Def Proper Time GR: Half or Integral Along Path?

[hawkdron496]

TL;DR Summary

How do we formally define "Proper time is the time measured by a clock travelling along the worldlike"?

In texts on General Relativity, the proper time $d\tau^2 = -ds^2$ (with an appropriate choice of metric signature) is commonly said that the time measured by a timelike observer traveling along a path is given by the integral of $d\tau$ along this path. Of course it's possible to construct a coordinate system (given some niceness assumptions about the path, presumably) where all the spatial coordinates are 0, and the timelike coordinate is the proper time. However, it seems to me that it would be equally easy to construct a coordinate system where the spatial coordinates are the same, and the time coordinate is halved, so in this system the coordinate time along the path would be half the proper time.

I'm wondering if there's a way to "single out" a coordinate system as corresponding to what an observer traveling along the path measures on their clock, or whether it's a postulate of GR that the proper time is what a clock traveling along the path would say.

 

[PeterDonis]

it seems to me that it would be equally easy to construct a coordinate system where the spatial coordinates are the same, and the time coordinate is halved

Doing this will also change the metric coefficients, so $d\tau$ will end up being the same. The arc length along any worldline is an invariant, and proper time is just arc length along a timelike worldline.

You are of course free to choose the units of time; for example, you could choose to measure time in milliseconds instead of seconds, while keeping your spatial units the same. This would make arc lengths along timelike worldlines numerically greater by 1000, but physically the numbers would still have the same meaning: 1000 milliseconds is the same physical time interval as 1 second.

I'm wondering if there's a way to "single out" a coordinate system as corresponding to what an observer traveling along the path measures on their clock

Once you've made a choice of units of time, this is just a matter of constructing your coordinates so the observer in question is always at rest at the same spatial coordinates for all time. But if the observer is not inertial, these coordinates will not form an inertial frame.

whether it's a postulate of GR that the proper time is what a clock traveling along the path would say.

It's a postulate in the sense that arc length along timelike worldlines is how GR models the readings of clocks mathematically.

 

[Dale]

Summary: How do we formally define "Proper time is the time measured by a clock traveling along the worldlike"?

it seems to me that it would be equally easy to construct a coordinate system where the spatial coordinates are the same, and the time coordinate is halved, so in this system the coordinate time along the path would be half the proper time.

Proper time is an affine parameter. So you can always scale it and shift it and the scales and shifted parameter will still solve the geodesic equation. As @PeterDonis mentioned, that amounts to choosing a different unit of time and shifting your choice of $t=0$

 

[hawkdron496]

Doing this will also change the metric coefficients, so $d\tau$ will end up being the same. The arc length along any worldline is an invariant, and proper time is just arc length along a timelike worldline.

I understand that much. My question is more about *coordinate* time, in the sense that I can imagine constructing a coordinate system in which the observer is spacially at rest, and coordinate time = proper time. I can also imagine constructing a coordinate system in which the observer is spacially at rest and coordinate time is half of proper time (take the coordinate system where the observer measures proper time and make a change of coordinates $t' = 2t$). The arclength of the worldline is of course the same, but only in one of those frames does coordinate time correspond to what an observer will see on a clock.

Of course the arc length along the worldline will be the same in either coordinate system, but what make the first frame "special" in the sense that coordinate time corresponds to what the observer measures on their clock?

 

[PeterDonis]

I can imagine constructing a coordinate system in which the observer is spacially at rest, and coordinate time = proper time.

Yes, you can always do this for a single observer.

I can also imagine constructing a coordinate system in which the observer is spacially at rest and coordinate time is half of proper time (take the coordinate system where the observer measures proper time and make a change of coordinates $t' = 2t$).

This just means you are measuring coordinate time and proper time in different units. And, as I said, the metric coefficients will have to change to correspond.

Of course the arc length along the worldline will be the same in either coordinate system, but what make the first frame "special" in the sense that coordinate time corresponds to what the observer measures on their clock?

Only if you insist on defining "what the observer measures on their clock" using a particular choice of units for time, and saying that, if coordinate time uses a different choice of units, it does not "correspond" to the time read on the clock. But this difference is trivial and does not reflect anything physical; it just reflects a weird choice of units for coordinate time as compared to proper time.

 

[hawkdron496]

Only if you insist on defining "what the observer measures on their clock" using a particular choice of units for time, and saying that, if coordinate time uses a different choice of units, it does not "correspond" to the time read on the clock. But this difference is trivial and does not reflect anything physical; it just reflects a weird choice of units for coordinate time as compared to proper time.

Is what you're saying that the proper time quantity won't be equal to the coordinate time quantity in the $t'$ frame, because in the $t'$ frame you're essentially measuring time in half-units, and once you do the unit conversion it will stop being an issue? Surely though you could construct a new frame $t'' = t'^2$ (and I guess work in a subset of $\mathbb{R}^4$ where $t' > 0$ so there are no issues with the transformation having derivative 0 at 0), at which point the coordinate time and proper time (which hasn't changed after this change of coordinates because the metric coeffs change to compensate) are no longer related by a simple unit conversion.

