Recall from elementary physics that the rate at which work is done on an electric charge by an electromagnetic field is: P = ·= q·= ·q If one follows the usual method of constructing a current density made up of many charges, it is easy to show that this generalizes to: dPdV = · for the rate at which an electric field does work on a current density throughout a volume. The magnetic field, of course, does no work because the force it creates is always perpendicular to or .

If we use AL to eliminate = - t and integrate over a volume of space to compute the rate the electromagnetic field is doing work within that volume: P = &int#int;· d^3x_0 = &int#int;{·() - ·t} d^3x_0

Using: (×) = ·() - ·() (which can be easily shown to be true as an identity by distributing the derivatives) and then use FL to eliminate , one gets: &int#int;· d^3x_0 = - &int#int;{(×) + ·t + ·t } d^3x_0

It is easy to see that:
&epsi#epsilon;·t & = & 2 ·t

1&mu#mu; ·t & = & 2
·t
from which we see that these terms are the time derivative of the *electromagnetic field energy density*:
&eta#eta;= 12&epsi#epsilon;·+ 12&mu#mu;·=
12(·+ ·)
Moving the sign to the other side of the power equation above, we get:
- &int#int;_V · d^3x_0 = &int#int;_V { (×) +
&eta#eta;t } d^3x_0
as the rate at which power flows *out* of the volume
(which is
arbitrary).

Equating the terms under the integral:
&eta#eta;t + = - ·
where we introduce the *Poynting vector*
= ×

This has the precise appearance of conservation law. If we apply the
divergence theorem to the integral form to change the volume integral of
the divergence of
into a surface integral of its flux:
&conint#oint;_&sigma#sigma;· dA + t &int#int;_V/&sigma#sigma; &eta#eta;
dV + &int#int;_V/&sigma#sigma; · dV = 0
where
is the closed surface that bounds the volume
.
Either the differential or integral forms constitute the *Poynting
Theorem*.

In words, the sum of the work done by all fields on charges in the
volume, plus the changes in the field energy within the volume, plus the
energy that flows out of the volume carried by the field must *balance* - this is a version of the *work-energy theorem*, but one
expressed in terms of the *fields*.

In this interpretation, we see that
must be the *vector
intensity* of the electromagnetic field - the energy per unit area per
unit time - since the flux of the Poynting vector through the surface
is the power passing through it. It's magnitude is the intensity
proper, but it also tells us the *direction* of energy flow.

With this said, there is at least one assumption in the equations above
that is not strictly justified, as we are assuming that the medium is
dispersionless and has no resistance. We do not allow for energy to
appear as heat, in other words, which surely would happen if we drive
currents with the electric field. We also used the *macroscopic*
field equations and energy densities, which involve a coarse-grained
average over the microscopic particles that matter is actually made up
of - it is their random motion that *is* the missing heat.

It seems, then, that Poynting's theorem is likely to be applicable in a
*microscopic* description of particles moving in a vacuum, where
their individual energies can be tracked and tallied:
&eta#eta;t + & = & - ·

& = & 1&mu#mu;_0×

&eta#eta;& = & 12&epsi#epsilon;_0 E^2 + 12&mu#mu;_0B^2
but not necessarily so useful in macroscopic media with dynamical
dispersion that we do not yet understand. There we can identify the
term as the rate at which the mechanical energy of the
charged particles that make up
changes and write:
dEdt = ddt (E_field + E_
mechanical) = -&conint#oint;_&sigma#sigma;· dA
(where
is, recall, an *outward* directed normal) so that this
says that the rate at which energy flows *into* the volume carried
by the electromagnetic field equals the rate at which the total
mechanical plus field energy in the volume increases. This is a
marvelous result!

Momentum can similarly be considered, again in a microscopic description. There we start with Newton's second law and the Lorentz force law: = q(+ ×) = ddt summing with coarse graining into an integral as usual: d_mechdt = &int#int;_V (&rho#rho;+ ×)d^3x As before, we eliminate sources using the inhomogeneous MEs (this time starting from the beginning with the vacuum forms): d_mechdt = &int#int;_V (&epsi#epsilon;_0() - &epsi#epsilon;_0 t ×+ 1&mu#mu;_0 ( ) ×)d^3x or &rho#rho;+ ×= &epsi#epsilon;_0{() + ×t - c^2 ×() }.

Again, we distribute:
t(×) = t ×+
×t
or
×t = - t(×) +
×t
substitute it in above, and add
:
&rho#rho;+ ×& = & &epsi#epsilon;_0{() +
c^2() .

& & - t(×) +
×t

& & . - c^2 ×() }.

Finally, substituting in FL:
&rho#rho;+ ×& = & &epsi#epsilon;_0{() +
c^2() .

& & . - ×() - c^2 ×() }

& &
- t(×)

Reassembling and rearranging:
d_mechdt + ddt &int#int;_V (×
)dV & = & &int#int;_V { () - ×(
) + .

& & . c^2() - c^2 ×()
} dV

The quantity under the integral on the left has units of momentum
density. We define:
= (×) = (×) =
1c^2(×) = 1c^2
to be the field momentum density. Proving that the *right* hand
side of this interpretation is consistent with this is actually
amazingly difficult. It is simpler to just define the *Maxwell
Stress Tensor*:
T_&alpha#alpha;&beta#beta; = {E_&alpha#alpha;E_&beta#beta;+ c^2 B_&alpha#alpha;B_&beta#beta;-
12(·+ c^2·)&delta#delta;_&alpha#alpha;&beta#beta;}
In terms of this, with a little work one can show that:
ddt(_field + _mechanical)_&alpha#alpha;= &conint#oint;_S
&sum#sum;_&beta#beta;T_&alpha#alpha;&beta#beta; ^_&beta#beta;dA
That is, for *each* component, the time rate of change of the total
momentum (field plus mechanical) within the volume equals the flux of
the field momentum through the closed surface that contains the volume.

I wish that I could do better with this, but analyzing the Maxwell Stress Tensor termwise to understand how it is related to field momentum flow is simply difficult. It will actually make more sense, and be easier to derive, when we formulate electrodynamics relativistically so we will wait until then to discuss this further.