# Tossing a clock

1. Aug 12, 2008

### snoopies622

This seems counter-intuitive to me, so I wondering if someone could please confirm or deny it:

I stand on the surface of the Earth holding two identical clocks, each set to the same time. I toss one straight up. When it comes down, it reads a later time than the one that stayed with me because while in flight it traveled on a geodesic, which maximizes proper time, while the one that stayed in my hand did not.

I guess the counter-intuitive part is that in space, a geodesic minimizes distance, but in space-time it maximizes distance. Is this correct?

2. Aug 12, 2008

### Fredrik

Staff Emeritus
Yes, it's correct.

3. Aug 12, 2008

### George Jones

Staff Emeritus
I think (would have to do the calculation to be sure) it's correct in this case, but it's not always true in general relativity!

4. Aug 12, 2008

### Fredrik

Staff Emeritus
Hmm...I think this sort of thing is interesting enough, and non-trivial enough, to be worth discussing in some detail. Right now I don't understand all the details myself. I'm going to have to think about it some more.

Consider this scenario: Suppose that we drill a hole along the rotational axis of a solid spherical planet without atmosphere. (We can imagine that it's a spherical diamond that's slightly smaller than the size that would make the hole collapse). Now we drop clock A down the hole from some altitude, so that it oscillates back and forth along the rotational axis. We also put clock B in a circular orbit so that it will "meet" clock A again when it has completed a full orbit. (It should be possible to adjust the height from which we drop clock A and the height of the orbit of clock B so that this will happen). Finally, we hold clock C stationary over the hole, at the height were A and B will meet.

At the event where the three clocks meet the first time, we set them all to zero. What times will they show when they meet again?

It's interesting that in this case we have two geodesics that connect the same two events. Both of them should have a longer proper time than the non-geodesic path of clock C, so $t_A>t_C$ and $t_B>t_C$, but is $t_A=t_B$ or is one of them bigger than the other?

I don't see the solution immediately. If someone else does, feel free to post it.

Edit: Is it possible that $t_C$ is actually bigger than one of $t_A$ and $t_B$? If the answer is no in this case, are there other situations where there are multiple geodesics connecting the same two events and there are non-geodesic timelike curves connecting the two events that have a longer proper time than some of the geodesics? George, is that what you're saying?

Last edited: Aug 12, 2008
5. Aug 12, 2008

### George Jones

Staff Emeritus
Yes, see

https://www.physicsforums.com/showthread.php?p=861892#post861892

6. Aug 12, 2008

### JesseM

Is my memory correct that there's some sort of variational principle where a timelike geodesic will at least be a local maximum in the sense that infinitesimally "nearby" paths will always have a smaller proper time?

7. Aug 12, 2008

### Fredrik

Staff Emeritus
That's what my memory says too.

8. Aug 13, 2008

### A.T.

I have a question on that scenario too:
Consider A is droped just from the planets surface into the tunnel, and a clock D is resting in the center of the planet. Will both clocks measure the same proper time between their meetings?

L. C. Epstein claims so his book "Relativity Visualized". He provides no math, but his argument goes like this: Spacetime is curved spherically inside the planet (interior Schwarzschild solution?). Both clocks are free falling so they travel on geodesics -> great circles. And since all great circles have the same arc length between their intersection points, the proper time along this world lines is also the same.

In other terms this would mean, that the greater gravitational time dilatation of clock D, and the greater velocity time dilatation of clock A would cancel out each other between the meetings.

Is there a calculation of that scenario somewhere?

9. Aug 13, 2008

### George Jones

Staff Emeritus
Epstein is wrong. I have done the calculation, which involves numerical integration.

10. Aug 13, 2008

### Ich

What about one orbiting the planet at surface level and on falling through?

11. Aug 13, 2008

### A.T.

Thanks. So which clock is faster? The oscillating one or the one resting at the center? And how big is the difference? Since you mention numerical integration, I assume there is no analytical formula. Could outline the computation?

12. