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Poynting's Theorem, Work and Energy

Recall from elementary physics that the rate at which work is done on an electric charge by an electromagnetic field is:

\begin{displaymath}
P = \mbox{\boldmath$F$}\cdot \mbox{\boldmath$v$}= q\mbox{\b...
...{\boldmath$v$}= \mbox{\boldmath$E$}\cdot q\mbox{\boldmath$v$}
\end{displaymath} (8.81)

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:
\begin{displaymath}
\frac{dP}{dV} = \mbox{\boldmath$E$}\cdot \mbox{\boldmath$J$}
\end{displaymath} (8.82)

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 $\mbox{\boldmath$v$}$ or $\mbox{\boldmath$J$}$.

If we use AL to eliminate

\begin{displaymath}
\mbox{\boldmath$J$}= \mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$}- \frac{\partial \mbox{\boldmath$D$}}{\partial t}
\end{displaymath} (8.83)

and integrate over a volume of space to compute the rate the electromagnetic field is doing work within that volume:
\begin{displaymath}
P = \int \mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}\ d^3x_...
...frac{\partial \mbox{\boldmath$D$}}{\partial t}\right\} d^3x_0
\end{displaymath} (8.84)

Using:

\begin{displaymath}
\mbox{\boldmath$\nabla$}\cdot (\mbox{\boldmath$E$}\times \mb...
...E$}\cdot (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$H$})
\end{displaymath} (8.85)

(which can be easily shown to be true as an identity by distributing the derivatives) and then use FL to eliminate $\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$E$}$, one gets:
\begin{displaymath}
\int \mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}\ d^3x_0 = ...
...rac{\partial \mbox{\boldmath$B$}}{\partial t} \right\} d^3x_0
\end{displaymath} (8.86)

It is easy to see that:

$\displaystyle \epsilon \frac{\partial \mbox{\boldmath$E$}\cdot \mbox{\boldmath$E$}}{\partial t}$ $\textstyle =$ $\displaystyle 2 \mbox{\boldmath$E$}\cdot\frac{\partial \mbox{\boldmath$D$}}{\partial t}$ (8.87)
$\displaystyle \frac{1}{\mu} \frac{\partial \mbox{\boldmath$B$}\cdot \mbox{\boldmath$B$}}{\partial t}$ $\textstyle =$ $\displaystyle 2 \mbox{\boldmath$H$}
\cdot\frac{\partial \mbox{\boldmath$B$}}{\partial t}$ (8.88)

from which we see that these terms are the time derivative of the electromagnetic field energy density:
\begin{displaymath}
\eta = \frac{1}{2}\epsilon \mbox{\boldmath$E$}\cdot\mbox{\bo...
...{\boldmath$D$}+ \mbox{\boldmath$B$}\cdot \mbox{\boldmath$H$})
\end{displaymath} (8.89)

Moving the sign to the other side of the power equation above, we get:
\begin{displaymath}
- \int_V \mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}\ d^3x_...
...dmath$H$}) +
\frac{\partial \eta}{\partial t} \right\} d^3x_0
\end{displaymath} (8.90)

as the rate at which power flows out of the volume $V$ (which is arbitrary).

Equating the terms under the integral:

\begin{displaymath}
\frac{\partial \eta}{\partial t} + \mbox{\boldmath$\nabla$}...
...\boldmath$S$}= - \mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}
\end{displaymath} (8.91)

where we introduce the Poynting vector
\begin{displaymath}
\mbox{\boldmath$S$}= \mbox{\boldmath$E$}\times \mbox{\boldmath$H$}
\end{displaymath} (8.92)

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 $\mbox{\boldmath$S$}$ into a surface integral of its flux:

\begin{displaymath}
\oint_\sigma \mbox{\boldmath$S$}\cdot \hat{\mbox{\boldmath$n...
.../\sigma} \mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}\ dV = 0
\end{displaymath} (8.93)

where $\sigma$ is the closed surface that bounds the volume $V$. 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 $\mbox{\boldmath$S$}$ 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:

$\displaystyle \frac{\partial \eta}{\partial t} + \mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$S$}$ $\textstyle =$ $\displaystyle - \mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}$ (8.94)
$\displaystyle \mbox{\boldmath$S$}$ $\textstyle =$ $\displaystyle \frac{1}{\mu_0}\mbox{\boldmath$E$}\times \mbox{\boldmath$B$}$ (8.95)
$\displaystyle \eta$ $\textstyle =$ $\displaystyle \frac{1}{2}\epsilon_0 E^2 + \frac{1}{2\mu_0}B^2$ (8.96)

but not necessarily so useful in macroscopic media with dynamical dispersion that we do not yet understand. There we can identify the $\mbox{\boldmath$J$}\cdot \mbox{\boldmath$E$}$ term as the rate at which the mechanical energy of the charged particles that make up $\mbox{\boldmath$J$}$ changes and write:
\begin{displaymath}
\frac{dE}{dt} = \frac{d}{dt} \left(E_{\rm field} + E_{\rm
me...
..._\sigma \mbox{\boldmath$S$}\cdot\hat{\mbox{\boldmath$n$}}\ dA
\end{displaymath} (8.97)

