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Momentum of light sans Maxwell

  1. Sep 3, 2013 #1
    I have an easy question but nonetheless I have not been able to find the answer: within the context of special relativity, in particular with light clocks, where the two opposite mirrors of a hypothetical light clock are traveling along very fast, they nonetheless do not leave the emitted photon behind, where it would miss the opposite mirror completely because the mirror has sped away, but the light instead, due to it's momentum, "vectors" forward to just the right position to reflect off the opposite mirror and then in turn angle back to the first mirror. How is this momentum of light demonstrated experimentally?
  2. jcsd
  3. Sep 3, 2013 #2


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    Staff: Mentor

    Well, there's the way that light pressure makes the tail of a comet point away from the Sun (although solar wind also contributes to that effect). There's the photoelectric effect, in which light striking a metal surface will impart momentum to electrons in the metal, knocking them free. You'll find plenty of other experiments if you google around.

    However, you don't need this momentum to explain the behavior of the light in the light clock. If you start with Maxwell's equations and solve for the behavior of the light waves, you'll get the same path between the mirrors whether they're moving or not; this works because Maxwell's equations transform the E and B fields properly to allow for the motion.
  4. Sep 3, 2013 #3


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    In the rest frame of the mirrors, you aim the photon to hit the second mirror, and it does. Intuitively then, it must hit the mirror in all frames. The fact that it hits the mirror cannot possibly depend on which frame you use to describe it!

    Mathematically, it's called aberration, aka the headlight effect. The path of a photon in a moving frame becomes angled forward.

    Let's say in the mirror's rest frame, the photon is moving at an angle θ wrt the x axis. The wave vector of the photon is then

    k = (kx, ky, kz, kt) = (ω cos θ, ω sin θ, 0, ω).

    The Lorentz transformation is

    kx' = γ(kx + (v/c) kt) = γω(cos θ + (v/c))
    ky' = ky = ω sin θ
    kz' = kz = 0
    kt' = γ(kt + (v/c) kx) = γω(1 + (v/c) cos θ)

    from which we can read off the relationship between ω and ω' (the Doppler effect), and the relationship between θ and θ' (the aberration)

    kx' = ω' cos θ' = γω(cos θ + (v/c))
    ky' = ω' sin θ' = ω sin θ
    kt' = ω' = γω(1 + (v/c) cos θ)

    In our particular case, in the original frame the light beam is perpendicular to the relative motion, so cos θ is zero, but cos θ' is not.
  5. Sep 3, 2013 #4
    Thanks Nugatory and Bill_K, it's back to the books for me. I guess I have been too pre-occupied looking for an experiment that basically duplicated a light clock. I see now there are other angles from which I might consider the answer to my question.
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