Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 4 9D Solution Created 2026-09-24 Updated 2026-10-14
For an autonomous Lagrangian, the canonical momentum and energy areAlong a solution of the Euler-Lagrange equation,The absence of explicit time dependence is essential to this conservation law.
The classical action is . If the final endpoint is moved along one fixed classical trajectory, while the initial endpoint is held fixed, the fundamental theorem of calculus givesThis total derivative is different from the partial derivative with fixed final position.
For a family of nearby classical trajectories, integration by parts gives the endpoint variationTo obtain the final term, the endpoint change in the variation at fixed parameter time is , while the changing upper integration limit contributes . The interior integral vanishes by the Euler-Lagrange equation. Therefore, on any smooth branch of classical boundary-value solutions,The chain rule along the endpoint trajectory now reads , consistently with the direct calculation.
Past exam of the mathematics course of the University of Cambridge 2011 ii Paper 1 9C i Solution Created 2026-09-24 Updated 2026-10-14
The canonical momentum is . Its time derivative contains the total derivative along the trajectory: . The Euler-Lagrange equation therefore givesThe magnetic field is , so the last term is . With the electric field , this is precisely the Lorentz force law:
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 46 3 Solution Created 2026-10-03 Updated 2026-10-07
For fields with a first-derivative Lagrangian density, suppose an infinitesimal global transformation at fixed coordinates is and , with constant infinitesimal . Noether theorem gives the on shell conserved currentThis follows by varying the Lagrangian, integrating the terms with by parts, and setting the Euler-Lagrange expressions to zero. If spatial boundary flux vanishes, is a conserved Noether charge. Spacetime symmetries are covered by including the induced field variation at fixed coordinates and the corresponding total derivative .
An example of a theory with a Lorentz four-vector is the source-free Maxwell field with electromagnetic four-potential . Under an Abelian gauge transformation,The change in is , so the Maxwell Lagrangian is gauge-invariant. For define and . With signature ,The minus sign in gives the positive electric kinetic term and positive physical field energy; reversing it would reverse the Hamiltonian's sign for the physical modes. The normalization accounts for antisymmetry and yields the conventional equations. Varying gives .
For time translation at fixed coordinates take . Since is constant, , and hence . Thus . The momentum derivative isNoether theorem therefore gives the canonical stress-energy tensor's time-translation currentHere was used. The canonical momentum of vanishes, while that of the lower spatial component is : acts as a multiplier for Gauss law, not a propagating degree of freedom. The density above is not manifestly gauge-invariant, but its charge agrees with the physical energy on the constraint surface and with appropriate boundary conditions.
The gauge-covariant time translation instead has . It is time translation combined with a gauge transformation of parameter . Directly,using the definition of , equivalently its Bianchi identity. The total derivative in is unchanged. The improved current isIt is the gauge-invariant Maxwell stress-energy tensor. In particular,The spatial current is the Poynting vector , so conservation gives . This is the familiar positive local electromagnetic field energy density, expressed entirely in measurable fields.
To compare the currents,The last term vanishes on the source-free Maxwell equations, leaving a stress-energy tensor improvement generated by a stress-energy superpotential. At the charge level,when the surface term vanishes and Gauss law holds. Thus the symmetries yield the same conserved energy under these conditions while the second supplies a gauge-invariant local density. Boundary terms would have to be retained if those boundary conditions were not imposed.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 47 2 Solution Created 2026-10-03 Updated 2026-10-07
Take and use the complex sine-Gordon theory reduced density , with physical lengths and times . Physical rest energy is , and the dimensionless Noether charge is obtained from the reduced action. Put , , , so . Initially take , avoiding the singular target-space boundary.
The canonical Hamiltonian density isHere , , and . Since , the kinetic contribution is simply . ConsequentlyTo fix the charge sign, take the Noether current directly from , :Thus positive internal rotation has positive charge in this convention. With and the supplied integral,Reversing the definition of the generator would reverse all charge labels but not any mass. The limit is the vacuum with zero mass and charge. At the field reaches at its center and the current's coordinate expression becomes singular. The mass still has the finite limiting value , butThe displayed local density does not assign a unique endpoint charge merely by substituting .
