Why do clocks show proper time?

[George Keeling]

TL;DR

I often see the assertion that clocks show proper time. I don't think ever seen a justification in GR. What is it?

I often see the assertion that clocks show proper time. (E.g. this thread). But I don't think that I have seen a good reason for the assertion in GR. It's not a problem in SR because one can always find a coordinate system that is stationary relative to a clock so, from the metric, $d\tau=dt$. But in GR you would get $d\tau=\sqrt{-g_{00}}dt$ or worse if the metric were not diagonal. The only reason I can think of to justify the assertion is that $dt$ is different in different coordinate systems and $d\tau$ is not. So it's, sort of, the only sensible choice left. Surely there must be better than this?

 

[Orodruin]

If it does not measure proper time it is not a clock because that is how a clock is defined in relativity.

In fact, even in SR proper time is more fundamental than coordinate time and a clock is something that measures proper time, not coordinate time. You can then use the proper times shown by family of clocks to define your coordinate time. The argumentation that $d\tau = dt$ in the instantaneous rest frame is just a heuristic that shows that you can in essence recover your old time concept.

 

[PeroK]

Summary: I often see the assertion that clocks show proper time. I don't think ever seen a justification in GR. What is it?

I often see the assertion that clocks show proper time. (E.g. this thread). But I don't think that I have seen a good reason for the assertion in GR. It's not a problem in SR because one can always find a coordinate system that is stationary relative to a clock so, from the metric, $d\tau=dt$. But in GR you would get $d\tau=\sqrt{-g_{00}}dt$ or worse if the metric were not diagonal. The only reason I can think of to justify the assertion is that $dt$ is different in different coordinate systems and $d\tau$ is not. So it's, sort of, the only sensible choice left. Surely there must be better than this?

Take a clock, defined by some physical process, like oscillations of a quartz crystal. Measure $d\tau$, as defined above for the path through spacetime of your clock. You get an exact correspondence between $d \tau$ and the number of oscillations, whatever the path. You can't do better than that!

 

[PeterDonis]

It's not a problem in SR because one can always find a coordinate system that is stationary relative to a clock

You can always do this in GR as well.

 

[Tendex]

It's not a problem in SR because one can always find a coordinate system that is stationary relative to a clock

You can always do this in GR as well.

Indeed, you can look up momentarily comoving frames of general relativity. This is prompted by the fact that GR is a generalization of SR that is still locally Minkowskian by the principle of equivalence.
This comes mathematically by the general covariance of the theory. A form of local gauge invariance whose group is non-compact, the group of all diffeomorphisms, GL(4), that contains as subgroups the compact gauge groups you might have heard of in QFT's Standard model of particles when restricting to SR and Minkowski base space.

 

[PeterDonis]

you can look up momentarily comoving frames of general relativity

I wasn't just talking about those. You can set up Fermi normal coordinates centered on any timelike worldline. In those coordinates, the metric is Minkowski everywhere on the worldline, so coordinate time equals proper time on the worldline.

 

[kent davidge]

so coordinate time equals proper time on the worldline

but they are disconnected along the worldline, correct?

 

[Dale]

Summary: I often see the assertion that clocks show proper time. I don't think ever seen a justification in GR. What is it?

I don't think that I have seen a good reason for the assertion in GR.

I would take the approach mentioned by @Orodruin and just say that in GR clocks are defined as devices that measure proper time.

Then it is an experimental question to determine if a given device is a good clock.

 

[PeterDonis]

they are disconnected along the worldline

What do you think is disconnected along the worldline?

 

[pervect]

Summary: I often see the assertion that clocks show proper time. I don't think ever seen a justification in GR. What is it?

I often see the assertion that clocks show proper time. (E.g. this thread). But I don't think that I have seen a good reason for the assertion in GR. It's not a problem in SR because one can always find a coordinate system that is stationary relative to a clock so, from the metric, $d\tau=dt$. But in GR you would get $d\tau=\sqrt{-g_{00}}dt$ or worse if the metric were not diagonal. The only reason I can think of to justify the assertion is that $dt$ is different in different coordinate systems and $d\tau$ is not. So it's, sort of, the only sensible choice left. Surely there must be better than this?

Because $d\tau$ is the same in all coordinate systems, it's what MTW calls "a geometric object". Being a scalar, proper time is one of the simplest geometric objects possible.

Tensors in general are regarded as "geometric objects", because they can be interpreted as having an existence that is independent of the coordinate choice. A certain amount of interpretation is required in general, because the components of tensors do vary when one chooses changes coordinates, but they change according to a strictly defined set of rules. The interpretation of the tensor is that it has an "existence", which is a philosophical idea, that is independent of the choice of coordinates.

