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For a constant electromagnetic field tensor, the relativistic Lorentz force is a constant-coefficient linear system on four-velocity, with solutionThe electric-type blocks of $F$ generate hyperbolic rotations, while magnetic-type blocks generate ordinary rotations.

Electromagnetic minimal coupling replaces the free kinetic momentum in an action or Hamiltonian by a combination involving the electromagnetic potential. For $L=-m\sqrt{{\dot{x}}^a{\dot{x}}_a}-q{A}_a{\dot{x}}^a$, the canonical covector is $P_a=-m{\dot{x}}_a/\sqrt{{\dot{x}}^2}-q{A}_a$, so $P_a+q{A}_a$ recovers the potential-independent covector. Under $A_a\mapsto A_a+{\partial }_a\chi$, the action changes by an endpoint term. The fixed-endpoint equations are the relativistic Lorentz force equation and are gauge invariant.

Let a dot denote differentiation with respect to an affine geodesic parameter $s$. The five-dimensional geodesic Lagrangian isBecause $z$ is cyclic, its Euler-Lagrange equation gives the conserved quantityThis is the conserved momentum per unit five-dimensional mass associated with the Killing vector field ${\partial }_z$. For the other coordinates, differentiate $\partial L/\partial {\dot{x}}^i={\gamma }_{ij}{\dot{x}}^j+k{A}_i$. Using $\dot{k}=0$ givesThe first two terms combine into ${\gamma }_{ij}({\ddot{x}}^j+{\Gamma }^j{}_{lm}{\dot{x}}^l{\dot{x}}^m)$, with the Christoffel symbols of $\gamma$. Define the four-dimensional electromagnetic field tensor $F_{ij}={\partial }_i{A}_j-{\partial }_j{A}_i$ and raise its index with $\gamma$. The complete reduced geodesic equations areThis is the Kaluza-Klein geodesic Lorentz force. The Christoffel symbols describe the gravitational motion in general relativity in the four-dimensional metric tensor, while the right-hand side is antisymmetric in its lowered indices and gives the relativistic Lorentz force. In particular,Thus the force preserves the four-dimensional velocity norm, as a Lorentz force must. A particle with $k=0$ is neutral and follows an ordinary four-dimensional geodesic. Nonzero compact momentum gives electric charge, with its sign determined by the direction of motion around the circle.

To identify the actual charge-to-mass ratio, suppose the five-dimensional particle has mass $M>0$ and $s$ is its five-dimensional proper time, so $g_5(\dot{X},\dot{X})=-1$. Then $\gamma (\dot{x},\dot{x})=-(1+k^2)$. Four-dimensional proper time obeys $d\tau /ds=\sqrt{1+k^2}$; since $k$ is constant, $u^i=d{x}^i/d\tau$ satisfiesHere $q$ is in the geometric normalization of the supplied potential $A$. The relation $m^2=M^2+p_z^2$ also follows directly from the five-dimensional mass shell, and continues to identify the effective mass when the higher-dimensional particle is massless. The distinction between the two proper times prevents mistaking $k$ itself for the physical four-dimensional charge-to-mass ratio. A rescaling of $A$ to a canonically normalized electromagnetic potential correspondingly rescales $q$.

For the wave calculation, take a scalar field of five-dimensional mass $M$ in units $\hslash =c=1$, obeying $({\square }_5-M^2)\Psi =0$. In the coordinate order $(x^i,z)$, direct inversion of the metric tensor givesThe determinant follows by the triangular change of coframe from $(d{x}^i,dz)$ to $(d{x}^i,dz+A_{i}d{x}^i)$, whose determinant is one. Write $J=\sqrt{|\det \gamma |}$. Since all coefficients are $z$-independent, the scalar covariant wave operator isExpanding the right-hand side verifies both cross terms $-2A^i{\partial }_i{\partial }_z\Psi$, the term $-J^{-1}{\partial }_i(J{A}^i){\partial }_z\Psi$, and the coefficient $1+A_i{A}^i$ of ${\partial }_z^2\Psi$.

Single-valuedness under the compact coordinate period requires the Fourier seriesFor the $n$th Kaluza-Klein mode, put $q_n=n/R$ and $D_i={\partial }_i-i{q}_n{A}_i$. The five-dimensional Klein-Gordon equation reduces toThis is the charged four-dimensional Klein-Gordon equation: the divergence expression includes both the spacetime connection and the gauge connection. Its interpretation as electric charge can also be checked directly. Under the fibre coordinate change $z^{\prime }=z-\chi (x)$, the metric retains its form with $A^{\prime }=A+d\chi$, and the scalar Fourier coefficient changes by ${\psi }_n^{\prime }=e^{i{q}_n\chi }{\psi }_n$. Hence $D_i^{\prime }{\psi }_n^{\prime }=e^{i{q}_n\chi }{D}_i{\psi }_n$, exactly the transformation law for a field of charge $q_n$.