The four collective parameters. Translation, a Lorentz boost of rapidity , and the global phase symmetry produceThe independent real parameters are , with periodic modulo . A further time translation is absorbed into these parameters. The energy and momentum are and , while is invariant. The internal angle is the collective coordinate whose canonical momentum is this Noether charge.
Semiclassical quantization. For one rest-frame internal rotation, the period in dimensionless time is . The action around the orbit isEquivalently this is , where is the reduced rest energy. Bohr-Sommerfeld quantization with gives . Because the internal orbit is a cyclic angle, not a one-dimensional oscillator between turning points, there is no half-integer Maslov index shift. HenceOn the nonsingular branch . The elementary charge-one mass approaches at weak coupling. For positive with , , where , . Thus : the charge- soliton cannot fragment into smaller like-charge particles. For a general charge-conserving partition, a constituent with already has mass at least ; if every , the decreasing ratio gives , strictly for nontrivial fragmentation. Thus opposite-charge constituents cannot open a lower-energy decay either. This is charged rather than topological stability, and the semiclassical spectrum is reliable at weak coupling.
Counting states and the endpoint. Write . If is not an integer, the regular branch givesIf is an integer, the local nonsingular calculation gives only regular states. The formal closed-branch list has charge labels, but its last two labels are limits of the same singular static solution. Counting them needs a global completion; it cannot be decided from a regular current integral at .
In the additional integer-level completion, at leading semiclassical order and charges are identified modulo integer . There are then nonzero charge sectors. For odd , positive and negative labels are distinct; for even , the endpoint labels describe one self-conjugate state. This supplies states when , rather than endpoint labels. This global interpretation, and the distinction between classical and renormalized couplings, are discussed in Dorey and Hollowood, section 2. A generic classical does not automatically define an integer or specify this quantum completion.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 57 1 i Solution Created 2026-10-03 Updated 2026-10-07
Use signature and write for the spatial induced metric, for its spatial covariant derivative, and . The negative shift vector convention gives and . Decomposing the scalar field derivative givesVarying the scalar field matter action with respect to the inverse metric tensor, including the variation of , gives the stress-energy tensorRaising both indices gives the requested . Contracting with the unit normal twice, or once with the spatial projection tensor, givesProjecting both indices of the stress-energy tensor onto the spatial hypersurface givesTwo printed identifications require correction. The canonical momentum of the displayed density is , not : explicitly and differentiation with respect to supplies . Thus is the normal scalar-field momentum. Also, the printed trace-free stress has an extra factor and uses where the general spatial induced metric requires . In an orthonormal spatial frame the latter becomes , but the coefficient is still one. For a gradient in that frame, direct projection gives , rather than the printed . The PDF's energy-density gradient is spatial ; the TeX aid's spacetime index there is an OCR defect.
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 57 3 a Solution Created 2026-10-03 Updated 2026-10-07
Let , with an inverse spatial Laplacian defined by a boundary or zero-mode prescription. It depends on , not on its time derivative. Differentiation gives the canonical momentumCount as first order. Inverting perturbatively givesFor the Legendre transform in mechanics, write with second order. The quadratic kinetic contribution is , whose correction is fourth order; in the cubic terms one may consequently replace by . The Hamiltonian density is thereforeThe cubic interaction Hamiltonian density isIn the free interaction picture, . Thus here with the time derivatives interpreted as free fields. This equality follows from the cubic Legendre transform for scalar derivative interactions; at higher orders additional terms can appear.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 40 4 Solution Created 2026-10-03 Updated 2026-10-06
For the real scalar field, the mostly-plus Minkowski metric gives a positive kinetic energy. Its canonical momentum and Hamiltonian operator areThe Euler-Lagrange field equation is , or . Canonical quantization of a real scalar field promotes the fields to Hermitian operators and replaces the equal-time Poisson brackets by canonical commutation relations:The Heisenberg equation of motion then gives and , so the classical field equation remains an operator identity. Products at coincident points need a regulator; equivalently, the field is an operator-valued distribution.
