Past exam of the relativity course of the University of Cambridge 2013 Paper 1 1 b Solution 2026-10-20
For geodesic motion of uncharged dust, use the pressureless stress-energy form, and . stress-energy conservation expands toContracting with kills the second term, since . Hence , and wherever ,On a flow line , so this becomes . Each relativistic dust element therefore follows a timelike geodesic with proper time as an affine parameter. pressure gradients or external nongravitational forces would add an acceleration term.
Past exam of the relativity course of the University of Cambridge 2013 Paper 1 3 Solution Created 2026-10-20 Updated 2026-10-20
Throughout this solution and .
Null metric tensor and boost. Inverting the coordinate definitions gives , , and henceFor a frame moving at speed along positive , use , . Their difference and sum giveThe reciprocal scale factors preserve . They are in rapidity notation.
acceleration invariants. Proper-time normalization gives , so . The four-acceleration has squared norm ; thereforeFor future motion write and . Then the second equation is . acceleration toward positive selects .
hyperbolic motion from an initial event and rapidity. The stated initial event fixes , , but not the initial rapidity . Integrating with positive signed acceleration gives the general familyThe initial event alone does not fix the initial velocity or signed acceleration. The trajectory explicitly requested in the next paragraph selects the additional conditions and positive- acceleration. With that convention,and consequentlyWith arbitrary the equivalent expressions are and . Thus position data alone do not select the displayed special hyperbola. The following calculations use that intended hyperbola, which is also explicitly given by the paper.
Diagram and horizon. The selected world line is the right-hand branch . Its future asymptote is a reception horizon: rightmoving signals have constant , whereas the observer has and approaches zero only as . Events with cannot send a future signal to this observer if the acceleration continues indefinitely. This is an observer-dependent causal restriction in flat spacetime, not a curvature singularity.
photon reception by a uniformly accelerated observer. A rightmoving photon from the stationary source has four-momentum . Contracting with the observer's four-velocity givesThe same result follows by differentiating its emission-time relation. A photon emitted at and time has , so intersection requiresThus is the supremum of receivable emission times, reached only in the limit of infinite receiver proper time. A photon emitted exactly then lies on the horizon and is never received at finite time; later photons cannot catch the ship. Restoring dimensions gives the limiting emission time for this chosen origin and trajectory.
Rindler coordinates. Substitution of the selected trajectory into the coordinate transformation givesDifferentiating the transformation also gives . Curves of fixed are hyperbolae, while curves of fixed are straight rays within the right wedge. The coordinate is the selected ship's proper clock; at another fixed , proper time obeys . The spatial coordinate is logarithmic in , rather than being proper distance itself. Fixed- observers have proper acceleration .
Nearby freely falling particles can accelerate relative to one another even though each follows a geodesic. A local inertial frame removes the common gravitational acceleration at one event; it cannot remove the variation across an extended laboratory. That variation is a tidal force, measured covariantly by geodesic deviation.
Let be the reference four-velocity and an infinitesimal separation to a neighboring geodesic, with matching affine parameters. With the convention ,This relates measured relative acceleration to the Riemann curvature tensor. Reversing the curvature-tensor convention reverses the displayed curvature sign, without changing the physical relative motion.
In a small freely falling frame outside a spherical mass, the weak-field tidal tensor has radial and transverse components proportional to . ThusAn initially spherical cloud stretches radially and compresses in both transverse directions. Two separated test masses, with their changing separation measured by light, form the basic idea of a gravity gradiometer; the common falling motion drops out. A uniform apparent acceleration in an accelerated laboratory can occur even in flat Minkowski spacetime, so it is the separation-dependent tidal response that diagnoses curvature.
Radial stretching and transverse compression from geodesic deviation
. Original local schematic of the weak-field tidal tensor around a freely falling reference particle. Arrows show relative acceleration, with radial direction horizontal; the cloud deformation is exaggerated and is not a trajectory calculation. Details. Past exam of the relativity course of the University of Cambridge 2018 Section B B2 a Solution 2026-10-18
Let denote coordinate acceleration and use for four-acceleration, to distinguish these two objects. Applying the chain rule to the Lorentz factor givesSince , differentiation of the four-velocity with respect to proper time yieldsThis is also , as required.
The invariant Minkowski norm of the four-velocity is . Differentiating this identity gives . In the instantaneous rest frame, , so and the spatial components equal the physically experienced proper acceleration . Therefore in the metric signature.
