# Two watches, one thrown in the air.

1. Dec 18, 2006

### Wallace

I was looking at this archived thread Two watches, one thrown in the air., that was also mentioned in the 'hard questions' thread recently.

Reading the long and detailed discussion it struck me that there was a simple approach that seemed to have been overlooked.

To briefly describe the situation, two clocks are held on the surface of the earth (or on a constantly accelerating spaceship, the two situations are equivalent). One is thrown up (ahead) and once it has fallen back to the clock that stayed 'stationary' (with respect to the thrower on the earths surface or ship) the question is asked about which, if either, clock has run slower.

I would say, very simply that the thrown clock is following a geo-desic path and hence by definition maximizes the proper time experienced between two points $$(t_1,x_1) \to (t_2,x_2)$$ that the paths of both clocks intersect.

Of course this dosn't give you the quantitative result straight out, but I think it's the best conceptual place to start from.

Last edited: Dec 18, 2006
2. Dec 18, 2006

### JesseM

The point about the thrown watch following a geodesic was actually brought up by pervect in the earlier thread, in post #27. But as he says, the argument is really only airtight if you're talking about a uniform gravitational field in flat spacetime, in curved spacetime a geodesic just has "extremal" proper time, which is usually the maximum proper time but can be the minimum (I don't know in what situation the geodesic would represent minimal proper time though, it might only happen in some unusual choices of spacetimes).

Last edited: Dec 18, 2006
3. Dec 18, 2006

### MeJennifer

My problem, surely due to my lack of understanding, with Carl's explanation is:

The idea of tranfering "exactly half" from an accelerated frame to an intertial frame seems incorrect to me.

4. Dec 18, 2006

### Chris Hillman

Oh, I see, well, disregard position modeled using Newtonian concepts, obviously.

The point of Carl's thought experiment is that held watch is being accelerated (by your hand), while the thrown watch is not (neglecting air resistance). In principle, the thrown watch should therefore be a bit behind when you catch it (this sudden decceleration doesn't affect this miraculous timepiece, since we assume it is an ideal clock.)

Chris Hillman

5. Dec 18, 2006

### JesseM

Shouldn't the thrown watch be ahead, not behind? I thought the point about geodesics was that this situation is analogous to the twin paradox where the twin who follows a geodesic ages more (so his watch is ahead) than the twin who accelerates.

6. Dec 19, 2006

### Wallace

I missed pervect's discussion of geodesic paths, oops .

However I'm not so sure about how limited the validity of this approach is. I may be misunderstanding your statement above, but I don't see why this needs to apply to a uniform gravitational field only. Say the thrown clock goes far enough away from the Earth during its journey that a Newtonian would say the gravitational 'acceleration' is significantly less than $$9.8 ms^{-2}$$, then this scenario is no longer taking place in a 'uniform gravitational field' and my simple reasoning may not apply. Is this what you mean? Apologies if I've misunderstood your statement.

Anyway, assuming this is what you mean, I don't think it changes the outcome, since the accelerated clock remains accelerated at a constant rate, since its position is not changing, and the other clock follows a geodesic. A geodesic is a geodesic is a geodesic regardless of the specifics of the space-time it moves through, it's still going to maximize the proper time experienced relative to the constantly accelerated clock.

At least that's my understanding, if I'm wrong please point out why. It seems no matter how much I think I understand relativity, something always seems to come along to rock the boat

7. Dec 19, 2006

### Chris Hillman

No pun intended

Yes, I misspoke. Let me try again:

The point of the twin paradox and of Carl's watch-juggling paradox is that accelerating an observer is equivalent to bending his world line, but in a Lorentzian manifold, a line (timelike geodesic arc) is the longest curve between sufficiently nearby two points, at least for sufficiently "nearby" competing arcs. So, the bent world line (the world line of the held watch) is shorter than the unbent world line (the world line of the thrown watch). So the elapsed time between toss and catch, as measured by the thrown watch, is longer than this time as measured by the held watch. So, the watch which was thrown is now running a bit ahead of the watch which was held, assuming they were synchronized before beginning the experiment.

Good catch!

8. Dec 19, 2006

### JesseM

I wasn't saying it would only apply to a uniform gravitational field only, I was just saying you need to be careful, because although in almost all normal situations it's true that the geodesic path maximizes the proper time (this is true of geodesic paths in the curved spacetime around the Earth, for example), from what I understand there are occasional situations in GR where the geodesic path would actually minimize the proper time, although I'm not sure what these situations would look like. So a totally rigorous argument would first need to prove that this is one of those situations where the geodesic path does maximize proper time (which, in fact, it does) rather than just assuming it.