The Kaluza-Klein charge quantization is thereforePositive and negative integers give opposite charges, and the zero mode is neutral. Restoring $\hslash$ gives compact momenta $p_z=n\hslash /R$, so their fundamental spacing is $\hslash /R$. These are charges in the metric's geometric normalization. If $A_i=\kappa {\mathcal{A}}_i$ for a physically normalized electromagnetic potential ${\mathcal{A}}_i$, the physical charge unit is $\kappa /R$ in natural units, or $\kappa \hslash /R$ with $\hslash$ restored. Determining $\kappa$ requires an electromagnetic action normalization not specified by the metric ansatz alone.

Use the Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$ and define the electromagnetic field tensor byAs a rank-two Lorentz tensor, it transforms under a Lorentz transformation asTwo independent quadratic electromagnetic field invariants are

With $A^{\mu }=(\varphi /c,\mathbf{A})$, $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$, and $\mathbf{B}=\nabla \times \mathbf{A}$, the components areDirect contraction giveswhere the sign of $I_2$ changes if the opposite spacetime orientation is chosen. Since both are complete tensor contractions, $S$ and $T$ are Lorentz scalars.

For a constant uniform field, the relativistic Lorentz force is a constant-coefficient linear ordinary differential equation. Its solution is the matrix exponential

Now $S=T=0$ implies that the field is a null electromagnetic field:The stated axes allow us to choosePut $\Omega =qB/m$. On the $(ct,x,y)$ coordinates, the mixed tensor divided by $B$ isThe particle starts with four-velocity $u(0)=(c,0,0,0)$, so the exponential terminates after its quadratic term and givesIntegrating with the particle initially at the origin gives the relativistic trajectory in a constant null crossed fieldEliminating $\tau$ yieldsIf the original positive $x$ direction is opposite to $\mathbf{E}$, both the sign of $x$ and the displayed signed coefficient reverse.

For a constant electromagnetic field tensor, the relativistic Lorentz force is a linear system with constant coefficients, so

PutThe initial four-velocity is $(u^0,u^1,u^2,u^3)={\gamma }_0(c,0,v_0,0)$. For parallel fields along $x$, the field tensor has one hyperbolic block in the $(0,1)$ plane and one rotation block in the $(2,3)$ plane. With the orientation convention in the question,Taking the initial spacetime point to be the origin and integrating $u^{\mu }=d{x}^{\mu }/d\tau$ gives the relativistic motion in parallel electric and magnetic fieldsThe expressions have their continuous $a\to 0$ or $b\to 0$ limits when one field vanishes; reversing the sign of $q$ reverses the electric acceleration and magnetic rotation.

Eliminating proper time between the first two equations yieldsAt early coordinate time,which is uniformly accelerated motion with the transverse relativistic inertia factor ${\gamma }_0$. At late positive time,so the longitudinal speed approaches the speed of light.

The $y$-$z$ projection satisfiesIt is therefore a circle of radius and proper-time period

For the pitch magnitude assume $v_0>0$ and put $\alpha =|q|E/(mc)$, $\Omega =|q|B/m$, andNumber the $n$th proper-time period as $[(n-1)\Delta \tau ,n\Delta \tau ]$. Its longitudinal displacement isConsequently, as $n\to \infty$,where

Choose, for example,The fields are perpendicular and satisfy $E_0^2=c^2{B}_0^2$, so both invariants vanish. Every nonzero constant background with vanishing invariants can be brought to this form by a spatial rotation, up to signs.

In this frame, and with a convention consistent with the relativistic Lorentz force, the mixed tensor isCalling the displayed dimensionless matrix $N$, direct multiplication givesConsequentlyThis is a tensor equation. Since it holds in the canonical frame, it holds in every Lorentz frame for any null electromagnetic field.

Let $X^{\mu }(\tau )$ be the worldline, $u^{\mu }=d{X}^{\mu }/d\tau$ its four-velocity, andThe relativistic Lorentz force in the constant background isBecause $F^3=0$, its exponential terminates:The particle starts from rest, soUsing the matrix from part (c) givesIntegrating from the origin and using $X^0=ct$ yields the parametric trajectoryThis is the relativistic trajectory in a constant null crossed field. The real cubic relation for $t(\tau )$ is strictly increasing, so it determines $\tau$ uniquely as a function of coordinate time if an explicit $\mathbf{x}(t)$ is desired.

With $u^{\mu }=\gamma (c,\mathbf{v})$ and $d/d\tau =\gamma d/dt$, the spatial components of the relativistic Lorentz force giveThe time component gives $m{c}^{2}d\gamma /dt=q\mathbf{E}\cdot \mathbf{v}$. It says magnetic fields do no work and electric power changes relativistic energy.

