I What Is the Proper Time in Curved Space?

[Kashmir]

In special relativity we've the invariant $\begin{aligned} d s^2=&-d t^2 \\ &+d x^2 \\+d y^2+d z \end{aligned}$.

For a clock moving along a worldline the above equation reduces to $\begin{aligned} d s^2=&-d t^2\end{aligned}$ , hence we can say that the time measured by the clock moving along the world line reads time dt such that $\begin{aligned} d s^2=&-d t^2\end{aligned}$ which is called proper time.

Then Hartle gravity pg 126 while motivating curvature of space gives an example of geometry such that $d s^2=-\left(1+\frac{2 \Phi\left(x^t\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x^i\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$

In this space what should the proper time be? I think that for a clock moving along a worldlines the spatial differential are zero thus the proper time should be $\frac{d \tau^2}{c}=\frac{-d s^2}{c\left(1+\frac{2 \phi}{c^2}\right)}$ but the author says that the proper time is
$d \tau^2=-d s^2 / c^2$.

Whats wrong with my thinking? Please help.

 

[Nugatory]

For a clock moving along a worldline...

It's always moving along a worldline, everything is.

Whats wrong with my thinking? Please help.

Basically what's going on here is that $ds^2=-dt^2$ is a special case that applies only when:
a) the coordinate system we're using is such that there is one timelike and three spacelike coordinates, and along the worldline we're considering the three spacelike coordinates are constant.
b) the coordinate system we have chosen is such that there are no off-diagonal terms in the metric.
c) the coordinate system we're using is such that the metric coefficient $g_{tt}$ is equal to one along that worldline.
(probably some more precise way of describing the special conditions, but also some danger of getting into nitpicking)

All three conditions when applying Minkowski coordinates to an object at rest in an inertial frame using those coordinates, and that's why you'll see $ds^2=-dt^2$ in special relativity textbooks.

In Hartle's example #c does not hold.

 

[vela]

Reread page 60 where Hartle introduces proper time. In particular, note that he says a clock moving along a timeline world line measures the distance $\tau$ along it.

If you apply the definition $d\tau^2 = -ds^2/c^2$ to a particle at rest in special relativity, you recover $d\tau = dt$. But this result is a special case that follows from the general definition. Your mistake is taking this result of a special case and trying to apply it to a completely different situation.

 

[PeterDonis]

what should the proper time be?

The proper time along any timelike curve in any curved spacetime is the integral of the metric along the curve.

I think that for a clock moving along a worldlines the spatial differential are zero thus the proper time should be $\frac{d \tau^2}{c}=\frac{-d s^2}{c\left(1+\frac{2 \phi}{c^2}\right)}$ but the author says that the proper time is
$d \tau^2=-d s^2 / c^2$.

Whats wrong with my thinking?

What's wrong is that you just pulled your equation for $d\tau^2$ out of thin air, instead of asking what the physical meaning of $ds^2$ is. The answer is the $ds^2$ is always the arc length along the curve. That means, for a timelike curve, $d\tau^2 = - ds^2 / c^2$ (assuming that you are using a spacelike signature convention for $ds^2$) is always true by definition, for any timelike curve and any curved spacetime.

 

[Dale]

but the author says that the proper time is
$d \tau^2=-d s^2 / c^2$.

This is the correct general definition. The other one is only a special case.

 

[vanhees71]

To make it more explicit. A general time-like worldline is described by a function $x^{\mu}(\lambda)$ of the spacetime coordinates of an arbitrary parameter of this worldline. Neither this parameter nor the spacetime coordinates a priori have a physical meaning. They are just labels for points in the four-dimensional spacetime manifold (in GR a pseudo-Riemannian manifold). This space time is specified by the pseudo-metric tensor which has components $g_{\mu \nu}$ with respect to the coordinate basis $\mathrm{d} x^{\mu}$.

The worldline is "time-like", and only such time-like worldlines can be worldlines of massive particles (or a clock), if $g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}<0$, where the dot indicates differentiation wrt. $\lambda$. Now the proper time is a physical quantity, i.e., it is independent of the choice of coordinates and also the parameter $\lambda$ of the worldline. indeed by definition
$$\mathrm{d} \tau =\frac{1}{c} \mathrm{d} \lambda \sqrt{g_{\mu \nu} \dot{x}^{\mu} \dot{x}^{\nu}}.$$

 

[Kashmir]

Reread page 60 where Hartle introduces proper time. In particular, note that he says a clock moving along a timeline world line measures the distance $\tau$ along it.

