence give and . This is the first-order transition in a cubic-quartic Landau potential; the symmetry restriction on the expansion is therefore part of the prediction.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 42 2 Solution Created 2026-10-03 Updated 2026-10-06
Work in a translation-invariant pure thermodynamic phase, and write to distinguish inverse temperature from the order-parameter exponent. The connected correlation function isAway from criticality in the massive scalar order channel, it decays exponentially, possibly multiplied by an algebraic factor. An exponential correlation length is defined by when that limit exists. At a continuous transition diverges and the critical decay is algebraic. In an ordered symmetric mixture, a nondecaying contribution can remain; selecting a pure branch prevents confusing it with connected critical fluctuations.
Let and . Differentiating the partition function with the energy term gives the correlation-function susceptibility sum ruleIf the source is instead the dimensionless , the explicit inverse-temperature factor is absent. In continuum physical coordinates the lattice sum is approximately .
Choose a normalized blocking kernel such that , replacing the sum by an integral for continuous variables. A deterministic example is a product of delta functions imposing that each coarse variable equals the average of spins in a block of sites, with its field normalization included consistently. Define the effective Boltzmann weight byKernel normalization makes the blocked partition function exactly equal to the original. Its coarse-grained variables retain the long-distance observables through their defining relation to the original fields. An exact RG step generally generates all symmetry-allowed operators; keeping only a small coupling set is an approximation, not an exact closure assumption. The geometrical coarse lattice hasso its physical volume is unchanged. A subsequent coordinate rescaling may restore the numerical lattice spacing to its initial value, which is the equivalent rescaled-coordinate convention.
Here define as free energy per lattice site, . Exact partition-function invariance givesBecause multiplies the identity operator, a blocking step has , where is the generated constant per blocked site. With this impliesThe identity-operator contribution to renormalization-group free energy includes the eliminated modes' entropy, connected vacuum terms and field-measure normalization. It cannot be discarded when calculating an absolute free energy, even though it cancels out of normalized correlation functions.
There is a useful convention behind the word “singular” in this inhomogeneous equation. After removing , still contains a regular coupling-dependent background. If and , then the genuinely nonanalytic part obeys . Thus the printed inhomogeneous representative and homogeneous singular scaling below are consistent after background subtraction. At resonances or marginal points this subtraction may leave additive or multiplicative logarithms; a strictly homogeneous pure-power form is then qualified accordingly.
A renormalization-group fixed point satisfies . Diagonalizing the linearized map gives scaling coordinates . For increasing coarse-graining length, defines a relevant operator, an irrelevant operator, and a marginal operator, whose fate needs nonlinear analysis. Here the are logarithmic scaling exponents: the eigenvalues of the discrete Jacobian matrix are , not the numbers themselves. The critical surface is the stable manifold flowing into the critical fixed point after every relevant scaling field is tuned to zero. A repulsive renormalization-group trajectory leaves the fixed point along relevant directions and flows toward a different long-distance phase.
For the requested two relevant fields, take and , with . Ignoring nonsingular irrelevant corrections and choosing a total blocking scale , the homogeneous singular free-energy density obeysChoose so the renormalized thermal field is . This provesMetric factors and regular analytic redefinitions of the scaling fields only change amplitudes. Likewise , so at the correlation-length critical exponent is . Two thermal derivatives of give the heat-capacity critical exponent , proving the hyperscaling relationTwo field derivatives also give ; one field derivative gives , and using to set the blocking scale on the critical isotherm gives . These statements assume that no dangerously irrelevant coupling makes the scaling function singular as it is removed. In particular the naive hyperscaling form need not describe the quartic ordered phase above four dimensions, where its stabilizing quartic coupling is dangerously irrelevant. That restriction reconciles this derivation with the mean-field exponents and Question 3.
The scaling form of the connected correlation function is in the continuum long-distance regime. For , tends to a finite nonzero constant. For , it decays exponentially with a possible algebraic prefactor; an isolated massive pole gives the familiar Ornstein--Zernike correlation function tail with a temperature-dependent amplitude. Exponential decay is the essential feature; the same critical algebraic power need not remain the exact large-distance prefactor.
Using the susceptibility sum rule and spherical integration givesFor and an ordinary exponentially decaying scaling function the integral tends to a finite constant: it converges at zero because is finite, and at infinity because of the exponential decay. Microscopic distances add a regular background. Thus , establishing the Fisher scaling relation
For a continuous field, field scaling and anomalous dimension giveThis field renormalization is additional to the canonical engineering factor . It makes . Invariance of the uniform source coupling then gives , agreeing with the susceptibility exponent above. With a convention , the fixed-point choice is ; the sign of the logarithmic derivative depends on this stated convention.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 75 1 a Solution Created 2026-10-03 Updated 2026-10-06
An acoustic analogy is an exact rearrangement of the fluid equations into a chosen linear propagation operator acting on an acoustic variable, with everything left over placed on the right as effective forcing. The rearrangement becomes a sound-prediction method only after a reference medium, boundary conditions and approximations to the forcing are specified. In particular, a right-hand-side term need not represent independently generated sound.
