Past exam of the optics and electrodynamics course of the University of Cambridge / 2021 / Section B / B6

The upper index identifies $P^{\alpha }$ as a contravariant Lorentz four-vector, obeying the rank-one contravariant tensor transformation law. Its components transform with the coordinate Jacobian,It is not invariant component by component. Its contracted square $P_{\alpha }{P}^{\alpha }=m^2{c}^2$, for metric signature $(+,-,-,-)$, is a Lorentz invariant. This distinguishes a contravariant transformation law from an invariant scalar.

First lower the index of the four-current with the Minkowski metric: $J_{\beta }=g_{\beta \alpha }{J}^{\alpha }$. The displayed expression is thenThereforeIts one free lower index transforms with the inverse coordinate Jacobian. The contraction with the metric here lowers an index; it does not make the current a scalar because the index $\delta$ remains free.

The chain rule givesThusDifferentiation with respect to a contravariant coordinate has a lower transformation index. In inertial Cartesian coordinates in Minkowski spacetime, no connection term is needed for this coordinate transformation of a derivative acting on a scalar; Lorentz transformations have constant Jacobians.

The lower derivative index and the upper electromagnetic four-potential index are fully contracted. For a constant Lorentz transformation,HenceThis is a statement about Lorentz transformations, not electromagnetic gauge transformations. The quantity is not generally gauge invariant; setting it to zero is the Lorenz gauge condition, which is Lorentz covariant because zero remains zero in every inertial frame.

Use coordinates $(ct,x,y,z)$ and put $\beta =v/c$, $\gamma =(1-{\beta }^2{)}^{-1/2}$. For an observer moving at $+v\hat{\mathbf{x}}$, the coordinate Lorentz boost isThe given electromagnetic field tensor transforms as $F^{\prime }=\Lambda F{\Lambda }^{\mathsf{T}}$. For example,whileThe $01$ component is unchanged because its boost determinant is ${\gamma }^2(1-{\beta }^2)=1$. The $23$ component is unchanged because neither index lies in the boost plane. Reading off all six components givesThe signs follow the stated positive observer velocity and the supplied tensor convention. In vector form for a boost in any direction, ${\mathbf{E}}_{\parallel }^{\prime }={\mathbf{E}}_{\parallel }$, ${\mathbf{E}}_{\perp }^{\prime }=\gamma ({\mathbf{E}}_{\perp }+\mathbf{v}\times \mathbf{B})$, with analogous ${\mathbf{B}}_{\perp }^{\prime }=\gamma ({\mathbf{B}}_{\perp }-\mathbf{v}\times \mathbf{E}/c^2)$.

A uniform Lorentz boost preserves the plane wave form, while changing both its frequency and its vector wave amplitude. Decompose vectors parallel and perpendicular to $\mathbf{v}$, so $\mathbf{k}={\mathbf{k}}_{\parallel }+{\mathbf{k}}_{\perp }$. In the observer's primed coordinates,The transformed wave amplitude is specified in part (d). The unprimed “rest frame” in this question is the original coordinate frame, not a rest frame of a vacuum light wave, which has no inertial rest frame.

Use the inverse coordinate transformation,Substitution into the wave phase $\mathbf{k}\cdot \mathbf{r}-\omega t$ and comparison of the coefficients of ${\mathbf{r}}^{\prime }$ and $t^{\prime }$ givesThe first is the relativistic Doppler effect; the changed direction of the wavevector describes relativistic aberration. Equivalently,The electric-field wave amplitude follows from part (b), after rotating the boost axis to $\mathbf{v}$:For a vacuum plane wave, Maxwell equations give ${\mathbf{B}}_0=\mathbf{k}\times {\mathbf{E}}_0/\omega$ and $\mathbf{k}\cdot {\mathbf{E}}_0=0$. Thus an expression using only the given electric wave amplitude and wavevector isEquivalently, ${\mathbf{E}}_0^{\prime }=\gamma ({\mathbf{E}}_0+\mathbf{v}\times {\mathbf{B}}_0)-[{\gamma }^2/(\gamma +1)]\mathbf{v}(\mathbf{v}\cdot {\mathbf{E}}_0)/c^2$. The wave amplitude is not a scalar carried unchanged by the wave phase transformation; its direction must remain transverse to the transformed wavevector.

The frequency and longitudinal wavevector formulas are exactly the boost of a contravariant Lorentz four-vector:where the matrix is understood on the time component and the signed scalar component along the boost direction. Hence the four-wavevector obeysFor the metric signature $(+,-,-,-)$, lowering the position index gives $R_{\alpha }=(ct,-\mathbf{r})$, soIt is a dimensionless Lorentz scalar, equal to minus the wave phase used in the stated complex wave. Invariance follows by contracting the mutually inverse position and wavevector transformations. Reversing metric signature reverses this displayed scalar's sign, but not its invariance or any observable phase difference. In vacuum, $K_{\alpha }{K}^{\alpha }=0$ is an additional null-vector check; the wave remains lightlike in every inertial frame.

Yes: the same physical resonant cavity and its field remain resonant in every inertial description. A Lorentz transformation changes the mirror worldlines and the field together. The boundary phase matching and the constructive-interference condition are statements about the same reflection events, and the contracted wave phase is a Lorentz scalar. A boosted resonant solution therefore still satisfies the moving-mirror boundary conditions.

It would be incorrect merely to insert a Lorentz-contracted length into the formula for two stationary mirrors while leaving both propagation frequencies unchanged. For example, let the mirrors be separated by $L$ in their rest frame and let ${\omega }_{0}L/c=m\pi$, including the usual equal-mirror reflection-phase convention. For an observer boosted at $+v$ along the cavity axis, the cavity moves at $-v$ and its two travelling components have frequenciesThey need not have one common laboratory frequency. The simultaneous mirror separation is $L^{\prime }=L/\gamma$, while the forward and return flight times are $L^{\prime }/(c+v)$ and $L^{\prime }/(c-v)$. The round-trip interval between events on the same mirror is therefore $2\gamma L/c$.

Along that moving mirror the wave phase rate of either wave component is ${\omega }^{\prime }-k^{\prime }{v}_{\mathrm{mirror}}={\omega }_0/\gamma$. Its advance over the round trip isThus the resonance condition is unchanged even though the observed travelling frequencies, lengths and flight times all change. For a boost in a general direction the beam directions also undergo relativistic aberration; the event-by-event phase-invariance argument still applies. A separate stationary cavity placed in the new frame, or an external laser whose frequency is not transformed with the original setup, is a different physical system.