Smooth formal matter, EM tensors and observer quantities — Examples

Let $M={\mathbb{R}}^4$ with length coordinates $(x^0,x^1,x^2,x^3)$, Minkowski metric $\eta =\mathrm{diag}(-1,1,1,1)$, orientation $d{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$, future time-orientation $N={\mathrm{\partial }}_0$, and $x^0=ct$. In SI bookkeeping, assign $[{\rho }_0]=\mathrm{k}\mathrm{g}{\mathrm{m}}^{-3}$, $[L]=\mathrm{m}$, and $[{\alpha }_{\ast }]=\mathrm{C}\mathrm{k}{\mathrm{g}}^{-1}$. Choose ${\rho }_0>0$, $L>0$, and ${\alpha }_{\ast }\mathrm{\ne }0$. Set

$U=c{\mathrm{\partial }}_0,\rho ={\rho }_0,\alpha (x)={\alpha }_{\ast }(1+(x^1\mathrm{/}L{)}^2),\sigma ={\rho }_0\alpha ,F=0.$

The fields are smooth and the ratio $\alpha =\sigma \mathrm{/}\rho$ is nonconstant in space. This is a mathematical example of the matter balance and transport equations; the sourced Maxwell equation is not imposed.

The example is not a coupled Einstein–Maxwell solution and makes no claim about physical existence. It shows exactly why constant $\sigma \mathrm{/}\rho$ is not a consequence of the dust exchange identity.

Let $M={\mathbb{R}}^4$ with global length coordinates $(x^0,x^1,x^2,x^3)$ and metric components $g_{ab}={\eta }_{ab}=\mathrm{diag}(-1,1,1,1)$ in the coordinate frame ${\mathrm{\partial }}_a:=\mathrm{\partial }\mathrm{/}\mathrm{\partial }{x}^a$. Use this metric to raise and lower indices and its Levi–Civita connection $\mathrm{\nabla }$. Take orientation $d{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$, future time-orientation $N={\mathrm{\partial }}_0$, and the positive parameters $c,{\mu }_0$ of Smooth formal matter, EM tensors and observer quantities, with $x^0=ct$. Choose constants with SI units $[E_0]=\mathrm{V}{\mathrm{m}}^{-1}$, $[\chi ]=\mathrm{S}{\mathrm{m}}^{-1}$, and $[{ϵ}_i]=[p_0]=\mathrm{J}{\mathrm{m}}^{-3}$, with $E_0\mathrm{\ne }0$, $\chi >0$, ${ϵ}_i>0$, and $p_0\in \mathbb{R}$. Let $I=(-{ϵ}_i\mathrm{/}(2\chi E_0^2),{ϵ}_i\mathrm{/}(2\chi E_0^2))$, which has units of seconds; put $t=x^0\mathrm{/}c$ and restrict all fields to the open spacetime domain $\mathcal{D}:=\{x\in {\mathbb{R}}^4:t(x)\in I\}\cong {\mathbb{R}}^3\times I$. The maps $\mathbf{E},\mathbf{B}:\mathcal{D}\to {\mathbb{R}}^3$ below give the spatial components relative to the positively oriented orthonormal frame $({\mathrm{\partial }}_0,{\mathrm{\partial }}_1,{\mathrm{\partial }}_2,{\mathrm{\partial }}_3)$, with ${\mathbf{e}}_x=(1,0,0)$. Define

$U=c{\mathrm{\partial }}_0,\mathbf{E}=E_0{\mathbf{e}}_x,\mathbf{B}=0,ϵ(t)={ϵ}_i+\chi E_0^{2}t,p=p_0,\sigma =0,q^a=(0,\chi E_0,0,0)\mathrm{.}$

Let $F$ be the antisymmetric covariant two-tensor with $F_{01}=-E_0\mathrm{/}c$ and all other independent components zero, and set $j=q$. All the displayed scalar, vector and tensor fields are smooth on $\mathcal{D}$. The interval gives ${ϵ}_i\mathrm{/}2<ϵ(t)<3{ϵ}_i\mathrm{/}2$ throughout $\mathcal{D}$. The imposed field is a fixed background for the matter calculation; the nonzero current is not claimed to solve its Maxwell equation.

The calculation is local and includes no equation of state, entropy balance, electromagnetic backreaction, or global fluid solution. The units $[\chi ]=\mathrm{S}{\mathrm{m}}^{-1}$ and $[E_0]=\mathrm{V}{\mathrm{m}}^{-1}$ make $\chi E_0^2$ a power density, but the example reports no measurement.

Work pointwise on Minkowski space with length coordinates $x^0=ct$, signature $(-,+,+,+)$, orientation $d{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$, and future time-orientation $N={\mathrm{\partial }}_0$. In SI bookkeeping assign $[B_0]=\mathrm{T}$ and $[E_0]=\mathrm{V}{\mathrm{m}}^{-1}$. First take $\mathbf{E}=0$, $\mathbf{B}=B_0{\mathbf{e}}_z$ with $B_0\mathrm{\ne }0$. Then take $\mathbf{E}=E_0{\mathbf{e}}_x$, $\mathbf{B}=(E_0\mathrm{/}c){\mathbf{e}}_y$ with $E_0\mathrm{\ne }0$. These are explicit field-component calculations; no coupled Einstein solution is asserted.

The first tensor is anisotropic even though its rest-frame spatial energy-current component vanishes. In the crossed-field case, the observer energy-current vector is future null, while the physical SI energy-flux vector is the nonzero spatial vector $(E_0^2\mathrm{/}({\mu }_{0}c)){\mathbf{e}}_z$ with magnitude $cu$. These are exact algebraic consequences of the displayed tensors and do not describe measured fields.

Let $\mathrm{\Omega }\subseteq {\mathbb{R}}^3$ be open and let $\varphi \in C^{\mathrm{\infty }}(\mathrm{\Omega },\mathbb{R})$, assigned volts in SI bookkeeping. On $M={\mathbb{R}}_{x^0}\times \mathrm{\Omega }$ with Minkowski metric, orientation $d{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$, future time-orientation $N={\mathrm{\partial }}_0$, and length coordinate $x^0=ct$, take the static one-form $A=-(\varphi \mathrm{/}c)d{x}^0$ and $F=dA$. Let $I\subseteq \mathbb{R}$ be an open interval with nonempty interior (Intervals of $\mathbb{R}$: the nine order-convex forms, nondegeneracy, and length) and let $z:I\to M$ be an existing $C^2$ future-timelike curve parametrized by proper time $\tau$ in the sense of Timelike worldlines, proper time, length and four-velocity on a Lorentzian four-manifold, with $U=dz\mathrm{/}d\tau$, $g(U,U)=-c^2$, and $m{\mathrm{\nabla }}_{U}U=q{F}^{\mathrm{♯}}(U)$ for constants $m>0$ and $q\in \mathbb{R}$, assigned kilograms and coulombs respectively. Write $\mathbf{w}=d\mathbf{x}\mathrm{/}dt$, $\mathrm{\mid }\mathbf{w}\mathrm{\mid }<c$, and ${\gamma }_L=(1-\mathrm{\mid }\mathbf{w}{\mathrm{\mid }}^2\mathrm{/}{c}^2{)}^{-1\mathrm{/}2}$.

This is a test-particle first integral in a prescribed static background. It does not assert that Minkowski space with nonzero $F$ solves the coupled Einstein equation. The expression uses kinetic momentum and the stated potential gauge; general gauge changes are covered by the compensator in the supplier theorem.