Why is the frame where $t$ corresponds to the arc length and all spatial coordinates are 0 somehow the "physical" one, over any other frame where all spatial coordinates are 0?

 

[PeroK]

$t^2$ has dimensions of time squared. That's like trying to measure length in $m^2$.

 

[Ibix]

It's like having a map with kilometre grid squares. It's easiest to give distances in kilometres, but it's not difficult to give them in miles. You have two options - one is to measure distance in kilometres then divide by 1.6, and the other is to get the grid rescaled to miles. The first option is analogous to $d\tau^2=-Ads^2$ where $A$ is some constant. The second option is changing from your $t$ to $t'$. Which is more convenient depends on application.

There's also nothing wrong with using non-uniform grid squares, which is what I think you are trying for with your $t''$. You'd end up absorbing the variable units into the metric - this is kind of analogous to the angular coordinate in polars, where a coordinate change of 1 means different distances depending on $r$.

 

[Nugatory]

Why is the frame where t corresponds to the arc length and all spatial coordinates are 0 somehow the "physical" one, over any other frame where all spatial coordinates are 0?

It’s not.
No frame is special and more “physical” than any other, and there is nothing special about those frames in which the spatial coordinates along a particular timelike curve are zero. The physical thing is the amount of time a clock following that curve will measure, and that is unaffected by our choice of coordinates. There’s a number on the face of the clock at one end of the curve, there’s a number on the face of the clock after it has reached the other end, the difference is the amount of time that passed between the two clock readings (by definition - “time is what a clock measures” said Einstein), and no coordinates are involved anywhere.

We often choose our coordinates so that the coordinate time has a simple relationship to what a clock following some worldline would read. That doesn’t make those coordinates any more physical than any others.

 

[Orodruin]

Why is the frame where t corresponds to the arc length and all spatial coordinates are 0 somehow the "physical" one, over any other frame where all spatial coordinates are 0?

It is not. The description of the physics is the same. That coordinate time = proper time is just a coincidence due to your choice of coordinates for a particular observer.

 

[Dale]

The arclength of the worldline is of course the same, but only in one of those frames does coordinate time correspond to what an observer will see on a clock

This is a false statement. In both the coordinate time will correspond to what an observer will see on a clock using some ad hoc unit of time. Again, proper time is an affine parameter.

what make the first frame "special" in the sense that coordinate time corresponds to what the observer measures on their clock?

Nothing. You are making an incorrect statement about its “special”-ness.

Surely though you could construct a new frame t″=t′2 (and I guess work in a subset of R4 where t′>0 so there are no issues with the transformation having derivative 0 at 0), at which point the coordinate time and proper time (which hasn't changed after this change of coordinates because the metric coeffs change to compensate) are no longer related by a simple unit conversion.

Yes, that would give you some non-inertial coordinates. That is not an affine transformation. So your resulting time coordinate would no longer represent the proper time of any clock and the metric in these coordinates would not be the Minkowski metric.

Why is the frame where t corresponds to the arc length and all spatial coordinates are 0 somehow the "physical" one, over any other frame where all spatial coordinates are 0?

Well, IMO no coordinates should be considered “physical”. But here the issue is obvious and by construction. In one there can be physical clocks that keep good time that match the coordinate time and in the other there are not. Or, more mathematically, one has Christoffel symbols and the other does not.

 

[hawkdron496]

This is a false statement. In both the coordinate time will correspond to what an observer will see on a clock using some ad hoc unit of time. Again, proper time is an affine parameter.

Surely though if in one frame the endpoint of the worldline is $(\tau, 0,0,0)$, in the frame transformed according to $t' = 2t$ the worldlike would end at $(2\tau, 0,0,0)$, meaning someone using the second set of coordinates wouldn't measure proper time along the path?

I think the root of my question may be this: when I hear "proper time is the time measured by a clock traveling along the worldline", it makes me think that there must be some "special" coordinate system corresponding to what a clock traveling along the worldline "actually" sees. It seems that this is not the case (you can set up a coordinate system where coordinate time = proper time, but it seems that there's nothing "special" about it, because you can't do it without a prior notion of proper time), so how is "time measured by a clock traveling along the worldline" formally defined? Is it equal to the proper time by definition? In that case, is it a postulate of GR that "Clock along the path sees proper time"? Or is there some other, deeper definition, from which we can derive "proper time is time seen by a clock along the path" as a theorem?

 

[Ibix]

Surely though if in one frame the endpoint of the worldline is $(\tau, 0,0,0)$, in the frame transformed according to $t' = 2t$ the worldlike would end at $(2\tau, 0,0,0)$, meaning someone using the second set of coordinates wouldn't measure proper time along the path?

No. It means one's measuring proper time in units of 2s. It might make more sense if you use $t'=60t$. In that case, one frame measures a coordinate difference of one minute, the other sixty seconds. They're both the same thing!

