# How is this approximation of gravity involving Rindler coordinates valid?

I have seen that it is possible to approximate the metric in the presence of a gravitational field by the Rindler metric:

Now as some answers in the links have pointed out, this doesn't quite describe a uniform gravitational, because the acceleration described by the coordinates depends on one of the spatial coordinates.

My question is, how can we refer to this as a gravitational field at all? The Rindler metric is derived from a coordinate transformation on inertial coordinates on Minkowski spacetime, so we know a priori the Riemann curvature tensor will be exactly zero in the Rindler metric.

In doing this "Rindler approximation" to the gravitational field, say, near the surface of the Earth, we started out with a nonzero Riemann curvature tensor (indicating spacetime curvature exists), and then we obtained a situation in which the curvature tensor vanishes everywhere in the region we're approximating. Doesn't this render the approximation invalid?

Even if you argue that the region of approximation is small (which makes sense), there is no sense in which we can make the curvature tensor outright vanish (because the Ricci scalar, which is a contraction of the curvature tensor, is supposed to be invariant).

Accompanying this strange change in the curvature, objects that "fall into the Earth" followed geodesic paths prior to the approximation, and under the new approximation, the same objects are now undergoing proper accelerations, meaning they are no longer following geodesic paths. Is there a physical interpretation or the mathematical reasoning behind this change?

It seems like we are replacing spacetimes outright as opposed to approximating them. Is this understanding correct?

• the Ricci scalar, which is a contraction of the curvature tensor, is supposed to be invariant” - The Ricci tensor and therefore the scalar are both zero in the Schwarzschild spacetime. The only curvature there is Weyl that defines the magnitude of the time dilation and length contraction, which are equal (or reciprocal depending on the view) to each other similarly to Special Relativity. Thus locally and radially the Schwarzschild spacetime is equivalent to Rindler, as it should be according to the Equivalence Principle. Commented May 27 at 6:26

Here’s a very simple thing to keep in mind: Rindler gives you something isometric to a portion of the Minkowski spacetime, so no gravitational fields. Of course, there can be curves which have non-zero acceleration, $$\nabla_{\dot{\gamma}}\dot{\gamma}\neq 0$$, but this does not mean we’re in a curved spacetime. And I would never use the term “gravitational field” to refer to anything in a flat spacetime.
What the Rindler observers are, are a 1-parameter family of curves (which can then be used to define a new local coordinate system) in Minkowski spacetime. Consider a 2-dimensional (for simplicity) Minkowski spacetime $$(\Bbb{M}^2,\eta)$$, where in global coordinates $$(T,X)$$ we have $$\eta=-dT^2+dX^2$$. For each $$\alpha>0$$, one can define a curve $$\gamma_{\alpha}:\Bbb{R}\to\Bbb{M}^2$$ by setting \begin{align} (T\circ\gamma_{\alpha})(\tau)=\frac{1}{\alpha}\sinh(\alpha\tau), \quad\text{and}\quad (X\circ\gamma_{\alpha})(\tau)=\frac{1}{\alpha}\cosh(\alpha\tau). \end{align} It then follows that
• these curves ‘fill up’ the open set $$U=\{p\in\Bbb{M}^2: |T(p)|, in the sense that for each $$p\in U$$, there is a unique $$\alpha>0$$ and $$\tau\in\Bbb{R}$$ such that $$\gamma_{\alpha}(\tau)=p$$.
• $$\eta(\dot{\gamma}_{\alpha},\dot{\gamma}_{\alpha})=-1$$, so each $$\gamma_{\alpha}$$ is parametrized by proper time.
• $$\eta(\nabla_{\dot{\gamma}_{\alpha}}\dot{\gamma}_{\alpha}, \nabla_{\dot{\gamma}_{\alpha}}\dot{\gamma}_{\alpha}) = \alpha^2>0$$, meaning the acceleration of $$\gamma_{\alpha}$$ has constant magnitude $$\alpha$$.
• $$\eta(\dot{\gamma}_{\alpha},\nabla_{\dot{\gamma}_{\alpha}}\dot{\gamma}_{\alpha})=0$$, so the acceleration is $$\eta$$-orthogonal to the velocity.