The partial of tensor is equal
$$\partial_{\mu}T^{\mu j} = \frac{\partial}{\partial t}T^{0j} + \nabla_{j} T^{ij} = -(\rho \vec{E} + \vec{j}\; \times \vec{B})^{j}$$
where does it come from?
In the equation above the term
$$T^{0j} = \vec{S} = \frac{1}{4 \pi}\vec{E} \; \times \vec{B}$$
I recognize as Poynting vector.
From momentum conservation
$$\frac{d\vec{P}^{j}}{dt}=0$$
follows equation
$$\frac{d\vec{P}^{j}}{dt}= - \int_{V} d^{3} x \rho E^{j} - \int_{V} d^{3} x (\vec{j} \; \times \vec{B})^{j} - \int_{V} d \sigma_{i} T^{ij}$$
where $$P = \int_{V} d^{3} x \vec{S}$$
I wanted to know what right hand side presents?
$$RHS = - \int_{V} d^{3} x \rho E^{j} - \int_{V} d^{3} x (\vec{j} \; \times \vec{B})^{j} - \int_{V} d \sigma_{i} T^{ij}$$
I've read that it is the Poynting vector evaluating in time, but I didn't see that.
Thank you for your explanation with mathematical description.