For a direct calculation, put and expand the temporal and spatial squares:The last step uses . This proper acceleration from coordinate velocity and acceleration also reads , providing the familiar one-dimensional limit . Each expression has units .
Past exam of the relativity course of the University of Cambridge 2018 Section B B4 Solution 2026-10-18
Local equivalence and tides. In Newtonian mechanics, inertial mass measures the response to a force, whereas passive gravitational mass couples to an imposed Newtonian gravitational field. A freely falling body obeysThe observed composition independence of free fall says that this ratio is universal. With a common choice of units it is one. This is the Newtonian content of the weak equivalence principle for test particles; it does not by itself specify all nongravitational physics.
The Einstein equivalence principle further says that local nongravitational experiments in a sufficiently small freely falling laboratory have the form predicted by special relativity, independently of the laboratory's location and motion. In such a local inertial frame, the metric tensor is locally the Minkowski metric and the Christoffel symbols vanish at the selected event. The strong equivalence principle extends the universality claim to local gravitational experiments and bodies with significant gravitational self-energy. None of these statements removes measurable tidal forces across a finite apparatus.
In an ideal small freely falling lift, a released ball initially floats relative to the cabin, because both undergo the same free fall. Two separated balls, however, need not retain their separation: they sample different parts of the gravitational field. For a point source, linearizing across a small separation gives the tidal tensor predictionA radial separation stretches with coefficient , whereas a transverse separation is compressed with coefficient . Initially parallel velocities therefore develop relative motion even though the individual bodies experience zero proper acceleration.
The relativistic statement is geodesic deviation. With the Riemann curvature tensor conventiona connecting vector between neighboring freely falling worldlines satisfiesHere is the covariant derivative along the reference geodesic, and its four-velocity. Curvature can be nonzero where the Christoffel symbols have been made zero. In the weak, static, slow-motion limit this equation reduces to , recovering the tidal tensor above. A finite falling lift can detect gravity through these relative accelerations; local equivalence does not imply a globally uniform field.
Weak-field connection and potential. Take and the metric signature. There is an index error in the printed formula for the Christoffel symbols: its first differentiated metric component must be , not . The correctly contracted expression isThe offending term vanishes under the stipulated time independence, so this correction does not change the intended calculation. Write and treat both and its derivatives as first order. To that order , , and . A stationary metric tensor givesFor the temporal component the inverse-metric coefficient is already first order, soFor a spatial component the zeroth-order inverse metric contributes:The time component of the approximate geodesic equation says that is constant to the retained order. In its spatial component, convert the proper time derivatives to coordinate-time derivatives and cancel :Comparison with the Newtonian equation identifies . For an isolated source, choose the gravitational potential to tend to zero at infinity and normalize the time coordinate there, giving . ThusThe quantity is the Newtonian gravitational potential, or gravitational potential energy per unit mass, in . For a point source, . A static clock has , connecting the Newtonian limit with gravitational time dilation.
Frequency of the orbiting source. Treat the gas blob as a test particle on a timelike circular Schwarzschild geodesic, with pressure and other nongravitational forces negligible. Treat our very distant location as infinity, so the observer's own gravitational potential is negligible. These are the assumptions that make a unique answer in terms of possible.
Put and use a local static observer at the emission point as an intermediate measuring frame. The remote observer lies on the symmetry axis of the orbit. A ray arriving on that axis has zero conserved axial angular momentum, so its local direction is perpendicular to the emitter's azimuthal velocity. Bending within a meridional plane does not alter this orthogonality. The transverse relativistic Doppler effect therefore giveswhere is the orbital speed measured by that local static observer. The gravitational redshift between static Schwarzschild observers then givesA gravitational circular orbit has in the weak-field limit, so . Expanding both factors,Solving for the emitted frequency givesThe gravitational part is and the transverse kinematic time dilation contributes another ; the absence of line-of-sight motion does not remove this second effect.
As an exact check, the local speed on a timelike circular Schwarzschild geodesic satisfies , where . Therefore the face-on frequency shift of a circular Schwarzschild source iswhose expansion agrees with the boxed answer. If the gas is supported or accelerated by pressure or magnetic stresses rather than following a geodesic, its orbital speed is not fixed by the mass and radius alone; the appropriate first-order result is instead , with supplied independently.


