Past exam of the symmetry and relativity course of the University of Oxford / 2016 (A13481W1, B2)

Source paper: www.whatdotheyknow.com/request/physics_past_exam_papers_2015_16/response/973997/attach/3/Cowell%20Attachment.zip?cookie_passthrough=1

All four numbered questions in the held source are covered. Letters on separately marked paragraphs are inferred; the source's nested letters and Roman numerals are retained. Use the Minkowski metric $(+---)$ and $X^{\mu }=(ct,\mathbf{x})$. Electromagnetic formulas use SI units unless explicitly specialized.

For a massive particle on a timelike world line, proper time satisfies $d\tau =dt\sqrt{1-v^2/c^2}=dt/\gamma$. Its four-velocity and four-acceleration are $U=dX/d\tau$ and $A=dU/d\tau$. Differentiating $U=\gamma (c,\mathbf{v})$ givesThe invariant Minkowski inner products areFor the explicitly requested contraction,Thus four-acceleration is orthogonal to four-velocity. Equivalently this follows by differentiating the constant Lorentz scalar $U^2=c^2$ with respect to proper time.

The event separation is timelike precisely whenA simultaneous pair in an inertial frame would instead have $(D-B{)}^2=-|\Delta {\mathbf{x}}^{\prime }{|}^2\le 0$. Since a Lorentz transformation preserves this interval, no such simultaneous frame exists. A timelike pair can be brought to the same spatial position: the required velocity is $\mathbf{u}=\Delta \mathbf{x}/\Delta t$, whose magnitude is less than $c$.

The separate condition $D\cdot B=0$ says that the two event-position four-vectors, measured from the chosen spacetime origin, are orthogonal in the Minkowski inner product. It is invariant under Lorentz transformations about that origin but not under arbitrary translations. It is not a general criterion for their causal separation.

For a concrete example, put $B=(cT,0,0,0)$ and $D=(0,R,0,0)$. Their inner product vanishes, but their separation has squared interval $c^2{T}^2-R^2$, which can be positive, zero or negative. If $B$ is timelike, a rest frame for $B$ makes $D^0=0$; this means that $D$ and the chosen origin are simultaneous in that frame, not that $D$ and $B$ are simultaneous.

The proper acceleration is $\alpha =\sqrt{-A^2}$, the magnitude of the instantaneous rest-frame three-acceleration. A pure force is a four-force preserving rest mass, or equivalently satisfying $F\cdot U=0$. Indeed $F=d(mU)/d\tau$ gives $F\cdot U=c^{2}dm/d\tau$.

For the right-hand branch of the hyperbola, use $c=1$, $x>0$ and $x=\sqrt{L^2+t^2}$, taking $L>0$. Direct differentiation givesThe sign of $v$ follows the sign of $t$, whereas the coordinate acceleration on this branch is positive. ConsequentlyThe proper-time parameterization $t=L\sinh (\tau /L)$, $x=L\cosh (\tau /L)$ independently gives constant proper acceleration. Restoring dimensions, a hyperbola $x^2-c^2{t}^2=L^2$ has $\alpha =c^2/L$. Constant proper acceleration does not mean that all inertial-frame components of the four-force are constant.

Let $f=|\mathbf{f}|$ and $\hat{\mathbf{f}}=\mathbf{f}/f$. A constant laboratory three-force means $d\mathbf{p}/dt=\mathbf{f}$, hence $\mathbf{p}=\mathbf{f}t$ from rest. Using $E^2=m^2{c}^4+p^2{c}^2$ givesThe dimensionless time $s=ft/(mc)$ makes the shapes transparent: $\gamma =\sqrt{1+s^2}$ starts at one with zero slope and becomes asymptotically linear; $\beta =s/\sqrt{1+s^2}$ starts linearly and approaches one without reaching it at finite time. The acceleration decreases as $f/(m{\gamma }^3)$.

Integrating the velocity gives the distance along the force,Thus a constant three-force produces hyperbolic motion with constant proper acceleration $f/m$, although its velocity is never Newtonian at large time. The following original curves supply the requested sketches.

