= Solution
Normalize the five-dimensional tangent as in part (c), and let $s=\sqrt{1-\xi^2}>0$. Since $u=c\tau$, one has $d\tau/d\lambda=s/c$, so
$$
\dot x^a=\frac{s}{c}u^a,\qquad
\ddot x^a=\frac{s^2}{c^2}\frac{du^a}{d\tau}.
$$
Substitution into the corrected equation in part (d) gives
$$
\boxed{\frac{du^a}{d\tau}=\frac{c\xi}{\sqrt{1-\xi^2}}F^a{}_b u^b.}
$$
This 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 $d(\eta_{ab}u^au^b)/d\tau=0$.
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.
Back to article page