Use Einstein summation convention and write . Differentiating the continuity equation in time and taking the divergence of the momentum equation eliminates :Set and . Since , the constant has zero Laplacian. The prescribed reference mass density and reference speed of sound are independent of time, soCombining the two identities givesOnly conservation of mass and conservation of momentum have been used; no equation of state or energy equation was needed. Spatial variation of creates no omitted derivative in this identity, because it multiplies a time derivative. The double divergence of the momentum flux tensor has the structure of an acoustic quadrupole.
For a localized flow, choose the reference fields to match the stationary surrounding medium: and outside the flow, with chosen so there. A uniform surrounding fluid permits ambient constant mass density and adiabatic sound speed, giving the familiar homogeneous wave equation. A nonuniform surrounding fluid calls for its actual stationary reference profiles, extended sensibly through the flow region. This makes the acoustic variable vanish in the unperturbed exterior and minimizes artificial contrast terms. In a uniform isentropic exterior the leading acoustic relation also makes vanish to first order. In a stratified exterior, propagation and entropy-advection effects can remain in , as the next part demonstrates.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 75 1 b Solution Created 2026-10-03 Updated 2026-10-06
At rest, all time derivatives and advective terms vanish. The continuity equation and the stated energy equation are then identities, while the momentum equation reduces to . The reference state need not have uniform mass density or entropy. Assume smooth positive reference pressure and mass density, and retain only first-order perturbations.
The linearized continuity equation, momentum balance and adiabatic energy relation areThe perturbation of multiplies a vanishing reference material derivative, so it does not enter at first order. Differentiate the last equation in time, then substitute the linearized momentum equation:Expanding the divergence yields the stratified acoustic pressure equationFor a perfect gas, . Put . ThenThus the divergence-form stratified acoustic pressure equation isThis is homogeneous linear propagation through the static medium, including its inhomogeneity. In the acoustic analogy of part (a), has no first-order contribution when viscosity is neglected and the reference velocity is zero. Nevertheless the chosen left-hand operator lacks the gradient terms in the genuine propagation equation. Those effects must consequently appear in . They describe propagation of an existing disturbance, rather than an independent source. Since reference-field choices also change the division between the operator and forcing, cannot be identified unambiguously with newly generated noise. Its presence in the right-hand side is a consequence of the chosen analogy, not a source-classification theorem.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 75 3 b Solution Created 2026-10-03 Updated 2026-10-06
Let , with , and set . Direct differentiation givesThe heat equation implies , so the bracket vanishes. Conversely its being independent of can be absorbed into a -dependent multiplicative normalization of , leaving unchanged. This proves the Cole-Hopf transformation with the positive sign appropriate to the negative-flux Burgers convention. Although the algebra works for nonzero , the given forward Gaussian function diffusion kernel and a physical vanishing-viscosity limit require .
Take for the stated Burgers N-wave. Integrating and normalizing the exterior value to one givesThis function is continuous at ; its logarithmic derivative has the specified jumps. Convolution with the heat kernel is positive and solves the heat equation for . Splitting the integral into the exterior baseline and the interior correction yieldsPut . Completing the square givesChanging the interior integration variable to changes its limits to and the Gaussian function width to . The Jacobian and normalization leave the factor . Thus the Cole-Hopf solution for a Burgers N-wave isHere is the normalized Gaussian function mass of its indicated interval. Its explicit error function representation isFor fixed , the Gaussian function concentrates at as . Consequentlyaway from the endpoints; at the limit is . The transition layer has width . This is an approximate identity argument, not a uniform step approximation across the endpoints.
For fixed and , both interval masses tend to one, and the exponentially large positive dominates the exterior correction. Its logarithmic derivative therefore givesFor , both interval masses are exponentially small and the weighted interior integral is also negligible, while the exterior contribution tends to one. Therefore in the specified far exterior.
