Gravitational time dilation, proper time and spacetime interval

[PAllen]

Hi.

Forget about $d\tau^2$ and use $ds^2$. This is a simple subscription. Using $d\tau^2$ means using co-moving coordinate of the body where the one stays still at its origin of space coordinate and the rest of the world move. I am afraid that co-moving coordinate of the body is variable and not easy to handle in GR. Best.

Not the way I've seen. dτ² is just -1/c² times ds², I.e. just a different metric convention, according to all GR books I own.

 

[sweet springs]

All your books allow imaginary $\tau$ for space interval?

 

[vanhees71]

No, the notion of proper time of course only makes sense for space-like trajectories in spacetime (in both SR and GR).

 

[PAllen]

All your books allow imaginary $\tau$ for space interval?

No, by pure convention, you use proper time differential for timelike or null, and for ds, any type of curve, adjusting the sign as needed. But there is no strong reason for this - it is perfectly common to use time as a unit for distance.

The main point is using proper time says nothing about coordinates in use, or even using coordinates at all. It is perfectly well defined in coordinate free conventions.

 

[vanhees71]

How do you apply it for null trajectories at all, and which physics sense should it make for space-like ones?

 

[PeterDonis]

All your books allow imaginary $\tau$ for space interval?

With the timelike signature convention, yes. With the spacelike convention, where the interval is typically written as $ds^2$, $s$ is imaginary for a timelike interval. If you insist on taking the square root of $ds^2$ or $d\tau^2$, imaginary numbers are unavoidable for one of the two types (spacelike or timelike).

More important, your statement that @PAllen responded to, about $ds^2$ somehow being more general while $d\tau^2$ requires using a "comoving coordinate", is incorrect. Both forms are equally general; it's just a matter of which signature convention you prefer for the metric.

 

[vanhees71]

Well, for the east-coast afficionados it's the very same as for the westcoast ones. Physics doesn't change under conventions. Proper time is defined for time-like trajectories and it's always real variable defined by
$$\mathrm{d} \tau^2 = \pm g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu} > 0.$$
The upper (lower) sign is for west- (east-) coast convention.

 

[PAllen]

How do you apply it for null trajectories at all, and which physics sense should it make for space-like ones?

For null trajectories, it is zero, and the significance is that this defines a null trajectory. For a spacelike interval it just gives distance in units of time, imaginary if you don't flip the sign, else not if you do.

 

[sweet springs]

In the case #26 what kind of clock ticks $\tau$?

 

[PeterDonis]

In the case #26 what kind of clock ticks $\tau$?

The interval we are discussing is always the arc length along a curve. If the interval is timelike, it's arc length along a timelike curve. In post #26, that timelike curve would be the worldline of an observer at rest relative to the gravitating body. That is the case whether you write the interval as $d\tau^2$ or $ds^2$ and whether you use one signature convention or the other.

 

[vanhees71]

Proper time, if used in the proper sense to time-like trajectories, is a clock comoving with the particle on that trajectory. I've no clue, what the difference between $\mathrm{d} \tau^2$ and $\mathrm{d} s^2$ might be (except a factor of $c^2$ when "unnatural units" are used).

 

[sweet springs]

Like a popped up baseball, free moving or falling clock that pass and come back to the same fixed point at its coordinate clock t1 and t2, ticks $\tau$ ,I assume. No?

 

[PeterDonis]

Like a popped up baseball, free moving or falling clock that pass and come back to the same fixed point at its coordinate clock t1 and t2, ticks $\tau$ ,I assume. No?

$\tau$ is arc length along a timelike worldline, as has already been said several times. That should give you the answer to this question.

 

[PeterDonis]

I've no clue, what the difference between $\mathrm{d} \tau^2$ and $\mathrm{d} s^2$ might be (except a factor of c2c2c^2 when "unnatural units" are used).

There isn't one; it's just two different notation conventions.

 

[sweet springs]

τ\tau is arc length along a timelike worldline, as has already been said several times. That should give you the answer to this question.

Hm... Blockhead am I. Coming back to #1 #26 and #27,
under stationary gravitation Let us see two events both at a fixed point,
Event O: Coordinate clock ticks 0 and the clock C ticks 0
Event A: Coordinate clock ticks t and the clock C ticks $\tau$
where $g_{00}c^2t^2 =c^2\tau^2=s^2$. t and $\tau$ are either infinitesimal or finite.

Is there such a clock C? If so what is the motion between or "arc" of clock C?

 

[PeterDonis]

Is there such a clock C?

Of course; it's the clock sitting at the fixed point in space where the two events happen.

what is the motion between or "arc" of clock C?

See above.

The mistake you appear to be making here is thinking of the "coordinate clock", the one ticking $t$ instead of $\tau$, as an actual clock. It isn't. It's just a bookkeeping device, a way of labeling events. The only actual clock is clock C, and it ticks $\tau$, not $t$.

