In space, a photon with momentum $P$ is reflected off a previously at rest mirror, imparting momentum to the mirror. Even if the wavelength of the reflected photon is much longer than the that of the incident, the reflected photon still has some momentum vector pointing the opposite direction, meaning that the momentum of the mirror must be greater than $P$ in order to follow the conservation of momentum of the entire system. But this would all mean that the sum of the magnitudes of the momentum vectors has increased, so somehow the energy of the system has increased? Where did I go wrong?
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$\begingroup$ Let me know if I should clarify $\endgroup$– Joey PelukaCommented Aug 5, 2021 at 8:43
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2 Answers
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The wavelength of the reflected photon will be slightly larger, so that its energy and momentum have decreased. So not all of the momentum vector magnitudes increase. The photon loses energy to provide it to the mirror. This effect is tiny, which is why usually we don't notice mirrors moving or wavelengths changing after reflecting off mirrors.
Related:
Energy conservation in reflection of light from a perfect mirror
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$\begingroup$ Sorry I meant to say longer instead of shorter. But this helps anyway thx $\endgroup$ Commented Aug 5, 2021 at 9:16
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Let's look at this one by one:
I'm assuming here that the mirror does not have infinite mass, and that it is perfectly reflecting (only absorbs energy by the momentum change of the photon, doesn't absorb any energy from the photon directly).
Now,
1. Momentum Conservation: We know that the momentum of a photon is p=h/λ. When the photon is reflected, it changes direction. If the mirror was large enough, we could assume that the photon didn't give any energy to the mirror, but this is not the case here.
So, some of the photon's energy (and hence, momentum) goes to the mirror.
Let P(1) and P(2) be the initial and final momenta of the photon, P(m) be the final momentum of the mirror, λ(1) and λ(2) be the initial and final wavelength of the photon.
By momentum conservation: P(1)=P(m)+P(2)
Taking direction of P(1) as positive, h/λ(1)=P(m) - h/λ(2) (Since some momentum is lost, P(1)>P(2), and so λ(1)<λ(2). So yes, the momentum of mirror is indeed greater than the initial momentum of photon.
2. Energy Conservation: We know that the energy of a photon is hc/λ. Since we've established that the mirror absorbs some energy (only due to the momentum change of photon), we can say that the photon loses energy, and is redshifted. Again, E(1) and E(2) are the initial and final energies of photon, and E(m) is the final energy of mirror.
By energy conservation: E(1)=E(m)+E(2), and so hc/λ(1)=hcλ(2)+E(m).
So, the mirror does absorb energy(and momentum), but no conservation laws are violated.
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$\begingroup$ So, the momentum of the mirror is greater than than the initial photon’s, but it’s energy is less than the initial photon’s? $\endgroup$ Commented Aug 5, 2021 at 9:49
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$\begingroup$ Yeah. It's seeming a bit counterintuitive to you because you may have developed a false intuition of momentum. Let me present it like this if I throw a ball at a stationary box, and (for the sake of simplicity), the ball rebounds off the box with the same velocity, the momentum the ball will impart on the box will be twice the initial momentum of the ball. While, the energy will obviously not be more than the ball. The reason for this is energy can't be negative, while momentum can be. So, a body can impart more momentum on an object than it has, if the body reverses direction. $\endgroup$– VVidyanCommented Aug 5, 2021 at 11:12
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$\begingroup$ Ok, sure, but now the ball is moving at the same speed, but the box is now also moving. So the total kinetic energy has increased somehow. $\endgroup$ Commented Aug 5, 2021 at 11:31
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1$\begingroup$ No, that's why I added for the sake of simplicity. In reality, the final speed of the ball will be less than the initial speed, based on the masses of the ball and the box. So, net kinetic energy will be constant. However, since the direction is reversed, the momentum of the box will be greater than the initial momentum of the ball. $\endgroup$– VVidyanCommented Aug 5, 2021 at 21:18