Resolve the free field into positive- and negative-frequency plane waves. With and ,Reality pairs the two terms by Hermitian conjugation. The equal-time canonical commutation relations are equivalent toFor example, the two mixed terms in each supply half the Dirac delta function. Thus the normalization is fixed, not optional. Each independent mode is a quantum harmonic oscillator, with and its annihilation operator and creation operator.
Choose the Fock vacuum by . Repeated creation operators construct the bosonic Fock space. Substitution into the Hamiltonian operator yieldswhere a finite volume can be used as a regulator. Normal ordering sets the flat-space vacuum reference to zero. The spatial momentum operator isIn particular, and . A one-particle state therefore has positive energy, momentum , and invariant mass . A scalar field transforms in the trivial spin representation, so its particles have spin zero. Because the field is real there is only one set of oscillators: the particle is its own antiparticle, with no distinct conserved particle-minus-antiparticle charge. The free theory does conserve its occupation-number sum, though generic scalar interactions need not.
A momentum eigenstate extends throughout space. A localized particle is described by a superposition such asIts wave packet evolves through the phase ; a narrow packet has group velocity , whose magnitude is at most one. Localization and dispersion concern these superpositions, rather than classical trajectories attached to individual field modes.
Relativistic spacetime behavior is encoded by microcausality. Directly from the oscillator commutators,This is a Lorentz-invariant distribution and vanishes at equal time. Any spacelike separation can be transformed to an equal-time separation, so for spacelike . Thus local operations at spacelike-separated points commute. The vacuum Wightman function need not vanish outside the light cone, but that correlation does not transmit a controllable signal. The Feynman propagator iswhich propagates positive energy forward and negative energy backward in the time-ordered product. For causal response one uses the retarded Green function, whose support lies in or on the future light cone.
Finally, commuting creation operators implyThe multiparticle state is symmetric under the particle exchange operator, which is Bose statistics. For a normalized single mode, the occupation states are with : any number of identical bosons can occupy it. There is no Pauli exclusion principle for these particles. This construction agrees with the Spin-statistics theorem relating integer spin to bosonic exchange symmetry under relativistic locality and positive-energy assumptions. It demonstrates the required scalar case without assuming that theorem as the quantization prescription.
The quantized real scalar field describes positive-energy, mass-, spin-zero bosons that are their own antiparticles; commuting creation operators give Bose statistics, and local field commutators enforce microcausality.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 43 1 Solution Created 2026-10-03 Updated 2026-10-06
Use natural units and the Minkowski metric . A real scalar field assigns a real variable to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation . With , this is the Klein-Gordon equation. An additional nonlinear part of describes interactions.
The canonical momentum is . The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations and recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.
To see its origin, first consider one coordinate with . Split a time interval into steps of length and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains . This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring replaces the weight by ; stationary phase explains the emergence of classical trajectories.
For the field, use scalar field configuration eigenstates , satisfying . Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral . The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.
For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: , so after normalization only the lowest-energy component remains as . This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With , the Euclidean path integral has the weight , whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.
Introduce a classical source and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down . The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.
The free theory illustrates the method. Its quadratic kernel is with the vacuum pole prescription, and completing the square gives the Gaussian evaluation of a free scalar generating functionalTwo source derivatives give . Higher derivatives give all pairings, the content of Wick theorem. For an interaction , one may use path-integral perturbation by source derivatives:Expanding this expression generates Feynman diagrams and their Wick contractions. The connected generating functional retains connected contributions; in particular . These functionals turn the computation of field-operator expectations into source differentiation of an ordinary regulated integral.
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 2 12C a Solution Created 2026-09-24 Updated 2026-10-06
The canonical momentum is , and a regular Legendre transform in mechanics defines after solving for . Its differential isbecause the terms cancel. With the Euler-Lagrange equation , this gives Hamilton's equationsIn particular at corresponding velocities, so a cyclic coordinate for the Lagrangian is also cyclic for the Hamiltonian.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 301 1 Solution Created 2026-10-03 Updated 2026-10-14
Use units , the Minkowski metric , and the Fourier transform convention . Here gamma matrices satisfy , and is the Dirac adjoint. A single Dirac field describes both Electrons and Positrons; these are the particle and antiparticle sectors of that field.