If you apply the definition $d\tau^2 = -ds^2/c^2$ to a particle at rest in special relativity, you recover $d\tau = dt$. But this result is a special case that follows from the general definition. Your mistake is taking this result of a special case and trying to apply it to a completely different situation.

Suppose in one frame I measure two events $A$ and $B$. I calculate the interval $ds$ using $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$Any other frame will measure the same interval.
Now suppose we take a clock moving along the worldline of our events. In the clocks frame the spatial differentials $dx, dy, dz$ for the two events $A$ and $B$ are zero. Suppose in the moving clocks frame we measure time $dt$.

This time $dt$ is also the proper time $d\tau$To find proper time we put $dx, dy, dz=0$ in $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$ which gives me
the proper time as $d{ \tau^2}=\frac{-d s^2}{c^2\left(1+\frac{2 \phi}{c^2}\right)}$

 

[PeterDonis]

This time $dt$ is also the proper time $d\tau$

No, it isn't. You have already been told this; see posts #4, #5, and #6.

 

[PeterDonis]

Suppose in one frame I measure two events $A$ and $B$. I calculate the interval $ds$ using $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$

This isn't a calculation of anything, it's just writing down the general metric. To calculate an interval (more precisely, an arc length along a curve) between events $A$ and $B$, you need to first pick a curve that connects the two events, and then integrate $ds$ along the curve.

Any other frame will measure the same interval.

It is true that the arc length along a given curve between two given events will be the same no matter what frame you use to calculate it, yes.

Now suppose we take a clock moving along the worldline of our events.

There is no such thing as " the worldline of our events". What you have done here is to assume that, in the coordinates you are using, both events, A and B, have the same spatial coordinates, and you have then picked a curve that goes from the coordinate time $t_A$ of event A to the coordinate time $t_B$ of event B, keeping the spatial coordinates the same.

And since the spatial coordinates are the same everywhere along the curve, the potential $\Phi$ is constant along the curve and you can pull the $1 + 2 \Phi / c^2$ factor out of the integral. Then you get $\tau = - \int ds = \sqrt{1 + 2 \Phi / c^2} \int_{t_A}^{t_B} dt^2 = \sqrt{1 + 2 \Phi / c^2} \left( t_B - t_A \right)$.

 

[Orodruin]

This time dt is also the proper time

No, it is not. The coordinate time t is nothing but a coordinate. You are ascribing it physical significance when it has none.

 

[Kashmir]

No, it is not. The coordinate time t is nothing but a coordinate. You are ascribing it physical significance when it has none.

Wikipedia : "proper time is defined as the time as measured by a clock following that line"

In our case the clock is our frame of reference, if it measures time $dt$ then it should be the proper time according to above definition.

 

[PeterDonis]

Wikipedia : "proper time is defined as the time as measured by a clock following that line"

Yes. That is $d\tau^2 = - ds^2$.

In our case the clock is our frame of reference

No, it's not. The clock's time is not the same as coordinate time. For the clock's worldline, $d\tau \neq dt$. That has already been shown.

if it measures time $dt$

It doesn't. See above.

@Kashmir, you have repeated the same wrong statement several times now, and been corrected several times now. If we are just going to keep going around in circles, there is no point in continuing this thread.

 

[Orodruin]

Wikipedia : "proper time is defined as the time as measured by a clock following that line"

In our case the clock is our frame of reference, if it measures time $dt$ then it should be the proper time according to above definition.

No, you are misunderstanding what coordinate time is. Proper time is the time measured by a clock. Coordinate time isn’t except in particular cases.

 

[Nugatory]

In our case the clock is our frame of reference, if it measures time $dt$ then it should be the proper time according to above definition.

The clock is measuring proper time $d\tau$, not coordinate time $dt$.

In this context ”our frame of reference” is just the convention that we use to assign x, y, z, and t coordinates to events. We can choose this convention in such a way that $dt$ is equal to $d\tau$ along the worldline of that one particular clock but we don’t have to - and in this case Hartle has not.

