2
$\begingroup$

Electromagnetic waves follows superposition principle. So that we can simply add the fields of waves to calculate the final field. Then let's think there are two waves that are moving through x-axis and y-axis. If we control the phase difference well, we can make a situation like figure.

Two electromagnetic waves

In the figure with some $t=t_0$, you can see that there is a poynting vector has a z-axis component. In initial condition, there was no z-axis component of momentum. This doesn't seem right.

What did I miss?(or miscalculated) Why there is a strange momentum out there?

edit: I thought that if two coherent lights have identical frequency, the direction of poynting vector will be always +z. So that the z-component doesn't vanish in average. Is there something wrong with my thoughts?

$\endgroup$
0
$\begingroup$

Your observation is correct. For two coherent waves there is a z-component of momentum density. Note that it vanishes on average.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Don't quite understand why it vanishes on average. Assuming the waves have the same frequency, when the phase changes by $\pi$ all the E and B arrows reverse. Doesn't that leave the direction of the Poynting vector unchanged? $\endgroup$ – Philip Wood Jul 31 at 21:01
  • $\begingroup$ I thought exactly same. Can you explain me why it vanishes on average? $\endgroup$ – littlegiant Aug 1 at 9:19
  • 1
    $\begingroup$ @ littlegiant I don't think it does vanish on average. I'm thinking, though, that there may be a loop-hole in your argument... For e-m waves we know that 𝑑𝐸/𝑑𝑑=𝑐 𝑑𝑝/𝑑𝑑 in which 𝑑𝐸/𝑑𝑑 is the rate of propagation of energy through a surface and 𝑑𝑝/𝑑𝑑 is the rate of propagation of momentum through that surface. But can we apply this to an arbitrary superposition of fields in which there is propagation of energy, as given by the Poynting vector? The waves, as we know, pass through each other; neither is propagating in the direction of the Poynting vector. Your thoughts? $\endgroup$ – Philip Wood Aug 1 at 17:08
  • $\begingroup$ @PhilipWood I could be really wrong here, but as you pointed out, there's just superposition at one single point in space - I don't think this superposition is therefore a wave and doesn't itself transport energy or momentum. To derive Poynting's Theorem, you can integrate the energy density in the EM field over a finite volume V - not sure the derivation will work for a single point. Therefore, I'm not sure the Poynting vector has much meaning for interference just at a single point? $\endgroup$ – Shrey Aug 2 at 0:59
  • $\begingroup$ @PhilipWood Yes, as you said it is not propagating. But I think even it is not propagating, it still can have energy and momentum. Well, actually static electromagnetic field also has momentum, though its momentum is cancelled out by "hidden momentum". But in this case, there is no particles to create the "hidden momentum". So I thought that something is weird, and posted my question. $\endgroup$ – littlegiant Aug 2 at 13:55

Your Answer

By clicking β€œPost Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.