The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along , so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is . It is independent of and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.
The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If is the photon quadratic operator and , completing the square givesIn the last expression and are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to , and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, , the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.
The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.
The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor ; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.
To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative obeys . Replacing by in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current is conserved by the matter equations, and the Electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.
With , expand in powers of . A term of order contains spacetime integrations and . Applying the Wick theorem pairs the free fields: an - Wick contraction supplies a photon propagator, and a - Wick contraction supplies an oriented Dirac propagator. Each insertion supplies an interaction vertex. The permutations of contractions cancel the expansion factorials except for the Feynman-diagram symmetry factor. Interchanging Grassmann fields produces the fermionic sign, including a minus sign for every closed fermion loop. This is how Feynman diagrams arise from expectation values, rather than an extra dynamical assumption.
The normalized vacuum generating functional removes components with no external insertions. In the operator formalism, for an interacting-vacuum expectation value of an inserted product , the same cancellation appears asVacuum projection and the Feynman i-epsilon prescription are implicit. The linked-cluster theorem exponentiates all connected vacuum bubbles into the same factor in numerator and denominator, so it cancels. Normalize by to remove every vacuum component. This cancellation of vacuum bubbles still leaves products of disconnected diagrams that each contain external insertions. If only connected correlators are wanted, differentiate the connected generating functional .
The resulting momentum-space QED Feynman rules for the bare theory can be stated in Feynman gauge as follows:
- An internal oriented fermion line of four-momentum contributes .
- An internal photon line contributes .
- A one-photon two-fermion interaction vertex contributes , with its spinor and photon indices attached to the incident lines.
- Each interaction vertex conserves four-momentum. With all incident momenta treated as incoming, include ; after using these constraints, integrate every independent loop four-momentum as .
- Contract the gamma matrix and Dirac propagator factors in their order along the fermion line. A closed fermion loop has a trace and a factor . A relative odd permutation of external fermions also contributes .
- Divide each labelled diagram by its Feynman-diagram symmetry factor and sum the allowed diagrams. Omit vacuum components as explained above.
- For an amputated scattering amplitude, an incoming Electron has and an outgoing Electron has ; an incoming Positron has and an outgoing Positron has . An incoming photon has and an outgoing photon has . The external polarization vectors are physical and transverse, and external four-momenta are on shell. This last step follows from the LSZ reduction formula and amputation of external propagators; an unamputated correlator keeps its external quantum field theory propagators.
The illustrated interaction vertex has incoming fermion momentum , incoming photon momentum , and outgoing fermion momentum . Its solid-line arrows indicate fermion flow. The photon wavy line has no fermion-flow arrow. The propagators, the vertex , momentum conservation, and the fermionic signs determine the perturbative amplitudes.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 301 3 Solution Created 2026-10-03 Updated 2026-10-06
Use and . The free real scalar field has actionThe canonical momentum is , and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes and to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives , , hence the Klein-Gordon equation .
Let and . The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts . Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor , the corresponding coefficient is a difference of those square roots and multiplies ; it vanishes because . Taking the adjoint gives . Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies , and normal ordering gives , after removing the constant zero-point energy. A one-particle excitation has energy and spin zero.
The Feynman propagator is the vacuum expectation value of the time-ordered product of two field insertions. For , the field at creates a one-particle excitation from the vacuum and the field at annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the - contraction survives, so with and ,Here is the Heaviside step function, and changing to in the second term gives the last line.
For the Fourier transform convention , insert a positive damping factor and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for , and above for , reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, has derivative jump at , soIt is the expectation-value convention for , including the numerator , that determines this source normalization.
The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 301 1 b Solution Created 2026-10-03 Updated 2026-10-05
Write . The additional gauge fixing variation is . Integration by parts gives the Euler-Lagrange field equation , hence . This equation follows from the gauge-fixed density without imposing as a separate operator identity.
