# Derivation of Proper Time in General Relativity

by paultsui
Tags: derivation, proper, relativity, time
 P: 13 In relativity, proper time along a world-line is be defined by $d\tau^{2} = ds^{2} / c^{2}$ However, proper time can also be understood as the time lapsed by an observer who carries a clock along the world-line. In special relativity, this can easily be proven: The line element in special relativity is given by $ds^{2} = (cdt)^{2} - dx^{2} - dy^{2} - dz^{2}$, therefore in a frame that moves along the world line, we have $dx^{2} = dy^{2} = dz^{2} = 0$, giving us $ds^{2} = (cd\tau)^{2}$ Tn general relativity, things seem to be a little tricker because of the metric element $g_{tt}$. Repeating the derivation ends up with $ds^{2} = g_{tt}(cd\tau)^{2}$ instead. I found a proof here: http://arxiv.org/pdf/gr-qc/0005039v3.pdf However, on p.2, the author states that $g_{t't'}$ can always be chosen as 1, hence completing the proof. This baffles me as I always think that $g_{t't'}$ is defined by the geometry of space-time, which cannot be chosen arbitrarily. Can anyone give me a hint on where my logic go wrong? Thank you!
 P: 260 GR is valid in any type of coordinates, so it shouldn't be surprising that you'd be able to find some that make g00=1.
 Quote by paultsui In relativity, proper time along a world-line is be defined by $d\tau^{2} = ds^{2} / c^{2}$ However, proper time can also be understood as the time lapsed by an observer who carries a clock along the world-line. In special relativity, this can easily be proven: The line element in special relativity is given by $ds^{2} = (cdt)^{2} - dx^{2} - dy^{2} - dz^{2}$, therefore in a frame that moves along the world line, we have $dx^{2} = dy^{2} = dz^{2} = 0$, giving us $ds^{2} = (cd\tau)^{2}$ Tn general relativity, things seem to be a little tricker because of the metric element $g_{tt}$. Repeating the derivation ends up with $ds^{2} = g_{tt}(cd\tau)^{2}$ instead. I found a proof here: http://arxiv.org/pdf/gr-qc/0005039v3.pdf However, on p.2, the author states that $g_{t't'}$ can always be chosen as 1, hence completing the proof. This baffles me as I always think that $g_{t't'}$ is defined by the geometry of space-time, which cannot be chosen arbitrarily. Can anyone give me a hint on where my logic go wrong? Thank you!
 P: 13 Derivation of Proper Time in General Relativity Thank you for replying! It is true that we can always choose a coordinates system such that $g_{00}$ to be 1 at the point of interest, but why do we have to pick $g_{00} = 1$ instead of ,say, $g_{00} = 2?$ Consider the following: Imagine person A is at infinity while person B is at a point near a massive object. Both of them have a clock. For person A, after time $dt$ (measured with his own clock), he would see that the person B to have travelled a certain distance ds in space-time. In addition, he would also see that the clock of B has lapsed a certain amount, which is the proper time. Therefore there must be a unique relation between $d\tau$ and $ds$. However, if we were allowed to choose $g_{00}$ to be 1, we can also pick another coordinates system such that $g_{00} = 2$. But since there is unique relation between $d\tau$ and $ds$, using a coordinates system such that $g_{00} = 2$ must be wrong. There must be something fundamental about HAVING to pick $g_{00} = 1$... but what is this?