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I did mean area. And I thought the number of photons was proportional to the square of the area because the probability amplitude is proportional to the area.
by grondilu 10y ago
I did mean area. And I thought the number of photons was proportional to the square of the area because the probability amplitude is proportional to the area. Therefore the probability should be proportional to the square of the area, shoudn't it?
It also seemed to make sense to me otherwise we would not need to build large telescopes, we could just build lots of small ones and fuse the images.
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- titanomachy 10y ago> the probability should be proportional to the square of the area, shoudn't it? No.
- grondilu 10y agoWhat's wrong with the reasoning, then? Isn't the probability amplitude proportional to the area?
- analognoise 10y agoIt's proportional to the area, which is already a term encompassing length^2.
- grondilu 10y agoIf the probability amplitude is proportional to the area, it doesn't matter if that term already encompasses a length² term. QM tells you then that the probability should be proportional to the square of that amplitude, even if that means it would encompass a length^4 term.
- candiodari 10y agoAnd if we change to "proportional to the area of the lens, not the sensor" ?
- davrosthedalek 10y agoNo, the probability is proportional to the area, not the amplitude. If a sensor of size A sees N photons, a sensor of the same size next to it sees another N. Fusing these two sensors together see 2N, not 4N. Otherwise, you violate energy/momentum conservation.
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- TeMPOraL 10y agoI love this kind of explanations. You can skip a lot of hard math with a simple counterexample that shows the absurdity of proposed theory.
- greglindahl 10y agoWe do build small(er) ones and fuse the images (using interferometry), that's what many large telescopes are these days. In the radio, we've been mostly doing it that way since 1978 (VLA).
- grondilu 10y agoInterferometry, as its name indicates, uses interferences. So it adds probability amplitudes.
- greglindahl 10y ago... we use interferomers because we want the added resolution, not just the additional photons. If you don't want extra resolution, you can just add up the images from all of the smaller telescopes. Light is a wave and a particle, and if you are getting wildly different answers from thinking about it as a wave and as a particle, and you're looking at a macro and not micro scale, then you're doing it wrong. That's why I answered your wave question with a photon count answer.
- grondilu 10y ago> That's why I answered your wave question with a photon count answer. But how do you count photons without using probability amplitudes? If you count them by using a classical reasoning of photons being small particles falling from the sky, I'd say you're doing it wrong, because photons are not classical particles.
- greglindahl 10y agoCCDs count photons in a particular fashion, and it happens to involve individual photons doing things. You might think of it as photons knocking electrons off of atoms, but it's actually semiconductors with a narrow bandgap, so CCDs work at much lower energies than needed for ionizing radiation.