# Misconception in the implications of the gravitational redshift experiments

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.