Let’s say there is a perfect mirror at rest. A photon is directed at it and bounces off.
Since the photon bounced and did not get absorbed, its energy/frequency remains the same as it was before the bounce. Because of conservation of momentum, though, if the photon is bounced off perpendicularly, the mirror gains double the initial momentum of the photon. The mirror’s momentum, I presume, is the classical m*v, so its velocity increases (I really don’t see how its mass would), thus its energy increases. But if the photon’s energy is the same and the mirror’s energy increased, that would violate the conservation of energy law.
How? Let’s say that the substance off which the photon is bouncing doesn’t have the quantum level that would absorb the photon and re-emit it. That’s what I meant by “perfect mirror”.
You are trying to apply a “Classical” Physics solution to an “Modern” Physics problem. In the Classical Physics case, the wall is infinitely massive and rigid, so it doesn’t experience any change in velocity, hence ∆mv = 0;
Photons do have momentum, and transfer it just like anything else can transfer momentum. And, in transferring momentum, they (like anything else) change their own momentum, also.
The OP’s problem might be a bit simpler to understand in the zero-momentum frame. Here, instead of the mirror starting off at rest, we must assume the mirror is moving slightly towards the photon, such that the momentum of the mirror cancels out the momentum of the photon, and the total momentum of the whole system is zero. Then, the collision happens, and the photon bounces off. After the collision, the photon is moving away from the mirror with the same energy as before, but momentum in the opposite direction, and the mirror is moving away from the photon, at the same speed as before but in the opposite direction.
True, in such a system, the problem with conservation of energy doesn’t arise. But in a system where the mirror is at rest to begin with, the problem does arise - basically for the reason that the photon cannot move at anything lower than c.
So in the mirror-at-rest system, how does the energy get conserved?
Are you saying that the act of reflection always lowers the light’s frequency? If so, can you explain the quantum mechanism by which this reduction in frequency happens?
To add on another question - does that mean that our reflection in a mirror is red-shifted slightly, to such a small degree that we don’t notice the change?
I knew that there’s no such thing as a perfect mirror (except for total internal reflection?) but I thought that just affected the overall intensity of the reflected light, not the wavelength.
For a metallic mirror, the electrons reflecting the photon are in the conduction band. I don’t think “absorbs and re-emits” a photon is a good description. That would be more appropriate for scattering from an atom.
For a real mirror, I would expect that the photon frequencies could actually be spread a little bit, both up and down, due to the non-zero temperature of the mirror. If the mirror were cooled to absolute zero, a photon would be red-shifted slightly, but I don’t know if it would be measurable.
In physics, everything is an approximation at some level, if you dig down far enough.
Yes, and if you take that solution and transform into the reference frame in which the mirror is initially stationary, it becomes clear that the frequency of the photon is different before and after the collision.
No. A mirror that isn’t free to move wouldn’t lower the frequency of the reflected light.
I think the issue you’ve pointed out can be resolved by considering the frames of reference carefully. The photon of frequency f is absorbed by the stationary mirror, which begins moving to the left (say). The photon emitted by the mirror has the frequency f in a reference frame moving with the mirror, but in the original reference frame of the stationary mirror, the frequency of the reflected photon is f’<f.
It’s the same system, just looked at in a different reference frame. In the reference frame where the mirror is initially stationary, the incoming photon is slightly blueshifted relative to the zero-momentum frame, and the outgoing photon is slightly redshifted. In other words, it loses energy.
Reflection need not always result in the photon losing energy. If you start out with the mirror approaching the photon at high speed (such that it has more momentum than the photon does), then the reflected photon will have higher energy than the incoming one.
As other responses have indicated (or at least implied) - it’s impossible for a photon to change direction and have its energy conserved in all reference frames. It may happen to have the same energy before/after the reflection in a particular reference frame, but looking at the same event from any other reference frame, the energy changes.
And all of this works just as well with bouncing a rubber ball off a wall, too. Even if the bounce is perfectly elastic (it won’t be, of course, but this can sometimes be a good approximation), if the wall is free to move, the ball will bounce off with a little less energy.