 

[sweet springs]

The mistake you appear to be making here is thinking of the "coordinate clock", the one ticking tt instead of τ\tau, as an actual clock. It isn't. It's just a bookkeeping device, a way of labeling events. The only actual clock is clock C, and it ticks τ\tau, not t.

Yea, you catch my idea fully. I thought both the coordinate clock and the clock C are actual ones. Let me think of your teaching. Thanks now.

 

[haushofer]

The mistake you appear to be making here is thinking of the "coordinate clock", the one ticking $t$ instead of $\tau$, as an actual clock. It isn't. It's just a bookkeeping device, a way of labeling events. The only actual clock is clock C, and it ticks $\tau$, not $t$.

I agree that the coordinate time is a bookkeeping device, but you can stick an observer to it. I'd say the coordinate t is measured as the time difference between the two events by an observer very far away from the black hole where spacetime can be considered to be flat.

 

[vanhees71]

You have
$$\mathrm{d} \tau^2 = g_{\mu \nu} \mathrm{d} x^{\mu} \mathrm{d} x^{\nu}.$$
The coordinate system defines a (local) reference frame, and by definition for an observer at rest $\mathrm{d} x^j=0$ for $j \in \{1,2,3 \}$. This implies that for such an observer the proper time is given by
$$\mathrm{d} \tau = \sqrt{g_{00}} \mathrm{d} x^0.$$
That's the relation between proper time $\tau$ (a physical quantity) and coordinate time (a sheer book-keeping device).

 

[Dale]

I've no clue, what the difference between $\mathrm{d} \tau^2$ and $\mathrm{d} s^2$ might be

Personally, I use $\mathrm{d} \tau^2$ for timelike intervals and $\mathrm{d} s^2$ for all other intervals (spacelike, null, or unknown). That is just a personal convention as far as I know.

 

[Ibix]

I agree that the coordinate time is a bookkeeping device, but you can stick an observer to it. I'd say the coordinate t is measured as the time difference between the two events by an observer very far away from the black hole where spacetime can be considered to be flat.

But that's a special case of coordinates that have a sensible physical interpretation, isn't it? You can replace $t$ with $t'=2t$ without breaking anything, and then there's no direct interpretation of the time coordinate in terms of anything much.

 

[vanhees71]

Yes, and that's why proper time has a direct physical meaning, while coordinates generally don't. It's as in electrodynamics: The four-potential is no direct physical meaning, because it's gauge-dependent. Only the field-strength tensor has direct physical meaning, i.e., is related to measurable quantities which are well defined by an equivalence class of appropriate measurement procedures.

 

[sweet springs]

Re:#47
$\tau$ is larger than t in SR but smaller than t in GR. It is an ambiguous saying but reveals my misunderstanding. Best.

 

[haushofer]

But that's a special case of coordinates that have a sensible physical interpretation, isn't it? You can replace $t$ with $t'=2t$ without breaking anything, and then there's no direct interpretation of the time coordinate in terms of anything much.

I interpret that physically as the same observer using a different time scale.

 

[PeterDonis]

I'd say the coordinate t is measured as the time difference between the two events by an observer very far away from the black hole where spacetime can be considered to be flat.

But such an observer cannot measure the time difference between two events that are not on his worldline. He can only assign such times as a convention, which is what a coordinate chart is.

 

[haushofer]

But such an observer cannot measure the time difference between two events that are not on his worldline. He can only assign such times as a convention, which is what a coordinate chart is.

Yes. That's the whole thing of coordinate time; it's not proper time :P But I'd still call it measuring with his clock.

 

[vanhees71]

Yes. That's the whole thing of coordinate time; it's not proper time :P But I'd still call it measuring with his clock.

But an ideal clock measures precisely its proper time!

 

[Orodruin]

The point is that measurements on a clock only are unambiguous for events on the world line of said clock (as already pointed out). The time coordinates of other events depend on an arbitrary foliation of space-time. Fine, the Schwarzschild coordinates are somewhat special in the sense that its time coordinate is chosen such that the corresponding basis vector field is a time-like Killing vector field showing that the space-time is stationary, but it is not much more than that.

 

[haushofer]

But an ideal clock measures precisely its proper time!

Yes, but coordinate times are not scary diseases, they 're just coordinate dependent. I agree with Orodruin that this particular case is special.

 

[vanhees71]

In any affine space, the parallel postulate holds. You don't need a metric or fundamental form. That's very easy to see. Just fix a point, and you can describe anything with vectors. A straight line is defined then by
$$g: \quad \vec{x}(t)=\vec{v} t+\vec{x}_0,$$
and two straight lines $g_1$ and $g_2$ are called parallel, iff $\vec{v}_1=\vec{v}_2$. Then either they are identical and have all points in common or they don't have any points in common. Nowhere did I need any reference to a scalar product or fundamental form.

If the affine space is two-dimensional then it's also easy to see that to any straight line and a given point not on this line there's exactly one parallel to the straight line going through the given point. Again you don't need any reference to a fundamental form.