# Does a nonzero Poynting vector mean that there is propagation of energy?

I don't know how this "paradox" can be solved. I'm given the following system: A permanent magnet with a magnetic field given by ($\hat{a}$ are unit vectors in the x and y directions)

$$\vec{H}=H_0\hat{a}_y$$

and a parallel plane capacitor with an electric field

$$\vec{E}=E_0\hat{a}_x$$

Poynting's vector is given by:

$$\vec{S}=\vec{E}\times\vec{H}=H_0E_0\hat{a}_z \neq 0$$

The funny part comes when the professor told that in a system like that "clearly" there is not propagation (wich I know will imply some short of energy flux) in the $z$ direction, hence the "paradox". Is there or is there not propagation of energy?

Any hint will be appreciated, thank you for your time.

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It's a static situation and div(S) = 0. There doesn't seem to be any physical paradox. –  Mark Eichenlaub Nov 17 '11 at 23:06
Hi so you mean that as $\nabla \cdot \vec{S}=0$ then the energy balance in the volume of my system is zero no? Thank you, I'm sorry because it was an easy question. –  Jorge Nov 17 '11 at 23:11
yeah, that's what I meant –  Mark Eichenlaub Nov 17 '11 at 23:22
If you discharge the capacitor, the magnet starts rotating from the angular momentum of the formerly circulating energy, correct? –  endolith Nov 22 '11 at 4:17

It depends on what you mean by "propagation of energy." Usually when we think of energy propagating, there is an non-constant distribution of energy in space, and so you can follow features of that distribution to see which way the energy is moving. For example, in an electromagnetic wave, the energy density as a function of position takes a sinusoidal form, $u\sim\sin^2 kx$, and you can follow the peaks and troughs in this distribution over time to see the movement of the wave.

But if you really think about it, energy is not a vector, so it doesn't have an inherent direction, and so in general there may not be any propagation to speak of. Without some feature in the energy distribution to follow, the whole idea of propagation becomes essentially inapplicable. In that case, all you can say about energy is how the amount of it in some particular region changes with time. That quantity is related to the divergence of the Poynting vector,

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

(in empty space), so as far as the energy distribution is concerned, it's only the divergence of $\vec{S}$ that matters, not its value. From Wikipedia:

The Poynting vector is usually interpreted as an energy flux, but this is only strictly correct for electromagnetic radiation. The more general case is described by Poynting's theorem above, where it occurs as a divergence, which means that it can only describe the change of energy density in space, rather than the flow.

However, the Poynting vector does have a different interpretation as the momentum density of the EM field. Momentum is a vector, so it does have a direction, and you can meaningfully talk about a "momentum propagation" (loosely speaking). When you calculate that $\vec{S} \neq 0$, what you're finding is that the EM field of this configuration does have a nonzero momentum density. However, the momentum normally has no effect because it never gets transferred to anything. If you put something that interacts with the EM field in its way, though, the object will experience radiation pressure.

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Thank you for your time. So, the Poynting vector can be viewed as a momentum per unit of volume, ok, I didn't knew. More things, your last paragraph meant that if I put something "near" my system, it will experience radiation pressure but there will be not momentum transfer between the field and the thing, because $\nabla \cdot \vec{S}=0$ is this? –  Jorge Nov 18 '11 at 0:54
No, the thing will experience radiation pressure and there will be momentum transfer between the field and the thing. The radiation pressure is momentum transfer. The momentum transfer is unrelated to $\nabla\cdot\vec{S}$, it is related to $\vec{S}$ itself. –  David Z Nov 18 '11 at 1:07

Yes there is energy-flux which is the same as momentum density because the stress-energy tensor is symmetric.

But the energy-flux will simply just loop around for any practical (non infinitely sized field) and the amount of energy which leaves a small volume of space at one side enters the same volume from the other side so the energy density does not change.

There are several examples of this in the second volume of Feynman's lectures on physics (section 27-5)

Regards, Hans

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Thanks for your answer. I'm afraid I'm not familiar with the stress-energy tensor. Of course, I will check Feynman's lectures for those examples! –  Jorge Nov 18 '11 at 0:56