You are allowing yourself to be misled by the way that in flat spacetime we can choose to use Minkowski coordinates. These coordinates have the nice property that $dt=d\tau$ along the worldlines of all objects whose spatial coordinates are constant (“at rest in our frame”), but that nice property is unique to those coordinates in that spacetime.

 

[Ibix]

@Kashmir - $d\tau$ is the time measured by a clock moving from coordinates $(t,r,\theta,\phi)$ to coordinates $(t+dt,r+dr,\theta+d\theta,\phi+d\phi)$. In the case where you are at rest in your chosen coordinate system, $dr=d\theta=d\phi=0$, and the line element you quoted simplifies to $$\begin{eqnarray*}
d\tau^2&=&-ds^2\\
&=&\left(1+\frac{2\Phi}{c^2}\right)dt^2
\end{eqnarray*}$$which tells you that $dt\neq d\tau$.

The rule that a clock's time matches coordinate time if it's at rest is a special case. It only applies where the $dt^2$ term in the line element is -1, like in SR.

 

[vanhees71]

Suppose in one frame I measure two events $A$ and $B$. I calculate the interval $ds$ using $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$Any other frame will measure the same interval.
Now suppose we take a clock moving along the worldline of our events. In the clocks frame the spatial differentials $dx, dy, dz$ for the two events $A$ and $B$ are zero. Suppose in the moving clocks frame we measure time $dt$.

This time $dt$ is also the proper time $d\tau$To find proper time we put $dx, dy, dz=0$ in $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$ which gives me
the proper time as $d{ \tau^2}=\frac{-d s^2}{c^2\left(1+\frac{2 \phi}{c^2}\right)}$

It doesn't make sense to say two events have some "distance" or "interval". You always have to specify a worldline connecting the two events. One way is to look for a geodesic connecting these two events. Then the invariant "length" of this worldline can be defined as a "distance".

In your example of the Newtonian approximation the coordinate time is the time of an observer at rest wrt. the chosen reference frame given by the coordinates at infinity, if you follow the usual convention that the Newtonian gravitational potential $\Phi(x) \rightarrow 0$ for $x \rightarrow \infty$. Coordinates only become a physical meaning when expressed in invariant, scalar terms, and that's the proper time of such an observer at rest for very far distances from the source of the gravitational field.

 

[Dale]

Wikipedia : "proper time is defined as the time as measured by a clock following that line"

In our case the clock is our frame of reference, if it measures time $dt$ then it should be the proper time according to above definition.

See example 4 in that Wikipedia section. That example is worked out and shows explicitly and clearly that $d \tau \ne dt$ for exactly this case.

It also shows in the section In Special Relativity explicitly $d\tau=ds/c$ which is not your $d\tau=dt$ and differs from our $d\tau^2=-ds^2/c^2$ only by the completely arbitrary choice of sign convention on $ds^2$.

 

[Ibix]

It doesn't make sense to say two events have some "distance" or "interval". You always have to specify a worldline connecting the two events. One way is to look for a geodesic connecting these two events. Then the invariant "length" of this worldline can be defined as a "distance".

Just to add, this comes naturally in SR because you can always draw a unique straight line between two events. That's why we can talk about "the interval between two events" in SR, and the absence of a unique path between pairs of events in curved spacetime is why we need to be picky about paths all of a sudden.

 

[vanhees71]

Since Minkowski space is a flat space (an affine pseudo-Euclidean manifold) the straight lines are the geodesics of this spacetime, and in this sense to define the pseudo-distance between two events is indeed "straight forward" ;-)).

 

[Kashmir]

No, it is not. The coordinate time t is nothing but a coordinate. You are ascribing it physical significance when it has none.

Ohh. That was the source of my confusion.

So basically the coordinates (t, x, y, z) are nothing but four numbers which label a point in spacetime .

So suppose I'm an observer, then to locate each point in spacetime I've to choose a way to represent each point by four numbers . Is that right?

 

[PeroK]

So suppose I'm an observer, then to locate each point in spacetime I've to choose a way to represent each point by four numbers . Is that right?

Hartle makes the point somewhere that in the general spacetimes of GR observers only make local measurements and coordinate systems (that do not neatly align with any "observer") are used to describe the spacetime and make the predictions about local measurements.

