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Consider a flat, thin, perfectly reflective square mirror of mass $m$ lying on edge on a frictionless, horizontal surface that is met by an electromagnetic wave with Poynting vector parallel to the normal to the square. Suppose the EM wave recoils from the square with the same outgoing momentum (and thus energy). That is, suppose the incoming wave has momentum $p$ and recoils with momentum $-p$. To conserve momentum, the square recoils with momentum $2p$, and thus has kinetic energy $\frac{2p^2}{m}$. Where did the kinetic energy of the square come from?

Thoughts: I suppose the square must lose heat or electron motion must slow or something of the sort. Obviously, something like this is purely hypothetical; is it even possible from a thermodynamic standpoint?

Dale
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Arjun Puri
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The problem is related with the concept of different coordinate systems: If you argue that the photon has Energy $E_0 = \hbar \omega_0$ and momentum $\vec p_0 = E_0/c \; \vec e_z$ before the collision and $E_1 = E_0$ and momentum $\vec p_1 = -\vec p_0$ after the collision, you are observing the process in the rest frame of the mirror. In this coordinate frame mirror stays at rest. Hence, after the collision the kinetic energy remains zero. However, if you are in the lab frame, where the mirror rests before the collision and moves after the collision, the momentum and energy of the photon must have changes.

In order to give you an idea about the line of argument, let's take a classical description and consider the absorption and the emission as two separated processes:

  • Before the absorption (=collision) the situation is described above.
  • After the absorption, but prior to the emission the mirror acquired the momentum $\vec p_0$. Thus, it is already moving to the right. If it now emits a wave with wavelength $\lambda_0 = 2\pi c/\omega_0$ in its rest frame, the observer in the lab frame will detect light with a larger wavelength $\lambda_1^{(lab)} > \lambda_0$ (=smaller energy, and therefore smaller momentum) due to the Doppler shift. Hence, the observer in the lab frame to see light which has a smaller energy, $E_1^{(lab)}<E_0$.

I don't know where to find additional material, which describes the light as a classical wave and discusses your problem. However, the Compton effect describes the scattering between a single photon and a single electron. It is well understood and the set-up from your question can be easily transferred: You simply replace the mirror by the electron. You will easily find material on wiki about the Compton effect.

NotMe
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Here is a discussion of elastic scattering and how energy and momentum conservation appear:

An elastic collision is defined as one in which both conservation of momentum and conservation of kinetic energy are observed. This implies that there is no dissipative force acting during the collision and that all of the kinetic energy of the objects before the collision is still in the form of kinetic energy afterward.

For macroscopic objects which come into contact in a collision, there is always some dissipation and they are never perfectly elastic. Collisions between hard steel balls as in the swinging balls apparatus are nearly elastic.

Momentum is always conserved, energy is a conserved quantity in total, as a scalar the addition of all types of energy make the total energy.

Suppose the EM wave recoils from the square with the same outgoing momentum (and thus energy).

The system you propose is un-physical because it violates total energy conservation. The -2p given to the square has to be in the total energy balance also, so the mass of the square is needed in calculations, and the energy is taken from the light beam. At the level of photons, the frequency of the light should become smaller .

See this discussion for further points.

Community
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anna v
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That is, suppose the incoming wave has momentum and recoils with momentum −. To conserve momentum, the square recoils with momentum 2, and thus has kinetic energy 2^2/. Where did the kinetic energy of the square come from?

Since the energy of the light is proportional to its momentum the light has the same energy coming and going. By the conservation of energy this implies that the KE of the mirror is $KE=\frac{2p^2}{m}=0$ which in turn implies $m=\infty$

If you wish to specify some $m<\infty$ then you cannot demand that the outgoing light have momentum $-p$. Instead, the chosen $m$ will determine the outgoing momentum based on conservation of energy and momentum.

Dale
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