Energy being destroyed

1. Dec 6, 2007

tomprice

Ok say we have two identical lasers firing beams perpendicular to each other such that the beams intersect at some point.
A semisilvered mirror is placed at this intersection, at 45 degrees to both of the laser beams, so that each beam is split into two beams of half the intensity, one going in the same direction as the original beam and one perpendicular to it.
Say that one of the lasers is placed so that, at the point of intersection, one of the beams has a phase shift of half a wavelength relative to the other.
I would expect the two beams to exactly cancel each other out from destructive interference, but this would be an example of energy being destroyed, would it not?

Where is the error in my thinking?

Thank you very much.

2. Dec 7, 2007

Demystifier

Either there will be places at which the interference is constructive rather than destructive; or the interference is destructive everywhere, so the laser beams are not even fired at the first place. In each case, the total energy is not destroyed.

3. Dec 7, 2007

dst

Let's say it's completely destructive interference. Where does the energy from flicking the laser power switch go? Into frying the laser?

4. Dec 7, 2007

Demystifier

I should know more technical details on how the laser works to answer that one. But I think this question is not essential. The essential point is that in that case the atoms will stay in the excited level and nothing will stimulate their decay.

5. Dec 7, 2007

dst

I was thinking of a diode laser. But I guess you're right.

Last edited: Dec 7, 2007
6. Dec 7, 2007

tomprice

Ok so if I understand correctly, having the semi-silvered mirror placed in the manner described will alter the functioning of the lasers so that no light is actually generated by them. But how does placing the semi-silvered mirror have this effect... in other words how do the atoms in the laser "know" that a semi-silvered mirror has been placed like that?

Also, something I just thought of: is it possible that having the light reflect off the semi-silvered mirror gives it a different phase shift than passing straight through, so the described situation is impossible? Maybe that would explain it. Thanks again.

7. Dec 7, 2007

HallsofIvy

I think you are misunderstanding "destructive interference". If you have the waves set so that, at a specific point, at a specific time, the waves sum to 0, then at a half-wavelength away (or at a half period in time) those same waves will be adding. At a given instant in time, you have the waves canceling at one point but adding at another. At a given point in space, you have the waves canceling at one time, adding at another. In neither case is any energy lost.

8. Dec 7, 2007

dst

We're talking exactly out of phase? Or do light waves behave differently? What I mean, is that suppose one is described by sin x, the other would be described by cos x. Adding amplitudes would always sum to zero.

Maybe I'm out of phase with light :)

Is there any way to construct that given scenario (above)?

9. Dec 7, 2007

tomprice

Let's say that the electric field caused by a given laser at a distance d from the mirror is given by sin(d).
Let's say that the electric field caused by the the other laser at a distance d from the mirror is given by sin(d + pi), which must be possible with correct positioning of the laser.

Then the total electric field at distance d from the laser is given by:
sin(d) + sin(d + pi)
=sin(d) - sin(d)
= 0

So the electric field is 0 for all d.

This is the waves being "destructive everywhere", is it not?

10. Dec 7, 2007

Sojourner01

You're confusing the 'wave nature' of light (which is no such thing) and the quantum nature of the photon. When you have destructive interference, what you're actually saying is that the electromagnetic field amplitudes of the photon wavefunction are zero in that region - making the probability of interaction infinitesimally small. However, the photon's probability distribution propagates through free space at the speed of light and is modified by the electromagnetic fields it passes through so that, in all space the photon could possibly be, the cumulative total of its interaction probabilities is 1. All you're doing in destructive interference is making it extremely improbable that the photon will transfer its energy in that region.

11. Dec 7, 2007

tomprice

Ok Sojourner, thanks for that answer, I think it's a little over my head though.
First of all, could you please elaborate on what you mean by:
a. "probability of interaction" - interaction with what?
b. "Transfer its energy" - transfer energy to what?

+ A little off-topic point for dst: adding the functions sin x and cos x does not give you 0, it gives a third wave with an amplitude of the square root of two.

12. Dec 7, 2007

mda

The source beams disappearing would imply entanglement (since there is no retroreflected beam), which may have some additional requirements.

One thing nobody has mentioned so far is the phase difference between reflection and transmission... for destructive interference this has to be an integer multiple of pi.

For a metal splitter this will be close to pi/2 so destructive interference cannot occur for both beams...

If we phase shift the sources by pi/2 we get destructive on one exit and constructive on the other, and power is conserved (total power from both sources appears on the constructive side).

I haven't considered other splitters but I suspect similar conditions hold.

13. Dec 7, 2007

mda

Same is true for a frustrated total internal reflection (typical cube type)...

14. Dec 7, 2007

pervect

Staff Emeritus
I'm pretty sure that there *must* be difference in phase between the reflected and transmitted beams in any lossless beam splitter.

This is for simple energy conservation reasons. Let the beams be plane waves, with a peak E-field of Emax. The power of the beam is proportional to E_max^2.

Then if you have a lossless beam splitter, you start out with a beam of E_max (power =1) and split it into two beams of .707 E_max (power = 1/2). You split one beam into two parts, each with half the power, and nothing gets lost.

This isn't possible if the two split beams are in phase where they overlap, but is perfectly possible if the split beams are 90 degrees out of phase where they overlap.

If you split a beam into .5 E_max and .5_E_max, you've split a beam of unit power into two beams of quarter power. This is a lossy beam splitter, not a lossless one.

