Confusion with certain aspects of the Equivalence Principle

Question

I took a course on GR and I am revising it after a while. I am heavily confused about the equivalence principle. Consider the following two statements:

1. On a Riemannian manifold, we can always choose coordinates such that the metric looks like the Minkowski metric ($\eta_{\mu\nu}$) locally.

2. From the principle of equivalence, we know that locally one cannot distinguish between free fall and absence of gravity. Another way to see this is that one can always choose accelerated coordinates locally to mimic the effects of a uniform gravitational field.

Now, it might be obvious to some, but I do not see how even point #1 comes from the equivalence principle. How is me being able to choose locally accelerated coordinates to mimic the effects of a uniform gravitational field have to do with the space-time being locally flat? Is it because we call these accelerated coordinates w.r.t. Minkowski metric ($\eta_{\mu\nu}a^{\mu}a^{\nu}=$ constant)? Or is it that these locally accelerated coordinates are precisely the coordinates that make the metric look like $\eta^{\mu\nu}$ locally?

Any help would be appreciated.

Answer 1

I think your question has a deep connection with Riemann Normal Coordinates.

The basic idea behind Riemann normal coordinates is to use the geodesics through a given point to define the coordinates for nearby points. Let the given point be $O$ and consider some nearby point $P$. If $P$ is close enough to $O$ then there exists a unique geodesic joining $O$ to $P$. Let $a^\mu$ be the components of the unit tangent vector to this geodesic at $O$ and let $s$ be the geodesic arc length measured from $O$ to $P$. Then the Riemann normal coordinates of $P$ are defined to be $x^\mu = sa^\mu$.

One trivial consequence of this definition is that all geodesics through O are of the form $x^\mu(s) = sa^\mu$ and that the $a^\mu$ are constant along each geodesic. By direct substitution into the geodesic equation $$\frac{d^2x^\mu}{ds^2} + \Gamma^\mu_{\alpha \beta}\frac{dx^\alpha}{ds} \frac{dx^\beta}{ds}=0$$

and its derivatives, one obtains, at the origin $O$ $$\Gamma^\mu_{\alpha \beta} = 0$$ $$\partial_\nu \Gamma^\mu_{\alpha \beta} + \partial_\alpha \Gamma^\mu_{\nu \beta} + \partial_\beta \Gamma^\mu_{\alpha \nu} = 0$$

From this results it is easy to see that $$\partial_\nu \Gamma^\mu_{\alpha \beta} = -\frac{1}{3}(R^\mu_{\alpha \beta \nu} + R^\nu_{\beta \alpha \nu})$$

from which it follows that $$\partial_\alpha \partial_\beta g_{\mu \nu} = -\frac{1}{3}(R_{\mu \alpha \nu \beta} + R_{\mu \beta \nu \alpha})$$

and finally $$R_{\mu \nu \alpha \beta} = \partial_\mu \partial_\beta g_{\alpha \nu} - \partial_\nu \partial_\beta g_{\alpha \mu}$$

This result can finally be used to reexpress the Taylor expansion of the metric in terms of the curvature i.e. The Riemann tensor. The Taylor expansion reads as follows $$g_{\mu \nu}(x)= \eta_{\mu \nu} + \partial_\alpha \partial_\beta g_{\mu \nu} \frac{x^\alpha x^\beta}{2} + O(\epsilon^2)$$

Therefore $$\boxed{g_{\mu \nu}(x)=\eta_{\mu \nu} - \frac{1}{3}R_{\mu \alpha \nu \beta} x^\alpha x^\beta}$$

Thus the leading terms of the metric could be expressed as a sum of a constant part plus a curvature part for nearby points. See that this result is not valid for points that are further away, where the Taylor expansion of the metric is not applicable and the unit tangent vector is not constant along the geodesic joining $O$ to $P$.

Physically, the real gravitational field is not the acceleration but the tidal forces that can not be vanished with any coordinate change. Tidal forces are usually measured as a geodesic deviation breaking the locality of two observers far enough apart.

As an example of this coordinates we have the Schwarzschild black hole, where near the event horizon can be shown that the metric of two nearby points has the same form than Rindler coordinates of an accelerated observer in a flat spacetime.

More information in Riemann coordinates.

Answer 2

We first need to separate coordinate systems from the manifold itself. The manifold itself has the same shape regardless of the coordinates you choose. Consider a free-falling particle displaying proper time $\tau$. In a coordinate system, its motion is given by the geodesic equation $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = -\Gamma^\rho_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$ This equation is coordinate-dependent. It is possible to change coordinates such that all $\Gamma^\rho_{\mu\nu} = 0$ at a single point, such that $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = 0$$ and by definition this is the motion of free-falling particles in flat spacetime with $\eta_{\mu\nu}$. If we use some other coordinate system instead, you will find that $\text{d}^2 x^\rho /\text{d}\tau^2 \neq 0$ which means that the free-falling particle is accelerating with respect to the coordinates, which is precisely what an observer using these coordinates perceive as "gravity".

More generally, the acceleration of a particle with respect to a coordinate system is the sum of proper acceleration $a^\rho$ and coordinate-induced (geometric) acceleration: $$\frac{\text{d}^2x^\rho}{\text{d}\tau^2} = a^\rho - \Gamma^\rho_{\mu\nu}\frac{\text{d}x^\mu}{\text{d}\tau}\frac{\text{d}x^\nu}{\text{d}\tau}$$ What this means is that the proper acceleration of an object, which is the acceleration measured by an accelerometer on the object itself, is coordinate-independent (although its components, given by the above equation, do depend on the choice of coordinate system). In GR, gravity is a result of spacetime curvature and not a force, so it is not a source of proper acceleration. Therefore, your term "locally accelerated coordinates" sounds vague to me. An observer either has proper acceleration or doesn't.

Answer 3

Surely 'absence of gravity' (assertion 2) is none other than 'metric is locally Minkowski' (assertion 1). Another way to state it: the locally Minkowski metric gives all connection coefficients zero at a point, and gravitational phenomena arise from non-zero connection coefficients.

Answer 4

How is me being able to choose locally accelerated coordinates to mimic the effects of a uniform gravitational field have to do with the space-time being locally flat?

If you mimic effects of a gravitational field, it means there is really no gravitational field, but it only appears to be so. If there is no gravitational field, the spacetime is flat.

Curvature is coordinate independent concept. If something has curvature, it does so in every coordinate system whatsoever. So if you can mimic local properties of metric in curved spacetime by choosing special coordinates in flat spacetime, this means, the two spacetimes have locally the same curvature, i.e. curved spacetime must locally be flat.