# Poynting Vector and Sources

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1. Apr 4, 2015

### devd

Suppose we have a charge, $q$ and a magnetic dipole moment, $\vec m$. They don't move, nothing changes with time, in short, a static situation.

Now, we have at least some regions of space where both the electric field, $\vec E$ and the magnetic field, $\vec B$ are non-zero. That means in those regions, the poynting vector, $\vec S$ is non-zero.

Now according to the interpretation of $\vec S$ as energy flux density, that is energy per unit area per unit time, energy is flowing between the regions. How is this true? What is the source of this energy? Am i missing something trivial?

2. Apr 5, 2015

### vanhees71

This is a very clever question! In the literature, somewhat misleading, it's called "hidden momentum". If you take the full relativistic (!) balance equation for momentum, including the mechanical momentum "hidden" in the magnetic moment, you get a 0 total momentum as it must be for a static situation.

The "hidden momentum" is, something very obvious: You have to use an external force to keep the charge-current-field distribution in the static arrangement, i.e., there are mechanical external stresses, keeping the momentum balance correct, giving zero total momentum for the system as a whole. You have to use the correct relativistic expressions for all kinds of momenta, and that's why some textbooks make a mystery out of this balance equation, because they think it's nice to make relativity as something mysterious, although that's not the case, and it doesn't help understanding the issue!

For a very good paper (published in Am. J. Phys.) making that issue very clear (including your example of a magnetic moment in a electrostatic field), see

http://arxiv.org/abs/1302.3880

3. Apr 5, 2015

### devd

Does the "hidden momentum" refer to the fact that to keep the charge, magnetic dipole from moving, the current from changing (under their mutual effects), there needs to be some external agency? When calculating electro or magnetostatic energy, we leave out this part, right? Or is it a purely relativistic effect?

I am going to read the paper. Thanks for the reply and the reference!

4. Apr 5, 2015

### vanhees71

Both! Sometimes "paradoxes" in this question come from the application of non-relativistic expressions in the mechanical-momentum piece and then looking at the electromagnetic momentum at a accuracy, which includes the terms left out in the non-relativistic expansion (which is a formal expansion in powers of $1/c$ with $c$ the "speed of light", which should rather be called the "limiting speed in relativistic space-time", but that's another story).

5. Apr 9, 2015

### DaPi

Am I setting myself up to be shot down?

Oh dear! It was a loooong time ago that I learned of Poynting's theorem in the integral form. We were given the health warning then that S being
is an unwarranted assumption and only the integral over a closed surface was physically meaningful.

In the static case I can imagine S being non-zero, but the surface integral and divS (for the differential form of the theorem) both being zero.

6. Apr 10, 2015

### vanhees71

This is also not entirely right! The energy-momentum-stress tensor has a physical meaning in General Relativity. It adds to the sources of the gravitational field in the Einstein-Hilbert field equations of motion! This makes it also clear, how to fix the ambiguity caused by the ambiguity of gauge dependence of the canonical energy-momentum tensor: The physical energy-momentum tensor (whose time-time and time-space components are the energy and momentum densities) is given by the variation of the Hilbert-Einstein action wrt. the metric components, leading to a gauge invariant energy-momentum tensor for the em. field as it must be for a physically relevant quantity.

7. Apr 15, 2015

### Jano L.

I like to think of it not as an unwarranted assumption, but as a definition. Provided we have explicit work-energy theorem where $\mathbf E\cdot \mathbf j$ gives work done on matter, we may choose any definition of EM energy and flux that fits this theorem; the usual (Poynting-like) expressions are the simplest, so we often use them.