4 ms·

I don't know what you mean by "surface", but the number of photos collected depends on the area of the lens. No quantum mechanics or interference involved.

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 ago
    What's wrong with the reasoning, then? Isn't the probability amplitude proportional to the area?
    - analognoise 10y ago
      It's proportional to the area, which is already a term encompassing length^2.
      - grondilu 10y ago
        If 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 ago
    And if we change to "proportional to the area of the lens, not the sensor" ?
- davrosthedalek 10y ago
  No, 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 ago
    I 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 ago
  We 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 ago
    Interferometry, 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 ago
        CCDs 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.
tigershark 10y ago
Nope, nothing to do with that. A lens with a 1m diameter f/10 will be less bright than a lens with a 10 cm diameter f/1.4
- greglindahl 10y ago
  You might want to read https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telescopes https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telesc...
  - tigershark 10y ago
    That is valid only for point sources. Stars are not points when seen in a telescope because of airy disks, atmospherical scattering and imperfections in the optics. If you have perfect optics, you are outside of the atmosphere and the only limit is the airy disk then yes, for point sources you don't have that problem. Now, are you saying that a valley is a point source?
    - greglindahl 10y ago
      If you look back, I made a statement about the number of photons collected. I totally understand that camera people, who have a shitload of photons available most of the time, don't think about them like astronomers do. But if it's very dark, it becomes astronomy.
  - tigershark 10y ago
    Not sure why I cannot reply to your other comment. Anyway, astronomy has nothing to do with the scarcity of photons. If you take a picture of M31 you have a relative abundance of photons, with most of them not being point sources. If you look at any planet or globular cluster or nebula is the same. Only when you are observing single stars you are having that. And single star observations are for sure not the whole astronomy.
    - greglindahl 10y ago
      HN requires a certain delay before responding, to encourage people to think first. Many distant things, like galaxies, are very dim. M31 is almost visible to the naked eye, which is not dim to an astronomer. You might want to study astronomy before having strong opinions about it.

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 ago
  What's wrong with the reasoning, then? Isn't the probability amplitude proportional to the area?
  - analognoise 10y ago
    It's proportional to the area, which is already a term encompassing length^2.
    - grondilu 10y ago
      If 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 ago
  And if we change to "proportional to the area of the lens, not the sensor" ?
davrosthedalek 10y ago
No, 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.
- deleted 10y ago
  [deleted]
- TeMPOraL 10y ago
  I 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 ago
We 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 ago
  Interferometry, 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 ago
        CCDs 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.

tigershark 10y ago

Nope, nothing to do with that. A lens with a 1m diameter f/10 will be less bright than a lens with a 10 cm diameter f/1.4

greglindahl 10y ago
You might want to read https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telescopes https://en.wikipedia.org/wiki/F-number#Focal_ratio_in_telesc...
- tigershark 10y ago
  That is valid only for point sources. Stars are not points when seen in a telescope because of airy disks, atmospherical scattering and imperfections in the optics. If you have perfect optics, you are outside of the atmosphere and the only limit is the airy disk then yes, for point sources you don't have that problem. Now, are you saying that a valley is a point source?
  - greglindahl 10y ago
    If you look back, I made a statement about the number of photons collected. I totally understand that camera people, who have a shitload of photons available most of the time, don't think about them like astronomers do. But if it's very dark, it becomes astronomy.
- tigershark 10y ago
  Not sure why I cannot reply to your other comment. Anyway, astronomy has nothing to do with the scarcity of photons. If you take a picture of M31 you have a relative abundance of photons, with most of them not being point sources. If you look at any planet or globular cluster or nebula is the same. Only when you are observing single stars you are having that. And single star observations are for sure not the whole astronomy.
  - greglindahl 10y ago
    HN requires a certain delay before responding, to encourage people to think first. Many distant things, like galaxies, are very dim. M31 is almost visible to the naked eye, which is not dim to an astronomer. You might want to study astronomy before having strong opinions about it.