# Are clocks used in general relativity?

1. Mar 29, 2013

### nortonian

In special relativity clocks are used to record events and proper time gives the time in an inertial frame so two times are used. In general relativity I only see proper time being used. Is there an implicit understanding about clocks?

2. Mar 29, 2013

### Passionflower

All clocks always record proper time, unless they are defect.

3. Mar 29, 2013

### Staff: Mentor

GR does not only use proper time. The time parameter in the Schwartzschild metric is not proper time, except in the limit of large distances from the central mass.

4. Mar 29, 2013

### bcrowell

Staff Emeritus
Proper time is the same thing as clock time in both SR and GR.

This is incorrect. Proper time plays the same role in GR that it does in SR. Your example of the time in the Schwarzschild metric (presumably expressed in Schwarzschild coordinates) is an example of a timelike coordinate in GR. Timelike coordinates exist in GR and are not the same thing as proper time.

5. Mar 29, 2013

### Passionflower

Then what is incorrect about Chestermiller's response?

6. Mar 29, 2013

### WannabeNewton

The clocks still measure proper time regardless (as you and Ben and others noted). Coordinate time is irrelevant. In fact coordinate time is present in SR too and is not the same as proper time there as well in general (unless of course we are in the rest frame of the observer whose worldline actually passes through the two events in between which the proper time is being measured) so there is no value in pointing out there are "other kinds of time" in GR as if this was special to GR. Clocks measure proper time - doesn't matter if it's GR or SR.

7. Mar 29, 2013

### Passionflower

I must be missing something, what do you think is wrong about Chestermiller's response?

8. Mar 29, 2013

### WannabeNewton

There is nothing wrong with it as it stands of course but it doesn't really relate to the OP's question. The notion / existence of coordinate time doesn't change the fact that the time measured on an observer's clock between two events in space - time is the proper time.

9. Mar 30, 2013

### nortonian

So in a time dilation experiment we have an atomic clock in an airplane which has a proper time and as it moves it passes through events that use the earth's proper time as a coordinate value in order to measure time dilation. The two times are not the same. So how can you have an event with two time coordinates?

You are saying that you can use either proper time or coordinate time depending on where you use it? Is there a convention to know when?

I realize that proper time is measured between two events, but earth time is a continuum so that shouldn't matter when they are being compared.

Last edited: Mar 30, 2013
10. Mar 30, 2013

### ghwellsjr

Every coordinate system assigns a different time to each event. Just like you can refer to the rover landing on Mars in Eastern, Mountain, Pacific Time or any of many other coordinate systems. If a clock is stationary in an Inertial Reference Frame (IRF) then its Proper Time will tick at the same rate as the Coordinate Time for that IRF. The significance of Coordinate Time is that it ticks everywhere simultaneously whereas individual clocks, say those on earth, even if they tick at the same rate as the Coordinate Time only tick at the location where they are at.
You always use Coordinate Time when specifying the time coordinate of an event in a particular IRF.
You can only use the Proper Time on a real clock to measure the Coordinate Time between two events if those two events are at the same location in the selected IRF.

Another subject is the spacetime interval between two events. If that interval is timelike, then you can use a real clock that is inertial to measure the spacetime interval if it is present at those two events. In other words, in the IRF in which those two events are at the same location, a clock that is also stationary in that IRF and present at the same location will tick at the same rate as the Coordinate Time in that IRF between those two events.

11. Mar 30, 2013

### WannabeNewton

Coordinate time is different from proper time. Lets say we are in the rest frame of an observer $O$ whose worldline passes through two events $A$ and $B$. Furthermore lets say $O$ has set up his frame with coordinates $(t,x,y,z)$. In the rest frame of $O$ he measures no spatial displacement when he goes from $A$ to $B$ of course so $\Delta \tau_{AB} = \Delta t_{AB}$ so the time measured on a clock held by $O$ coincides the coordinate time in the coordinates set up by $O$ with the proper time, between the two events.

Let's say now we have another observer $O'$ who is moving with velocity $v$ in the $x$ direction relative to $O$ and say $O'$ has set up in his frame coordinates $(t',x',y',z')$ (assume they have synchronized their clocks etc.). According to $O'$ the proper time for $O$ to go from $A$ to $B$ will be $\Delta \tau'_{AB} = \sqrt{\Delta t_{AB}'^{2} - \Delta x_{AB}'^2}$.

