0
$\begingroup$

When a photon is reflected by a mirror, it transfers momentum to the mirror, causing the mirror to gain kinetic energy. However, the photon's energy does not change. This seems to imply that the mirror must lose its rest mass energy.

I understand this scenario through the principle of four-momentum conservation.

But how can this be possible? My guess is that the mirror might lose heat energy, as an increase in heat energy could increase the mirror's mass.

$\endgroup$
8
  • 3
    $\begingroup$ The photon energy does change. It is red shifted by the reflection. $\endgroup$ Commented Dec 5 at 12:52
  • $\begingroup$ @JohnRennie However, in the exercises of Schutz's relativity book, redshift is not considered. Is this simply a situation created by ignoring certain conditions for the purpose of the problem, and is it something that wouldn't actually occur in reality? $\endgroup$
    – Kim
    Commented Dec 5 at 13:03
  • 2
    $\begingroup$ Because to a good approximation the mirror is infinitely massive and does not move when the photon reflects off it. $\endgroup$ Commented Dec 5 at 13:17
  • $\begingroup$ @JohnRennie That would fundamentally ignore both my guess and the conditions of the problem. Both the problem and my guess assume that the mirror's momentum increase is a basic condition. $\endgroup$
    – Kim
    Commented Dec 5 at 13:27
  • $\begingroup$ Which problem in Schutz? $\endgroup$ Commented Dec 5 at 16:14

1 Answer 1

0
$\begingroup$

The phenomenon of reflection is closely related to the phenomenon of absorption, since the reflected intensity depends on the imaginary part of the complex index of refraction, as Feynman explains in The Feynman lectures vol. II chapter 33 and it in turns is related to the absorption coefficient chapter 32. We could then consider that in order of light to be reflected, it should be absorbed and emitted by an atom at the surface of the mirror.

During the absorption and emission, the atom suffers recoil due to conservation of momentum. This recoil manifests as a Doppler shift of light frequency, and it is more noticeable when we are talking about high frequency emission\absorption, like in nuclear phenomena. This is well explained by Mössbauer in his Nobel lecture and this is what turns the Mossbauer effect interesting (the absence of such recoil).

If we neglect the Doppler shift, as you said, we are basically ignoring such recoil. It means that we are considering the photon's momentum negligible compared to atom's momentum, or because the later is too heavy or the first has very low frequency.

In any case, one should not expect to have momentum conservation neglecting the atom recoil. But if you are talking about the Mössbauer effect, the explanation is given by Mössbauer in his lecture

In such a recoilless emission process, the entire excitation energy is transferred to the emitted quantum, and the same holds for the recoilless absorption. Here, the term "recoilless" relates only to the recoil energy transferred in a nuclear transition, and not to the transferred momentum. The value of this transferred momentum is determined by the energy of the gamma quantum and is essentially constant, independent of any change in the internal state of motion of the crystal. This momentum is, therefore, transferred to the lattice in all emission or absorption processes, even in the recoilless processes. It is always absorbed by the crystal as a whole, resulting in a translational velocity that is negligibly small.

So in this case, the mirror velocity changes, but it is negligible.

$\endgroup$
4
  • $\begingroup$ In this problem, the frame of reference is not attached to the mirror. Since the analysis is conducted from a third-party reference frame, the Doppler effect does not occur. $\endgroup$
    – Kim
    Commented Dec 5 at 15:17
  • $\begingroup$ That is impossible. Consider for simplicity that the mirror is at rest. If the observer is in the same frame of the mirror, it observes the same frequency of light before absorption, but the mirror adquire velocity after absorption, so there will be doppler shift. If the observer is moving in relation to the atom initially, it will observe a difference frequency that absorbed by the atom, and again there will be Doppler shift. $\endgroup$
    – Ruffolo
    Commented Dec 5 at 16:41
  • $\begingroup$ Reading the fundamental works of science is the basis for independent and creative thinking. Ruffolo how do you think about Is it possible to make a light follow a non-geodesic path? $\endgroup$ Commented Dec 6 at 21:00
  • $\begingroup$ I don't know if I understand this question, but for me, the geodesic path only makes sense if we are in the geometric limit of optics. Diffraction could be understood as a deviation from geodesic propagation, since it is a wave feature of light. $\endgroup$
    – Ruffolo
    Commented Dec 7 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.