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In my lecture on General relativity, it is said that, by taking the following metric :

$$g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x),~ {\rm with}\ h_{\mu\nu}\ll1$$

one has the definition below of proper time :

$$\text{d}\tau^2 = \text{d}t^2 \left(g_{00}+2g_{0i}\dfrac{\text{d}x^{i}}{c\,\text{d}t}+g_{ij}\dfrac{\text{d}x^i}{c\,\text{d}t}\dfrac{\text{d}x^{j}}{c\,\text{d}t}\right)$$

I don't know how to demonstrate this definition or equation, if someone could help me this would be nice.

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Working with the signature $(+,-,-,-)$, the general definition of proper time (its differential) can be expressed as: $$ d \tau ^2 = ds^2 /c^2 = \frac{1}{c^2}g_{\mu \nu}dx^\mu dx^\nu $$ Remember that when we have repeated indices in a term they are to be summed (this is the Einstein summation convention) for all coordinates. Thus we can write: $$ d \tau^2 = \frac{1}{c^2}(g_{00}c^2 dt^2 +2g_{0 i}c dt dx^i + g_{ij}dx^i dx^j) $$ where we have explicitly expanded the sum, and which can be directly rearranged to obtain your equation. Note here that when latin (greek) indices are used they denote spatial (all) coordinates, so the last term in this expression is a sum over $x$, $y$ and $z$.

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    $\begingroup$ didn't you forget the factor 2 into $g_{0 i}c dt dx^i$ ? $\endgroup$
    – youpilat13
    May 15 '19 at 21:15
  • $\begingroup$ @faya13, yes, I forgot about it, sorry! It's now edited. $\endgroup$
    – Lith
    May 16 '19 at 15:30

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