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I know about special relativity and Lorentz transformations. However, there is also the general coordinate transformation (GCT) which is allegedly used in general theory of relativity. What is GCT and how is it different from the Lorentz transformation?

I know how do Lorentz transformations look like. I want to see how does a GCT look like. Please include that in the answer. Moreover, Lorentz transformation keeps $ds^2$ unchanged. Does GCT also keep $ds^2$ unchanged?

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  • $\begingroup$ en.wikipedia.org/wiki/Diffeomorphism $\endgroup$
    – user4552
    Mar 18, 2018 at 17:36
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    $\begingroup$ Wikipedia is impenetrable. I'm looking for a simpler version of explanation, if possible. $\endgroup$ Mar 18, 2018 at 17:43

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A Lorentz transformation is a coordinate transformation between inertial reference frames. The transformation is linear in the coordinates and the metric easy. Instead a GCT (general coordinate transformation) is a map between differentiable manifolds, the transformation is not linear and the metric much more complicated. It is the realm of curved geometry, which allows for a mathematical description of GR (general relativity). To operate in that context you need years spent in acquiring the theoretical and technical skills.

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  • $\begingroup$ Dear @MicheleGrosso thanks for the answer. I think I know metric, in a curved spacetime: $ds^2=g_{ab}dx^a dx^b$ where $a,b$ are spacetime indices. I needed one clarification. Can we define GCT as those transformations (map) $x^a\to y^a(x^a)$ which leave $ds^2$ invariant? Any other criterion to be obeyed? I have very little knowledge of GR. By the way, you defined GCT as a map between two differentiable manifolds. These two different manifolds are two different frames? $\endgroup$ Nov 8, 2018 at 13:24
  • $\begingroup$ $ds^2$ is the spacetime interval squared. Being a distance, it is an invariant independently of the coordinates system. Mathematically the invariance is guaranteed as it is the composition of two vectors (the differencials of the coordinates). Differentiable manifolds are different reference frames. $\endgroup$ Nov 9, 2018 at 17:42

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