The magnitude of the electric force is $f=e\mathcal{E}$, directed so that the electron gains energy. The work across the gap is $e\mathcal{E}L=50\mathrm{MeV}$. With the source precision $m{c}^2=0.511\mathrm{MeV}$,Reducing the field by 20 percent gives kinetic energy $40\mathrm{MeV}$ andA large fractional energy change therefore produces only a small speed change, since the original electron is already highly relativistic.

For the transit time, combine the distance formula from solution with ${\gamma }^2-1=[ft/(mc){]}^2$:This is slightly longer than $L/c$, since the electron starts from rest. For the reduced field it becomes $3.37798\times 10^{-8}\mathrm{s}$.

For a Lorentz boost of speed $v$ along $x$, the velocity-addition formula formulas areSubstitute $u_x=u\cos \theta$, $u_y=u\sin \theta$ and take the ratio. The common time-transformation denominator cancels, yieldingTo specify the direction rather than just its tangent, use ${\theta }^{\prime }=atan2(u\sin \theta /{\gamma }_v,u\cos \theta -v)$. This keeps the correct quadrant when the boost reverses the sign of the longitudinal velocity. For physical subluminal speeds the canceled denominator is positive. At $u_x=v$ and nonzero $u_y$, the transformed direction is perpendicular to the boost; if the particle is exactly comoving, its transformed velocity vanishes and the angle is undefined.

The initial momentum magnitude satisfies $p^2/(m^2{c}^2)={\gamma }_i^2-1=288$. Equality of longitudinal and transverse magnitudes therefore givesAn electric field along $z$ changes $p_z$, while the magnetic Lorentz force only rotates the transverse momentum. Its magnitude $p_{\perp }$ remains fixed. At the longitudinal turning point $p_z=0$, henceFor an electron initially moving toward positive $z$, the decelerating field points toward positive $z$ because its charge is $-e$. The electric work over $\Delta z=1\mathrm{m}$ givesThe radius is determined by transverse momentum rather than total energy:so the requested radius plot is horizontal. The cyclotron angular frequency changes as $e|B|/(\gamma m)$ and the helix pitch shrinks to zero. These statements neglect radiation, as stipulated.

Each photon has energy $\epsilon =\hslash \omega$. Their total four-momentum isIts squared invariant is $P^2=2{\epsilon }^2(1-\cos \theta )/c^2$. Therefore the system's rest energy isTwo individually massless particles can thus have positive total invariant mass. Back-to-back photons give rest energy $2\hslash \omega$, whereas exactly parallel photons have zero invariant mass and no massive-system rest frame.

The center-of-momentum frame moves with velocity $\mathbf{V}=c^2{\mathbf{p}}_{\mathrm{tot}}/E_{\mathrm{tot}}$. Using the total four-momentum in solution givesFor $0<\theta \le \pi$, its direction bisects the two photon directions, and its speed decreases from the limiting value $c$ to zero. At $\theta =\pi$, the lab already is the center-of-momentum frame. At $\theta =0$, the formal speed is $c$ and no inertial rest frame exists; this endpoint is a limit, not an allowable boost.

Let $\epsilon =\hslash \omega$ be the incident laser-photon energy and $\gamma =W/(m{c}^2)$. The maximum in the usual soft-laser geometry occurs for a head-on encounter followed by a forward outgoing photon along the electron beam. Boost into the initial electron rest frame: the incident energy becomes ${\epsilon }_{\ast }=\gamma (1+\beta )\epsilon$. Backscattering there gives the Compton scattering recoil relation ${\epsilon }_{\ast ,s}={\epsilon }_{\ast }/[1+2{\epsilon }_{\ast }/(m{c}^2)]$. Boosting this outgoing photon forward to the lab givesThe familiar $4{\gamma }^2\epsilon$ limit additionally requires $\gamma \gg 1$ and ${\epsilon }_{\ast }\ll m{c}^2$. The denominator is the recoil correction, not an optional factor at arbitrary incident energy.

