Zero divergence with decreasing slice charge in an open region

Let $M={\mathbb{R}}^4$ with coordinates $(x^0,x^1,x^2,x^3)$, metric $\eta =\mathrm{diag}(-1,1,1,1)$, orientation $d{x}^0\wedge d{x}^1\wedge d{x}^2\wedge d{x}^3$, and future direction ${\mathrm{\partial }}_0$. Let $\mathrm{\Omega }=(0,L)\times (0,L_2)\times (0,L_3)$ and $K=[0,T]\times \mathrm{\Omega }$, with compact closure $\bar{K}=[0,T]\times [0,L]\times [0,L_2]\times [0,L_3]$. Define the constant vector $u=(\gamma ,\gamma \beta ,0,0)$ and scalar $s=s_0[1+a(x^1-\beta x^0)]$. Since $0<\beta <1$ and $aT<1$, $1+a(x^1-\beta x^0)\ge 1-a\beta T>0$ on $\bar{K}$, so $s$ is positive there and extends smoothly to a neighborhood of $\bar{K}$. Also $\eta (u,u)=-{\gamma }^2(1-{\beta }^2)=-1$ and $u$ is future-directed.

Let $Q$ be the closed rectangular hull of either $\mathrm{\Omega }$ or a side-face parameter rectangle ${\mathrm{\Omega }}_i$, and let $R$ be the corresponding endpoint-variant product. For a continuous density $f:Q\to \mathbb{R}$, the zero-extension convention in [F3] represents ${\int }_{R}f$ by a Riemann integral on $Q$. The extension differs from $f$ only on $\mathrm{\partial }Q$, which has content zero by [F5]. If $\mathrm{\mid }f\mathrm{\mid }\le M$, finite box covers of $\mathrm{\partial }Q$ with arbitrarily small total volume bound the change in the upper and lower Darboux sums by $2M$ times that volume; hence the two integrals agree as the volume tends to zero.

Put $J=su$, so $J^0=\gamma s$ and $J^1=\gamma \beta s$. In this chart $w=1$, and [F2] gives ${\mathrm{\nabla }}_a{J}^a={\mathrm{\partial }}_0{J}^0+{\mathrm{\partial }}_1{J}^1=\gamma (-s_{0}a\beta )+\gamma \beta (s_{0}a)=0$ throughout the slab; the remaining components vanish. Thus local production is zero everywhere.

By step 1.2, [F3] and [F5], the integral of the slice density over the open rectangle equals its ordinary integral over the closed rectangular hull. Integrating the affine density gives $Q_J(t)=\gamma s_{0}L{L}_2{L}_3(1+\frac{aL}{2}-a\beta t),$ so $Q_J(T)-Q_J(0)=-\gamma s_{0}a\beta TL{L}_2{L}_3<0.$ The slice charge strictly decreases.

Only the $x^1$ side faces carry flux. Their upper-minus-lower density is $J^1(t,L,x^2,x^3)-J^1(t,0,x^2,x^3)=\gamma \beta s_{0}aL$, independent of $t,x^2,x^3$. By step 1.2, the integral over the mixed-endpoint face parameter rectangle equals the integral over its closed rectangular hull, so the integrated net outward side flux is ${\mathrm{\Phi }}_J(B)=\gamma s_{0}a\beta TL{L}_2{L}_3>0.$ The factor $T$ is the duration of the slab. By [F4], $Q_J(T)-Q_J(0)+{\mathrm{\Phi }}_J(B)=0$, in agreement with the zero production calculation. The example refutes the unrestricted monotonicity assertion and shows that the side-flux hypothesis is essential.