I think the time has come for you to leave the notion of "observer" behind. It's overused in SR, IMHO, and when it comes to GR, it's problematic to describe spacetime in terms of "observers". Try to think in terms of coordinate systems instead.

 

[Ibix]

So suppose I'm an observer, then to locate each point in spacetime I've to choose a way to represent each point by four numbers . Is that right?

Kind of. The problem with observers is that there are often things they can't observe - inside event horizons or outside the observable universe, for example. Coordinate systems can cover regions you can't observe, and can be very helpful for predicting physics there.

But you are correct that all you are doing with a coordinate system is proposing a systematic labelling of all events (possibly only in a limited region). The coordinates don't have to have a trivial relationship to anything physical.

 

[Kashmir]

Hartle makes the point somewhere that in the general spacetimes of GR observers only make local measurements and coordinate systems (that do not neatly align with any "observer") are used to describe the spacetime and make the predictions about local measurements.

I think the time has come for you to leave the notion of "observer" behind. It's overused in SR, IMHO, and when it comes to GR, it's problematic to describe spacetime in terms of "observers". Try to think in terms of coordinate systems instead.

Thank you so much.
Can you please give me an example which illustrates the use of coordinates.

 

[PeroK]

Thank you so much.
Can you please give me an example which illustrates the use of coordinates.

Your previous thread on the metric with the gravitational potential. That's the spacetime described using a certain coordinate system that does not correspond to a single observer.

Part of your confusion over proper time in GR may be due to trying to relate coordinate time at a location to the proper time of a clock at that location.

 

[Kashmir]

Kind of. The problem with observers is that there are often things they can't observe - inside event horizons or outside the observable universe, for example. Coordinate systems can cover regions you can't observe, and can be very helpful for predicting physics there.

But you are correct that all you are doing with a coordinate system is proposing a systematic labelling of all events (possibly only in a limited region). The coordinates don't have to have a trivial relationship to anything physical.

Thank you.
So suppose I've a coordinate system now, then how will these coordinates tell me what's going on?
How do we interpret these coordinates to understand what is happening?

 

[PeroK]

Thank you.
So suppose I've a coordinate system now, then how will these coordinates tell me what's going on?
How do we interpret these coordinates to understand what is happening?

That's all covered in Hartle's book. Essentially you make predictions about local measurements. E.g. you cannot directly measure a light ray traveling through space. But, you can predict the angle of incidence when it is detected on Earth. From that you infer that your model of spacetime is accurate.

 

[Kashmir]

Part of your confusion over proper time in GR may be due to trying to relate coordinate time at a location to the proper time of a clock at that location.

Yes confusion did arise from it.

What i thought was this:
We've an observer , he sets up a set of three mutually perpendicular axis and has a clock.
The three axis and the clock are the coordinates (t, x, y, z).

However as I've understood since then is that coordinates can be anything which uniquely determine a point in spacetime and
(x, y, z) in the coordinate set (t, x, y, z)
aren't necessarily spatial coordinates measured on three perpendicular meter rods. And the t appearing in the set similarly isn't necessarily connected trivially to any clock.

Also there is a connection between these coordinates such that the spacetime interval $ds^2$ remains invariant upon choosing another coordinate system.

 

[PeroK]

Yes confusion did arise from it.

What i thought was this:
We've an observer , he sets up a set of three mutually perpendicular axis and has a clock.
The three axis and the clock are the coordinates (t, x, y, z).

That's a local process in curved spacetime.

However as I've understood since then is that coordinates can be anything which uniquely determine a point in spacetime and
(x, y, z) in the coordinate set (t, x, y, z)
aren't necessarily spatial coordinates measured on three perpendicular meter rods. And the t appearing in the set similarly isn't necessarily connected trivially to any clock.

Yes, exactly. E.g. spherical (polar) coodinates.

Also there is a connection between these coordinates such that the spacetime interval $ds^2$ remains invariant upon choosing another coordinate system.

Yes.

 

[Kashmir]

That's a local process in curved spacetime.

Can you please tell me What's a local process? Does it mean for a small neighborhood this is a valid way of assigning coordinates?

 

[Kashmir]

Suppose we are in Minkowski space, there i choose my coordinates { a, b, c, d} to represent spacetime.

