So in wikipedia for the Poynting's theorem we have:

$-\frac{\partial u}{\partial t}= \nabla \cdot\vec S + \vec j \cdot\vec E$


$-\frac{\partial u}{\partial t}$ is the rate of change of energy density per unit volume.

$\nabla\cdot \vec S$ is energy flow out of the volume, given by the divergence of the Poynting vector $\vec S$.

$\vec j \cdot\vec E$ is the rate at which the fields do work on a charges in the volume ($\vec J$ is the current density corresponding to the motion of charge, $\vec E$ is the electric field, and $\cdot$ is the dot product).

I don't understand this 3rd component in our equation.

  1. How is rate of work connected with the dot product of the current density and Electric field?

  2. Since the Poynting theorem considers electromagnetic fields, why do we observe only the realtionship between the current density and electric filed $\vec E$ but also not the relationship between $\vec j$ and $\vec B$?

  3. What is the physical meaning of doing this scalar product?

  • 1
    $\begingroup$ What about the explanation in the derivation section of the Wiki article is unclear to you? $\endgroup$
    – ACuriousMind
    Commented Dec 17, 2021 at 15:14
  • $\begingroup$ The third term is the rate of change of the EMF energy density per unit volume. It may describe dissipation (due to resistance, for example) or gain (amplifying the field by the charges, like in lasers with "free" electrons). $\endgroup$ Commented Dec 17, 2021 at 15:49
  • $\begingroup$ Also "\cdot" is the command for a dot product :) $\endgroup$ Commented Dec 17, 2021 at 16:16

1 Answer 1


Poyntings theorem can be derived with 0 assumptions of energy conservation.It can be entirely derived from maxwell equations and work = $f.dr$

Consider I have a charge distribution.

Then the force element on a volume dv is given by

$df= (\vec{E}+\vec{v}×\vec{B})\rho dv$

$df= \vec{E}\rho dv+(\vec{v}×\vec{B})\rho dv$

an infinitesimal work element, is then given b

$df \cdot dr = \vec{E} \cdot \vec{dr} \rho dv+(\vec{v}×\vec{B})\cdot \vec{dr} \rho dv$

$dw = \vec{E} \cdot \vec{dr} \rho dv+(\vec{v}×\vec{B})\cdot \vec{dr} \rho dv$

dr isn't a line element in the form r(t), it is instead a position vector field line element representing the paths that each charge $\rho dv$ takes,

this is equivelent to $ dr = \vec{V} dt$ thus,

$dw = \rho\vec{E} \cdot \vec{V}dt dv+\rho(\vec{v}×\vec{B})\cdot \vec{V}dt dv$

the second term is zero since a vector by definition perpendicular to V, dotted with V is zero ( aka, magnetic fields do no work)

So the qauntity

$dw = \rho\vec{E} \cdot \vec{V}dt dv$

Represents the infinitesimal amount of work done by a single DQ of charge moving a tiny infinitesimal distance dr, ( or the distance moved by that charge with a velocity "V" in time "dt")

Meaning the rate at which work is done on a charge DQ(divided by dt) must be given by

$\frac{dw}{dt} = \rho\vec{E} \cdot \vec{V} dv$

Which, substituting in the definition of J is equivelant to

$\frac{dw}{dt} = \vec{E} \cdot \vec{J} dv$

This is the rate at which work is being done on an infinitesimal charge DQ,

lastly, to find the TOTAL rate at which work is being on charges in a volume, you simply integrate this about some volume

$\frac{dW}{dt} = \iiint \vec{E} \cdot \vec{J} dv$

I have chosen to work with rate of work on a single dq and then integrate last, you could have done the same integrating first

From here you can then derive poyntings theorem by eliminating J in favour of the fields using amperes law

To answer those last questions, E dot J intiluatively is the component of the velocity of my charge in the direction of the field, which is indicative of the work expression. There is also no relationship between J and B since magnetic fields do no work


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