With Cartesian Minkowski spacetime coordinates $x^{\mu }=(ct,x,y,z)$, introduce the four-current $J^{\mu }=(c\rho ,\mathbf{J})$ and electromagnetic four-potential ${\mathcal{A}}^{\mu }=(\varphi /c,\mathbf{A})$. The antisymmetric electromagnetic field tensor $F^{\mu \nu }={\partial }^{\mu }{\mathcal{A}}^{\nu }-{\partial }^{\nu }{\mathcal{A}}^{\mu }$ has components $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}^{ijk}{B}_k$. Maxwell equations becomeThe time and space components reproduce Gauss's law and the Ampère-Maxwell equation; the dual equation gives absence of magnetic monopoles and Faraday's law of induction. Antisymmetry gives charge conservation ${\partial }_{\nu }{J}^{\nu }=0$. The relativistic Lorentz force is $d{p}^{\mu }/d\tau =q{F}^{\mu \nu }{u}_{\nu }$. Every displayed equation transforms tensorially under a Lorentz transformation; the electric field and magnetic field are different components of the same object.

Two independent algebraic electromagnetic field invariants areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations but changes sign under spatial reflection. For perpendicular fields, the sign of the first distinguishes a frame with purely magnetic field from one with purely electric field. Both invariants vanish for a plane electromagnetic wave with $\mathbf{E}\perp \mathbf{B}$ and $E=cB$, although the field itself is nonzero.

For the electric-dominated crossed-field orbit, choose a boost $V=c^{2}B/E$ along $x$. Since $|E|>c|B|$, it is subluminal. The Lorentz transformation of electromagnetic fields gives $B_z^{\prime }=0$ and $E_y^{\prime }=E/\Gamma$, where $\Gamma =(1-V^2/c^2{)}^{-1/2}$. The initially stationary electron has $U^{\prime }(0)=(\Gamma c,-\Gamma V,0,0)$, not zero velocity in this frame.

Let $q=-e$, $a=q{E}_y^{\prime }/m_e$ and $s=a\tau /c$. Neglect radiation reaction, as in the prescribed external-field model. Integrating the relativistic Lorentz force in this purely electric frame gives the exact trajectoryIndeed $U^{\prime 0}=\Gamma c\cosh s$, $U^{\prime x}=-\Gamma V$, and $U^{\prime y}=\Gamma c\sinh s$ satisfy both the force equation and $U^{\prime 2}=c^2$. The constant transverse momentum must be retained.

The inverse boost gives a useful exact laboratory parametrization as well:For $E,B>0$, the lab electron starts accelerating toward negative $y$ and bends toward positive $x$, the direction of $\mathbf{E}\times \mathbf{B}$. It never acquires a cyclotron orbit in this electric-dominated case. At late times, $v_x\to V$ and $v_y\to \mathrm{sgn}(q{E}^{\prime })c/\Gamma$, so its speed approaches $c$ along an oblique asymptote. Reversing field signs reverses the appropriate directions. The zero-$B$ limit is the ordinary electric-field hyperbolic motion.

Choose Minkowski metric $(+,-,-,-)$, so $A_{\mu }=(\varphi /c,-\mathbf{A})$ and $d\tau =dt\sqrt{1-v^2/c^2}$. In laboratory time the relativistic charged-particle action has LagrangianHere $e$ denotes the signed particle charge in the action. The canonical momentum and the position derivative areThe Euler-Lagrange equation therefore givesThe last term is $(\mathbf{v}\times \mathbf{B}{)}_i$, and the first two terms form the electric field. Thus the relativistic Lorentz force equation isIts power relation and acceleration form areThe power relation follows by taking the scalar product with $\mathbf{v}$ and using the relativistic energy-momentum relation. It also shows why the magnetic field alone cannot change the particle's energy.

With no electric field, the power relation from part (a) givesConsequentlyThe magnetic part of the relativistic Lorentz force changes the direction of mechanical momentum, but performs no work. This conclusion does not require the magnetic field to be spatially uniform: it follows from the local orthogonality of $\mathbf{v}$ and $\mathbf{v}\times \mathbf{B}$.

Let $\mathbf{b}=\mathbf{B}/B$ for a nonzero constant magnetic field. Since $\gamma$ is constant, the relativistic Lorentz force reduces toThe projection along $\mathbf{b}$ is constant. In the perpendicular plane define the signed cyclotron frequency $\Omega =eB/(m{\gamma }_0)$; thenThe explicit rotating velocity isThis is the relativistic cyclotron orbit frequency in laboratory time, reduced from the nonrelativistic frequency by $1/{\gamma }_0$. With nonzero parallel velocity, the trajectory is a helix. If ${\mathbf{v}}_{\perp 0}=0$, that perpendicular vector stays zero and has no nontrivial fundamental period; the displayed $T$ is the period of a nonzero transverse orbit.

When uniform $\mathbf{E}$ and $\mathbf{B}$ are parallel, the relativistic Lorentz force splits into independent hyperbolic motion along the fields and circular motion in the perpendicular plane. The circle is periodic in proper time, while electric acceleration makes the helix's longitudinal pitch grow exponentially from one proper-time cyclotron period to the next.