However I note that one coordinate is special from the other three because it gets a -1 sign when writing the spacetime interval $ds$.
How does one know which of { a, b, c, d} is that one?

 

[vanhees71]

That's convention, i.e., it depends on how you order your coordinates. When you use a pseudo-Cartesian basis usually you take the time-like basis vector to be labelled with the index 0 and the space-like ones with 1, 2, 3. Then you get $g_{\mu \nu} = \boldsymbol{e}_{\mu} \cdot \boldsymbol{e}_{\nu}=\mathrm{diag}(-1,1,1,1)$ (when using the east-coast convention concerning the metric, i.e., the signature $(3,1)$).

 

[PeroK]

Can you please tell me What's a local process? Does it mean for a small neighborhood this is a valid way of assigning coordinates?

Having rigid rods that spread out along three spatial axes is a local idea in generally curved spacetimes. Mathematically, this manifests itself in the loss of the concept of a position vector. In a curved spacetime, vectors are a local concept (technically vectors are defined in the tangent space at each point). They are not defined globally.

 

[Kashmir]

In SR we set up a set of three mutually perpendicular axis and has a clock. The three axis and the clock define the coordinates (t, x, y, z) . Dont they?
Why isn't that sufficient in GR?

 

[Orodruin]

In SR we set up a set of three mutually perpendicular axis and has a clock. The three axis and the clock define the coordinates (t, x, y, z) . Dont they?
Why isn't that sufficient in GR?

Because in GR spacetime is curved.

 

[PeterDonis]

In SR we set up a set of three mutually perpendicular axis and has a clock. The three axis and the clock define the coordinates (t, x, y, z) . Dont they?

To make this a coordinate chart, you need to do it everywhere in spacetime: there needs to be a family of worldlines, one for each clock, that fills the entire spacetime, and at every point along each worldline, there needs to be a set of three mutually perpendicular axes, and the clock readings and mutually perpendicular axes at neighboring events need to match up. All of this is necessary for the coordinates (t, x, y, z) to have the properties they have in SR. And all of this is only possible in flat spacetime. In curved spacetime, i.e., in the presence of gravity, it cannot be done.

 

[Ibix]

Why isn't that sufficient in GR?

For the same reason you can't set up a Cartesian coordinate system covering all of Earth - it won't work on a curved surface because initially parallel lines will not generally remain so.

 

[DrGreg]

For the same reason you can't set up a Cartesian coordinate system covering all of Earth - it won't work on a curved surface because initially parallel lines will not generally remain so.

To clarify, Ibix is referring to a 2D $(x, y)$ Cartesian coordinate system covering all of the Earth's surface, not a 3D $(x, y, z)$ Cartesian coordinate system covering all of the Earth including its interior.

Flat 2D maps of a large part of the Earth's surface have to include some form of distortion. But there's a metric to calculate Earth-distances from map coordinates, such as $(latitude, longitude)$.

 

[Kashmir]

Thank you everyone..
So we can't have a 'Three perpendicular rods and a clock' as a coordinate system throughout a curved spacetime , although we may use it in a small region.

Now below is an example from *Hartle,pg 127* .
Here we've a metric given as $d s^2=-\left(1+\frac{2 \Phi\left(x^t\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x^i\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$
The figure explains what's going on

Author says :
"The coordinate separations between the two emissions at location $x_A$ are

$\Delta_t$ and $x= y =z=0$".Now my questions are:

1) with respect to whome are A and B stationary?

2) in this particular example do coordinates (x, y, z) have any physical significance?

I'm confused to be honest.

 

[PeroK]

1) with respect to whom are A and B stationary?

"Stationary" has no physical meaning. In general, stationary means with constant spatial coordinates.

However, this is not just an abstract example of a metric. It's intended to represent the gravitational field outside the Earth (with the appropriate potential function, as specified by Hartle). So, stationary suggests with respect to the Earth.

You would, of course, need some physical process to determine the value of $(x, y, z)$ at the locations where the experiment takes place. In Newtonian physics, because there is only one possible space and time and everything is Euclidean, this is simple. In curved spacetimes, this is more complicated. Nevertheless, you can imagine that given any location outside the Earth, you can have a consistent physical process to determine its coordinates.

