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Dale
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Is proper time equal to the Invariant Interval or the time elapsed in the Rest Frame

In SR consider two time like separated events - In some frame \begin{equation}ds^2= dt^2 - dx^2\end{equation}

In a frame where the events occur at the same place ( rest frame; $dx' =0$) then according to what I know proper time is the time elapsed in that frame i.e. $d\tau=dt'$.

Hence \begin{equation}ds'^2 = d\tau^2 =dt'^2\end{equation} ( since $d\tau =dt'$ in that frame)and since the interval is invariant; \begin{equation} d\tau^2=dt'^2= dt^2 - dx^2.\end{equation}

Consider the same event in GR, in some frame ( coordinate system) \begin{equation}ds^2= g_{00}dt^2 - g_{11}dx^2....\end{equation} In the frame ( coordinate system) where the events occur at the same place ( $dx'^2=0$) then according to what I know proper time is the time elapsed in that frame i.e. $d\tau=dt'$, and we should have, \begin{equation} ds'^2= g_{00}dt'^2 = g_{00}d\tau^2\end{equation} ( by the same analogy as in SR, $d\tau =dt'$) and since the interval is invariant, we should have, \begin{equation} g_{00}d\tau^2 = g_{00}dt^2 - g_{11}dx^2....\end{equation}.

But from time dilation formula in GR, I know this wrong.

Precisely, according to what I have unsertood, proper time is the time elapsed in the rest frame of the particle, just like in SR, so for the interval\begin{equation} ds^2= g_{00}dt^2 - g_{11}dx^2....\end{equation}, if two events happen at the same place \begin{equation}ds^2 = g_{00}dt'=g_{00}d\tau^2\end{equation} ( by definition just as in the SR case).

Why is this wrong. Is my reasoning that proper time is the time measured in the rest frame wrong

Or is that the time coordinate in a general metric not represent time measured by any clock.

Shashaank
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