In both Sean Carroll's book and Misner,Wheeler,Thorne, the authors take the gravitational redshift experiments to show that spacetime has a curved geometry. There is a paper by Harvey R Brown(published in American Journal of Physics) that challenges this viewpoint by pointing out how this is a misconception. A single redshift experiment does not indicate curvature. Rather, different such experiments in different places is the main reason. (https://arxiv.org/abs/1512.09253) It presents 3 misconceptions, and the one I am asking about is the second one.

What I understood from the presentations of Carroll and Misner, Wheeler,Thorne book is as follows.

  1. Take two non inertial accelerated observers, they observe a difference in the emission and absorption interval ∆t0 and ∆t1. ∆t0 is the time interval two successive crests of the emitted light wave, ∆t1 is that for the absorption case.

  2. The same experiment is done in the tower case, and we get the same result. This time we used a frame where the flat metric(in the coordinate system we use) must be diag(-1,1,1,1) had SR been strictly true, and we would then have ∆t0=∆t1. But in reality, we have ∆t0<∆t1. And we conclude that diag(-1,1,1,1) is not the true metric. Rather the true metric is like that of the accelerated case. But since this is the inertial frame and we use a Cartesian coordinate, here the metric being like that means the geometry is curved.

This is where Dr Brown has his objection. He derives the relationship between ∆t0 and ∆t1 by using the accelerated frame metric and shows that the redshift relation is because of the fact that in the accelerated coordinate system, the minkowski metric does NOT have the simple diag(-1,1,1,1) form. And this is the main reason behind the relationship between ∆t0 and ∆t1. The same relationship between ∆t0 and ∆t1 holds in the gravitational case due to the Weak Equivalence Principle. And this implies that the metric in the gravitational case is also not diag(-1,1,1,1). Rather the metric is that of an accelerated frame in Minkowski spacetime. But this metric is a flat spacetime metric and the components of the Riemann tensor still vanishes even here. So no curvature is involved.

Question: Is my understanding of the argument correct? And if it is correct, it makes me wonder why from Misner to Carroll so many prominent experts on GR simply missed this issue.


MTW certainly didn't miss this issue. Carroll may have. The paper you linked quotes Carroll's textbook as saying of the Pound-Rebka experiment:

simple geometry seems to imply that the [emission and reception intervals] must be the same. But of course they are not; the gravitational redshift implies that the elevated experimenters observe fewer wavelengths per second [...] What went wrong? Simple geometry – the spacetime through which the photons traveled was curved.

That does seem to be an error.

GR is widely misunderstood even by professional physicists, not because it's especially difficult but because it's almost useless, so few people are motivated to study it in any depth. But there always have been people who do understand it in depth. Nothing in Brown and Read's paper is new, and they aren't claiming that it is.

I don't think that it's necessary to do the Pound-Rebka experiment at multiple locations on the earth. It's enough to argue by symmetry that the result would be the same (or at least similar) elsewhere. It's hard to analyze, though, because I don't know of any rival to GR that predicts a different result for this experiment. Nordström's theory predicted no bending of starlight by the sun, and I think no precession of Mercury, but even it had a gravitational redshift.

| cite | improve this answer | |
  • $\begingroup$ It's almost useless. Ouch .Is it so? I had the impression given by books that GR is one of the most beautiful theories with applications from black hole physics to GPS. $\endgroup$ – Nesar Aug 26 at 17:47
  • 1
    $\begingroup$ Perhaps it would be better to write that it was almost useless (few practical applications) , but that is no longer true as @Nesar pointed out $\endgroup$ – magma Aug 27 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.