So where does this leave coordinate time? It leaves coordinate time as a convention, a label. It's a human choice. It's like giving someone a map of the countryside, and saying "the cabin is at B4 on the map". "B4" is just a label, a convention. The creator of the map gave the particular location a name or label, that name being "B4", but the name isn't part of the physics, it's a communication tool to tell someone where the cabin is. To have physical meaning, one needs to have access to the map. A person with a different map might give the cablin a different label, such as E8. Thus, the label (coordinates) don't tell you anything physical, unless you also know the labeling system (coordinate system) that is being used.

 

[George Keeling]

Thanks everybody especially @PeroK and @Orodruin

 

[George Keeling]

I came back to this and found a much more satisfying answer on page 9 of Carroll's book. It does not involve super accurate clocks (PeroK ) or saying "it is so" (Orodruin , Dale). I've added it in case other poor souls come here!

Using his conventions, in special relativity proper time is the invariant interval $\Delta\tau$ given by$$
\left(\Delta\tau\right)^2=-\left(\Delta s\right)^2=-\eta_{\mu\nu}\Delta x^\mu\Delta x^\nu
$$Just under equation his 1.17 he says, his italics, the proper time between two events measures the elapsed time as seen by an observer moving on a straight line between the two events.

The reason is that we define coordinate time as the time measured by stationary clocks at fixed spatial coordinate positions. Duh! Therefore for a stationary observer ($\Delta x^i=0$) we have $$
\left(\Delta\tau\right)^2=\left(\Delta t\right)^2
$$Therefore the elapsed proper time is the elapsed clock time between two events measured by a stationary observer.

If the observer is moving with a constant velocity they measure elapsed time between two events with a clock which they carry with them, not by noting the times on the clocks passing by at fixed spatial coordinate positions.

In another coordinate system where the observer is stationary, the elapsed time will still be the proper time. The proper time is invariant. You could have measured the proper time in any coordinate system and got the same answer for the elapsed time between the two events.

This carries over to general relativity, I think: By the equivalence principle a small enough region of spacetime is flat. So our clock and observer's journey can be chopped up into little journeys where the above argument still applies.

 

[Dale]

Well, I am glad that you like that explanation. Certainly Carroll is an excellent source. I don’t like the explanation because coordinate time doesn’t need to be equal to the time on a physical clock, and in most cases it is not. So to me it doesn’t work, but then again I am no Carroll.

 

[vanhees71]

It doesn't need to be a "straight line" though. Proper time is defined by the said equation for any time-like worldline. The same equation of course also holds in GR,
$$\mathrm{d} \tau^2 = g_{\mu \nu} \mathrm{d} q^{\mu} \mathrm{d} q^{\nu}$$
(using the signature (+---)).

Whether or not a specific clock shows "proper time" is of course a question of its construction. I doubt that, e.g., a pendulum clock shows proper time if it's accelerated ;-)).

 

[Orodruin]

Well, I am glad that you like that explanation. Certainly Carroll is an excellent source. I don’t like the explanation because coordinate time doesn’t need to be equal to the time on a physical clock, and in most cases it is not. So to me it doesn’t work, but then again I am no Carroll.

To add to this, the explanation in Carroll tells you why coordinate time in an inertial frame is equal to the time elapsed on a clock at rest in that frame. There is no reason a coordinate time would hold any special meaning or actual relation to a clock. It is therefore actually more of a reverse argument of why we pick that coordinate time rather than why a clock following a particular world line shows the proper time along that world line. The latter is true by definition.

The argument in Carroll is however good for a heuristic argument for why proper time is something related to what we are used to calling ”time”.

 

[vanhees71]

Well, locally you can make any coordinate time equal to an observer's coordinate time when you choose coordinates, where
$$\mathrm{d} s^2 = \mathrm{d} t^2 +2 g_{0 k} \mathrm{d} t \mathrm{d}q^k -g_{jk} \mathrm{d} q^j \mathrm{d} q^k,$$
where latin indices run from 1 to 3 only. Then for the observers at rest, i.e., $\mathrm{d} q^j=0$ you have $\mathrm{d} s=\mathrm{d} t$.

 

[PeterDonis]

The reason is that we define coordinate time as the time measured by stationary clocks at fixed spatial coordinate positions.

Unfortunately, no, this is not always true. It is only true for certain coordinate systems in certain spacetimes. One of those cases is global inertial coordinates in the flat spacetime of SR, which is the case most people intuitively think of, and that's the case you focus on in your post, so it looks like what you are proposing works in your post. But unfortunately it doesn't generalize.

