Past exam of the relativity course of the University of Cambridge / 2021 / Section B / 4 / e / Solution

Normalize the five-dimensional tangent as in part (c), and let $s=\sqrt{1-{\xi }^2}>0$. Since $u=c\tau$, one has $d\tau /d\lambda =s/c$, soSubstitution into the corrected equation in part (d) givesThis is the Lorentz force form in four-dimensional Minkowski spacetime. The effective charge-to-mass coefficient is proportional to the conserved fifth-direction momentum: in $c=1$ units and a potential normalization identifying $A$ with the electromagnetic four-potential, $q/m=\xi /\sqrt{1-{\xi }^2}$. The factor converting the geometrical $A$ to an SI potential fixes the physical charge units. Opposite signs of $\xi$ give opposite effective charges, and $\xi =0$ gives a neutral inertial projection. Antisymmetry gives $d({\eta }_{ab}{u}^a{u}^b)/d\tau =0$.

This is a Kaluza-Klein theory mechanism: a higher-dimensional geodesic appears as forced motion after projection. With the explicitly supplied $+---$ base signature, the positive fibre term gives a second positive direction in the five-dimensional metric. The derivation is a valid geometrical reduction of that metric; a conventional single-time spacetime would require a spacelike extra direction and corresponding sign changes. If one instead retains the sign printed in part (d), the inferred effective charge reverses, but that version is not the geodesic equation of the metric supplied here.