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 2 p 2 m . Where did the kinetic energy of the square come from?

Question

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             2              p        2              m  . Where did the kinetic energy of the square come from?

acorazarxf · Accepted Answer

The problem is related with the concept of different coordinate systems: If you argue that the photon has Energy       E    0    =  ℏ      ω    0   and momentum                     p        →              0    =      E    0        /    c                      e        →              z   before the collision and       E    1    =      E    0   and momentum                     p        →              1    =  −                    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&#039;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                     p        →              0  . Thus, it is already moving to the right. If it now emits a wave with wavelength       λ    0    =  2  π  c      /        ω    0   in its rest frame, the observer in the lab frame will detect light with a larger wavelength       λ    1          (      l      a      b      )        &amp;gt;      λ    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          (      l      a      b      )        &amp;lt;      E    0

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