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Different masses don't move the same way without any forces? But that's ok?

  1. May 10, 2010 #1
    Since it is the case in classical physics, I had always thought that objects with different mass acted the exact same way if they are affected only by geometry. In fact, I had taken that to be a statement of what it means for an effect to be geometrical.

    However, I recently discovered the COW experiment, where it is demonstrated that the phase difference along 2 paths in a gravitational field depends on the mass of the particle. This is taken by some to indicate a departure from the strong equivalence principle where gravity is seen as equivalent to an "acceleration field" that acts the same on all particles.

    But http://arxiv.org/abs/gr-qc/0206030 shows that acceleration alone produces phase differences that depend on the mass.

    We can even determine the mass of a particle by just changing our reference frame.
    Say we have a beam of non-relativistic particles with average momentum [tex]p = \hbar k[/tex]. If we then move at a (non-relativistic) speed v in the same direction as p, in my new frame, p' is different and given by [tex]p' = \hbar k'[/tex]. The mass of our particle is then given by [tex]m = \frac{\hbar \Delta k}{v}[/tex]
    We can measure the wave length in either frames by sending the particles through a double slit. Of course this doesn't produce a force on our particles because the ones that make it through the double slit are those that we didn't touch. So I see that I can distinguish between different masses even without pushing on them.

    So we should expect that objects of different mass are affected differently by gravity when we consider quantum effects. I can tell their difference when I accelerate!


    That's very cool, but I'm quite confused about this last equation [tex]m = \frac{\hbar \Delta k}{v}[/tex]
    I don't know where it comes from! Ehrenfest's Theorem gives us [tex]\frac{\mtext{d}\langle x \rangle}{\mtext{dt}}=\frac{\langle p \rangle}{m}[/tex], and I guess it should follow from that, but I'm very unused to changing frames in Quantum Mechanics. I'm uncomfortable with the idea of moving at a definite velocity v at all.

    On the other hand, even though I'm uncomfortable with some of the specifics of the derivation (In particular, I worry that the uncertainty principle could spoil this for small masses), it has a strong ring of truth to it. After all, I can in fact measure the difference in wave length as I change frames, so [tex]\frac{\hbar \Delta k}{v}[/tex] is a measurable quantity, and its average value would be the mass if p=mv were approximately true, which is the case in the classical limit.
  2. jcsd
  3. May 10, 2010 #2
    Of course we can't do the same thing in classical physics because [tex]p[/tex] isn't as easily measured. I don't know any classical experiment that will determine the momentum of an object without pushing on it.

    Any thoughts? I was completely shocked by this.

    Has any experiment looked for this?
    Last edited: May 10, 2010
  4. May 10, 2010 #3
    well now as you mentioned it I am also looking for such thing and hope to reply soon with a good argument.
  5. May 12, 2010 #4
    How does this contradict GR in any sense? Even without gravity, quantum-mechanical objects with different masses behave differently. You can do Young's double slit experiment with electrons. But if you change to neutrons with the same velocity, the fringe spacing will change, because electrons and neutrons have different masses, and therefore they have different de-Broglie wavelengths when travelling at the same speed.

    So mass does make a difference in experiments sensitive to the de-Broglie wavelength, even in the absence of gravity! Why should such a difference disappear in an experiment involving gravity? Your reasoning should be reversed - it's only in the case these differences disappear that the equivalence principle is violated.

    The short answer to the problem is as follows. "Different masses behave in the same way under gravity" is a statement that is only valid for test masses governed exclusively by the geodesic equation. Remember that the geodesic equation is the direct generalization of Newton's 1st law in general relativity, and it only applies to perfect point-like objects with no internal dynamics at all. The identical behaviour of objects with different masses is a consequence of the fact that the geodesic equation, just like Newton's 1st law, is explicitly independent of mass. When we go to the quantum theory, the wavefunction (as in Schrodinger's equation) or field (as in quantum field theory) are extended continuum objects with their internal dynamics coupled to gravity, rather than free-falling objects. In particular, the de-Broglie length is explicitly dependent on the mass (unlike the geodesic equation which does not), so the very assumption that different masses behave identically has already broken down, and it makes no sense to consider it as a test of the equivalence principle. Perhaps it's more clear if we just look at the time-dependent Schrodinger's equation which governs the evolution of a wavefunction. The equation involves the Hamiltonian which has explicit mass dependence p^2/2m. Obviously there's no hope that different masses can still be equivalent.

    (An additional explanation which is not directly related to the discussion: Though Newton's 1st law applies to extended objects, the counterpart in GR, the geodesic equation, is only valid for point-like objects, because an extended object can sense the difference in curvature across its size when spacetime is curved, and a geodesic path can only be precisely defined if its exact starting point is known.)
    Last edited: May 12, 2010
  6. May 12, 2010 #5
    IMHO "objects with different masses fall at the same rate" is not a very fundamental statement of the equivalence principle. The fundamental statement is that the laws of physics are the same whether in an inertial frame without gravity or in a free-falling local reference frame.

    Now we shall use the more fundamental version to prove the former statement. Part 1: First consider non-interacting particles A and B placed at the same point at rest in an inertial frame without gravity. They will continue to stay at rest, regardless of their mass, so they will never separate from each other. Part 2: Now let's switch to the free-falling frame of A in the presence of gravity. Since the laws of physics are the same here, again A and B can never separate each other. Therefore A and B must fall at the same velocity even though they have different masses.

    Now suppose we let some mass-dependent physics, such as the de-Brogle wavelength, enter the scene. Now the proof breaks at Part 1, because mass is no longer an irrelevant issue. Therefore we can't advance to Part 2 and complete our proof.
  7. May 12, 2010 #6
    It doesn't contradict GR. Sorry if I made it seem that way. My example doesn't even use GR at all, it just uses a velocity transform (Galilean since I was thinking of non-relativistic QM)

    I was just pointing out the important fact that in QM, inertial motion depends on mass and that there are measurable consequences. This is something that hadn't really sunk in for me before, and it seems to be a common misconception. (and this has nothing to do with internal structure or extended bodies, just mass, whatever that means to you)

    Also, it seems that the Geodesic equation needs some addition qualifications in Quantum Mechanics because it doesn't give a path that a point particle will travel irrespective of mass. Mass affects the trajectory of a point particle even when you neglect the gravitational reaction from the mass. Of course, I'm sure people know how to do this.
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