Energy

editor · 2005 October 10th, 16:26

If two EM waves 180 deg. out-of-phase meet in a vacuum they would cancel out.
1. Where would the energy go ?
2. How would you explain the result using photons ?

_mark_ · 2005 October 10th, 16:49

You are already treating light as a wave when it interacts to go out of phase, i dont know if the result can be explained using photons.
Generally I have only ever treated light as a wave when its traveling and as a photon when it interacts with matter. I'd also like to know the answer to 2.

As for 1, has the energy gone? The two waves are still there.

Brinx · 2005 October 10th, 17:34

I think Xerxes pointed that out to me a while ago. Allegedly, if two EM-waves interfere destructively, only the E-field gets to be zero at that point. The B-field still oscillates strongly there. I don't know the finer details of this interaction though, so maybe Xerxes could lend a hand in explaining this.

editor · 2005 October 10th, 21:43

The energies in the E and B fields are equal, so even if only the E fields cancel, either their energy is added to double the B fields or it goes somewhere else. You can't destroy energy.
2. Even if one photon is the other photon's antiparticle and they annihilate - what would the product be ??

Brinx · 2005 October 11th, 04:23

Quote:

Originally Posted by editor

...either their energy is added to double the B fields...

I thought that was the gist of it, yes. Again, I'm not sure of this at all - I'd expect there to be a decent page about this online somewhere. :)

kublai · 2005 October 11th, 04:58

Quote:

Originally Posted by editor

2. Even if one photon is the other photon's antiparticle and they annihilate - what would the product be ??

Photons are their own antiparticle, they don't annihilate when they collide, just bounce off.

Xerxes314 · 2005 October 11th, 15:39

No, I think my objection was that the situation is impossible to set up. If you take one emitter and fire it at another emitter located half a wavelength away, the second emitter will not emit. It will just absorb the energy of the other emitter. You'll see an addition of out of phase waves summing to zero beyond the second emitter, but the paradox of energy loss is easily resolved by noting that the second emitter is actually an absorber.

Xerxes

Brinx · 2005 October 11th, 15:50

Well, the beams need not be fired in opposite directions to achieve fully destructive interference. The two-slit setup will do just as nicely. What happens, field-wise, at the positions of the dark bands?

editor · 2005 October 11th, 16:20

My post vanished ! It went like this: What decides which emitter, if they are equal in all respects, is the emitter and which the absorber.
If this could happen in the macro world, e.g. two radio wave transmitters, you could measure the energy going into each transmission.
How about sound cancellation in cars, there's a working example.
In other words - not convinced by that explanation.

**Fernanda** · 2005 October 11th, 19:00

you don't have to have to emmitters pointing at each other... you can simply have it hit a coating that is 1/2 a wavelength apart. That's how anti-reflective coatings work... the difference there is that the light hitting your lenses are not coherent. But a coherent laser with the right wave width would anhialate each other upon rebound. back to his question above

Xerxes314 · 2005 October 11th, 21:45

It's still easy. The coating is absorbing the energy.

Xerxes

editor · 2005 October 11th, 21:51

Yes and no. The reflection from the front of the coating has a 1/2 wave phase shift. the reflection from the back of the coating has no phase shift if the lens is a lesser refractive index than the coating, and a 1/2 wave phase shift if the lens is a higher ref. index. So the thickness of the coating has to be adjusted, and can be arranged to give a black appearance or a bright appearance. try explaining that with photons. Feynman ties himself in knots in his book QED trying to do just that. pages 60 to 70.
With waves, as shown on the attachment, it's easy to visualise the two-hole interference.

editor · 2005 October 11th, 21:56

I originally said 'in a vacuum' to specifically avoid that sort of answer.

EnergRecruitm · 2008 November 12th, 06:51

Expressions are obtained for all α³Ry contributions to the energy levels of the two-fermion system in electrodynamics. These expressions are evaluated from a relativistic two-body equation which takes binding into account in its interaction operator. They are specifically calculated for the n=2 levels of the system. Corrections arise from three sources: improved treatment of pair effects of the Coulomb field and of the exchange of transverse photons, self-energy and vacuum polarization terms, and, in second-order annihilation processes.

editor · 2008 November 12th, 21:55

Can you do that again in simpler language ?