I just finished reading Einstein's "Relativity: The special and the General Theory", and if I understand Einstein correctly, he says that an accelerated reference frame in Minkowski space (i.e., empty space with no mass) will have a metric $g_{ik}$ which is a non-constant function of the coordinates in the accelerated reference frame, but there will exist a transformation of coordinates to an inertial reference frame where the $g_{ik}$ transform back to the plus or minus 1s (and zeros) of the flat metric on Minkowski space. The observer in the accelerated reference frame then interprets the non-constant $g_{ik}$ as some sort of gravitational field. However, in the presence of mass, a pure gravitational field is induced which alters the local metric of spacetime $g_{ik}$ in a way that no coordinate transformation will transform the $g_{ik}$ back to the zeros and plus or minus 1s of the Minkowski metric, and this is due to the fact that the $g_{ik}$ are encoding a non-euclidean curved spacetime, so that its metric is never euclidean in any coordinate system.

Now if my interpretation of what Einstein is saying is essentially correct, I don't understand how an accelerated frame in Minkowski space is indistinguishable from that of a gravitational field, since even though the coordinates in the accelerated reference frame will induce non-constant $g_{ik}$, they are still the $g_{ik}$ of a flat Minkowski metric. In the case of a gravitational field induced by mass however, the $g_{ik}$ are of a different nature, as they encode a non-euclidean curved spacetime.

As such, something must be wrong with my understanding of the situation.

  • $\begingroup$ The equivalence principle states that local measurements cannot distinguish acceleration and gravity. $\endgroup$ – Danu Oct 27 '15 at 9:19
  • $\begingroup$ The $g_{ik}$ are only ever local expressions of the spacetime metric. $\endgroup$ – dezign Oct 27 '15 at 10:04
  • $\begingroup$ Tidal forces (i.e. non-local effects) can clearly distinguish gravity from uniform acceleration. $\endgroup$ – Danu Oct 27 '15 at 10:05

There are local transformations that take the curved spacetime to a Minkowski spacetime, but they only work locally not globally.

For example the spacetime geometry of a uniformally accelerating observer is described by the Rindler metric. There exists a coordinate transformation that converts this to the Minkowski metric everywhere. So far so good.

The spacetime geometry for a black hole is described by the Schwarzschild metric. in the immediate vicinity of a freely falling observer this looks flat i.e. there are coordinates that locally make the spacetime look flat. However if you leave the immediate vicinity of the observer spacetime starts looking curved again. Unlike the accelerating observer there is no coordinate transformation that makes spacetime flat everywhere.

You could look at this a different way: if you're hovering a constant distance from the black hole you'll feel an acceleration and your spacetime will locally look just like Rindler spacetime. You can use the same coordinate transformation as in Rindler spacetime to transform the local region around you to flat spacetime. This is why acceleration is locally indistinguishable from gravity. However, as before, this only works locally. Look too far from your position and you'll find the geometry deviates from the Rindler geometry and your coordinate transformation no longer works.

  • $\begingroup$ If the spacetime manifold has non-zero curvature at a point, there are NO transformations which transform the metric in a neighborhood of the point to the flat Minkowski metric, even in an infinitesimal neighborhood. This is why you can't map even the smallest region of a sphere to the plane without distorting distances in some way, because the sphere has non-zero curvature everywhere. $\endgroup$ – dezign Mar 14 '16 at 1:11

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