# Accelerated reference frame in Minkowski space vs pure gravitational field

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.

• The equivalence principle states that local measurements cannot distinguish acceleration and gravity.
– Danu
Commented Oct 27, 2015 at 9:19
• The $g_{ik}$ are only ever local expressions of the spacetime metric. Commented Oct 27, 2015 at 10:04
• Tidal forces (i.e. non-local effects) can clearly distinguish gravity from uniform acceleration.
– Danu
Commented Oct 27, 2015 at 10:05