Quite easy. Atmospheric pressure = 14psi (pounds per square inch) = 100000 Pascal = 10 Newtons / centimeter^2.

The total force acting over a dime-sized hole is about 10 Newtons or the equivalent weight of a ~2-pound object on earth. Paper or thin card might tear but a piece of strong multilayer corrugated cardboard should be fine. Might need to give it some help sealing (glue or something) too.

Quantum mechanics is best understood with a solid grasp of classical wave mechanics.

If there were two water waves in different locations, it's easy to keep track of them, even if they momentarily cross. But if there were two travelling together in the same direction -- is that still two waves? If they then separate, which one is which? This is what indistinguishability means.

Photons are fundamentally just bumps in the global electromagnetic field. When the bumps are well-separated, we can say this is bump 'A' and this is bump 'B'. When they are close, or moving together, or interfering ... those labels are not possible.

> A quality answer and even some better added mystery. Nice! Well done. So, why do physicists fight about that? Is it actually unsettled science?

It's not as mysterious or unsettled as portrayed -- rather it is a case of using non-specific language when trying to simplify for students or laypeople, leading to confusion. The math is exact and well tested. Have a look at my direct reply.

As for interpretations of quantum mechanics -- that one is unsettled. We know the math of QM works very well, but we have little idea what physical meaning (if any) to assign to a lot of the intermediate operations we do in a calculation. Ars Technica recently published a decent article on the topic.

Right, and I didn't contradict you on that, I'm adding to your answer there. The only mistake you made here was you have confused GR and SR.

As far as I understand you, you're asking two different questions.

> So if photons travel through a non-vaccum medium by being absorbed and re-emitted, how the heck does the information travel through that medium? Who tells the emitting atom to generate photons of exactly this frequency and polarisation in exactly that direction? How does it actually generate that frequency, e.g. the 432.1THz of a ruby laser when passing through a pane of glas?

This is the wrong type of "absorbed and re-emitted". Photons are not completely absorbed and then re-emitted by a single atom, like you get when you cause fluorescence or something. See my longer explanation. So while you are correct to worry about random direction or energy in the case of classical particle absorption and re-emission, that's not what's happening.

> If one adds unspecific energy to the same piece of glass, i.e. melts it, it glows in yellow or white. Is there any way to make that glass emitting photons of a certain frequency except shining the right frequency into it?

If you had just a tiny amount, like a few atoms, of glass, you would pretty much only see photon emissions at their characteristic energy levels (associated with electron shells, vibrational modes, etc). But as you add more and more atoms, the modes get washed out and photons get absorbed and re-emitted within the glass itself many times before finally emerging (here I am talking about complete absorption and re-emission) such that the final spectrum always looks like black-body radiation. That's why objects of a certain temperature end up looking like certain colors.

> according to General Relativity, a massless particle must always travel at c.

That's special relativity, not general relativity (yes, GR includes SR, but you get what I mean).

The fundamental wrong assumption people make is that a wavepacket of the EM field -- a photon -- in a vacuum is somehow "the same" as the wavepacket of the EM field in a complex background of charged particles, i.e. a real material made of atoms.

To make an analogy (of limited range, please don't abuse it :)): no-one is surprised that a classical gravity wave in water and a wave in honey behave differently. Why is it surprising for an EM wave? /u/Keudn 's description is classical and doesn't explain how to calculate the speed in a medium, but aside from that missing depth, strictly correct.

I think you're on the right track, mostly. I can try to add some detail/clarification.

> 1) you emitted a single photon, there is a high probability that photon would be detected by the detector.

Yeah, more or less -- the probability depends on the coefficient of reflection of the surface of the glass. It could be high, or low, depending on the angle and refractive index of the glass, and will match the classical calculation of the relative amplitudes of transmitted and reflected waves (Fresnel equations), because fundamentally, the photon wavefunction is obeying pretty much the same wave propagation laws. Note you only get this behaviour if the photon wavefunction spatial extent (typically many photon wavelengths) means that it has the possibility to interact with many identical atoms which are spaced in a uniform way (relative to the wavelength of the photon). This is mostly true for optical wavelengths hitting normal materials (typical atom spacing 0.1nm, optical wavelength 500nm, any random variation is too small to be "seen" by a photon) but is not necessarily true at shorter wavelengths like X-rays, or if your "glass" is only a few atoms big. Those scenarios are more complicated and you don't always end up with the "normal" ray optics rules after summing those probabilities.