For completeness, if the incidence direction is prescribed rather than optimized, let its unit vector be $\mathbf{n}$ and the outgoing direction ${\mathbf{n}}_s$. Squaring the residual electron four-momentum on its mass shell gives the more general kinematic relationThus an unspecified crossing angle must not silently be identified with head-on incidence; the boxed result is the maximum for the stated head-on soft-photon arrangement.

Numerically $W=2\mathrm{GeV}$ gives $\gamma =3913.89$, and a $1\mathrm{cm}$ photon has energy $1.23984\times 10^{-4}\mathrm{eV}$. The maximum scattered energy is $7.5970\mathrm{keV}$, givingThe rest-frame incident photon is only about one electronvolt, so the small recoil correction is consistent with the amplification estimate.

Three useful observational phenomena are relativistic Doppler effect in spectral lines from approaching and receding ejecta, relativistic beaming through jet brightness asymmetry, and apparent superluminal motion inferred from successive radio images of moving knots. The first compares spectral frequencies with rest-frame lines; the second compares the two sides after accounting for intrinsic emission and absorption; the third tests the light-travel-time relation ${\beta }_{\mathrm{app}}=\beta \sin \vartheta /(1-\beta \cos \vartheta )$. NRAO's time sequence of Galactic microquasar ejecta gives a concrete example of the third measurement.^[ref]

These are tests of relativistic radiation kinematics when source geometry and environmental effects are controlled, rather than claims that any one redshift or brightness contrast by itself uniquely proves special relativity.

For the frequency transformation, a photon has wave four-vector $K=(\omega /c,\mathbf{k})$ with $|\mathbf{k}|=\omega /c$ in vacuum. The temporal component of a Lorentz boost isThis is the relativistic Doppler effect: frequency changes both because of relative motion along the ray and because clocks in relative motion are time dilated. The angle here is measured in S. A frame chasing the photon redshifts it, ${\omega }^{\prime }/\omega =\sqrt{(1-\beta )/(1+\beta )}$; a head-on frame blueshifts it. Transverse angles defined in different frames must not be interchanged.

The phase velocity is the speed of a constant-phase surface, $v_{\mathrm{ph}}=\omega /k$. The group velocity is the envelope speed of a narrow packet, $v_{\mathrm{gr}}=d\omega /dk$, requiring a local dispersion relation rather than only a single monochromatic wave. In an isotropic transparent medium with $k=n(\omega )\omega /c$,For a vacuum wave, $n\equiv 1$ over the packet bandwidth, so both speeds are $c$ and $\boxed{v_{\mathrm{ph}}{v}_{\mathrm{gr}}=c^2}$. No superluminal signalling is implied by a phase speed in a dispersive material.

Strictly, the value $n({\omega }_0)=1$ at one frequency alone does not prove the product identity. For example $n(\omega )=1+q(\omega -{\omega }_0)/{\omega }_0$ gives $v_{\mathrm{ph}}({\omega }_0)=c$ but $v_{\mathrm{gr}}({\omega }_0)=c/(1+q)$. The vacuum interpretation, or an equivalent nondispersive assumption, is necessary.

Keep the unequal masses: equal and opposite velocities are not equal and opposite momenta. Write the rest energies $r=m_e{c}^2$, $R=M_p{c}^2$, $S=r+R$, and let the incoming Lorentz factor of each particle be $\gamma$. ThenThe transverse photon has energy $\epsilon =hc/\lambda$ and no $z$ momentum. If the final composite has rest energy $Q$, squaring its four-momentum gives the exact constraintThe mixed term contains no longitudinal momentum contribution because the photon is transverse. For a specified final rest energy, the physical positive root isprovided it is real and at least one.

The source does not specify the composite rest mass or its binding energy. With the usual approximation that the lowest final rest energy is $Q=S$, the minimum incoming energy occurs at that lowest mass and isIf a hydrogen bound state is intended at precision sensitive to its binding, replace $Q$ by its actual rest energy in the preceding equation; the mass cannot be inferred exactly from the phrase “single particle.” Higher composite excitation raises the required incoming energy on this branch. Setting initial total momentum to zero would give a different answer and contradict the stipulated equal-speed electron and proton.