We know how the coordinates $(t,x,y,z)$ of $O$ will related to the coordinates $(t',x',y',z')$ of $O'$ through the lorentz transformations $t' = \gamma (t - vx), x' = \gamma (x - vt)$. We see that $\Delta t_{AB}'^{2} = \gamma^{2} \Delta t_{AB}^{2}, \Delta x_{AB}' = \gamma ^{2}v^2\Delta t_{AB}^{2}$ so $\Delta \tau'_{AB} = \sqrt{\gamma ^{2}\Delta t_{AB}^{2}(1 - v^2)} = \sqrt{\gamma ^{2}\Delta t_{AB}^{2}\gamma ^{-2}} = \Delta t_{AB} = \Delta \tau_{AB}$. Indeed, both $O$ and $O'$'s measurements yield the same proper time between events $A$ and $B$ even though $O$ has set up different coordinates in his frame from those of $O'$. Proper time is a frame invariant quantity in this sense whereas the coordinate times $t$ and $t'$ are not.

Last edited: Mar 30, 2013
12. Mar 30, 2013

### WannabeNewton

In order to avoid confusion that *might* arise, this is the same spatial location. There might be an ambiguity because "location" in SR can be interpreted as an event.

13. Mar 30, 2013

### ghwellsjr

Well then, they are the same event, aren't they?

14. Mar 30, 2013

### WannabeNewton

How do you mean? There is a temporal displacement between the two events even if there is no spatial displacement (I'm talking about two distinct events, both lying on the image of observer $O$'s wordline and doing measurements in observer $O$'s rest frame).

15. Mar 30, 2013

### ghwellsjr

This, to me, is very confusing. I can't tell if there is just one clock whose Proper Time interval is being "measured" in two different frames or if there are two clocks that both "measure" the same Proper Time interval between a pair of events.

16. Mar 30, 2013

### WannabeNewton

It's the first - a clock passing through the two events. Was the wording ambiguous? Perhaps: the measurements made by $O$ and $O'$ regarding the clock. I'll change it on account of what you said.

17. Mar 30, 2013

### nortonian

Just to make sure I am on the same page, coordinate time refers to the euclidean space tangent to curved space-time, right? and I assume Euclidean space would also be used to describe the stress-energy.

18. Mar 30, 2013

### Passionflower

No, coordinate time can be anything.

Imagine a 2 year old drawing horizontal and vertical wiggly (but not so much as to cross its direct neighbor) but smooth great circles on a football, then that is a perfectly valid coordinate system.

19. Mar 30, 2013

### WannabeNewton

This is getting into the more advanced realm of differential geometry and general relativity (GR) and away from the physics of special relativity (SR) but we can take a crack at it.

There is no a prioi relationship between the coordinates set up by an observer in a neighborhood of space - time and the tangent space to space - time at some point in this neighborhood (by the way this tangent space is not Euclidean space itself but is isomorphic to it in the category of vector spaces).

Let $M$ be a space - time and let $O$ be an observer. On an open subset $U\subseteq M$, observer $O$ can set up coordinates $(t,x^1,x^2,x^3)$ on $U$ so that he can represent events $p\in U$ in terms of his coordinates; this is a way of representing measurements made by $O$ using his measuring apparatus (three perpendicular meter sticks and a clock). The $t:U\rightarrow \mathbb{R}$ coordinate function is the coordinate time. As you can see, in its bare form it has nothing to do with the tangent space to $M$ at $p$ (denoted $T_p(M)$) which is, as its name suggests, the set of all vectors tangent to $M$ at $p$. I can go on to describe how to relate the coordinates set up by this observer to vectors in this tangent space but I fear it might get too abstract and it is not really related to the thread question anyways.

Your final sentence, "and I assume Euclidean space would also be used to describe the stress energy" doesn't make sense to me unfortunately.

20. Mar 31, 2013

### ghwellsjr

If you say that the word "location" can mean "event" and isn't specific enough because it might denote something more than "spatial location" then it seems that something more must be time. So if I say two events are at the same location, then it seems to me you are saying that my statement can be interpreted that the two events (both their spatial and temporal locations) are all the same.