Kaluza-Klein geodesic Lorentz force ($\frac{D{\dot{x}}^i}{ds}=k{F}^i{}_j{\dot{x}}^j$)

For $d{s}_5^2={\gamma }_{ij}d{x}^{i}d{x}^j+(dz+A_{i}d{x}^i{)}^2$ with $z$-independent coefficients, the geodesic Lagrangian makes $k=\dot{z}+A_i{\dot{x}}^i$ constant. Its remaining Euler-Lagrange equations give the displayed Lorentz force, where $F=dA$ and $D$ uses the Levi-Civita connection of $\gamma$. For a particle of five-dimensional mass $M$ with proper-time parameter $s$, the conserved momentum is $p_z=Mk$ and the four-dimensional rest mass is $m=M\sqrt{1+k^2}$. Reparametrizing by four-dimensional proper time gives $D{u}^i/d\tau =(p_z/m)F^i{}_j{u}^j$.

Take $L=[{\eta }_{ab}{\dot{x}}^a{\dot{x}}^b+W^2]/2$ with $W=\dot{y}+A_a{\dot{x}}^a=\xi$ constant. The Euler-Lagrange equation for $x^a$ isThus, with $F_{bc}={\partial }_b{A}_c-{\partial }_c{A}_b$,Reversing the explicit derivative order while retaining this metric and the same definition of $\xi$ changes the equation rather than merely changing a field-tensor naming convention. More generally, a fibre term $\sigma W^2$, $\sigma =\pm 1$, gives $\sigma \xi F^a{}_b{\dot{x}}^b$; its conserved canonical fibre momentum is $p_y=\sigma \xi$.

In the normalized positive-fibre case with $|\xi |<1$, four-dimensional proper time gives $d{u}^a/d\tau =c\xi F^a{}_b{u}^b/\sqrt{1-{\xi }^2}$. This has the Lorentz force form. The scale relating the geometrical $A$ to a physical electromagnetic four-potential fixes the conversion to an SI charge-to-mass ratio.

For $d{s}_5^2={\eta }_{ab}d{x}^{a}d{x}^b+(dy+A_{a}d{x}^a{)}^2$, with $A$ independent of $y$, an affine parameter $\lambda$ makes $\xi =dy/d\lambda +A_{a}d{x}^a/d\lambda$ and the tangent norm $\mathcal{N}=g_{AB}{\dot{x}}^A{\dot{x}}^B$ constant. The four-dimensional interval satisfiesConsequently the formula $d{\lambda }^2=d{u}^2/(1-{\xi }^2)$ assumes the additional normalization $\mathcal{N}=1$; affine parametrization alone does not set this value. With base metric signature $+---$ and $u=c\tau$, the projected curve is timelike only for $\mathcal{N}>{\xi }^2$. The normalized case gives $d\tau /d\lambda =\sqrt{1-{\xi }^2}/c$. The positive fibre term and the $+---$ base together have two positive directions; the same sign convention must be retained when interpreting the reduction.