# How does the Equivalence Principle imply that derivatives of the metric vanish in a freely falling frame?

Why do the first derivatives of $$g_{\mu\nu}$$ vanish in a freely falling coordinate system?

I would like to start from the Equivalence Principle that for any point in spacetime there exists a locally inertial frame Cartesian coordinate system. In these coordinates, and in a sufficiently small region around the spacetime point, the laws of physics are consistent with special relativity and there's no gravitational force. This is true for

I believe that these locally inertial coordinates are those that would be used in "freely falling" reference frame. Is that right? I.e. if you do any physics experiment in a little elevator in free fall, the results you get (in the frame of the elevator) are just those predicted by special relativity and E&M, with no gravitation. And the natural Cartesian coordinates you would set up in the elevator would be the locally inertial coordinates mentioned above (maybe modulo a boost and a spatial rotation).

In this picture the metric at a spacetime point $$x_P^\mu$$ is defined by the coordinate transformation between these local inertial coordinates $$\xi_P^\mu$$ and the "lab" coordinates $$x^\mu$$: $$g_{\mu\nu}(x_P) \equiv \eta_{\alpha \beta} \frac{\partial\xi_P^\alpha(x_P)}{\partial x^\mu}\frac{\partial\xi_P^\beta(x_P)}{\partial x^\nu},$$ where $$\eta_{\alpha \beta}=\mathrm{diag}(1,-1,-1,-1)$$.

The way I'm interpreting this, the metric is just an object you can use to calculate the special relativistic invariant interval $$ds^2$$ between two events that you would find in the freely falling frame.

In particular if we are in the free falling elevator (i.e. the coordinates $$x^\mu$$ are the $$\xi_P^\mu$$) then the metric is $$\eta_{\mu\nu}$$ as we expect since special relativity holds in the free fall system. But apparently the derivative of the metric in this system is also zero. Why?

Weinberg's Gravitation and Cosmology book says that non-zero derivatives of the metric would manifest themselves in a local experiment that would find that rates of different clocks would be different (violating special relativity). But he doesn't spell out exactly how this argument works.

• If the derivatives of the metric would not vanish locally in a freely falling frame, the Christofel symbols would not vanish, which you would experience as a net (inertial) force on the frame. – mmeent Feb 18 '19 at 19:46
• In my understanding, the "Christoffel symbols" you would define in order to calculate these net forces would be $\Gamma^\mu_{\alpha\beta} \equiv (\partial^2 \xi^\nu/\partial x^\alpha \partial x^\beta) (\partial x^\mu / \partial \xi^\nu)$, not the usual Christoffel symbols that use the metric derivatives. The equivalence of the two definitions for the Christoffel symbols uses the fact that the metric has no derivatives in a locally inertial frame I believe. – Alex Feb 18 '19 at 20:17
• @Alex Yes, great point. Weinberg actually makes this clear twice (well, abundantly clear once (pp. 101) and subtly clear the other time (pp. 74)). The fact which is true without using the vanishing of the first derivatives is that the difference between the affine connections and the Christoffel symbols is a tensor. That this tensor vanishes can only be proven using the vanishing of the first derivatives. – Dvij D.C. Feb 21 '19 at 21:39
• On p. 101 Weinberg says that the first derivatives of the metric tensor have to vanish in inertial coordinates because "there can be no gravitational red shift between infinitesimally separated points". I think this is the statement I want to understand more rigorously. – Alex Feb 25 '19 at 14:12

## 1 Answer

How does the Equivalence Principle imply that derivatives of the metric vanish in a freely falling frame?

What you're claiming is not true. The metric can have all kinds of forms depending on the coordinate system. For example, if you use polar coordinates, then the components of the metric are varying, and the Christoffel symbols are nonzero.

Also, coordinate systems do not correspond to frames of reference. See How do frames of reference work in general relativity, and are they described by coordinate systems?

In particular if we are in the free falling coordinate system (i.e. the coordinates $$x^\mu$$ are the $$\xi_P^\mu$$) then the metric is $$\eta_{\mu\nu}$$ as we expect since special relativity holds in the free fall system.

Not true. The metric can have various forms in SR depending on your coordinate system, and stating that you want to use a free-falling frame does not imply any particular coordinate system.

I believe that this locally inertial frame is the same as a "freely falling" reference frame. Is that right? I.e. if you do any physics experiment in a little elevator in free fall, the results you get (in the frame of the elevator) are just those predicted by special relativity and E&M, with no gravitation.

Yes, although there's not much point in saying "locally inertial frame," because frames of reference are always local in GR.

We could try to change your statement around so that it became true. If we wanted to do that, the first thing we'd have to decide is whether you want to restrict the discussion to flat spacetime. You never say anything about that in the question.

• Thanks Ben. I edited the question to be more precise. In particular, the local inertial coordinates that I'm talking about should always be Cartesian. And the question is dealing with an arbitrary gravitational field, not necessarily flat spacetime. – Alex Feb 18 '19 at 21:34