Divergence-free stress-energy with a nonconserved slice integral

The field. Define the smooth symmetric field by $T^{00}=x^0$, $T^{01}=T^{10}=-x^1$, $T^{11}=x^0$, all other components zero. By [F2] its force density has components $f^0={\mathrm{\partial }}_0{x}^0+{\mathrm{\partial }}_1(-x^1)=1-1=0$, $f^1={\mathrm{\partial }}_0(-x^1)+{\mathrm{\partial }}_1{x}^0=0$ and $f^2=f^3=0$; hence ${\mathrm{\partial }}_a{T}^{ab}=0$ on all of affine Minkowski space and $T$ is locally conserved.

The slice integral is not conserved. Let $L>0$, $c_2<d_2$, $c_3<d_3$, put $A=(d_2-c_2)(d_3-c_3)>0$ and $K=[0,L]\times [c_2,d_2]\times [c_3,d_3]$, and let $Q^0(s)={\int }_K{T}^{00}(s,x)d^{3}x=A{\int }_0^L{x}^0(s)d{x}^1=ALs$. For $s_1<s_2$, [F1] applied to the sub-box with $f=0$ gives $Q^0(s_2)-Q^0(s_1)=-({\int }_{[s_1,s_2]\times K_1^{+}}{T}^{10}-{\int }_{[s_1,s_2]\times K_1^{-}}{T}^{10})-0$, and $T^{10}=-x^1$ takes the constant values $-L$ on $K_1^{+}=\{x^1=L\}$ and $0$ on $K_1^{-}=\{x^1=0\}$, so the side-flux difference equals $-A(s_2-s_1)\cdot (L-0)=-AL(s_2-s_1)$ and $Q^0(s_2)-Q^0(s_1)=AL(s_2-s_1)\mathrm{\ne }0$ whenever $s_1<s_2$; the transverse fluxes vanish because $T^{2b}=T^{3b}=0$. For example $A=6$, $L=2$ and $s_2-s_1=3$ give the value $36$.

Thus zero local divergence alone does not conserve a finite slice integral: this field has $f=0$ everywhere yet its slice integral $Q^0$ varies linearly with the slice parameter, so the side-flux (isolation) hypothesis in the finite-flux conservation statement cannot be dropped. The prescribed tensor is not claimed to arise from any matter model of this page and no choice principle is used.