Role of Maxwell's stress tensor in the conservation of momentum of an EM wave

Question

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

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?

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?).