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.