If we look at a flat Minkowski space-time (without any gravitation) an choose an accelerated frame of reference, what happens to the metric tensor is it still in Minkowski coordinates or will he be affected by general relativity?

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    $\begingroup$ In accelerated frame the metric will not look like a flat one.It would be curved...So either you can think it as if accelerated frame in flat spacetime or as if there is no acceleration but the spacetime is curved...You can have a look at Rindler spacetime..The content of that is precisely what you are asking... $\endgroup$ – kau Nov 12 '15 at 13:50
  • $\begingroup$ More PSE posts on Rindler coordinates: physics.stackexchange.com/search?q=rindler+coordinates $\endgroup$ – Kyle Kanos Nov 12 '15 at 14:38
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    $\begingroup$ @kau While Minkowski geodesics are not straight lines in Rindler coordinates, Rindler "spacetime" is not curved. The Riemann tensor vanishes. It is therefore not equivalent to, say, the Schwarzschild spacetime, which is curved. $\endgroup$ – AGML Nov 12 '15 at 17:06
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    $\begingroup$ @AGML Sorry .I would try to correct my comment.I am not finding an edit option there. Rindler spacetime can't be truely curved(meaning nonvanishing curvature tensor). Since it's the Minkowski spacetime and just looking at it using Rindler's coordinates. If I transform it back I should get a flat spacetime back. Although it has horizons and pseudo gravity like effects but it's different from actual scene when a massive body is present there. I hope this is correct. $\endgroup$ – kau Nov 13 '15 at 7:41

A reference frame is equivalent to a choice of coordinates. So, choosing an accelerated frame in Minkowski space is equivalent to choosing a specific coordinate system on Minkowski space. Most importantly, this means that there is not genuine curvature in an accelerated frame, i.e. it is fundamentally different than gravity.

The equivalence principle amounts to the statement that the Chistoffel connection coefficients $\Gamma^{\mu}_{\rho\sigma}$ are not tensorial objects; that is, they have no instrinsic geometric meaning. In the same way that we can choose a local Lorentz frame in a general curved spacetime (i.e. we can choose coordinates so that locally the connection coefficients are zero), we can also make flat space "look" like it is curved by choosing coordinates where the connection coefficients do not vanish.

However, we cannot use the equivalence principle to mask the presence of genuine spacetime curvature (i.e. gravity). We have the coordinate freedom to make the metric tensor appear flat, and we can even make the first derivatives of the metric tensor vanish. We cannot, however, guarantee that the second derivatives of the metric tensor vanish, and so tensors which depend on the second derivatives of the metric (principally the Riemann curvature tensor) will be able to detect whether or not spacetime is genuinely curved (whether gravity is present).

The geodesic equation is \begin{equation} \frac{d^2 x^{\mu}}{ds^2}+\Gamma^{\mu}_{\rho\sigma}\frac{dx^{\rho}}{ds} \frac{dx^{\sigma}}{ds}=0. \end{equation} So, if we are in flat spacetime, we can choose a reference frame (coordinate system) in which some of the $\Gamma^{\mu}_{\rho\sigma}$ are non-zero. Then Newton's law \begin{equation} \frac{d^2 x^{\mu}}{ds^2}=0, \end{equation} is modified by the presence of fictitious forces.


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