 

[Dale]

Surely though if in one frame the endpoint of the worldline is $(\tau, 0,0,0)$, in the frame transformed according to $t' = 2t$ the worldlike would end at $(2\tau, 0,0,0)$, meaning someone using the second set of coordinates wouldn't measure proper time along the path?

No, it just means that one was measuring time in ticks and the other was measuring time in tocks where 1 tick = 2 tocks. Both are measurements of proper time.

If you disagree with that then what is the natural unit of time that is forced on us by nature and cannot be changed. It certainly isn't the second since that is set by the BIPM and has been changed multiple times.

it makes me think that there must be some "special" coordinate system corresponding to what a clock traveling along the worldline "actually" sees.

This is simply not the case. Coordinate systems are arbitrary labels and there is nothing physical about them at all. No object "actually" sees any coordinate system. They are completely oblivious to whatever coordinate system you use. The coordinate system is part of the analysis, not part of the physics.

how is "time measured by a clock traveling along the worldline" formally defined? Is it equal to the proper time by definition? In that case, is it a postulate of GR that "Clock along the path sees proper time"?

I would say "yes", but others might disagree. Here are my thoughts.

A theory consists of a mathematical framework and a mapping between parts of the framework and experimental measurements. The mapping between parts of the framework and experimental measures is called the minimal interpretation.

In GR, the mathematical framework is pseudo-Riemannian manifolds of signature (-+++) that are solutions to the Einstein field equations. In a pseudo-Riemannian manifold you can define a timelike path $P$, called a worldline and you can calculate $\tau = \int_P \sqrt{-g_{\mu\nu} dx^{\mu} dx^{\nu}}$, then part of the minimal interpretation is that $\tau$ is mapped to the reading on a clock that travels along the path $P$.

So that would put this concept in the theory itself, but in the minimal interpretation part of the theory rather than in the mathematical framework part of the theory.

 

[hawkdron496]

I would say "yes", but others might disagree. Here are my thoughts.

A theory consists of a mathematical framework and a mapping between parts of the framework and experimental measurements. The mapping between parts of the framework and experimental measures is called the minimal interpretation.

In GR, the mathematical framework is pseudo-Riemannian manifolds of signature (-+++) that are solutions to the Einstein field equations. In a pseudo-Riemannian manifold you can define a timelike path $P$, called a worldline and you can calculate $\tau = \int_P \sqrt{-g_{\mu\nu} dx^{\mu} dx^{\nu}}$, then part of the minimal interpretation is that $\tau$ is mapped to the reading on a clock that travels along the path $P$.

So that would put this concept in the theory itself, but in the minimal interpretation part of the theory rather than in the mathematical framework part of the theory.

I believe this clarifies things a lot, thank you for the help, I appreciate it.

 

[PeterDonis]

Is what you're saying that the proper time quantity won't be equal to the coordinate time quantity in the $t'$ frame, because in the $t'$ frame you're essentially measuring time in half-units, and once you do the unit conversion it will stop being an issue?

It doesn't have to stop being an issue because it's never an issue in the first place. You're just choosing to measure proper time and coordinate time in different units. This is weird, but it poses no problem for physics. It just makes your calculations more tedious.

Surely though you could construct a new frame $t'' = t'^2$ (and I guess work in a subset of $\mathbb{R}^4$ where $t' > 0$ so there are no issues with the transformation having derivative 0 at 0), at which point the coordinate time and proper time (which hasn't changed after this change of coordinates because the metric coeffs change to compensate) are no longer related by a simple unit conversion.

Sure they are, the conversion is just different for different coordinate times. And the metric coefficients will have to be coordinate time-dependent accordingly. This is weirder than your original proposal, but it still poses no problem for physics. It just makes your calculations even more tedious.

 

[PeterDonis]

$t^2$ has dimensions of time squared.

No, the $t$ coordinate in the second frame will have units equal to the square root of the units of the $t$ coordinate in the first frame. Again, this is weird, but poses no problem for physics; it just makes calculations more tedious.

 

[PeterDonis]

Surely though if in one frame the endpoint of the worldline is $(\tau, 0,0,0)$, in the frame transformed according to $t' = 2t$ the worldlike would end at $(2\tau, 0,0,0)$

In coordinate terms, yes. But the metric would be different in the two frames.

I assume that in the first frame you intend the metric to be the standard Minkowski metric. The only metric coefficient that matters for your specific case is $g_{tt}$, which is $1$ (if we use the timelike signature convention), so the arc length along the worldline from the origin to the endpoint you give is $\sqrt{g_{tt} dt^2} = \sqrt{\tau^2} = \tau$.

In the second frame, your transformation $t' = 2 t$ means that you will have $g_{t' t'} = 1/ 4$ in the second frame, so the arc length along the worldline from the origin to the endpoint you give is $\sqrt{4 \tau^2 / 4} = \tau$. The same as before. Coordinate transformations must preserve arc lengths to be valid.

meaning someone using the second set of coordinates wouldn't measure proper time along the path?

Yes, they would. See above.