2
$\begingroup$

We know that the energy in a wave goes as electric field amplitude squared($E_0^2$) as given by poynting vector but the amplitude of electric fields are added linearly. To see why this creates a paradox, consider 2 lasers both with linearly polarized lights with electric field $E_0$. The total energy coming out of these lasers is then proportional to $2*E_0^2$. Now, suppose you find a way to add the amplitude of these 2 waves coherently so that the crest and trough match exactly with each other giving a net amplitude as $2*E_0$. Now, the energy carried by this wave would turn out to be proportional to $4*E_0^2$ which is greater than the energy we started out with!!!. How is this possible?

$\endgroup$
3
$\begingroup$

Indeed, if you manage to superpose two electromagnetic waves, each with intensity $I$, so that they're traveling in the exact same direction with the exact same phase and frequency, then you'll get a wave of intensity $4I$. This isn't a violation of conservation of energy -- it's a proof that you can't do this without adding additional energy to the system.

For example, suppose we started with two waves traveling in different directions, then tried to superpose them using mirrors. This is impossible, because you would require a mirror that takes light rays traveling in two different directions and reflects them so that they exit in the same place in the same direction. This violates existence and uniqueness.

It's easy to have a mirror system that gets the two waves in the same place traveling in slightly different directions, but then you'll get interference fringes. The intensity will vary between $4I$ and zero, averaging out to $2I$.

As another example, suppose you sourced the first wave by vibrating a charge sinusoidally, doing total work $W$ in the process. You might think that by vibrating a second charge in the path of the wave emitted by the first, in the exact same way, with just the right phase, will create a wave with total energy more than $2W$. This is true, but that's simply because vibrating that second charge will take more energy than the first, since you will have to do work against the existing electric field.

This last example is precisely what is happening in your example with the laser. The second laser has to do extra work, which ultimately comes from the laser's power source. In all cases, energy is conserved.

$\endgroup$
  • 1
    $\begingroup$ knzhou, would you consider addressing the (idealized) case of combining two laser beams with a power beam splitter? $\endgroup$ – Hal Hollis Jan 25 '19 at 20:16
  • $\begingroup$ Does this answer the question? Nothing was said about adding more energy, just about what happens with what was given. Is there evidence that two light beams can just add in such a way in any case? $\endgroup$ – user234190 Jul 7 '19 at 10:08
  • $\begingroup$ @user47014 What, is there evidence that interference happens? That might actually be one of the most well-tested facts in all of physics. Literally thousands, maybe tens of thousands of experiments require it to work just as expected. $\endgroup$ – knzhou Jul 7 '19 at 10:20
  • $\begingroup$ @knzhou I mean can beams add together constructively only, without an interference pattern--i.e. without destructive interference also, that conserves energy so it evens out to just adding the intensities. If you have a link to a video where two beams add to a stronger intensity without creating any interference pattern, I would like to see it $\endgroup$ – user234190 Jul 7 '19 at 10:23
  • $\begingroup$ @user47014 The whole point of my answer is that it's basically impossible to do this in practice. $\endgroup$ – knzhou Jul 7 '19 at 10:28
-1
$\begingroup$

Is is the key point of interference in optics (Young double slits or Michelson interferometer...).

On a bright fringe, the intensity is $4I0$ but this is compensated by the dark fringes, with zero intensity. Energy is conserved but with a different spatial repartition.

Sorry, I understand the question a little better. (sorry for my poor english)

Indeed, if we double the field, the power is multiplied by 4.

It seems to me that the problem is to believe that the power to be supplied when we draw two antennas indefinitely near each other is simply the sum of the individual powers when they are alone. Very close, they will interact with one another and it will take more power to make them radiate ?

$\endgroup$
  • $\begingroup$ Yeah, I understand the redistribution of energy in the Young's Double slit experiment but in this case, there is no destructive interference by design. The lasers are made to coherently interfere exactly. $\endgroup$ – Rishabh Jain Jan 25 '19 at 13:17
  • $\begingroup$ The field of the laser is not $E0$ in all space, it is rather of the form ${{E}_{0}}\cos (\omega t-kx)$ in a simple model. $\endgroup$ – Vincent Fraticelli Jan 25 '19 at 13:20
  • $\begingroup$ In my question I have specified that the crests and troughs of the laser match. So, the field of both the lasers are considered to be E*cos(wt-kx) which are made to coherently intefere. $\endgroup$ – Rishabh Jain Jan 25 '19 at 13:24
  • $\begingroup$ Sorry ! I have completed my answer. Hope it can help. $\endgroup$ – Vincent Fraticelli Jan 25 '19 at 13:43

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