Defining the Wave:

Let's assume an electromagnetic wave exists in the form

$\vec{E}(x,y,z,t) = \vec{E}_0 \, cos(kx-\omega t)$

$\vec{B}(x,y,z,t) = \vec{B}_0 \, cos(kx-\omega t)$

How the wave interacts with particles:

Let's say for example my proton is at the origin $$(x,y,z) = (0,0,0)$$

What are the associated electric and magnetic fields attached to that specific point in space?

$\vec{E}(x,y,z,t) = \vec{E}_0 \, \cos(\omega t)$

$\vec{B}(x,y,z,t) = \vec{B}_0 \, \cos(\omega t)$

Force on the charge :

$\vec{F} = q(\vec{E} +\vec{V}×\vec{B})$

As $(\vec{V} × \vec{B})<< \vec{E}$

$\vec{F} = q\vec{E}$

$m\vec{a}= q \vec{E}_0 \, cos(\omega t)$

$\vec{a}= \frac{q}{m} \vec{E}_0 \, cos(\omega t)$

$\vec{ v}(0)=0$

$\vec{v}= \frac{q}{m\omega} \vec{E}_0 \, sin(\omega t)$

This shows that the EM wave can cause a oscillatory motion by just considering the electric component, in a direction that is perpendicular to the wave vector $\vec{k}$.


Momentum density of the electromagnetic field, can be shown to be: $\mu_0 \epsilon_0 \vec{S}$

For an electromagnetic wave of the form

$\vec{E}_0 \, cos( \vec{k} \cdot \vec{r} -\omega t)$

The momentum density at all points in space, is in the same direction as the vector $\vec{k}$, which is the direction that the wave is moving!

So the total momentum in the electromagnetic field has zero components perpendicular to $\vec{k}$.

So how is momentum conserved if there is a momentum transfer perpendicular to $\vec{k}$ ( the one that causes perpendicular osscilations)

Specifically I would like some insight on the role of the maxwell stress tensor, $\nabla \cdot \sigma$, in electromagnetic momentum conservation.

Momentum conservation in EM:

$$\nabla \cdot \sigma - \frac{\partial \vec{p}}{\partial t} = \mu_{0} \epsilon_{0} \frac{\partial \vec{S}}{\partial t}$$

I assume its fine that the EM field momentum isn't conserved as the total momentum is, but I am still unsure what the significance of $\sigma $ is. As ignoring sigma, the law states that an increase in mechanical momentum density at a point is equal to the decrease in field momentum. $\sigma$ is known to be the momentum transport. Could this term be indicative of the EM wave generated by the charge osscilating?

  • $\begingroup$ "As B⃗ <<E⃗ " is not true for an EM wave. $\endgroup$ Commented Mar 14, 2022 at 18:02
  • $\begingroup$ @JerroldFranklin it is better to say that the comparison simply does not make sense in this context, since E and B do not have the same units. Per OP's equations it is vB that should be compared to E. But, regardless, the B field is still perpendicular to k... So, OP's question remains. $\endgroup$
    – hft
    Commented Mar 14, 2022 at 18:11
  • $\begingroup$ @jensen paull, to see how momentum is conserved for a swarm of particles in an EM field, you can look at this answer: physics.stackexchange.com/questions/683773/…. You will see that the derivation requires the field to vanish at infinity. This is not true for a plane wave (the field does not vanish at infinity), but might nevertheless be helpful for you to see where the missing momentum might come from. $\endgroup$
    – hft
    Commented Mar 14, 2022 at 18:28
  • $\begingroup$ You leave $\vec S$ undefined. $\endgroup$
    – my2cts
    Commented Mar 14, 2022 at 20:35
  • $\begingroup$ Congratulations, this is an interesting paradox. $\endgroup$
    – my2cts
    Commented Mar 15, 2022 at 8:33

1 Answer 1


One way to keep the EM picture and simultaneously the photon picture is the following.

The electron can not follow the oscillation of E because they are too fast and the electron has mass (there is no particle with charge without mass). So in fact it doesn't move up down. Nevertheless (here comes QM magic - I can not explain this!) the currant j associated with the electron is not 0 but has small value. Then jxB gives a force which coincides with the momentum of photons (virtual?).

  • $\begingroup$ I'm not bothered at the moment about the photon picture, momentum of the EM field at a point in space can be derived independently to QM. If you simply look at the lorentz force you will see there is a forward momentum transfer without the need to evoke photons, one component of which IS a perpendicular to \vec{k}. Taking the mass of an object into account, as I have done, shows there is still sinusoidal behaviour. $\endgroup$ Commented Mar 17, 2022 at 11:50
  • $\begingroup$ The main question was on the stress tensor role in conservation. As although the field momentum is only on k's direction, the transfer of momentum to matter at a point in space, can still be perpendicular $\endgroup$ Commented Mar 17, 2022 at 11:52
  • $\begingroup$ The field momentum decrease at a point in space, doesn't have to be the increase in mechanical momentum at that point in space. There is an oscillatory motion , aswell as a forward momentum. The oscillatory component coming from the stress tensor I presume $\endgroup$ Commented Mar 17, 2022 at 12:47
  • $\begingroup$ Lets calculate approx. how much an electron will move. \begin{aligned}dp=F\cdot t\\ mv=eEt\\ v=\dfrac{e}{m}E\dfrac{T}{2}\end{aligned} $\endgroup$
    – Mercury
    Commented Mar 19, 2022 at 13:21
  • $\begingroup$ Your math is incorrect, look at my math in the question. Although small, it is still not zero. $\endgroup$ Commented Mar 19, 2022 at 13:25

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