2) in this particular example do coordinates (x, y, z) have any physical significance?

Yes, in the sense that we expect that metric to represent the spacetime outside the Earth. You can't allocate $x, y, z$ coordinates arbitraily.

 

[PeterDonis]

"Stationary" has no physical meaning.

Careful. There is a physical meaning to the term for a certain class of curved spacetimes. See my further comments below in response to the OP.

So we can't have a 'Three perpendicular rods and a clock' as a coordinate system throughout a curved spacetime with the same properties as a global inertial frame in flat spacetime , although we may use it in a small region.

See the bolded addition I made above. It's important. See below.

1) with respect to whome are A and B stationary?

With respect to the gravitating body. Note, though, that the term "stationary" has two different possible meanings:

(1) The object has constant spatial coordinates. This is, of course, coordinate dependent, and in general does not tell you anything useful physically. In fact, some coordinate charts don't even support this meaning, since not all coordinate charts have one timelike and three spacelike coordinates.

(2) The object is "hovering" at a constant distance from a gravitating body whose gravitational field does not change with time. (The more technical definition is that the object's worldline is an integral curve of a timelike Killing vector field.) Both A and B meet this definition in the scenario you give, and this definition, unlike the one above, does tell you something physically meaningful.

2) in this particular example do coordinates (x, y, z) have any physical significance?

In this particular example, yes, because the coordinates have been carefully chosen so that both meanings of "stationary" given above match: objects such as A and B, whose spatial coordinates are constant (note that this requires that the coordinate chart we chose has one timelike coordinate, $t$, and three spacelike coordinates, $x$, $y$, and $z$--but you can't just tell this from the names of the coordinates, you have to look at the metric), also are "hovering" at a constant distance from a gravitating body whose gravitational field is not changing with time. That is only possible in special scenarios like this one, where there is a gravitating body whose gravitational field is not changing with time.

 

[PeterDonis]

So we can't have a 'Three perpendicular rods and a clock' as a coordinate system throughout a curved spacetime

Expanding further on my previous post, with regard to this point:

In the scenario you describe, we do have "three perpendicular rods and a clock" at each event; but they do not match up with the coordinates t, x, y, z in the same way as they would in a global inertial frame in flat spacetime. Clocks at rest in these coordinates do not tick coordinate time $t$ (unless they are at spatial infinity, where the potential $\Phi(x)$ goes to zero), and rulers do not exactly measure coordinate intervals $x$, $y$, $z$ (unless, again, they are at spatial infinity, where $\Phi(x)$ goes to zero). At finite $x$, $y$, $z$, the coefficients in the metric give correction factors that you have to apply to convert coordinate intervals into proper intervals, i.e., to times ticked off by clocks and distances measured by rulers.

 

[Kashmir]

spatial coordinates.

You mean Spacelike coordinates ?

constant distance

What distance are we talking about here? $ds^2$?

gravitational field does not change with time

Which time? coordinate time?

 

[PeroK]

You mean Spacelike coordinates ?

What distance are we talking about here? $ds^2$?

Which time? coordinate time?

I think a lot of your questions are answered in Chapter 7. With something like GR perhaps you should try to press on instead of stopping in your tracks. Park these questions for now. No book can explain everything at once.

 

[PeterDonis]

You mean Spacelike coordinates ?

Any coordinate that is "spatial" will have to be spacelike, yes.

What distance are we talking about here? $ds^2$?

The integral of $ds^2$ along a spacelike curve of constant coordinate time.

Which time? coordinate time?

If the time coordinate is chosen properly, which it is in the example you give, yes. The more technical definition is that the spacetime is stationary if it has a timelike Killing vector field. One can always choose a time coordinate $t$ such that the metric is independent of $t$ in a spacetime that has a timelike Killing vector field (proving this is often an exercise in GR textbooks).

 

[PeroK]

If the time coordinate is chosen properly, which it is in the example you give, yes. The more technical definition is that the spacetime is stationary if it has a timelike Killing vector field.

Hartle introduces Killing vectors in Chapter 8 (Geodesics), which is why I think the OP needs to press on and see whether subsequent chapters answer these questions.

 

[Kashmir]

Ok. I'll move on and see if things get clear, right now I'm somewhat confused.

But thank you all for your help.