9. Dec 19, 2006

### Stingray

FYI, a timelike curve will locally maximize the proper time between two points only if it is a geodesic with no conjugate points along its length. See Wald theorem 9.3.3. There's also no guarantee that the local maximum is a global maximum. But these exceptions would really be very rare and have no relevance for tossing a watch on earth.

10. Dec 19, 2006

### pervect

Staff Emeritus
Working some examples out in this case is probably the most helpful to see where some of the pitfalls are.

Consider two objects - one at rest on the surface of the Earth at the equator, and the other in very low Earth orbit, at an altitude of 0 meters at the equator.

The two objects worldlines will intersect periodically. While the object in LEO is following a geodesic, and the object sitting on the Earth's surface is not a geodesic, it's fairly straightforward to calculate that the object sitting on the surface has a longer elapsed proper time than the object, in spite of the fact that it is not following a geodesic, when they re-unite.

For the orbiting object, dtau = g00 (dt^2 - r^2 d $\theta$^2)

For the object sitting on the Earth's surface, dtau = g00 dt^2

By inspection dtau for the second object is larger.

Note that this does not disagree with Stingray's very interesting observation about the theorem about conjugate points - i.e. we can show that conjugate points exist in this situation. (We need consider only two "nearby" geodesics, i.e. orbits, which intersect, ,such as a circular orbit and a slightly elliptical orbit, to demonstrate that a conjugate point exists via Wald's definition.)

Since the path of longest elapsed time must be a geodesic, and we've calculated that the LEO path isn't the path of longest elapsed time, there must be ANOTHER geodesic which is. The "true" maximum must also be a local maximum.

A little bit of thought shows that it is just the thrown watch (the example is due to Feynman, I believe) that has this longest elapsed time. One has to adjust the height of the watch throw so that it intersects the exact same two points to make a fair comparison. So we have three watches in this case, one orbiting, one standing still, one thrown up. All three re-unite at some point at the same time. The thrown watch reads the longest time of all three. The problem is somewhat degenerate, in that one can throw the watch higher so that it reunites with the others after they make multiple orbits.

(Eliminating the "conjugate points" problem by insisting that the two points be closer in time than an orbital period is another way to guide one's thinking to the correct answer. If you don't allow orbits, the thrown object becomes more apparent as the object following a geodesic that connects the two points at different times on the Earth's surface).

I don't recall Feynman's exact wording of this, but basically he observers that you want to get the watch as far away from the Earth's gravitational field as possible in the shortest time to maximize its elapsed time. But you don't want to move so fast that the SR time dilation due to speed becomes significant. It is this sort of thinking that leads to the picking the thrown watch as the correct solution for the global (and not just local) maximizing of proper time.

There are more interesting (and not necessarily intuitive) examples along this line. For instance, consider a watch at rest at the center of the Earth (assumed to be non-rotating and of constant density). It is following a geodesic. Now, consider a watch bobbing up and down in a tunnel, following another geodesic. Which watch reads longer? Is the assumption of constant density important to this answer?

I've only worked this problem approximately (to the first order) but the answer is very interesting and was initially not expected. I'll try and find the post, but meanwhile since we are/were talking about hard problems (I guess it was actually another thread) it might be an interesting one for people to consider. So I won't give the answer to the problem that I came up with earlier at this time. Perhaps someone will even work out a more exact solution than the one I came up with. (But attacking the problem using anything other than a PPN approximation would be very hard for little reward, IMO).

Last edited: Dec 19, 2006
11. Dec 19, 2006

### MeJennifer

It seems that the problem can be simplified by using the following equivalent situation as described by the following space-time diagram:

Two synchronized watches W1 and W2 having the same straight worldline in flat space-time. At a given event clock W1 changes direction instantly and follows in a straight line while clock W2 changes direction and follows a hyperbolic curve. The wordlines meet at a later event provided that W1's acceleration was not too high in relation to W2's.

Here is it obvious that the proper time for watch W2 is shorter, since an arc is by definition longer than a straight line.

Or am I making a mistake in thinking that this situation is equivalent?

Last edited: Dec 19, 2006