# Proper time in a curved space

• I
• Kashmir
A and B and calculate the time intervals between each of the two events. In summary, 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 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
Kashmir said:
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.

• vanhees71 and Orodruin
Ibix said:
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, vanhees71 and Ibix
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.

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Kashmir said:
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.
Kashmir said:
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.

PeroK said:
"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.

Kashmir said:
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.

Kashmir said:
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.

Kashmir said:
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.

Kashmir said:
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.

• PeroK
PeterDonis said:
spatial coordinates.
You mean Spacelike coordinates ?
PeterDonis said:
constant distance
What distance are we talking about here? ##ds^2##?
PeterDonis said:
gravitational field does not change with time
Which time? coordinate time?

Kashmir said:
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.

• Dale and Kashmir
Kashmir said:
You mean Spacelike coordinates ?
Any coordinate that is "spatial" will have to be spacelike, yes.

Kashmir said:
What distance are we talking about here? ##ds^2##?
The integral of ##ds^2## along a spacelike curve of constant coordinate time.

Kashmir said:
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
PeterDonis said:
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.

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.

• martinbn
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.

• Orodruin
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:

wiki said:
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).

• vanhees71 and Ibix