For , each adjacent pair of walkways has relative rapidity . There are such steps between and , so the equal-rapidity chain of collinear Lorentz boosts has total rapidity . Thus the actual velocity isIt is strictly below the speed of light for every finite . Adding the adjacent velocities as ordinary numbers ignores the relativistic velocity-addition formula.
For large , the Taylor series of the inverse hyperbolic tangent givesConsequentlyMore precisely, , so the limit is approached from above. The case would prescribe an adjacent relative velocity equal to the speed of light; it does not define an admissible walkway inertial frame.
Differentiate the worldline with respect to :Since and , this is a future-directed timelike curve. The proper time element satisfies , henceAlice's total proper time is thereforeThis definite integral is an exact answer; it is an elliptic integral of the second kind. In notation , it is . Although this parameter exceeds one, the integrand is real on the stated interval because its minimum squared value is .
The endpoint events have and . Sam stays at in his inertial frame, so his elapsed proper time is . For , the integrand is strictly less than except at , and .
Past exam of the symmetry and relativity course of the University of Oxford 2015 3 d Solution 2026-10-25
Take the kaon decay as the common origin. A pion traveling freely with constant momentum has lab decay time and position . The supplied proper survival times therefore giveConsequently the separation from the first decay event to the second isThe requested distance is the separation of the two events, not the sum of the two individual pion path lengths.
Here , so the separation is a spacelike separation. There is an inertial frame in which the events are simultaneous; in that frame their distance equals the invariant spacelike length:One such frame moves along with , of speed . The simultaneity condition is then .
The spacelike distance between daughter decay events provides an independent invariant check: the daughter four-velocity contraction givesIt produces the same distance without using the particular kaon laboratory boost. A simultaneous frame exists only when this expression is positive, as it is for these two survival times.
Past exam of the symmetry and relativity course of the University of Oxford 2019 3 a Solution 2026-10-25
An invariant has the same value when the same physical event or state is described in different inertial frames. A conserved quantity has the same value as the physical system evolves, subject to its equations and isolation conditions. These are different comparisons. The invariant mass of a heated object can change with time, while an isolated system's conserved energy has different numerical values in boosted frames.
Under a Lorentz transformation , a Lorentz four-vector transforms asA covector transforms by the inverse matrix, . A general tensor has one corresponding factor for each index; for example,Contraction with the Minkowski metric produces Lorentz scalars; it does not by itself make those scalars conserved.
Past exam of the symmetry and relativity course of the University of Oxford 2019 3 b Solution 2026-10-25
To prove the vanishing component theorem for a Lorentz four-vector, first suppose in every inertial frame. An infinitesimal boost along spatial direction gives . It too vanishes for every , so each . If instead a specified spatial component vanishes in every frame, spatial rotations first imply that all three spatial components vanish in a chosen frame, and a boost in the specified direction then forces . ThereforeA physical consequence is that conservation of energy in every inertial frame implies full four-momentum conservation: apply the result to the difference between outgoing and incoming total four-momenta.
The statement fails for general tensors. A nonzero antisymmetric rank-two tensor has in every frame, because antisymmetry is preserved. For an off-diagonal component, the nonzero Minkowski metric is another counterexample: in all orthonormal Lorentz frames, while is certainly not zero. The transformation law can preserve structural zero components without forcing the entire tensor to vanish.
Past exam of the symmetry and relativity course of the University of Oxford 2021 1 a Solution 2026-10-25
Let be the magnitude of the electron's electric charge, , , and . The Lorentz force is , so the electron and Positron acquire opposite momenta:Using the relativistic energy-momentum relation, , and integrating from rest gives the world linesFor , the world lines are vertical at on a spacetime diagram. They join smoothly to the accelerated branches with zero initial velocity. At small , and ; at large , their velocities approach , respectively. Each accelerated world line is a branch of a hyperbola, since and . The electric field separates the opposite electric charges in this orientation. The asymptotes are lightlike, while every finite-time tangent is timelike.
Past exam of the symmetry and relativity course of the University of Oxford 2021 1 b Solution 2026-10-25
Write for the Lorentz factor of the inertial frame , distinguishing it from the particles' changing Lorentz factors. Apply the Lorentz transformationto each world line in a. Both particles initially have velocity in . The switching events have coordinatesThus the Positron starts accelerating earlier in . Its velocity first becomes less negative, vanishes when its laboratory velocity reaches , and then becomes positive. The electron continues at until , then accelerates toward . The relativistic velocity-addition formula makes this precise:The electric field parallel to the Lorentz boost has unchanged magnitude, but its switching surface is , rather than everywhere.
