Past exam of the relativity course of the University of Cambridge / 2012 / Paper 1 / 1

In the rest frame, conservation of energy would require $E_{\gamma }=m{c}^2$, whereas conservation of momentum would require the photon momentum to vanish. These contradict $|{\mathbf{p}}_{\gamma }|=E_{\gamma }/c$. More generally, four-momentum conservation would equate the initial massive four-momentum to a null four-momentum, although their Minkowski norms differ:An isolated massive particle cannot decay into one photon in any inertial frame. Directly, a laboratory speed $u<c$ gives $E=\gamma m{c}^2$ and $|\mathbf{p}|=\gamma mu$; matching a photon would require $E=c|\mathbf{p}|$, hence $u=c$. Recoil against another object could change the kinematics, but would introduce an additional participant.

Write the four-velocity $U^{\mu }=\gamma (c,\mathbf{v})$ and four-acceleration $A^{\mu }=d{U}^{\mu }/d\tau$. Since $U_{\mu }{A}^{\mu }=0$, spatial orthogonality $\mathbf{U}\cdot \mathbf{A}=0$ implies $U^0{A}^0=0$. Thus $A^0=cd\gamma /d\tau =0$ and the speed is constant. If “acceleration” is read as the laboratory three-acceleration instead, $\gamma \mathbf{v}\cdot d\mathbf{v}/dt=0$ immediately gives the same conclusion.

For four-acceleration in uniform circular motion, the speed is $r|\omega |<c$, so $\gamma =(1-r^2{\omega }^2/c^2{)}^{-1/2}$ is constant. Differentiating the four-velocity using $d/d\tau =\gamma d/dt$ givesThe proper acceleration is $\sqrt{-A_{\mu }{A}^{\mu }}={\gamma }^{2}r{\omega }^2$. The vanishing time component reflects constant energy; the spatial components describe the changing momentum direction.

The geodesics of the product-weighted diagonal plane metric follow from the geodesic Lagrangian $L=(y^2{\dot{x}}^2+x^2{\dot{y}}^2)/2$. With dots denoting an affine parameter derivative, its Euler-Lagrange equations giveSubstituting $y=mx$ with constant nonzero $m$ reduces both equations to $x\ddot{x}+{\dot{x}}^2=0$. Thus $x\dot{x}$ is constant, and a suitable affine parametrization exists; explicitly, $x^2=C_1\lambda +C_2$ on a segment of fixed sign.

These straight rays are geodesics only in the regular domain $xy\ne 0$. The metric tensor determinant is $x^2{y}^2$. On the axes, including the origin, this is a degenerate metric tensor: the calculation does not establish a regular geodesic continuing through the origin. The slope-zero ray is outside the regular domain. Geometrically, on each quadrant the change $\rho =|xy|/\sqrt{2}$, $\vartheta =\log |y/x|$ gives $d{s}^2=d{\rho }^2+{\rho }^{2}d{\vartheta }^2$. A fixed slope has constant $\vartheta$ and is therefore locally a radial geodesic of a flat metric tensor.