The only real difference to the classical wave picture is in the moment of detection -- instead of measuring a classical wave, with some part reflected and some transmitted, we are set up for a discrete, quantum mechanical interaction in the detector. A 1 or a 0. This collapses* the wavefunction according to the probabilities mentioned before, to either interact with the detector (and so we say the photon was transmitted and then absorbed in the detector) or not (and here we can say the photon was reflected).

> 2) ... However, because they are all part of a wavefront, none of them would be emitted out the sides of the transparent material?

You can hopefully see how the previous explanation scales up to emitting many photons, the numbers of photons detected will match the probability calculations. But to address your last sentence -- in this toy example, I assumed that there was no internal absorption or scattering inside the glass. In such a case, the original+scattered wavefunctions sum up such that the only possibilities are transmission straight through or reflection right at the interface, and there is no chance for a "random" photon to emerge out of the side of the glass at some angle. This isn't to do with photons being "part of a wavefront", unless I misunderstand you. More fundamentally, photons are waves; or at least, they travel in the same ways that classical waves do, and waves moving through a uniform medium don't just randomly scatter.

Now if we make things a bit more realistic -- the glass is not going to be a perfect crystal, and the uniform background of atoms assumption is not going to be exactly true. There will now be a small possibility of the wavefunction scattering differently off some imperfection in the atomic structure. Because it's associated with an imperfection, this part of the wavefunction won't be cancelled out by all the neighbouring scattered amplitudes, and you will end up with a real (but small) probability to have a photon emerging out of the side of the glass in some random direction, in addition to the main probabilities of transmission or reflection at the surfaces.

* "The measurement problem", or what exactly wave-function collapse really means physically, is a very thorny issue to which there isn't a good answer. But we know that mathematically, it works.

> Sure, it's not individual photons being absorbed by individual charges,

You are correct overall but dismissing this specific point as if most people understand it easily is pedagogically dangerous -- based on my teaching experience that's exactly what a lot of people who hear "photon absorption and emission" end up thinking of.

> a mirror is also an absorption and re-emission situation

I already addressed the use of "absorption and re-emission" in a previous reply to you, so I won't repeat that -- but for anyone reading, this is another case where you must not confuse "classical particle absorption and re-emission" (photon comes in, is absorbed, is gone for a moment, is emitted) with the collective scattering of probabilities/wavefunctions that we calculate in QM.

The law of reflection does not (cannot) come from a classical particle scattering model. It can be derived from either a classical wave model or a properly quantum mechanical model (wavefunctions are waves after all).

> The answer comes down to conservation of momentum and interference. Since the incoming photon has momentum, there is a higher probability that the photon will be emitted in the same direction to conserve momentum.

We've already discussed why you should not imagine the photon being completely absorbed by an atom when trying to derive the laws of optics. But let me address the momentum part of this answer (interference part is fine) as well because it's also misleading. Why should an atom prefer to return exactly to its previous state of "zero momentum"? Answer: It doesn't. Further: First of all the momentum we are talking about is tiny when a photon is absorbed by an atom, so even the thermal motion of the atoms is already going to wash out any momentum change. Second the atom is free to exchange momentum with its neighbours via phonons, so there's no reason why it has to return to zero by itself. Third, we measure single photon/single atom absorption and re-emission behaviour experimentally and it truly is isotropic (well, kinda), while your answer effectively claims it is not (contradicting your very first statement). (Incidentally, when 1 and 2 above are not true any more, like for a very cold gas, we can do fun stuff like laser Doppler cooling.)