 

[Orodruin]

To summarize, things will generally fall into place better once you start realising that the coordinate t is nothing but a coordinate without direct physical significance. In many cases it can be chosen in such a way that it has a particular physical interpretation, but then you still have to be careful not to over extend that interpretation.

 

[vanhees71]

In addition one should say that the observables in GR are given by local scalar quantities. A reference frame refers to a local observer or a family of local observers in a small neighborhood around a point in spacetime, where and when the observer performs the measurement. Formally the reference frame is given by the (necessary timelike) worldline of the observer (or a local congruence of such timelike worldlines) and a set of tetrades along this worldline (congruence). The most convenient choice of the tetrads are those that are "non-rotating" from the point of view of the observer, i.e., given by Fermi-Walker transport along the worldline.

 

[pervect]

The OP might find a careful reading of Misner's "Precis of General Relativity", on arxiv as https://arxiv.org/abs/gr-qc/9508043, helpful. Possibly not, but it's worth a try. It's one of the few papers that talk about some of the underpinings of the theory that address the OP's question.

It's rather long, but I'll give a couple of quotes I think are relevant.

Omitting the motivations and historical connections, and also the
detailed calculations, I state succinctly the principles that determine
the relativistic idealization of a GPS system. These determine the
results that Ashby presents in his tutorial.
A method for making sure that the relativity effects are specified correctly
(according to Einstein’s General Relativity) can be described rather briefly.
It agrees with Ashby’s approach but omits all discussion of how, historically
or logically, this viewpoint was developed. It also omits all the detailed
calculations. It is merely a statement of principles.
One first banishes the idea of an “observer”. This idea aided Einstein
in building special relativity but it is confusing and ambiguous in general
relativity. Instead one divides the theoretical landscape into two categories.
One category is the mathematical/conceptual model of whatever is happen-
ing that merits our attention. The other category is measuring instruments
and the data tables they provide.

TL/DR. We divide GR into two parts, a conceptual model and physical measurements

The conceptual model for a relativistic system is a spacetime map or
diagram plus some rules for its interpretation. For GPS the attached Figure
is a simplified version of the map. The real spacetime map is a computer
program that assigns map locations xyzt to a variety of events.

Tl/dr. The conceptual model says that coordinate are just labels that we use to describe events in space-time.

Misner also discusses the measuring instruments. However, I won't quote that part, because understanding it properly rquies first that one realize the difference between coordinate time and proper time, which is the point ujnder discussion.

Misner proceeds to give an approximate metric of the space-time of the Earth as an example of a particular instance of the conceptual model.

dτ^2 = [1 + 2(V − Φ0)/c^2]dt^2 − [1 − 2V /c^2](dx2 + dy2 + dz2)/c^2

Misner points out that this metric is a conceptual model. The purpose of this model is to provide labels (coordinates) for events. The events are physical. The coordinates are just labels. The map is not the territory. If we say "the destination is at square A4 of the map", the destination is physical, the reference to grid "A4" on the map is a map reference, it's not physical. Some different map from a different atlas might assign different labels to the same destination.

The following quote from Misner is of particular relevance - it basically says the same thing everyone else has been trying to tell the OP.

The constant Φ0 is chosen so that a standard SI clock “on the geoid” (e.g.,
USNO were it at sea level) would give, inserting its world line x(t), y(t), z(t)
into equation (1), just dτ = dt where dτ is the physical proper time reading
of the clock.

The direct statement isn't perhaps clear, until one realizes the implications. Namely, a clock that is NOT on the geoid does not keep proper time.

A quote from Wiki, says the same thing:

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 the 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 the geoid (mean sea level).

This gives an example of the difference between coordinate time and proper time in the context of a simple and hopefully familiar example of GR - namely, time on the surface of the Earth.

So - a quick recap.

Coordinates are just labels, they don't have physical significance.
Proper time does have physical significance. Proper time is NOT the same as coordinate time - on the Earth, as an example, the two are the same for clocks at sea level, but any clock other than sea level will have a proper time $d\tau$ that is different from the coordinate time $dt$. Thus it is important to know which one is talking about - the physical clock, that keeps proper time, or the coordinate clock, which does not necessarily keep coordinate time, depending on it's location (specifically, it's height above sea level).