Direct differentiation of the density as printed gives the canonical momentum components conjugate to the lower-index fields:There is a boundary-term convention to reconcile with part (c). The Feynman-gauge Maxwell kinetic density after a boundary-term subtraction isIt has the same Euler-Lagrange field equations, but its canonical momentum components are . The mode expansion supplied in part (c) is the expansion of these latter momenta. Thus it is valid after the stated boundary-term subtraction; it is not the direct derivative of the original density. For example, a time-independent spatially varying with gives but .
The boundary-term shift of canonical field momenta is a canonical transformation. If spatial boundary terms vanish, , and . This explains why the two conventions have the same dynamics while their momentum formulas differ.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 320 4 Solution Created 2026-10-03 Updated 2026-10-14
For a regular Lagrangian define the canonical momentum components . Require the velocity Hessian to be invertible locally so that the velocities can be expressed in terms of . The Legendre transform in mechanics definesThe explicit time argument cannot generally be omitted when the given Lagrangian depends on time. Its differential is , since the terms cancel. The Euler-Lagrange equations give , and comparison yields Hamilton's equations:For the time-independent stellar gravitational potential below, is conserved. Its sign convention differs from the binding energy in question 2: here is the ordinary specific orbital energy.
The prescribed nondegenerate prolate spheroidal coordinates require ; equality collapses the coordinate family. Their physical velocity components are , and . Per unit stellar mass the Lagrangian is , so . Its Hamiltonian isThis uses acceleration , as specified, rather than importing the opposite gravitational potential convention.
For the Stäckel potential, use the Hamilton-Jacobi equation with principal function . Axisymmetry makes constant. Write and . Multiplying by and usingseparates it as , where . Each side depends on its own coordinate only and therefore equals a constant, chosen as . Here is the generating function of the gravitational potential, not the gravitational constant. HenceThis proves the conserved axisymmetric Stäckel third integral. Eliminating between the two expressions and using givesAt regular generic phase-space points the three separation constants are independent; special circular or degenerate orbits can reduce their local independence. Thus the model is an integrable Hamiltonian system.
For the intended regular, phase-mixed steady model, Jeans theorem permits a galactic distribution function . At any fixed point, and depend on and only through their squares, and depends only on . Thus is even separately in and , whether or not it is even in . Odd integration proves velocity-ellipsoid alignment in a Stäckel potential:The argument allows azimuthal streaming: it does not require zero mean . The means in the two meridional directions vanish, so the off-diagonal velocity covariance also vanishes. Second moments must exist for this conclusion.
There is a hypothesis behind the printed universal wording. On a generic nonresonant regular invariant torus, a continuous steady distribution is constant along a dense orbit and hence depends only on the torus integrals; this justifies the phase-mixed form. Bare stationarity alone is not sufficient in a degenerate resonant gravitational potential with additional integrals. For example is a Stäckel Hamiltonian: choose , up to a constant vacuum shift. The additional conserved quantity gives a positive normalizable steady tracer for . Its local . At azimuth zero and a regular point where the direction has a radial component, this gives nonzero . This tracer need not itself source the harmonic gravitational potential. It exhibits why the stronger, ordinary phase-mixed three-integral assumption is necessary instead of claiming that every imaginable steady distribution has the asserted alignment.
The geometry follows by multiplying the coordinate-defining equation and evaluating at :Constant surfaces are prolate spheroids, with equatorial semiaxis and polar semiaxis . Constant interior surfaces are two-sheeted hyperboloids. Their common foci lie at . The orbit volume in an axisymmetric Stäckel potential is determined byIts allowed coordinate intervals have turning boundaries or coordinate endpoints. A typical bound orbit with oscillates between inner and outer spheroids, oscillates vertically between hyperboloidal limits, and circulates about the symmetry axis with . The equatorial fold joins the positive and negative branches; it is not necessarily a physical turning point in .
