Physical meaning of $F^{0\nu}j_\nu$ In my problem I showed that $F^{\mu\nu}j_\nu$ is a contravariant 4-vector. Now the question is what is its 0-component's physical meaning, i.e. the meaning of $F^{0\nu}j_\nu$ if $F^{\mu\nu}$ is the electromagnetic tensor and $j$ the 4-current $(\rho,\vec{j})$. From the definition of $F^{\mu\nu}$ I know that $$F^{0\nu}j_\nu = -\vec{E}\cdot\vec{j},$$ but I can't figure out what the physical significance of this is.
 A: 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.

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) $\mathrm{dp}\boldsymbol{=}\mathrm{dw}/\mathrm{d}t\boldsymbol{=}\mathbf{F}\boldsymbol{\cdot}\mathrm{d}\mathbf{x}/\mathrm{d}t\boldsymbol{=}\mathbf{F}\boldsymbol{\cdot}\mathbf{u}$, 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 $\rho\boldsymbol{=}\mathrm{d}Q/\mathrm{d}V$ the electric charge volume density.
