Read this thread.
As far as I can ascertain from reading this thread, the answer is: “the photon’s frequency must be reduced because otherwise the law of conservation of energy would be violated”. Which is circular reasoning once again, same as on my previous Hawking radiation thread.
I believe Chronos and I have explained how the frequency appears to be reduced in some reference frames. It so happens that the way it happens satisfies the laws of conservation of energy and momentum.
I don’t see how it’s circular. The 1st Law of Thermodynamics is well established by numerous other observations. So it’s perfectly valid to use it as an explanation for why something happens. It would only be “circular” if the photon energy shift were the only proof for conservation of energy.
It is valid to use it to assume that the photon’s frequency changes in that case. But it is not valid to use it to explain why it changes. For that, you have to do more than to just say “well, magic happens in order to maintain the law”.
By causing it to bounce. Even if you want to gloss over the absorption and re-emission, you’re still granting that it bounces, right? Well, it’s in the bounce (however that happens) that the energy loss occurs.
That’s my question. HOW does it happen? What are the mechanics of the “bounce”?
The reflectivity comes from free electrons. An electron bound to a nucleus can absorb and emit light, but that isn’t what’s happening here. Note that a photon and a free electron can interact, and it’s these free electrons that are making the mirror reflective. You can tell that the reflection can’t be from atomic absorption because the mirror produces specular reflection. If atomic absorption/re-emission were the mechanism at work, you’d get diffuse reflection and you wouldn’t see an image (since an atom does not remember an incoming photon’s direction when it later emits a photon).
On a classical level, the EM wave is wiggling the conduction (read: “free”) electrons, and those wiggling electrons are emitting radiation due to their acceleration. On a quantum level, it’s electron/photon scattering.
When the photons excite those free electrons and get re-emitted, are there quantum levels of excitation / re-emission like there are in an atom?
They’re free electrons, so effectively, there are no discrete states (discrete energy levels only arise in bound systems).
Ok, so - when the photon hits a free electron and is absorbed, the electron gets excited then emits a photon - what determines the energy of the emitted photon?
Quantum electrodynamics. But that’s not what you’re really asking.
Let’s take a simpler example: billard balls. One billard ball impacts another, then they separate again, having undergone a certain change in momentum and energy each. In order to calculate what happens, one typically sets up conservation of momentum and energy for the problem, then solves the resulting equations. You’re saying that this is circular – that in the end, the two balls go each their own merry way with a total energy that equals the total energy before the impact is only true because we put it in. And that’s true. But there’s a reason why we put it in – that reason is that if we were to have two billard balls collide the same way tomorrow, they would undergo the same process, whatever that may be. So we want the physics of the collision to be independent from the date – we want, to use the right buzzword, time-translation invariance. This is a symmetry of the laws of physics; due to a famous result known as Noether’s theorem, a symmetry leads to a conserved quantity. Time-translation symmetry leads, in particular, to energy conservation.
So that’s the take-away message here: if you want the laws of physics to be independent of time, to work the same way today as they do tomorrow, you get energy conservation out of it without even having to ask. The problem is, while this is a clear result mathematically, it’s somewhat hard to make intuitive.
Ok but bear with me for a second: in one of the reference systems (specifically, the one with stationary mirror pre-collision) the photon that is reflected has less energy than the photon had before the reflection. Thus, if you go into the details of how reflection happened, the photon would get absorbed by a free electron, excite that electron by the amount of energy that the photon contains, but THEN the electron would emit a photon that is slightly less energetic than the absorbed one. Is that correct? That interaction is what is bothering me.
I understand that there is a law of conservation of energy. But what is unsettling is how that law is used like magic in explanations of physical processes. The free electron in the paragraph above magically knows to emit an electron that is slightly less energetic - in order to maintain that law. In the Hawking radiation/quantum foam explanation, the particle being torn away from the pair magically knows it has to have negative energy - in order to maintain that law.
In that reference frame, yes. But from the electron’s point of view, it’s the same energy. The difference arises because the electron changes reference frames (i.e., it starts moving when it wasn’t before).
Are you saying that the sequence of events is as follows:
- The photon is absorbed by the free electron.
- The free electron starts moving because it absorbed the momentum from the photon
- The free electron emits a photon in the opposite direction - because the electron is now moving, from the stationary reference frame the new photon is redshifted
- Now the free electron is moving even faster
I can kinda see how this works now. One more question: how does the electron “know” to emit the photon in the right direction?
This is easiest to answer if we treat light as a wave, rather than as a particle. The boundary conditions for the electric and magnetic fields only match up correctly for a wave reflected in the correct direction or directions.
This sequence is not correct. The photon is never absorbed by the free electron in the way you are thinking. It is a scattering (collision) process. There is no intermediate state like there is with an atomic transition.
If your incoming particle was an electron rather than a photon (i.e., if you had an electron colliding with an electron), the billiard-ball imagery might sit better. The photon-hits-electron case is not any different from the electron-hits-electron case in terms of the kinematics. In both cases, two particles come together and interact – transferring some momentum in the process – and two particles then leave.
In contrast, when a photon excites an atom, the photon is kaput. Initially, you have an atom and a photon. Then, you have an excited atom. Then, that excited atom decays into a de-excited atom and a photon.
That’s a bit like saying, if you have a bucket full of water, and you split the water up between two smaller buckets, how does the second bucket magically know how far to fill in order to respect ‘conservation of water’? Because that’s the amount of water left! We are entailed to use conservation of water in order to explain the filling height of the second bucket, just as much as we are entailed to use conservation of energy and momentum in order to determine scattering kinematics.
Related question: Are there any experimental results with momentum transfer to a reflective surface - maybe a light sail?
IKAROS is the big flashy one, but a mid-range laser has enough momentum to impart to demonstrate this on lab scales, too.