Thus for a beam splitter to be lossless, it must shift phase.

15. Dec 8, 2007

Sojourner01

OK, I think I may have muddled the issue by introducing quantum mechanics here...

With anything. A photon - or any other quanta (-um?) - is defined by its probability distribution, that is, the probability that it is 'found' there. Until you 'find' it, it isn't anywhere. To locate it, you have to interact with it, because that's what a photon actually is; a quantisation of the electromagnetic field.

To whatever's interacting with it. In the case of atomic orbitals - in all matter, for example - the photon has to be entirely absorbed by the atom because that's what quantisation is. It may be instantaneously re-emitted - but then it's effectively a different photon. So essentially, what you're saying is that the photon transfers all its energy - which is the crux of the original question - wherever you interact with it, and that location is probabilistically determined by the quantum nature of the photon.

What the destructive interference region does is to modify the probability distributions of the emitted photons so that the interaction probabilities are extremely small in the interference region; shifting the total probability (which, remember, has to sum to 1 over all space because you have to find it somewhere eventually) to elsewhere in the space surrounding the system - or even in the system - like inside the mirrors, or even in the laser. I can't say exactly how this would change without doing some quantitative thinking.

16. Dec 8, 2007

tomprice

Thanks Sojourner I think I get that...
So, if the two lasers, instead of being the exact same frequency, were of a very slightly different frequency, we would get both areas with nearly completely destructive constructive interference and with nearly completely destructive intereference (and everything in between). Since in the latter, the probability of interaction would be extremely small, could we actually place a wall there and have the beam pass through the wall? Cause that would be pretty awesome...

17. Dec 8, 2007

Sojourner01

Hmm...

OK, good question, you've stretched by brain a bit further than it wanted to go there!

I think - guesstimating, mind you - that the wall itself modifies the photon probability distributions such that the interaction probability at its surface is large. This leaves very little 'room' for the distribution beyond the wall so technically the beam can be seen to occasionally go straight through. However, this is true for any situation.

I think it would be best not to take my tautological model too far - I merely intended to convey that the destructive interference situation is a quantum mechanics problem. I have most likely made some logical errors somewhere, so don't quote me on this!

18. Dec 8, 2007

mda

I agree. As you say, it can be proven that a lossless beam-splitter must shift by exactly 90 degrees, otherwise energy is not conserved (could be created or destroyed depending on the phase of the sources).

Even more interesting is the case for lossy splitters... it turns out that interfering the beams causes resonant tunneling/anti-tunneling (depending on source phase), meaning you can get back up to 100% of the source energy despite absorption, or force the splitter to absorb all of the energy. Of course in the latter case the energy is not destroyed... it is converted to heat by the splitter.

Thanks to the OP for such an interesting problem!

19. Dec 9, 2007

Claude Bile

The suppression of radiation from emitters is a real phenomenon and has been observed since the early 1980s. The reason radiation is suppressed is because there is no mode for the radiation to emit into - to put that in more layman terms, the electric field can't find a way to wriggle around in such a way than energy is actually emitted and transferred in a particular direction, bound by the constraints of the laws of electromagnetism, i.e. Maxwell's equations. Sometimes, though this is much more rare, one can get a so-called "complete photonic bandgap" where the emitter cannot emit at all in which case the emitter will happily sit in an elevated energy state - a kind of induced metastability.

Photonic crystals are popular meta-materials that exploit this phenomenon, and oodles of information about them can be found online as well as in scientific literature.

Claude.

20. Dec 9, 2007

tomprice

whoa that's weird... so complete photonic bandgap will not necessarily happen here because of the relative phase shift between reflection and transmission; could you please give an example of a situation in which it would happen? Thanks.

21. Dec 9, 2007

Chronos

mda is getting hot [as are various system components]! 'Lost' energy has a way of finding itself in other places via scattering and absorption. Think about destructive interference in water waves. Is any energy 'lost'?

22. Dec 10, 2007

Claude Bile

The situation outlined in the original post is highly idealised and somewhat unphysical since it treats the beams as 1D entities and so on, but you would find a photonic bandgap in materials with a periodic refractive index. Opal for example gains its iridescent character because it is made of of countless silica nanospheres stacked in a lattice.

To obtain a complete photonic bandgap, you need, in addition to the periodic lattice, a strong refractive index contrast. One method is by infiltrating an opal-like structure with titanium oxide then dissolving the silica spheres to end up with a so called "inverse opal". The neat thing about this method is that the titanium oxide mixture can be doped to include rare earth ions such as Erbium of Ytterbium that act as light sources embedded within the inverse opal lattice.

There are other methods as well - this is a very active field of research.

Claude.

23. Dec 10, 2007

The energy is within the photons and not in their fields. Even if you were to create complete interference this would not mean that the photons ceased to exist. It would only mean that their wave nature would be constant rather than fluctuating.

24. Dec 11, 2007

mda

so how does one detect "constant" photons, and if they are not detectable, are they really existing as photons?

25. Dec 11, 2007

mda

I'm not really sure what you're getting at here.
The only similar situation I can think of is where one source is "behind" the other (otherwise they are not two separate sources with co-propagating waves). In that case the travelling waves outside the sources cancel (assuming infinite line sources), and a standing waves forms inside. Energy is not lost - the sources reflect the waves from each other, so the energy is confined to the counter-propagating region between the sources. Of course, in the ideal limit this is not the same as the beam-splitter since in that case "no" light will be reflected back to the sources.