Physical meaning of $F^{0nu}j_nu$

You must see your expression as $::mathrm{P}boldsymbol{=}mathbf{E}boldsymbol{cdot}mathbf{j}boldsymbol{=}mathbf{E}boldsymbol{cdot}rho,mathbf{u}::$ where $mathbf{E}$ the electric field intensity 3-vector, $mathbf{j}boldsymbol{=}rho,mathbf{u}$ the electric current density 3-vector, $rho$ the electric charge density and $mathbf{u}$ the charge velocity 3-vector. Then $mathrm{P}$ is the work done per unit time (that is power) per unit volume.

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This is rate of energy per unit volume produced or absorbed by the application of the electric force on the moving charges. The infinitesimal work done ($boldsymbol{+}$ if produced,$boldsymbol{-}$ if absorbed) by the application of a force $mathbf{F}$ on a particle moving infinetisimally by $mathrm{d}mathbf{x}$ is $mathrm{dw}boldsymbol{=}mathbf{F}boldsymbol{cdot}mathrm{d}mathbf{x}$. If this motion takes place in time $mathrm{d}t$ then we have work per unit time (that is power) $mathrm{dp}boldsymbol{=}mathrm{dw}/mathrm{d}tboldsymbol{=}mathbf{F}boldsymbol{cdot}mathrm{d}mathbf{x}/mathrm{d}tboldsymbol{=}mathbf{F}boldsymbol{cdot}mathbf{u}$, where $mathbf{u}boldsymbol{=}mathrm{d}mathbf{x}/mathrm{d}t$ the velocity of the particle.

For a group of particles of infinitesimal electric charge $mathrm{d}Q$ constrained in an infinitesimal volume $mathrm{d}V$ and moving in an electric field $mathbf{E}$ with velocity $mathbf{u}$ then $mathbf{F}boldsymbol{=}mathrm{d}Q,mathbf{E}$ and the work done per unit time (power) per unit volume is $mathrm{P}boldsymbol{=}mathrm{d}Q/mathrm{d}V,mathbf{E}boldsymbol{cdot}mathbf{u}boldsymbol{=}rho,mathbf{E}boldsymbol{cdot}mathbf{u}$ where $rhoboldsymbol{=}mathrm{d}Q/mathrm{d}V$ the electric charge volume density.