In that case, the photons have a uniform distribution, so they have the same amplitude at the edge as away from the edge. If you have something like a Gaussian beam, the amplitude falls off away from the beam axis. With that centered in the aperture, the amplitude near the edges will be small. While the amplitude of a Gaussian beam never goes precisely to zero, you could make it arbitrarily small if you’re allowed to use arbitrarily high frequency photons.
I’m going to weasel out of this one by pointing out that I said “significantly deflect the light”. You’ll get some amount of diffraction effects for any size hole and any wavelength, but if the hole is much larger than the wavelength, then the effects will be small. If the hole is comparable in size to the wavelength, or smaller, then the effects will be large.
The probability of an N particle event goes up like the Nth power of the density of particles. When I am talking about reaching the breakdown field of the vacuum, I’m explicitly talking about enormous energy densities. In reality, if you create this situation with lasers or other pratical light sources, talking about single photon states is not going to be an accurate representation. Better is to use a Poisson distribution of n-photon states, a.k.a. Glauber’s coherent states. Better still is to just think about classical fields. Then it becomes like Hawking’s analysis of radiation from the event horizon of a black hole. A virtual electron positron pair is created. If the electric field is strong enough, they will gain enough energy to pay for their rest mass before they’ve gone a compton wavelength. This is a very high order process if you insist on using the photon picture, but it seems to me you should still be able to calculate this in perturbation theory if you start with classical electric fields.
I’m not sure of my physics here, I am no QED jockey, so I am glad to bow to higher authority if this is incorrect.
what if enough photons are deflected by diffraction that they hit the hole and start jiggling it around?
Yes, but consider this – you can do that diffraction experiment where you reduce the intensity to such a point that only one photon at a time goes through the slit and do this for long enough that you get a consistent pattern equal to what you get if you let a bunch go through at the same time. the patterns will be the same.
The patterns for a 100 micron slit and a 200 micron slit are different, and have different-looking edge “ripples”. Yet the photons, going through one at a time “know” how wide the slit is. This clearly isn’t a case of “invariably some photons must be passing by the edge, and thus get deflected”, since the width of the slit influences exactly how the pattern appears. Clearly, each photon “senses” the slit width, which implies that the photon extends to the edges of the slit*. You could imagine, if you want a workable mental picture of a photon, that each of them isn’t like a pingpong ball, but is more like a dust bunny, with a probability distribution that “tails off” like that gaussian distribution you mentioned above. It’s never quite zero, even though it does get smaller.
I’m not saying that mental picture is what a photon really is like. It’s a model, but a better one for visualizing the photon’s actions. And other particles act more like fuzzballs that hard spheres, too.
*The edge effect fringes approach a limiting value for an edge with the other edge really far away, as Grimaldi saw around his objects, which is much like what you’d expect from a mental picture where the photon gradually tailed off.
Whether you think classically, or in terms of photons, isn’t the point. You can still arrange for the classical field strength, or for the low probability tail for each photon, to have arbitrarily small amplitude at the edge of the aperture. In your example of diffraction at the edge of an object, they were intentionally shooting photons at the edge. In your comparison of diffraction through 100 and 200 micron slits, the amplitude is uniform over the slit opening. Those examples don’t say anything about whether you can have a hole where photons won’t touch the edge, because they weren’t trying to avoid touching the edge. They were trying to hit the edge.
Nobody was trying to hit anything. The photons go where they will. It’s not a given that illumination is uniform over the field, although that’s a reasonable supposition. The point of the low-intensity photons is that the photons would have statistically carry them through all parts of the aperture, but the end results are always the same, so even if the center path of the photon is as far as it can be from one edge, you still see diffraction from that edge.
And you really can’t make the amplitude aribitrarily small. The non-zero wavelength of the light will place a realizable size and tail on your wavepacket, and you won’t be able to get below a certain amplitude for a given size. Loo at it as a case of Fourier analysis of your arbitrary wavefunction. The only way to completely fit your arbitrary function is with an infinite range of wavelengths, which you won’t have.
