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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.

The electromagnetic four-potential is $A^{\mu }=(\Phi /c,\mathbf{A})$, and the vacuum four-wavevector is $K^{\mu }=(\omega /c,0,0,k)$ with $k=\omega /c>0$. Define $\psi =K_{\mu }{X}^{\mu }=\omega t-kz$ and $C=(\omega k{)}^{-1/2}$. The given transverse potential is $A_y=-C\sin \psi$. Its physical fields areso the Poynting vector points along positive $z$. For our signature, ${\partial }^{\mu }=(c^{-1}{\partial }_t,-\nabla )$, $F^{0i}=-E_i/c$, and $F^{ij}=-{ϵ}_{ijk}{B}_k$. ThereforeThe square-root prefactor is simply the source's chosen wave normalization; an overall dimensional amplitude could multiply every field.

The four-current is $J^{\mu }=(c\rho ,\mathbf{j})$, with $J_{\mu }=(c\rho ,-\mathbf{j})$. Charge conservation takes the equivalent formsLowering both indices of $F$ changes the sign of its time-space entries: $F_{0i}=E_i/c$, while $F_{ij}=-{ϵ}_{ijk}{B}_k$. The temporal component of ${\partial }^{\mu }{F}_{\mu \nu }$ is $\nabla \cdot \mathbf{E}/c={\mu }_{0}c\rho$, by Gauss's law. Its spatial components are $c^{-2}{\partial }_t\mathbf{E}-\nabla \times \mathbf{B}=-{\mu }_0\mathbf{j}$, by the Ampère-Maxwell equation. HenceFor the displayed free wave the current vanishes because $k=\omega /c$; the component derivation also proves the sourced identity. The two homogeneous Maxwell equations follow separately from expressing $F$ in terms of a potential.

Force-sign qualification. With these definitions the physical Lorentz force density is $f_{\mu }^{\mathrm{phys}}=F_{\mu \nu }{J}^{\nu }$, whereas the source prints the reversed contraction $J^{\nu }{F}_{\nu \mu }=-f_{\mu }^{\mathrm{phys}}$. ConsequentlyThe physical time component is the power transferred to matter per volume divided by $c$, as in Poynting's theorem. For example parallel positive $j_x$ and $E_x$ do positive work, but the literal printed contraction has negative time component. This sign cannot be repaired while simultaneously retaining the source's tensor definition and the Maxwell identity above; reverse the force contraction to describe the physical force.

Use $c=1$ and the mostly-plus Minkowski metric fixed in part a. The electromagnetic four-potential and its covariant form areDefine the electromagnetic field tensor by $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. Since $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$,Raising both indices changes the signs of its time-space entries. The lower-index signs agree with ^[ref]; the signature and index placement must be kept consistent.

For a passive boost to a frame moving with $v\hat{\mathbf{x}}$, transform the tensor asFor example, $F^{\prime 02}=\gamma F^{02}-\gamma v{F}^{12}=\gamma (E_y-v{B}_z)$ and $F^{\prime 12}=\gamma (B_z-v{E}_y)$. Reading off all entries givesFor an arbitrary boost velocity $\mathbf{v}$, rotational covariance gives the Lorentz transformation of electromagnetic fieldsHere all velocities are in $c=1$ units. Equivalently the parallel components are unchanged and the perpendicular components have the cross-product expressions multiplied by $\gamma$.

The scalar contraction of the electromagnetic field tensor isWith orientation ${ϵ}^{0123}=+1$, define the dual electromagnetic field tensor by ${\tilde{F}}^{\mu \nu }=\frac{1}{2}{ϵ}^{\mu \nu \rho \sigma }{F}_{\rho \sigma }$. Direct substitution givesThe first is a Lorentz scalar; the second is a pseudoscalar, invariant under proper Lorentz transformations such as boosts and rotations. Under spatial inversion $\mathbf{E}$ changes sign while $\mathbf{B}$ does not, so $\mathbf{E}\cdot \mathbf{B}$ reverses sign. For example, $\mathbf{E}=\mathbf{B}=\hat{\mathbf{z}}$ has product $+1$ before inversion and $-1$ afterwards. Thus the unrestricted statement including improper transformations needs this qualification.

For the spacetime integral, assume a globally regular electromagnetic four-potential on ordinary Minkowski space and sufficient boundary decay. Using $\nabla \cdot \mathbf{B}=0$ and ${\partial }_t\mathbf{B}=-\nabla \times \mathbf{E}$,ConsequentlyThe magnetic helicity boundary term controls the integral. If its surface flux vanishes and the magnetic helicity $H_B(t)=\int \mathbf{A}\cdot \mathbf{B}{d}^{3}x$ vanishes at both temporal ends, thenCompactly supported smooth potentials are one sufficient version of the intended falloff assumption. The local identity is exact, but vanishing fields at spatial infinity alone is insufficient. For a concrete counterexample, take $\varphi =0$ and, in dimensionless coordinates,where $a$ changes smoothly from zero to one and $\dot{a}$ has compact temporal support. Both electric fields and magnetic fields vanish rapidly at spatial infinity, butThe counterexample leaves a final magnetic field and therefore fails the extra temporal boundary condition. It shows exactly why the printed integral cannot follow from a merely spatial reading of “at infinity”; vanishing of the full boundary term is the needed hypothesis.

Use Minkowski metric $g_{\mu \nu }=\mathrm{diag}(-1,1,1,1)$ and units with the speed of light equal to one. Thus $u^{\mu }{u}_{\mu }=-1$, and $h_{\mu \nu }=g_{\mu \nu }+u_{\mu }{u}_{\nu }$ is the spatial projector of a timelike congruence. Keep the medium velocity fixed when varying the electromagnetic four-potential. Since the electromagnetic field tensor is $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$, its variation givesBoth tensors are antisymmetric. Integration by parts, with variations vanishing at the boundary, therefore givesThe Euler-Lagrange field equation is ${\partial }_{\mu }(F^{\mu \nu }-M^{\mu \nu })=0$. Renaming the free index and using antisymmetry givesThese are the inhomogeneous Maxwell equations with the bound charge and bound current encoded by the polarization-magnetization tensor.

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.