For a forward wave, the Faraday's law of induction fixes the relative amplitudes and signs:Thus the two scalar amplitudes in the source cannot both be positive when the stated propagation direction and axes are used. The supplied product magnitude fixes only an overall amplitude in the chosen field units.

One convenient potential representative isIndeed $-{\partial }_t\mathbf{A}$ gives the electric field and $\nabla \times \mathbf{A}$ gives the signed magnetic field above. Moreover $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\phi =0$, the Lorenz gauge condition. The potentials are not unique because residual gauge transformations may preserve this condition.

Since $n>1$, the electromagnetic electromagnetic field invariants satisfy $\mathbf{E}\cdot \mathbf{B}=0$ and $B^2-E^2/c^2=(n^2-1)E^2/c^2>0$ away from field nodes. A boost along positive $z$ with speed $u=c/n$ givesHere ${\omega }^{\prime }={\gamma }_u(\omega -uk)=0$ and $k^{\prime }=(\omega /c)\sqrt{n^2-1}$. The wave becomes a static spatially varying magnetic pattern in this frame. This is possible because its medium wave four-vector is spacelike, $K^2=(1-n^2){\omega }^2/c^2<0$; it is not a vacuum-photon rest frame. The medium moves in the new frame, so its original rest-frame constitutive relation must not be reused there unchanged.

No inertial frame makes the magnetic field vanish throughout a nontrivial wave: that would turn the positive invariant $B^2-E^2/c^2$ into a nonpositive one. Both fields vanish at isolated nodes already, which is distinct from removing a field everywhere.

Take $A^{\mu }=(\phi /c,\mathbf{A})$, ${\partial }_{\mu }=(c^{-1}{\partial }_t,\nabla )$ and $J^{\mu }=(c\rho ,\mathbf{j})$. ThenExpanding the cyclic derivative of $F^{ab}={\partial }^a{A}^b-{\partial }^b{A}^a$ produces six second derivatives, which cancel in pairs because partial derivatives commute. HenceThis Bianchi identity contains $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{E}=-{\partial }_t\mathbf{B}$. The inhomogeneous Maxwell equations areThe temporal component is $\nabla \cdot \mathbf{E}=\rho /{ϵ}_0$, since ${ϵ}_0{\mu }_0{c}^2=1$; the spatial components give $\nabla \times \mathbf{B}-c^{-2}{\partial }_t\mathbf{E}={\mu }_0\mathbf{j}$. The source suppresses the coupling factor, as in units with ${\mu }_0=1$ or a rescaled current.

For the physical Lorentz force density on matter, the consistent tensor formula isIts temporal component is power transferred to matter per unit volume divided by $c$. The printed expression $J^{\nu }{F}_{\nu \mu }$ has the opposite sign for the stated definition of $F$ in this convention. A stationary positive charge in a positive $x$ electric field is a concrete check: the physical upper spatial force is $+\rho E_x$, whereas the printed expression, raised to an upper index, gives $-\rho E_x$. It can represent the opposite momentum transfer to the electromagnetic field, but must not be identified with the force on the charge without changing a convention.

In the massive particle's rest frame, $P=(m_{0}c,0)$ and the physical Pauli-Lubanski pseudovector is $W=(0,m_{0}c{\mathbf{s}}_{\mathrm{rest}})$, up to a common overall sign. Their Minkowski inner product is therefore zero. Since it is a Lorentz scalar,in every inertial frame, for any nonzero scalar normalization of $W$. This proves the requested orthogonality without a boosted component calculation.

The printed assertion that its particular normalization is always a unit vector needs qualification. For the displayed components $W=(\mathbf{s}\cdot \mathbf{p},E\mathbf{s}/c)$,Thus dividing by $m_{0}c|\mathbf{s}|$ is unit spacelike only when $\mathbf{s}\times \mathbf{p}=0$, or when $\mathbf{s}$ in that denominator is explicitly replaced by the proper rest-spin magnitude with correctly boosted components of $W$. As a counterexample to the literal displayed normalization, choose $\mathbf{s}\perp \mathbf{p}$ with nonzero momentum: it gives $U_W^2=-{\gamma }^2$, not $-1$. Orthogonality remains true, but it does not establish normalization.

