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Poynting's theorem is given by

$$\frac{\partial}{\partial t}\int_{v}Udv + \oint_{A}\vec S\cdot \vec {dA} +\int_{v}\vec E\cdot\vec J dv =0 $$


  • the total electromagnetic energy inside the volume v is $U = \frac 1 2 (\vec E\cdot\vec D+\vec B\cdot\vec H)$
  • the Poynting vector $\vec S=\vec E\times \vec H$

This equation is interpreted as the conservation of electromagnetic and mechanical energy for a volume of space with each term representing respectively the rate at which

  1. electromagnetic energy changes inside the volume
  2. electromagnetic energy crosses the boundary of the enclosing surface
  3. mechanical work is done on charges inside the volume

Now take the case of a charge accelerating from rest by a static electric field, and initially at the centre of a spherical volume fixed in space with radius cT where c is the speed of light and T the time taken for electromagnetic fields to propagate from the center to the spherical boundary. For 0 < t < T both the magnetic and mechanical energy inside the volume increases without electromagnetic energy crossing the boundary.

So during this time, where does the negative term come from to maintain the RHS = 0 in Poynting's theorem?

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I'm finding this easier to visualize if I consider a volume of radius greater than $cT$... –  David Z Aug 28 '12 at 0:37
Have you actually attempted to calculate the time derivative of the energy density taking into account the external field and the field of the charge? It would be a good exercise. –  user37496 Jun 24 '14 at 18:04
@user37496 yes, I realized this after asking the question so made it into a brief answer. –  John McVirgo Jun 26 '14 at 11:43

3 Answers 3

What makes you so sure the net electromagnetic energy inside the volume has increased? Yes, there is a radiation field, but there was also a pre-existing field which the source of acceleration on the charge, and that pre-existing field had an energy density, and electromagnetic energy is not linear in the fields. It must be that whatever mechanical work has been done is offset by a net reduction of EM energy when the pre-existing field density is included.

As the other commenter mentioned, there is the issue of the moving boundary. However one could rewrite the 'paradox' with a fixed boundary greater than CT at a specific time.

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For the case of a moving boundary your conservation law derived from the local densities should read

$$\int_{v}\frac{\partial U}{\partial t}dv + \oint_{A}\vec S\cdot \vec {dA} +\int_{v}\vec E\cdot\vec J dv =0$$

The negative term you are looking for comes then from

$$\int_{v}\frac{\partial U}{\partial t}dv = \frac{\partial}{\partial t} \int_{v} U dv - U[\partial v]$$

where the last term is positive and due to contributions from the moving boundary of $v$, $\partial v = A $.

It is instructive to review the visually appealing picture of the field distribution around an accelerated charge ( http://physics.weber.edu/schroeder/mrr/MRRtalk.html ). There is indeed a discontinuity in the field lines on the sphere of radius cT surrounding the charge.

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sorry, I should have made it clearer that the volume is fixed. –  John McVirgo Aug 28 '12 at 0:10

The negative term comes from the change in the electric energy inside the volume contributed by the charge itself, which is a maximum at t<=0 and diminishes as it moves away from the center.

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