The real answer is that we're not dealing with classical absorption/emission of photons by one atom, we're dealing with a collective behaviour and scattered probabilities/partial waves (see previous linked discussion, etc.) -- even in the case of few photons. So though all the scattered components are random in direction, we add them coherently (interference, as you said). The newly constructed wave/wavefunction has a clear direction of travel obeying the law of reflection.

/u/dack42

There's some good info here but also a lot of misleading or easily misread info.

You're correct on the fungibility of photons. Nothing to add there.

But you are dangerously vague for the rest.

> The easiest to understand model is the one you mentioned- and it does work.

Careful. The confusion you describe (even among physicists) comes at the root from not being specific on what we mean by absorption and emission, and unless you clarify you have not helped the situation. Let me try to be extremely clear about the possible models.

1. Classical scattering of waves. Here, there is a main EM wave which can partially scatter off atoms, since they are charged -- and importantly it's a collective scattering from all of the atoms in the path. The superposition (combination) of the original wave and the small scattered waves leads to a new, slower wave, with the same frequency as the original. Note that at no point is the incoming wave completely extinguished (absorbed), it's only ever a partial effect. This model works mathematically.

2. Classical scattering of particles. Here, you imagine a billiard ball photon travelling along that happens to hit an atom and be absorbed. For a moment there is no more photon anywhere. Then the atom relaxes, and a photon is re-emitted to continue on its journey. This model does not work mathematically in any way to explain the speed of light in a medium. You cannot assign the slowing down of light to a random time delay between classical particle absorption and emission nor to a particle taking a 'longer zigzag route' through the medium.
   When people ask this question on the internet, it's usually with a foundation of only classical mechanics, not quantum mechanics, and so this is usually what they are imagining when they say "absorption and re-emission" and it is wrong.

3. Quantum mechanical scattering. This one is tricky to understand without at least an intro to quantum mechanics but fundamentally we are scattering probabilities, not whole particles. There are several ways to do the maths -- you can consider the propagation, partial scattering, and interference of the photon wavefunction as it interacts with virtual energy levels in the atoms (looks very similar to the classical wave math -- superpositions of the original and partially scattered probability waves). Virtual energy levels are guaranteed to be unstable and exciting the atom to one is fundamentally different to exciting to a stable energy level. Or you can consider a path integral approach like that of Feynman (again, summing probable paths, not definite or discrete ones), or you can start calculating the collective excitation of the photon and all neighbouring atoms as a dressed quasiparticle with new properties (specifically, mass -- so travels less than c). The important thing here is that all the behaviour is collective, never discretely with a single atom -- and at no point is the photon (with its original frequency and direction) completely gone. When physicists say "absorption and re-emission" they are often thinking of this model because we use the same terms for classical scattering and scattering of a QM wavefunction -- but that does not mean it is the same model as 2) above. It is not, it is fundamentally different.

> The most common complaint is that an atom can only absorb very specific wavelengths, but light of all wavelengths is slowed down by materials. But, this is handled by understanding that collections of particles will have nearly an infinite number of modes of excitation

Your answer doesn't work. It is true that "excitations are always discrete" is an oversimplification -- the existence of black-body radiation proves that. But you don't get to claim that all materials absorb at all frequencies and therefore slow light through absorption and reemission, without also explaining why the transparency of a material doesn't have anything to do with the speed of light through it (cf. glass at optical wavelengths). Again this confusion you introduced comes from not clearly differentiating between the QM scattering of probabilities -- which may involve virtual energy levels and whole lot of behind-the-scenes stuff -- and complete, classical absorption of a photon to a new stable energy level. The real answer is simply that you don't need stable energy levels (which cause the material to become opaque) to exist to do the QM scattering math. Though transitions to virtual energy levels are low probability and necessarily temporary, their collective sum, including the original photon as well, gets us to where we want to go.

A "laser thermometer" as commonly sold is just a small FoV infrared thermometer with a laser pointer strapped on top. (The exception is very advanced thermometers which use calibrated lasers and optical ranging to measure reflectivity.)

Usually these thermometers accept emissivity as a setting. They don't measure it.