Amount of light through one hole

Is there a volume limit of light that can pass a hole? Let’s do this in vacuum and use lenses so that light beams through the hole without touching the walls of it.

If there is a limit, what does it suggest?

Photons are bosons* so no, there is no limit.

Fermions, not so much, you can’t cram them together like that. At some point you won’t be able to shove more than a certain number of them through a given hole.

  • Go on, look it up, it’s fun stuff. As usual it’s because of quantum.

Enough light in one place will collapse into a kugelblitz, a black hole. At that point it isn’t light anymore (a black hole is a black hole, regardless of what originally went into it), so that’s a limit to how much light you can aim through a particular hole.

Does “quantum state” define position as well? I thought it only defined energy, spin and whatnot.

[QUOTE=AaronX]
Does “quantum state” define position as well? I thought it only defined energy, spin and whatnot.
[/QUOTE]
For a certain meaning of “position” yes. One example, two electrons can’t be in the same state/place in an atom - which given their size* should be quite a roomy place. You can put one electron “on top” of another if you change another property, like its spin.

I’ve got to work right now (and IAMNAPP), I expect Stranger on a Train can wright a nice essay explaining this.

  • For a certain meaning of “size” - it depends how you look at them, literally.

Actually, you should be able to arrange the light in such a way as to circumvent this limit, and get an arbitrarily large intensity through the hole. You just need to make sure that all of your photons are traveling very close to parallel to each other.

If the intensity gets high enough, I would wonder if photon-photon scattering would start to cause a loss of intensity. Of course, if the incident photons were traveling pretty much in the same direction, so will the produced photons, so even with this taken into account, you wouldn’t lose too much intensity through the hole. (To first order, anyway.)

Photon-photon scattering should also go to zero as the angle between the incident photons approaches zero.

Could someone explain how an infinite number of photons could pass through the hole at the same time? Please use small words.

Simplest answer is, there’s nothing stopping them. We tend to think of “no two objects can occupy the same space” as being a fundamental law of everything, but it’s not: That law (or something roughly equivalent to it) actually only applies to one of the two classes of particles, the fermions. Most stuff that we’re familiar with happens to be made out of fermions, which is why it seems so universal to us. But there are also a number of particles in the other class, called bosons, for which that law does not apply. And photons are an example of that class.

And to add: Photons can occupy same space and time because they have no mass. And if you line em up properly, they won’t interact with each other and collapse into a black hole.

They can occupy the same space at the same time (and with the same quantum numbers [i.e., other properties]) because they are bosons, not because they are massless.

As you increase the intensity, the electric field gets higher. Eventually you reach the breakdown strength of the vacuum and start creating electron-hole pairs. This sets a limit on the flux through the hole.

Does all this mean there is no hole to small for a photon to pass through?

That’s a little more complicated. A photon can pass through an arbitrarily-small hole. However, if the size of the hole is small compared to the wavelength of the light, then you’ll get diffraction effects, which can in practice significantly deflect the light passing through in some direction or another.

JWT Kottekoe, my QFT is getting rustier than I’d like, but I don’t think that what you describe would actually happen, for a sufficiently well-collimated beam. That’d be another way of describing the photon-photon scattering MikeS mentioned. But there just wouldn’t be sufficient energy nor time in the center-of-mass frame.

Thought this question sounded familiar
http://boards.straightdope.com/sdmb/showthread.php?t=659900&highlight=Light

I’ve forgotten what little I ever knew of QFT, but I don’t understand why, if the intensity is high enough, I wouldn’t get electron/positron pairs. I can treat the electric field classically in this limit (but not the fermion fields).

A system of an arbitrary number of parallel photons still has zero rest-mass, but an electron-positron pair does not. Said another way, you can switch to a frame of reference where the photons have arbitrarily small energy, too low to make an electron-positron pair.

This assumes there are no CMB photons coming the other way, and that the parallel photons don’t interact with the edges of the hole.

I suppose that, strictly speaking, you could have interactions between larger number of photons to produce a single electron-positron pair. But even three-particle interactions are so rare as to be unheard-of, and they’re going to be that much less likely in a well-collimated photon beam, where the center-of-mass frame is going to be extremely time dilated.

This isn’t really correct – you’ll get diffraction effects when the size of the hole is significantly larger than what you’d think the photon can be. Virtually all diffraction experiments use situations where the slits are huge compared to the wavelength of light.

in fact, in response to the OP:

I don’t think it’s possible to have a hole where photons won’t “touch the walls”. The very first observaions recorded of diffraction effects, by Franchesco Grimaldi, were of diffraction at the edge of isolated objects, so the “holes” were effectively infinitely wide.
Light WILL pass through sub-wavelength apertures, in distinction to what you might expect from the behavior of radio waves and Faraday cages, but the behavior gets weird, especially when you have arrays of sub-wavelength apertures, where you can get enhanced transmission.