imagine two beams of light in a vacuum that are of the same wavelength arranged so that they interfere with each other destructively. They combined beams “cancel each other out”.
What happens to the energy? Thermodynamic law dictates it can’t be destroyed, so the energy must be transferred to something. I don’t think it gets converted to heat (makes no sense to me in a vacuum).
So; 'splain to me where the energy goes.
This was, believe it or not, a hot topic one day at the Institute of Optics, although in a slightly different form. What happens, someone asked, if I build an interferometer and adjust it to exactly cancel out one of the emitted arms? The answer is that the amplitude goes up in the other arm to compensate. (Just as, when you have two beams interfering at an angle, the peaks go higher when the “zeroes” are carved out. The energy is just redistributed, not destryoed).
“Do you mean that if I take a terawatt laser”, asked one profesor, talking about the interferometer case, “and I shine it into the interferometer, the two beams will exactly cancel out?”
“Yes,” answered another professor. “The energy just goes into the other arm.”
The first professor pondered this for a moment.
“I still wouldn’t want to stick my arm in there,” he said.
It’s impossible to have a spatially confined perfectly collimated beams of light (Uncertainty principle).
The energy of the two 180 degree out of phase beams of light completely cancel out at the exact center, but all the energy can still be found at angle deviations from this exact center.
Nothing behaves ideally.
This is one of my wife’s chief complaints.
Part of the key to understanding this is to realize that every source is also a receiver. Suppose, for instance, that you have a radio antenna, with a comfortably long wavelength. If I put another antenna of the same sort very close to the first one (i.e., much less than a wavelength distant), and drive it at 180 degrees out of phase with the first, I’ll get almost total cancellation, in all directions. So the total amount of energy coming out of the antennas will be very small. Where did the extra energy go? Nowhere: It didn’t go into the antennas to begin with.
To see this, suppose that you didn’t power the second antenna at all. It’d then be acting as a receiver, not as a transmitter. There would still be electrons moving in it, in response to the incoming radio wave. Those electron movements would, in fact, be exactly the same sort of movements (and only slightly smaller) you’d force into the antenna, if you wanted to produce a wave 180 degrees out of phase with the first one. So when you turn the power on to the second antenna, it’s already doing almost exactly what you want it to do: You only have to add a very tiny amount of additional energy to get it to do exactly what you want.
Meanwhile, the first antenna is also getting this same effect from the second antenna. So you also need only apply a tiny amount of the first antenna, to get the electron motions you want in it. And the total energy you’re putting into both antennas is, unsurprisingly, exactly the amount of energy the waves (what portion of them isn’t canceled out entirely) are carrying away.
Long ago, in a galaxy far away, I wondered the same thing - there are some quite good answers in this thread.
(although, unfortunately, the diagram I linked in that thread is no more)
Thanks for the radio antenna concept, I think I can follow that. But, I was thinking more along the lines (snerk) of using a diffraction grating. Can you fill me in on where the energy goes in the regions where they get cancelled out? Type slowly and use small words.
See my post above – the energy that disappears from the places where it goes dark gets added to the regions that have light – they get brighter.