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Derivation of Proper Time in General Relativity 
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#1
Apr312, 05:45 AM

P: 13

In relativity, proper time along a worldline is be defined by [itex]d\tau^{2} = ds^{2} / c^{2}[/itex]
However, proper time can also be understood as the time lapsed by an observer who carries a clock along the worldline. In special relativity, this can easily be proven: The line element in special relativity is given by [itex]ds^{2} = (cdt)^{2}  dx^{2}  dy^{2}  dz^{2}[/itex], therefore in a frame that moves along the world line, we have [itex]dx^{2} = dy^{2} = dz^{2} = 0[/itex], giving us [itex]ds^{2} = (cd\tau)^{2}[/itex] Tn general relativity, things seem to be a little tricker because of the metric element [itex]g_{tt}[/itex]. Repeating the derivation ends up with [itex]ds^{2} = g_{tt}(cd\tau)^{2}[/itex] instead. I found a proof here: http://arxiv.org/pdf/grqc/0005039v3.pdf However, on p.2, the author states that [itex]g_{t't'}[/itex] can always be chosen as 1, hence completing the proof. This baffles me as I always think that [itex]g_{t't'}[/itex] is defined by the geometry of spacetime, which cannot be chosen arbitrarily. Can anyone give me a hint on where my logic go wrong? Thank you! 


#2
Apr312, 09:11 AM

P: 261

GR is valid in any type of coordinates, so it shouldn't be surprising that you'd be able to find some that make g_{00}=1.



#3
Apr312, 09:11 AM

Mentor
P: 5,376

I hope this helps. Chet 


#4
Apr312, 11:48 AM

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 [itex]g_{00}[/itex] to be 1 at the point of interest, but why do we have to pick [itex]g_{00} = 1[/itex] instead of ,say, [itex]g_{00} = 2?[/itex] 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 [itex]dt[/itex] (measured with his own clock), he would see that the person B to have travelled a certain distance ds in spacetime. 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 [itex]d\tau[/itex] and [itex]ds[/itex]. However, if we were allowed to choose [itex]g_{00}[/itex] to be 1, we can also pick another coordinates system such that [itex]g_{00} = 2[/itex]. But since there is unique relation between [itex]d\tau[/itex] and [itex]ds[/itex], using a coordinates system such that [itex]g_{00} = 2[/itex] must be wrong. There must be something fundamental about HAVING to pick [itex]g_{00} = 1[/itex]... but what is this? 


#5
Apr312, 12:54 PM

Sci Advisor
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P: 5,083

In no way do you have to pick gtt=1.
Note the author in the reference insists on a diagonal metric. This already excludes many coordinate systems, and is only possible locally, in general (that is, there are spacetimes in GR such that no coordinate system can have diagonal metric everywhere). Actually, it seems their whole argument is local: at a given point or small region, we can put the metric in diagonal form. Note, in particular, use of frames which are strictly local in GR. But then, the whole thing seems trivial to me, because at one point or small region, you can make the metric (+1,1,1,1), always, after which most of their argument is irrelevant. My conclusion: a whole essay about something trivially true. However, playing their game... I see the comment you mention as just: we can make gtt = 1, then (13) becomes (5). But this choice isn't used for anything. In particular, (14) makes no assumption that gtt=1. (14) is what is taken as the 'general definition' of d tau. Upshot: no requirement for gtt=1, and the author's don't require it as I read this little paper. 


#6
Apr312, 01:26 PM

Sci Advisor
PF Gold
P: 5,083

I should also add that most people take it as simply a definition that:
ds^2 = c^2 d tau^2 and an axiom of GR, to be tested by experiment, that tau measures time experienced along a world line. 


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