You’re implicitly assuming a photon amplitude more-or-less uniform across the aperture. A single photon with a uniform amplitude across a circular aperture will certainly have a larger probability of scattering than a single photon whose amplitude is that of a Gaussian beam. The spread of the first photon will be governed by the aperture, but the spread of the second will be governed almost entirely by its Gaussian shape.
For arbitrarily high frequency, which is the qualification I used in my first post, you don’t have a limit due to a fixed wavelength, and you can make the amplitude arbitrarily small.
What I don’t understand now is how they can collapse into a black hole if they have no mass.
In air there is a limit to how much laser light you can put through an aperture before the field strength becomes so high that the air ionizes. This is a common way to damage laser optics, and limits how much power you can get out of diode lasers assuming you can deal with the heat. I don’t know if this applies to incoherent light though, as random photons would tend to have the electric fields of photons average near zero I would think.
One photon has no mass. A system of two or more photons can have mass, and probably does. Contrary to your expectations, the mass of a system is not equal to the sums of the masses of its constituents.
I know this is a backtrack from the peanut gallery, but why are you analyzing (I think you are) photons-are-massless in the context of, stupid analogy, how many clowns can be stuffed in a Volkswagon (which, of course, I’m sure you’re not, but that’s how I approached OP initially).
The photons are zipping through the hole, so, in motion, they have mass. Having claimed that–and of course I look toward to being corrected–does this alter anything in what I am attempting to follow in this conversation?
Just being in motion doesn’t give them mass. Mass is invariant: An object’s mass does not depend on its velocity at all. You’re probably confused because, at some point in your education, a teacher or book told you that mass increases for things that are moving quickly. But when that teacher or book said that, really all they meant was that energy increases, and they should have just said so to begin with. Much better to restrict the word “mass” to mean what that teacher or book would have called “rest mass”.
Is there a “small words” way of giving an example of how it happens with, say, a system of two photons?
Suppose you have two green photons travelling in the same direction towards you. They have a certain total energy. If you are in a different reference frame traveling towards them (say you accelerate towards them). they will be blue shifted, and will have a higher total energy. (Blue photons are higher frequency, and have more energy than green photons.) Conversely, if you accelerate away from them, they will be red-shifted, and have less total energy (red photons are lower frequency, and have less energy than green photons). Accelerate fast enough away from those photons, and their total energy in your frame of reference can become very small.
Now suppose you again have two green photons traveling towards you, but from opposite directions. If you accelerate away from one of them, its energy will be less, but the energy of the other will be more. No matter which way you move, you won’t be able to make them have less than some energy Emin. The mass of that system of two photons is then given by that famous equation as M = Emin/c^2. This holds, even though the mass of each photon individually is 0.
A single massive particle has energy and it’s energy and mass are connected by special relativity (E=mc[sup]2[/sup]), however when we talk about mass we mean rest mass which is directly proportional to the it’s total energy in the frame of reference where it has the least energy i.e. the frame where it is at rest.
Now take a single photon, it too has energy, but in all frames it is never at rest. This means that for any given frame (of reference) you can always find another frame where it has less energy by finding a frame where the photon is redshifted relative to the first frame. In fact you can keep on redshifting a photon so its energy is arbitrarily close to zero. This means that a single photon has no rest mass.
Now take two photons travelling antiparallel (i.e. along the same line, but in opposite directions) and consider the total energy of the two-photon system. Redshifting one photon by changing frame will blueshift the other photon increasing it’s energy and in fact photons can be blueshifted to arbitrarily high energies. So as one photon’s energy approaches zero by redshifting the other photon’s energy will approach infinity by blueshifting. The minimum energy frame is the frame in which both photons have the same energy and the rest mass of the two-photon system will be directly proportional to the sum of the photons’ energies in this frame.
Very loosely speaking the amount of gravitational attraction a system ‘causes’ is related to its rest mass. A system consisting of a single photon has no rest mass and causes no gravitational attraction, whereas a system of two non-parallel (but as we have seen possibly antiparallel) photons has rest mass and causes gravitational attraction.