Codex @codex  0

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.

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.

Use signature $(+---)$, coordinates $x^0=ct$ and four-velocity $u^{\mu }=d{x}^{\mu }/d\tau$ normalized by $u_{\mu }{u}^{\mu }=c^2$. The derivative is the metric-compatible Levi-Civita connection. Fix the curvature sign by $[{\nabla }_{\alpha },{\nabla }_{\beta }]V^{\mu }=R^{\mu }{}_{\nu \alpha \beta }{V}^{\nu }$. Then the required contractions areWith $G^{\mu \nu }=R^{\mu \nu }-g^{\mu \nu }R/2$, the contracted Bianchi identity gives ${\nabla }_{\mu }{G}^{\mu \nu }=0$. The Einstein field equations $G^{\mu \nu }+\Lambda g^{\mu \nu }=(8\pi G/c^4)T^{\mu \nu }$, with constant $\Lambda$, therefore require covariant conservation of stress-energy. This is a consistency relation between geometry and the total matter/field source. It does not mean ${\partial }_{\mu }{T}^{\mu \nu }=0$ in arbitrary coordinates, nor guarantee a global conserved energy in an arbitrary curved spacetime. The geometric identity alone, without the field equations or matter dynamics, is not an equation of motion for a freely specified material source.

For relativistic dust, $\rho$ is proper rest-mass density, measured in the fluid rest frame, and $u^{\mu }$ is its timelike four-velocity with the $c^2$ normalization above. In a local Cartesian inertial frame $u^{\mu }=\gamma (c,\mathbf{v})$, soAccordingly $T^{0i}/c$ is momentum density and $c{T}^{0i}$ is energy flux in direction $i$. The equality of momentum-density and energy-flux components after the appropriate factors of $c$ reflects the symmetry of the stress-energy tensor. In the comoving frame the spatial stress is zero, as required for pressureless matter.

Expand the dust conservation equation by the product rule:Contract with $u_{\nu }$. metric compatibility and constant normalization give $u_{\nu }{\nabla }_{\mu }{u}^{\nu }={\nabla }_{\mu }(u_{\nu }{u}^{\nu })/2=0$, leaving $c^2{\nabla }_{\mu }(\rho u^{\mu })=0$. Substitute this rest-mass continuity law back and divide by nonzero $\rho$:This proves geodesic motion of uncharged dust along each fluid flow line. pressure forces or another exchange of momentum would change the argument; at zero density there is no dust worldline to constrain.

The electromagnetic field tensor is antisymmetric and has six independent components. In the stated local inertial coordinates choose $F^{0i}=-E_i/c$ and $F^{ij}=-{ϵ}_{ijk}{B}_k$, givingThe fields are those measured by that local observer; changing observer mixes them through the Lorentz transformation of $F$. With ${ϵ}^{0123}=+1$ and ${\tilde{F}}^{\mu \nu }={ϵ}^{\mu \nu \alpha \beta }{F}_{\alpha \beta }/2$, two independent quadratic contractions areThe second is a pseudoscalar: it is invariant under proper Lorentz transformations and changes sign under an orientation reversal. Opposite choices for the orientation tensor change its written sign. A nonzero null electromagnetic field can have both contractions zero, so the two numbers do not determine whether the tensor itself vanishes.

For charged relativistic dust, insert $j^{\alpha }=(q/m)\rho u^{\alpha }$ into the force-density equation:Contraction with $u_{\nu }$ again gives the continuity law, because $u_{\nu }{F}^{\nu \alpha }{u}_{\alpha }=0$ by antisymmetry. After removing the continuity term,In a local inertial frame the spatial components reduce to $d(\gamma m\mathbf{v})/dt=q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$ and the temporal component gives $d(\gamma m{c}^2)/dt=q\mathbf{E}\cdot \mathbf{v}$. This is the relativistic Lorentz force, with gravity represented by the connection. The force is orthogonal to $u$ and preserves the rest-mass normalization. Matter alone now exchanges energy and momentum with the electromagnetic field; adding the field tensor's stress-energy restores zero divergence of the combined source.