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

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?

• "As B⃗ <<E⃗ " is not true for an EM wave. Commented Mar 14, 2022 at 18:02
• @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.
– hft
Commented Mar 14, 2022 at 18:11
• @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.
– hft
Commented Mar 14, 2022 at 18:28
• You leave $\vec S$ undefined. Commented Mar 14, 2022 at 20:35
• Congratulations, this is an interesting paradox. Commented Mar 15, 2022 at 8:33