Confusion regarding the equivalence principle

Question

In section V.2 of Prof. A. Zee's book Einstein Gravity in a Nutshell, it is given that to get the action of a point particle in a gravitational field from that of the action in SR, one just replaces $\eta_{\mu\nu}$ with $g_{\mu\nu}$. In other words,

$$ S = -m\int \sqrt{-\eta_{\mu\nu} dx^\mu dx^\nu} \Rightarrow -m\int \sqrt{-g_{\mu\nu}(x) dx^\mu dx^\nu} $$

This follows from the equivalence principle which states that one can mimic the effects of a uniform gravitational field by transforming to locally accelerated coordinate frames. But if only uniform gravitational fields can be mimicked in this manner, how does the action hold for the whole curved space-time? Shouldn't the above replacement be true only locally? How can we talk about global trajectories like geodesics when this only works locally? I think I am missing a fairly elementary point here.

Any help would be appreciated.

Answer 1

There are two main points

1. The trajectory on which particle moves extremizes the action. This means, that action is invariant under infinitesimal perturbations of the trajectory and not necessarily under finite perturbations.

2. the integral $\int\sqrt{-g_{\mu\nu}dx^\mu dx^nu}$ along the curve is really the proper time of the curve $\int_{\gamma(t_{init})}^{\gamma(t_{fin})} \sqrt{-g(\dot\gamma(t),\dot\gamma(t))}dt$. This is independent of the coordinates. If you then separate the integral on $N$ infinitesimal intervals:
   $$\sum_{i=0}^N\int_{\gamma(t_i)}^{\gamma(t_{i+1})} \sqrt{-g(\dot\gamma(t),\dot\gamma(t))}dt,$$ where $t_0=t_{init}$ and $t_N=t_{fin}$ you can actually use different locally inertial coordinates on each interval and this must give the same result as using any other coordinate system. But we know each interval gives the same result up to $o((t_{i+1}-t_i)^2)$ as corresponding STR formula, so after summing all intervals the result is still $o(\Delta t)$, i.e. the results are the same.

So the locality is actually there, hidden in variational principle.

Answer 2

I think the argument given by Umaxo is very nice, and it contributes a very helpful piece of understanding. However, it is not a proof that the replacement of $\eta_{\mu\nu}$ by $g_{\mu\nu}$ is the only possible thing to do when getting a theory that respects the equivalence principle and reduces to special relativity in flat spacetime. One could add to the formula for the action further terms involving the Riemann curvature tensor (or its contractions) and such an action would still reduce to special relativity in the case of zero curvature. I think the most one can claim, starting out from special relativity, is that the replacement of $\eta_{\mu\nu}$ by $g_{\mu\nu}$ here is the simplest way to get a scalar quantity which could possibly serve as Lagrangian. One then checks by analysis that it does indeed make sensible predictions and hence one is encouraged to propose it as a conjecture, or, if you prefer, as part of a physical theory. Then one must test it by experiments.

Having said that, an interesting aspect of general relativity is that one does not need to propose the action of the 'field' (i.e. spacetime and its curvature) on particles as an independent statement from the field equation (here the Einstein equation). This is because one can claim that a test particle ought to move in the same way as a local change or 'bump' in the field, and the field equation tells that local 'bump' how to move. This is explored in the MTW textbook, among others.