For three incommensurate frequencies, the regular trajectory densely samples its projected allowed volume, with nonuniform occupation determined by its speeds. It does not follow one coordinate surface, and it need not close. Resonant orbits may be closed or lie in a smaller projected set; confines azimuthal motion to a meridional plane (with the usual axis coordinate qualification), and circular or equatorial limits are also lower-dimensional. Smooth Stäckel separation removes the generic chaotic behavior of nonintegrable potentials; it does not assert that every orbit fills a three-dimensional volume.
Past exam of the mathematics course of the University of Cambridge 2018 iii Paper 306 1 ii Solution Created 2026-10-03 Updated 2026-10-05
The canonical momentum equation is . Eliminating from the phase-space action yields the Lagrangian densityThis is the Polyakov action after identifying its inverse metric tensor density asThe determinant of this matrix is , as required in two dimensions. The Weyl transformation leaves the matrix unchanged, so encode the metric modulo its conformal factor.
Define the induced worldsheet metric by . Varying the independent inverse metric tensor in the Polyakov action givesConsequently , where . For a nondegenerate Lorentzian string worldsheet with the usual compatible time orientation, take , or . The undetermined function is precisely Weyl invariance. Since , elimination of the independent metric tensor givesThus the Nambu–Goto action measures minus string tension times Lorentzian worldsheet area. The metric elimination holds in the nondegenerate interior; null endpoints are understood as boundary limits.
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 4 15B Solution Created 2026-09-24 Updated 2026-10-03
Define the canonical momentaWhen the velocity Hessian is nonsingular, these equations can be inverted for , and the Hamiltonian is the Legendre transformIts differential isbecause the terms cancel. The Euler-Lagrange equation identifies , so coefficient comparison gives Hamilton's equations
Both and are cyclic coordinates, so and are conserved. The Hamiltonian has no explicit time dependence and is also conserved. Since is independent of ,under the Poisson bracket. Hence three independent first integrals in involution are
Past exam of the mathematics course of the University of Cambridge 2018 ii Paper 4 8B Solution Created 2026-09-24 Updated 2026-10-03
Let a one-parameter infinitesimal transformation be and suppose its change in the Lagrangian is a total derivative,With canonical momentum , the chain rule and the Euler-Lagrange equations give, along a motion,Comparison with the assumed total derivative proves Noether theorem:
The given Lagrangian has no explicit time dependence, so its energy function is conserved:It is also invariant under the dilation : the velocities and all scale by the same factor, while is unchanged. The generator is and . SinceNoether's theorem supplies the second first integralThe integrals and are independent on an open dense subset of phase space.
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 321 2 b Solution Created 2026-10-03 Updated 2026-10-05
The Euler-Lagrange equations of the particle Lagrangian in a shearing sheet areThe terms coupling and are the Coriolis acceleration. The cyclic coordinate has conserved canonical momentumFor a Newtonian potential of a point mass, and . Set and substitute in the radial equation. It becomes , a harmonic oscillator equation about the epicyclic guiding center . Integration givesThe four real constants in and the two in account for the six initial position and velocity data.
Expanding the inertial specific angular momentum gives . Thus measures the angular-momentum offset from the reference circular orbit, and specifies the radius of its associated epicyclic guiding center.
The conserved horizontal energy in the rotating frame isIt is the horizontal part of the local Jacobi energy in a shearing sheet, rather than the inertial specific orbital energy. Its positive term is the epicyclic energy; its negative term is the energy of the background shear at guiding-center position . Independently, the vertical harmonic oscillator has conserved energy
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 1 8E a Solution Created 2026-09-24 Updated 2026-09-29
Letbe the infinitesimal generator of the continuous symmetry, and letbe the canonical momentum. Noether theorem states that along every trajectory satisfying the Euler-Lagrange equations,is conserved.
Indeed, differentiate the invariance of with respect to at zero:On an Euler-Lagrange trajectory, , soThis proves the Noether conserved quantity for a mechanical point symmetry.
Past exam of the mathematics course of the University of Cambridge 2019 ii Paper 2 14E b Solution Created 2026-09-24 Updated 2026-10-03
The canonical momentum is , and the Legendre transform givesIts Hamilton's equations areand, componentwise,Combining the second equation with recovers the Lorentz force law.

