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Derivation of Proper Time in General Relativity

by paultsui
Tags: derivation, proper, relativity, time
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paultsui
#1
Apr3-12, 05:45 AM
P: 13
In relativity, proper time along a world-line 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 world-line.

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/gr-qc/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 space-time, which cannot be chosen arbitrarily.

Can anyone give me a hint on where my logic go wrong?
Thank you!
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elfmotat
#2
Apr3-12, 09:11 AM
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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.
Chestermiller
#3
Apr3-12, 09:11 AM
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Quote Quote by paultsui View Post
In relativity, proper time along a world-line 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 world-line.

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/gr-qc/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 space-time, which cannot be chosen arbitrarily.

Can anyone give me a hint on where my logic go wrong?
Thank you!
In GR, for observers in a frame of reference at rest with respect to a massive body, the metrical properties of spacetime can be described in terms of a single time parameter t, which is a kind of "reference time" for the system. But equal increments in this time parameter do not represent equal increments in clock time for each of the observers. The rate at which the clock time varies with the reference time parameter t depends on location relative to the massive body. For example, with regard to the Schwartzchild metric, the time parameter t is equal to the clock time only at large distances from the massive body. For observers closer to the body, equal increments in reference time t correspond to smaller increments in (proper) clock time. So clocks near a massive body appear to be running slower. Increments in proper time are related to increments in reference time by dτ=sqrt(gtt)dt. So proper time is still given by dτ=ds/c.
I hope this helps.

Chet

paultsui
#4
Apr3-12, 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 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 [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?
PAllen
#5
Apr3-12, 12:54 PM
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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.
PAllen
#6
Apr3-12, 01:26 PM
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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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