(where $\hat{\mbox{\boldmath$n$}}$ 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:

\begin{displaymath}
\mbox{\boldmath$F$}= q(\mbox{\boldmath$E$}+ \mbox{\boldmath...
...\times \mbox{\boldmath$B$}) = \frac{d\mbox{\boldmath$p$}}{dt}
\end{displaymath} (8.98)

summing with coarse graining into an integral as usual:
\begin{displaymath}
\frac{d\mbox{\boldmath$P$}_{\rm mech}}{dt} = \int_V (\rho\mb...
...dmath$E$}+ \mbox{\boldmath$J$}\times \mbox{\boldmath$B$})d^3x
\end{displaymath} (8.99)

As before, we eliminate sources using the inhomogeneous MEs (this time starting from the beginning with the vacuum forms):
\begin{displaymath}
\frac{d\mbox{\boldmath$P$}_{\rm mech}}{dt} = \int_V \left(\e...
...s
\mbox{\boldmath$B$}) \times \mbox{\boldmath$B$}\right)d^3x
\end{displaymath} (8.100)

or
\begin{displaymath}
\rho\mbox{\boldmath$E$}+ \mbox{\boldmath$J$}\times \mbox{\bo...
...\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}) \right\}.
\end{displaymath} (8.101)

Again, we distribute:

\begin{displaymath}
\frac{\partial }{\partial t}(\mbox{\boldmath$E$}\times \mbox...
...th$E$}
\times \frac{\partial \mbox{\boldmath$B$}}{\partial t}
\end{displaymath} (8.102)

or
\begin{displaymath}
\mbox{\boldmath$B$}\times \frac{\partial \mbox{\boldmath$E$}...
...th$E$}
\times \frac{\partial \mbox{\boldmath$B$}}{\partial t}
\end{displaymath} (8.103)

substitute it in above, and add $c^2\mbox{\boldmath$B$}(\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$}) = 0$:
$\displaystyle \rho\mbox{\boldmath$E$}+ \mbox{\boldmath$J$}\times \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle \epsilon_0\left\{\mbox{\boldmath$E$}(\mbox{\boldmath$\nabla$}\cdo...
...2\mbox{\boldmath$B$}(\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$}) \right.$  
    $\displaystyle \quad - \frac{\partial }{\partial t}(\mbox{\boldmath$E$}\times \m...
...}) + \mbox{\boldmath$E$}
\times \frac{\partial \mbox{\boldmath$B$}}{\partial t}$  
    $\displaystyle \quad\quad \left. - c^2 \mbox{\boldmath$B$}\times (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}) \right\}.$ (8.104)

Finally, substituting in FL:

$\displaystyle \rho\mbox{\boldmath$E$}+ \mbox{\boldmath$J$}\times \mbox{\boldmath$B$}$ $\textstyle =$ $\displaystyle \epsilon_0\left\{\mbox{\boldmath$E$}(\mbox{\boldmath$\nabla$}\cdo...
...2\mbox{\boldmath$B$}(\mbox{\boldmath$\nabla$}\cdot \mbox{\boldmath$B$}) \right.$  
    $\displaystyle \quad \left. - \mbox{\boldmath$E$}\times (\mbox{\boldmath$\nabla$...
...oldmath$B$}\times (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$}) \right\}$  
    $\displaystyle \quad\quad - \epsilon_0 \frac{\partial }{\partial t}(\mbox{\boldmath$E$}\times \mbox{\boldmath$B$})$ (8.105)

Reassembling and rearranging:

$\displaystyle \frac{d\mbox{\boldmath$P$}_{\rm mech}}{dt} + \frac{d}{dt} \epsilon_0 \int_V (\mbox{\boldmath$E$}\times
\mbox{\boldmath$B$})dV$ $\textstyle =$ $\displaystyle \epsilon_0 \int_V \left\{ \mbox{\boldmath$E$}(\mbox{\boldmath$\na...
...dmath$E$}\times (\mbox{\boldmath$\nabla$}\times
\mbox{\boldmath$E$}) + \right.$  
    $\displaystyle \left. c^2\mbox{\boldmath$B$}(\mbox{\boldmath$\nabla$}\cdot \mbox...
...ath$B$}\times (\mbox{\boldmath$\nabla$}\times \mbox{\boldmath$B$})
\right\} dV$  

The quantity under the integral on the left has units of momentum density. We define:
\begin{displaymath}
\mbox{\boldmath$g$}= \epsilon_0 (\mbox{\boldmath$E$}\times \...
...times \mbox{\boldmath$H$}) = \frac{1}{c^2}\mbox{\boldmath$S$}
\end{displaymath} (8.106)

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:
\begin{displaymath}
T_{\alpha \beta} = \epsilon_0 \left\{E_\alpha E_\beta + c^2 ...
...math$B$}\cdot\mbox{\boldmath$B$})\delta_{\alpha\beta}\right\}
\end{displaymath} (8.107)

In terms of this, with a little work one can show that:
\begin{displaymath}
\frac{d}{dt}(\mbox{\boldmath$P$}_{\rm field} + \mbox{\boldma...
...ical})_\alpha = \oint_S
\sum_\beta T_{\alpha\beta} n_\beta dA
\end{displaymath} (8.108)

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.


next up previous contents
Next: Magnetic Monopoles Up: Maxwell's Equations Previous: The Coulomb or Transverse   Contents
Robert G. Brown 2014-08-19