An exponentially weighted Gaussian function tail should not be discarded solely because its unweighted interval mass tends to zero. In fact, comparing the order-one exterior term with gives the sharper inviscid Burgers N-wave fronts , not . Away from these fronts, the vanishing-viscosity limit isInside , this follows from the sign of in the exponential. Outside , the constrained Gaussian function maximum lies at an interval endpoint and has negative exponent. The right shock wave's speed is , equal both to the derivative of and to minus half the sum of its two limiting states; the left shock wave is its reflection. This checks consistency with the Rankine-Hugoniot condition and shows that the requested near-center and far-exterior approximations are compatible with the full entropy limit.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 58 2 a Solution Created 2026-10-03 Updated 2026-10-06
For a uniform-density stellar model, and . With zero surface pressure, integrate the stellar hydrostatic equation:Using the ideal gas equation of state, . ThereforeEquivalently and . The constant mass density is imposed as a structural approximation; a perfect gas can realize it hydrostatically by allowing its temperature and entropy to vary with radius.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 58 3 b Solution Created 2026-10-03 Updated 2026-10-06
The stellar adiabatic exponents are fixed-composition, constant-specific entropy derivatives:The chain rule gives . Define the pressure derivatives and . The first law of thermodynamics, with , givesCombining this with on an adiabat givesThus the specific-heat ratio is not generally equal to the three stellar adiabatic exponents.
For the mixture, and . The specific heat capacity at constant volume, obtained by differentiating at fixed mass density, isTherefore the adiabatic exponents of a monatomic gas-radiation mixture and its specific-heat ratio areIf a relation involving only the exponents is wanted, eliminate from :For a pure monatomic perfect gas, and . For any calorically perfect gas with constant heat capacities, the same equality holds with its own . In a genuine gas-radiation mixture, , while the exponents are given separately above.
In the radiation limit, and the adiabatic temperature gradient is . This follows independently from photon entropy: a comoving volume has , so an adiabat obeys and . However, is singular in the pure-radiation limit, not . This is the pure-radiation constant-pressure heat-capacity singularity. The equation fixes temperature whenever pressure is fixed, so an ordinary constant-pressure temperature derivative is not available; along the mixture limit and . For photons alone, mass-specific quantities additionally require a material mass label. The often quoted radiation index is its pressure-density adiabatic exponent, not a finite constant-pressure/constant-volume heat-capacity ratio.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 59 3 a Solution Created 2026-10-03 Updated 2026-10-06
For a fixed polytropic index and fixed equation-of-state constant , let be the first zero of the regular Lane-Emden equation solution. A finite-radius model requires such a zero; for the usual nonnegative indices this holds for . The surface radius and mass follow from and :The last equality integrates the Lane-Emden equation; define the positive Lane-Emden surface mass constant . Substituting givesThe central-density exponents cancel when the required powers are taken. Thus the polytropic mass-radius relation isIt is independent of , with and composition held fixed. For the radius is fixed; for the mass is fixed. These limiting powers need no division by a vanishing exponent.
For , the literal pressure-density power and the supplied expression for are singular. Interpret this case separately as an incompressible planetary interior with constant density. Then , or at fixed density. A weakly compressed rocky body, or a rough uniform-density approximation to Earth, is an example; realistic terrestrial planets are stratified and compressible.
For , . This describes a cold nonrelativistic degenerate electron gas of fixed composition, or a fully convective monatomic ideal gas at fixed entropy. Examples are a nonrelativistic white dwarf, a sufficiently cooled partly degenerate brown dwarf as an approximation, and an approximately fully convective low-mass star for the ideal-gas version. The fixed- scaling is . It must not be applied to an entire main-sequence stellar sequence with different entropies; the entropy dependence of a polytropic mass-radius relation explains the distinction.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 59 3 b Solution Created 2026-10-03 Updated 2026-10-06
The mass-radius curve of solar-composition substellar objects reflects the transition from weak compression to pressure ionization and electron degeneracy pressure, followed by sustained hydrogen burning. A schematic joining representative object classes is:
Schematic mass-radius sequence from ice giants through gas giants and brown dwarfs to low-mass stars
. The illustration is not an age-specific numerical evolutionary model. In particular, ice giants contain much more heavy material than a solar-composition giant, so a single uniform-composition equation of state does not describe the entire joined curve.
- At the low-mass, weak-compression end, a fixed-density or fixed-composition approximation gives . Ice giants such as Neptune and Uranus have substantial water/rock-rich interiors and modest H/He envelopes; changing envelope fraction changes the radius markedly. Their heat comes from retained formation energy, contraction and radiogenic heating of heavy material. Fluid interiors generally convect, while composition stratification can impede mixing; outer radiative layers release the heat.
- Ordinary gas giants reach radii of order over a broad range around Jovian masses: an effective polytrope explains the approximate segment. Increased mass compresses material enough to offset the added volume. Cooling and Kelvin-Helmholtz contraction, with additional differentiation energy such as helium settling in Saturn, supply the intrinsic luminosity. Their deep envelopes are usually convective, with radiative photospheres.
- More massive brown dwarfs become increasingly supported by electron degeneracy pressure. The cold nonrelativistic limit gives , but finite entropy and Coulomb effects flatten actual giant/brown-dwarf curves and their radii depend on age. They cool and contract; temporary deuterium fusion occurs above a composition-dependent deuterium-burning mass near . This threshold does not cause a sharp structural kink or permanent stellar luminosity. Interiors are largely convective and surface emission is radiative.