In the laboratory, the total three-momentum is zero for all : the two Lorentz forces cancel at equal laboratory times. In , each active force is still , since this is one-dimensional motion with a parallel electric field. Define . For the total three-momentum, evaluated at equal , isThis follows by integrating from its switching event and from its later switching event. During the interval between them only the Positron feels the prescribed electric field, so momentum conservation for the particle pair fails. Once both forces act, the total three-momentum is constant again.
The equal-time momentum of an externally driven particle pair illustrates a distinction. The sum of particle four-momenta at a fixed pair of events is always a four-vector. However, a fresh sum at equal time in each inertial frame generally uses different events. Here transforming an equal- laboratory sum would give , which varies with ; this differs from the equal- answer above. For example, with and , the late-time equal- sum is , whereas transforming the equal- sum gives . This is a concrete failure of treating the pair's equal-time sums as components of a single frame-independent four-vector.
A closed system's total four-momentum, including the electromagnetic field and the apparatus supplying it, is conserved and independent of the choice of spacelike integration surface under the usual isolated-system boundary conditions. The particles alone exchange energy and momentum with an external system, so neither that independence nor universal particle-only momentum conservation is guaranteed.
Past exam of the symmetry and relativity course of the University of Oxford 2021 2 a Solution 2026-10-25
Let be the point electric charge's world line, with proper time and four-velocity . For the field event , choose the retarded source event: is future-directed and null. Write , where . In the instantaneous rest inertial frame of that source event,Consequently the covariant expression matching this Coulomb potential isThe denominator is a Lorentz scalar, so the right-hand side is a four-vector with the required instantaneous-rest components. This describes the retarded potential in Lorenz gauge, with no added free electromagnetic wave. Lorentz covariance alone does not establish the boundary conditions or exclude additional terms depending on four-acceleration. The physical input is the retarded point-source solution of Maxwell equations: the potential at the retarded event depends on the instantaneous four-current, while its derivatives generate the acceleration field. Equivalently, evaluating the retarded source integral gives a Jacobian from the light-cone constraint; this fixes that input for arbitrary smooth subluminal motion.[ref]
Retarded Potentials by Richard Fitzpatrick. §1.7, equations (71)–(76): causal source integrals, with source quantities evaluated at retarded time
In any inertial frame, put and . Then , and the Liénard–Wiechert potential isDifferentiate using and , including the position and time dependence of the retarded time. For example, and . The resulting Liénard-Wiechert field isHere the dot denotes a derivative with respect to source coordinate time. The first term is the velocity or non-radiative field: it falls as and exists even without acceleration. The second is the acceleration or radiation field, falling as and transverse to . Its Poynting flux falls as , so its outward power integrated over a large sphere stays finite as its radius tends to infinity. The first term's squared-field flux instead vanishes in that limit. This large-distance distinction defines radiation; a moving near field can carry local energy without constituting radiation escaping to infinity.
Past exam of the symmetry and relativity course of the University of Oxford 2021 2 b Solution 2026-10-25
For uniform circular motion, the transverse Lorentz force supplies the change in direction of three-momentum:The magnetic rigidity relation gives . Using ,The coordinate centripetal acceleration is . For transverse acceleration, the magnitude of the four-acceleration, or proper acceleration, is . The supplied Larmor formula, in its proper acceleration form, therefore givesThe emitted power here is also the laboratory total emitted power: in the instantaneous rest inertial frame the dipole radiation pattern has zero net radiated three-momentum, so both emitted energy and elapsed coordinate time gain the same Lorentz factor on transformation. This does not assert that a directional detector flux is invariant.
Numerically , per electron, andThe synchrotron's radiofrequency accelerating system must replace this lost energy to keep and magnetic rigidity fixed over many revolutions. The fraction lost per turn is about , so a fixed-orbit calculation over one turn is self-consistent.[ref]
Synchrotron Radiation by Richard Fitzpatrick. §12.22, equations (1975)–(1978): transverse radiated power and energy per revolution
Past exam of the symmetry and relativity course of the University of Oxford 2021 3 a ii Solution 2026-10-25
In the magnetic-dominated case the suitable Lorentz boost instead hasThe hypothesis gives . The electromagnetic field invariants now have , allowing a purely magnetic inertial frame. The initial velocity is again , perpendicular to . Since a magnetic field does no work, the electron's relativistic energy and speed in this inertial frame remain constant. Its trajectory is a circle in the plane with Lorentz factor and cyclotron frequency .
For , set and . A parametrization with the correct initial velocity and Lorentz force isTransforming the world line back to the laboratory givesThe laboratory trajectory is a stretched cycloid, with cusps at where the electron is momentarily at rest. It first accelerates toward , then curves toward , returning to its original coordinate each cycle while advancing in . Its mean drift is , the E-cross-B drift. Its laboratory kinetic energy oscillates as the electric field does positive and then negative work; it does not grow indefinitely. The strict inequalities in the two branches exclude the null case , where neither field can be removed by a finite Lorentz boost.