An isolated free electron in vacuum cannot spontaneously emit a nonzero real photon. If $P=P_f+Q$, where $Q$ is the emitted photon four-momentum, $P^2=P_f^2=m^2{c}^2$ and $Q^2=0$, conservation would require $P_f\cdot Q=0$. A future-pointing timelike four-vector has strictly positive inner product with every nonzero future-pointing null four-vector, so this is impossible.

The same conclusion follows from the classical Cherenkov radiation condition $v>c/n$: in vacuum $n=1$, while the electron with $\gamma =100$ has $\beta =\sqrt{1-10^{-4}}<1$. Hence there is no vacuum emission angle for the specified free particle. Acceleration by an external field or scattering from another body would change the physical problem and can permit radiation.

In a transparent uniform medium, constructive phase interference along the trajectory requires $\omega -\mathbf{k}\cdot \mathbf{v}=0$. Since $k=n\omega /c$, the Cherenkov cone angle obeysHere $n\beta =2\sqrt{1-10^{-4}}>1$, so emission is allowed andThe electron is not an isolated vacuum particle: it interacts with the dielectric polarization of the medium. Here the angle describes the photon direction relative to the trajectory; the conical wavefront surface has the complementary angle.

This angle is the classical or negligible-recoil result. At finite photon energy, energy-momentum bookkeeping in a nondispersive medium gives a recoil correction proportional to $\hslash \omega /E$; no photon frequency is supplied for such a correction. In a dispersive medium use $n(\omega )$ separately for each transparent frequency band.

Model the wire as an infinite straight line along $z$. Let ${\lambda }_0$ be its rest-frame line charge density and $\lambda$ its laboratory line charge density. Length contraction gives $\lambda ={\gamma }_v{\lambda }_0$. If the symbol $\rho$ in the source is intended as volume charge density, replace these line densities by their integrals over the cross section; an unspecified wire radius does not determine an interior field.

In the comoving frame the wire is static, so Gauss's law and cylindrical symmetry give the exterior fieldsIf the source's $\rho$ means laboratory line density, this electric field is $\rho /(2\pi {ϵ}_0{\gamma }_{v}r)$; if it means proper line density, it is $\rho /(2\pi {ϵ}_{0}r)$. Stating the density's reference frame is necessary for an unambiguous factor of ${\gamma }_v$.

Boost the comoving electric field perpendicular to itself along the wire. The transverse electric field gains a factor ${\gamma }_v$, while a circulating magnetic field appears. Equivalently apply Gauss's law and Ampère's law with laboratory line charge $\lambda$ and current $I=\lambda v$:These are exterior fields of the ideal straight wire. For laboratory line density $\rho$, substitute $\lambda =\rho$; for proper line density $\rho$, substitute $\lambda ={\gamma }_v\rho$. The radial distance is transverse to the boost and unchanged between the two frames.

The final unnumbered force question has this inferred third subpart. For either wire's exterior field,This electromagnetic field invariants is positive in every inertial frame. A purely magnetic nonzero field would instead have $E^{\prime 2}-c^2{B}^{\prime 2}<0$. Thus no physical boost makes the interaction purely magnetic. There is a comoving frame with a purely electric interaction, which is consistent with the invariant's sign.

The convected charge current gives the same conclusion. Its effective line-current invariant is ${\lambda }^2-I^2/c^2={\lambda }_0^2>0$, so no inertial frame can neutralize the charge while leaving only a current. In particular for a longitudinal boost of speed $u$, ${\lambda }^{\prime }={\gamma }_u\lambda (1-uv/c^2)$ cannot vanish when $|u|,|v|<c$.

For two identical wires separated by $d$, the laboratory transverse force per unit length isElectric repulsion exceeds the magnetic attraction. This assumes each moving wire has uncompensated convected charge; a neutral wire carrying conduction current is a different charge-current system and has a different invariant classification.