# What does the Poynting flux represent?

I know that the Poynting flux is the cross product between $\vec E$ fields and $\vec B$ fields, but I'm wondering, are the field lines of the Poynting flux considered as electromagnetic energy? Let's say we have charged particles in this field, would the Poynting flux exert force on them and move them along the field lines?

I have a plot of a Poynting flux that looks like this: What would happen if there were charged particles in this field?

• It doesn't give you information about how the particles move. Just think about the magnitudes of $E$ and $B$. You can take $B$ larger and larger and $E$ smaller and smaller, and that will leave $E\times B$ the same.
– user12029
Apr 25 '16 at 22:24
• @NeuroFuzzy What does the poynting flux tell me then? how is it useful? Apr 25 '16 at 22:46
• it allows you to see how energy is flowing in space. In your plot there is a lot of energy flowing towards the center at that moment in time. The precise statement is in Poynting's theorem en.wikipedia.org/wiki/Poynting%27s_theorem (the vacuum case is most relevant, in which case $J_f=0$)
– user12029
Apr 25 '16 at 22:48
• @NeuroFuzzy Is this energy in form of electromagnetic waves? Apr 25 '16 at 22:53
• This question is answered by the first sentence of the relevant Wikipedia article (en.wikipedia.org/wiki/Poynting_vector): "In physics, the Poynting vector represents the directional energy flux density (the rate of energy transfer per unit area) of an electromagnetic field." Apr 25 '16 at 23:20

The Poyntings vector is given by

$${\textbf{S}=\frac{1}{\mu_0} (\textbf{E}\times \textbf{B})}$$

or $${\textbf{S}=\textbf{E}\times \textbf{H}}$$

It is known that electromagnetic waves carry energy with them. The purpose of the Poynting vector is well explained by the Poynting's theorem which is the work energy theorem in electrodynamics. According to Poynting's theorem, the rate at which work is done on a charge by the Lorentz force on a charge distribution is:

$${\frac{\mathrm dW}{\mathrm dt}=-\frac{1}{2}\frac{\partial}{\partial{t}}\int_V \left(\frac{1}{\mu_0}B^2+\epsilon_0E^2\right)\mathrm dV-\frac{1}{\mu_0}\int_S (\textbf{E}×\textbf{B})\cdot \mathrm d\textbf{S}}$$

This means that the power (energy) imparted on a charge by an electromagnetic wave is equal to the decrease in the energy stored in the fields over a volume V (first integral) minus the amount of energy radiated out through the surface S enclosing the volume V (second integral). If there is no charge present, then ${\dfrac{\mathrm dW}{\mathrm dt}=0} \;.$ In that case, the decrease in the energy stored in the field over a volume is equal to the energy radiated out through a surface enclosing the volume. This is the law of conservation of energy.

So, in that sense, the second integral (surface integral of Poynting vector) represents the rate at which energy flows out through the surface. So the Poynting vector $\textbf{S}$ is the rate per unit area at which the energy crosses a surface. That's why it is termed as the energy flux density of an electromagnetic field (apply Gauss's divergence theorem to the second integral). So the lines of Poynting vector of course represents the electromagnetic energy radiated out through a surface in the cost of the energy stored in the fields. It can also give us a continuity equation as it says that anything that flows out should be in the cost of what remaining inside. It is not the Poynting vector that do work on the charged particles to make it move through the lines you drawn. Look at the Poynting's theorem. The work done on the charge to make it move plus the energy flux radiated out will be equal to the decrease in the energy stored in the fields. Where does the decreased energy in the field go? A part of it do some work and the remaining radiates out through the surface. The charged particles move in the direction of the Lorentz force acting on the charge. The energy flux density lines are not lines of force. It has nothing to do with the motion of the charge. If there is charge or not, the energy always radiate which corresponds to a decrease in the energy stored in the fields.

What does the Poynting flux represent?

Energy flow associated with wave motion. See this Blaze Labs picture: The Poynting vector is pointing in the direction the wave is travelling, transporting E=hf energy. Or energy-momentum if you prefer.

I know that the Poynting flux is the cross product between $$\vec E$$ and $$\vec B$$

I'm afraid there aren't really any such fields, see Jackson section 11.10 where he says "one should properly speak of the electromagnetic field Fμν rather than E or B separately". Also see the Wikipedia electromagnetic radiation article and note this: "the curl operator on one side of these equations results in first-order spatial derivatives of the wave solution, while the time-derivative on the other side of the equations, which gives the other field, is first order in time”. If it was an ocean wave and you were in a canoe, E denotes the slope of your canoe and B denotes the rate of change of slope. There's aren't two waves present, just one, and it's an electromagnetic wave, which is an electromangetic field variation.

but I'm wondering, are the field lines of the Poynting flux considered as electromagnetic energy?

I'm not sure field lines is quite the right phrase, but I'd say the answer is yes. Think about pair production and electron spin and spin angular momentum, then take a look at the static fields section of the Wikipedia Poynting vector article. The Poynting vector is still orthogonal to both E and H, only now it it isn't linear, it goes round and round: Public domain image by Michael Lenz, see Wikipedia

Note this:

"Although there are only static electric and magnetic fields, the calculation of the Poynting vector produces a clockwise circular flow of electromagnetic energy, with no beginning or end. While the circulating energy flow may seem nonsensical or paradoxical, it is necessary to maintain conservation of momentum. Momentum density is proportional to energy flow density, so the circulating flow of energy contains an angular momentum".

This angular momentum is exhibited via the Einstein-de Haas effect which "demonstrates that spin angular momentum is indeed of the same nature as the angular momentum of rotating bodies as conceived in classical mechanics.".

Let's say we have charged particles in this field, would the Poynting flux exert force on them and move them along the field lines?

Yes. If you had two charged particles such as an electron and a positron (see the Wikipedia positronium article), they would move linearly and/or rotationally, something like this: But note that this is a simplified "flat" picture. The positron has the opposite chirality to the electron. Both are Dirac spinors, and a Dirac spinor is a bispinors. It's like you've have a steering-wheel rotation, and a coin-spin rotation both going on at the same time, one at twice the rate of the other, hence spin half.

I have a plot of a Poynting flux that looks like this: What would happen if there were charged particles in this field?

This field looks like it is the field of a charged particle such as a positron. If you placed an electron down near it, linear motion would happen. Counter-rotating vortices attract, co-rotating vortices repel. If you threw the electron past it, rotational motion would also occur.

All this might sound novel, but don't forget Maxwell's page title. Or that in atomic orbital electrons exist as standing waves. Standing wave, standing field. And when you have a standing wave in a mirror-box, when you drop one of the sides that wave is off like a shot from a standing start, because there always was motion at c, even though you couldn't see it. It's the same for electron-positron annihilation.