Past exam of the relativity course of the University of Cambridge / 2019 / Section A

The conserved initial four-momentum of a particle with positive rest mass satisfies $P_{\mu }{P}^{\mu }=m^2{c}^2>0$. A single photon instead has null four-momentum, $K_{\mu }{K}^{\mu }=0$. Four-momentum conservation for an isolated one-body decay would require $P^{\mu }=K^{\mu }$, contradicting these Lorentz invariants. Thus decay into a single photon is impossible in every inertial frame.

Equivalently, the incoming energy and momentum obey $E^2-c^2|\mathbf{p}{|}^2=m^2{c}^4$, whereas a single outgoing photon obeys $E=c|\mathbf{p}|$. Changing the laboratory speed cannot change this mismatch of the invariant mass shell.

For motion along one spatial axis, introduce the rapidity $\eta$ so that the four-velocity is $u^{\mu }=c(\cosh \eta ,\sinh \eta ,0,0)$. Its four-acceleration isA constant signed proper acceleration $g$ therefore gives $d\eta /d\tau =g/c$, hence $\eta ={\eta }_0+g\tau /c$. Differentiating the four-velocity twice yieldsThe zero transverse components also satisfy this equation. The constant acceleration measured on board is the proper acceleration, rather than a constant coordinate acceleration in the laboratory.

Use the geodesic Lagrangian $L=\frac{1}{2}(y^2{\dot{x}}^2+x^2{\dot{y}}^2)$, where dots denote an affine parameter derivative. The Euler-Lagrange equations giveThese are the geodesic equations wherever $xy\ne 0$, so the Riemannian metric is nondegenerate. Substitution of $y=mx$, $\dot{y}=m\dot{x}$ and $\ddot{y}=m\ddot{x}$ makes both equations proportional toThus $\boxed{x^2=A\lambda +B,y=mx}$ is a family of affinely parametrized geodesics, on intervals staying within one quadrant. In particular, $m\ne 0$ and $x\ne 0$ keep the metric nonsingular; the coordinate axes themselves are not regular points of this metric. The coordinate $x$ need not be linear in the affine parameter, despite the straight coordinate-space shape of these curves.