This carries over to general relativity, I think

No, it doesn't. It doesn't even carry over to non-inertial coordinates in SR (for example, Rindler coordinates).

 

[Dale]

locally you can make any coordinate time equal to an observer's coordinate time when you choose coordinates

Hmm, I wouldn’t say it this way. You cannot make any coordinate time equal to an observer’s coordinate time. For any observer you can define local coordinates as you describe but not just any coordinates will do, they have to satisfy the equation you posted which some coordinates do not. Not all coordinate systems even have a coordinate time.

 

[vanhees71]

Of course. I don't understand your last sentence though. The metric components should have everywhere the signature $(1,3)$ (or $(3,1)$). Of course the coordinates are in general only locally defined and you can have coordinate singularities which you need to cover with another map (other coordinates). So shouldn't any four coordinates have one time-like, which I label usually with $q^0$. Then the coordinate lines defined by $(q^1,q^2,q^3)=\text{const}$ should be time-like in some region of the described spacetime, or do you have a counter example of useful coordinates, where this is wrong?

 

[George Jones]

So shouldn't any four coordinates have one time-like

No. For example, it is possible to have four independent spacelike coordinates signature (1,3) spaceitme. It is also possible to have four independent timelike coordinates.

 

[PAllen]

No. For example, it is possible to have four independent spacelike coordinates signature (1,3) spaceitme. It is also possible to have four independent timelike coordinates.

or 4 independent lightlike coordinates, as will be covered in an insight I have in progress.

 

[PeterDonis]

shouldn't any four coordinates have one time-like

No. An obvious counterexample is null coordinates.

 

[Dale]

do you have a counter example of useful coordinates, where this is wrong?

Well, I don’t know about “useful” but I was thinking of examples like the previous posters mentioned. Especially light like coordinates

 

[George Jones]

it is possible to have four independent spacelike coordinates signature (1,3) spaceitme.

For example, Painleve-Gullstrand coordinates inside the event horizon of a Schwarzschild black hole.

 

[pervect]

Well, I am glad that you like that explanation. Certainly Carroll is an excellent source. I don’t like the explanation because coordinate time doesn’t need to be equal to the time on a physical clock, and in most cases it is not. So to me it doesn’t work, but then again I am no Carroll.

I'm going to expand on this point a bit. Note that it is a point that I agree with, nothing I say is critical of Dale's response, it was simply an inspiration for my own response.

To give a specific example, here on the Earth, to get coordinate time (TAI time, international atomic time, for example, or any of the other coordinate times derived from it, including UTC, Coordinated Universal Time), we need to rate adjust the physical atomic clocks for their altitude to correct for "gravitational time dilation".

As a consequence, only at mean sea level do atomic clocks (which keep proper time) correctly indicate coordinate time without further adjustment. Basically, coordinates are just labels, defined by convention. The sort of time that physical clocks read is proper time, i.e. the SI definition of the second is a definition of the proper second.

To give a wiki reference see the link [International Atomic Time].

In the 1970s, it became clear that the clocks participating in TAI were ticking at different rates due to gravitational time dilation, and the combined TAI scale therefore corresponded to an average of the altitudes of the various clocks. Starting from Julian Date 2443144.5 (1 January 1977 00:00:00), corrections were applied to the output of all participating clocks, so that TAI would correspond to proper time at mean sea level (the geoid). Because the clocks were, on average, well above sea level, this meant that TAI slowed down, by about one part in a trillion. The former uncorrected time scale continues to be published, under the name EAL (Echelle Atomique Libre, meaning Free Atomic Scale).[13]

 

[kent davidge]

It doesn't even carry over to non-inertial coordinates in SR (for example, Rindler coordinates)

Don't understand you here. A clock at rest in these coordinates marks a time which is directly proportional to the "coordinate time". For example, if we consider only one spatial dimension, it is $d\tau = \rho dt$, where $\rho$ is the proportionality constant (constant in the sense that it does not depend on $\tau$, as opposed to the coordinates).

Or am I getting this wrong?

 

[PeterDonis]

directly proportional to the "coordinate time"

Is not the same as "coordinate time". "Directly proportional to" is not the same as "equal to" here. The factor of proportionality is coordinate-dependent.

Or am I getting this wrong?

Yes. A "constant" proportionality factor would be independent of all coordinates, not just $\tau$.

 

[A.T.]

I came back to this and found a much more satisfying answer on page 9 of Carroll's book. It does not involve super accurate clocks (PeroK ) or saying "it is so" (Orodruin , Dale).

Note that there is a difference between "because it is so" and "because it was defined to be so".