Do photons bunch up at an interface? Can they lase (laze?)

[ImaO]

TL;DR Summary

If the light slows at an interface, what happens to the photons coming in after the slowed ones? Can these cohere with the leading (slowed) ones and create what amounts to a FEL?

If the light slows at an interface, what happens to the photons coming in after the slowed ones? Can these cohere with the leading (slowed) ones and create what amounts to a FEL?

 

[phinds]

Summary:: If the light slows at an interface, what happens to the photons coming in after the slowed ones? Can these cohere with the leading (slowed) ones and create what amounts to a FEL?

If the light slows at an interface, what happens to the photons coming in after the slowed ones? Can these cohere with the leading (slowed) ones and create what amounts to a FEL?

Are you talking about the Event Horizon of a black hole? If so, the light does NOT slow down as it moves into the BH. Only light trying to escape from just outside the EH travels at c locally but appears slowed down to an observer far away from the EH.

If you are not talking about the EH of a BH, then what are you talking about? Are you talking about light going from one medium to another? Hm ... re-reading your post I now think that IS what you are talking about and I answered the wrong question.

 

[Dale]

Lasing occurs with stimulated emission, which in turn requires a population inversion. Photons slowing down doesn't by itself lead to a population inversion.

 

[PeterDonis]

If the light slows at an interface

This is a description of what happens at the boundary between two media with different indexes of refraction that is (a) highly oversimplified, and (b) classical, i.e., not consistent with describing light using photons.

If you are going to use the photon description, photons do not "slow down" when crossing into a medium with higher index of refraction; they just get absorbed and re-emitted more often.

 

[ImaO]

Thank you, Peter. I take from your note that the photon view isn't really appropriate (or anyway isn't optimal) for describing refraction; from the photon point of view, the speed is a group phenomenon (i.e., the light travels between ab/em events at c (not slower). But, per Dale's note, my question re lasing still holds, I think. If there are more absorption events, then there could indeed be a population inversion, resulting in lasing. No?

 

[PeterDonis]

If there are more absorption events, then there could indeed be a population inversion, resulting in lasing. No?

No. Population inversion means something much more specific than "more photons closer together". You need very specific conditions for a population inversion to happen, which is why lasers have to be very precisely constructed.

 

[ImaO]

Yes, more atoms in an excited state than in the ground state, (or a lower state), right? At the interface you have higher absorption so excitation, and the photons that follow right behind should cause SE on those, no? So at least increased coherence if not lasing?

 

[PeterDonis]

more atoms in an excited state than in the ground state, (or a lower state), right?

Only momentarily, until the photons that got absorbed get re-emitted.

But the more fundamental problem is that the atoms aren't all in the same excited state. See below.

So at least increased coherence

No. Increased coherence would require all of the atoms to be put in the same excited state. But they aren't. Atoms in a transparent medium have all kinds of different excited states available to them.

 

[PeterDonis]

the photons that follow right behind should cause SE on those

No, because SE requires that the state the atoms are going to emit into is the same state that the incoming photons are in, which isn't the case.

 

[ImaO]

Okay. Good. This is useful, thank you all. Let me explain why I’m on about this.

I have always been bothered by the “wave train” explanation of why light bends at the interface between two media of different refractive indexes (indices?) - even Feynman gives this seeming crazy explanation about wave trains having to be matched at the interface, and so they bend to match. But phonons don’t really have spatially extended wave trains in the sense that the ocean does. So the only sense I was able to make of this explanation was if they was a cluster of coherent photons at the interface, or something.

But probably the whole line of reasoning is misguided.

 

[PeterDonis]

I have always been bothered by the “wave train” explanation of why light bends at the interface between two media of different refractive indexes (indices?) - even Feynman gives this seeming crazy explanation about wave trains having to be matched at the interface, and so they bend to match.

This is (obviously ) an explanation using a wave model of light. But photons are not waves, so you should not expect this explanation to make sense if you are using a photon model of light.

 

[Nugatory]

If the light slows at an interface, what happens to the photons coming in after the slowed ones?

Photons don't work the way you're thinking: a beam of light is not a stream of photons the way a river is a stream of water molecules moving by.

The frequency $\omega$, speed $v$, and wavelength $\lambda$ of a wave are related by the equation $v=\omega\lambda$. In a vacuum, the relationship between frequency and wavelength is such that the speed of an electromagnetic wave always comes out to be $c$. It will be less when passing through a medium because the interactios between the oscillating electromagnetic field and the charged particles making up the medium reduces the wavelength at any given frequency.

The interface that you're describing is between a region where $\omega\lambda$ is larger and a region where it is smaller. However, the frequency $\omega$ is unaffected: for every wave crest that arrives at the interface another wave crest leaves it. The outgong peaks are more closely spaced ($\lambda$ is smaller) so the speed of the outgoing wave is less, but nothing is piling up at the interface.

 

[ImaO]

> The interface that you're describing is between a region where ωλωλ is larger and a region
> where it is smaller. However, the frequency ωω is unaffected: for every wave crest that
> arrives at the interface another wave crest leaves it. The [outgoing] peaks are more closely
> spaced (λλ is smaller) so the speed of the outgoing wave is less,...

Good. Great. Thank you. So, then, then what I'll call (for lack of knowing the right phrase) the "lambda-delta" model of differential (per wavelength) refraction, as explained everywhere (incl. Feynman!) is sensible under this model. Just to be clear, so then the reason for the reduced λλ is NOT increased absorption and re-emission, unless by absorption (and emission) we mean something other than the process wherein an electron changes its energy state. (Perhaps it is mathematically equivalent, and so one could say that there is "virtual" absorption/emission, in the same sense that there are virtual particles per vacuum fluctuation?)

 

[PeterDonis]

so then the reason for the reduced $\lambda$ is NOT increased absorption and re-emission

You are mixing up different models. In the model with $\lambda$ there are no photons, just classical waves, so there is no absorption and re-emission. In the model with absorption and re-emission, there are no waves and therefore no $\lambda$, there are just photons. Both models are approximations, and which one you use depends on the specific problem you are trying to solve.

Perhaps it is mathematically equivalent, and so one could say that there is "virtual" absorption/emission

I believe the "absorption and re-emission" in the photon model does involve what are, somewhat misleadingly, called "virtual particles", yes.

 

[ImaO]

Thank you! Should I (everyone) just forget about the particle and wave models, and just figure out how field theory describes refraction (everything)?

 

[PeterDonis]

Should I (everyone) just forget about the particle and wave models, and just figure out how field theory describes refraction (everything)?

Not unless you want to do the full computations of field theory every time instead of using the much simpler approximations that work well for a wide variety of problems.

 

[ImaO]

Thanks. I guess I care more about understanding how the world works than about calculation efficiency. (I'd probably have a different view if I was trying to get something actually done. :-)

 

[Dale]

Thanks. I guess I care more about understanding how the world works than about calculation efficiency. (I'd probably have a different view if I was trying to get something actually done. :-)

We don't know that the more advanced calculations are "how the world works" either. It is likely that there is an unknown and even more difficult calculation required. In fact, we can never know that our most advanced calculation is the final one, so it is always best to use the easiest calculation to adequately describe the phenomenon.