Past exam of the symmetry and relativity course of the University of Oxford / 2016 / 3

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.