Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 4

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.