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.(adsbygoogle = window.adsbygoogle || []).push({});

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

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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.

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

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