- Near the hydrogen-burning minimum mass, roughly – or – for near-solar composition, sustained fusion prevents indefinite cooling into a degenerate object. The low-mass main-sequence star branch turns upward, with approximately over the illustrative interval. Its entropy is not constant across masses, so this branch is compatible with an approximately internal profile. Hydrogen fusion through the proton–proton chain provides energy; the lowest-mass main-sequence stars are fully convective, capped by radiative atmospheres.
Planet/brown-dwarf naming conventions and deuterium burning do not define a universal discontinuity in the mass-radius relation. Composition, age and irradiation move the curves; the hydrogen-burning transition changes the long-term energy source more fundamentally.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 59 4 j Solution Created 2026-10-03 Updated 2026-10-06
The hot-Jupiter radius inflation problem is that some strongly irradiated giant planets have radii much larger than standard age-, mass- and composition-dependent cooling models predict. Greater internal entropy generally means a larger radius at fixed mass. The two broad classes of explanation are retaining existing heat by delaying cooling and depositing additional energy into the deep planet.
- Enhanced atmospheric opacity slows radiative leakage and keeps the deep interior hot. Required enrichment or persistent cloud opacity must be compatible with composition and spectra; adding heavy material also tends to increase density. Insulation can preserve initial heat but cannot necessarily reinflate an already cooled planet.
- Layered convection in a giant planet uses a stabilizing composition gradient and double-diffusive layers to reduce heat transport. The needed gradient and layer structure must survive mixing, and their efficiency is model-dependent; very inefficient transport cannot simply be assumed for every planet.
- Tidal heating dissipates orbital or spin energy, often requiring maintained eccentricity or obliquity. Nearly circular, synchronized planets have little of the simplest eccentricity-tide power, so a pumping mechanism or different tidal configuration is needed to explain them.
- Ohmic heating of a giant planet dissipates wind-induced electric currents in a conducting, magnetized atmosphere and interior. It requires appropriate ionization, conductivity, magnetic field, wind speeds and sufficiently deep deposition; magnetic drag can limit the available mechanical power.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 303 1 a Solution Created 2026-10-03 Updated 2026-10-06
Use inverse temperature , reserving without a subscript for the critical exponent. Write ; each undirected bond occurs once, so the number of bonds is . In the mean-field theory of the Ising model, write each Ising spin as and neglect the product of fluctuations:The resulting independent-spin statistical Hamiltonian isThe constant corrects the double counting of interaction energy. The Ising spin sums now factorize and can all be evaluated:This is the approximate partition function at an assumed mean field; equilibrium fixes self-consistently. The corresponding Ising auxiliary mean-field free energy isIts equilibrium value gives the mean-field Helmholtz free energy in the imposed field. Away from a stationary point its parameter is an assumed field variable, not necessarily the actual mean Ising spin of that independent-spin distribution.
For the full small- expansion at fixed , set and . Differentiating gives successive derivatives , , , and at . ThereforeAt zero field, spin inversion symmetry eliminates odd powers and this simplifies toThe quadratic coefficient changes sign and the quartic coefficient is positive atThus the zero-field mean-field prediction is a continuous, continuous phase transition: the stable zero spin magnetization develops two symmetry-related nonzero minima continuously below .
Both requested routes give the same mean-field self-consistency equation. First, the one-spin expectation in the effective field isSecond, differentiating the Ising auxiliary mean-field free energy giveswhose stationary condition is precisely that mean-field self-consistency equation. Choose its stable, lowest-free-energy branch rather than every algebraic solution.
An equally useful mean-field approximation parametrizes the trial distribution by its actual mean Ising spin. Its probabilities are , giving the Bragg-Williams free energy of the Ising modelThis energy-minus-entropy function has expansionIts stationary equation is , again equivalent to the mean-field self-consistency equation. The two functions differ away from equilibrium, but agree on stationary branches: for and , the entropy bracket equals . Their small- coefficients therefore need not agree at arbitrary temperature; at they have the same leading critical quartic coefficient . This distinction prevents confusing the auxiliary-field expansion with the physical-magnetization variational expansion.
For the order-parameter critical exponent, expand the equation of state at :On a nonzero stable branch,Hence . The spontaneous magnetization is understood by selecting a branch with an infinitesimal field after the thermodynamic limit; a finite symmetric sample has zero exact zero-field mean Ising spin.
For the magnetic susceptibility per site, define with in energy units. Implicit differentiation givesAbove , and . Below , evaluated on a selected ordered branch, the expansion of gives . Thus the magnetic-susceptibility critical exponent is on both sides, with different amplitudes. At , the equation of state becomesso the critical-isotherm exponent is . These are mean-field critical exponents, not a claim that neglecting fluctuations gives the exact Ising transition in every dimension.


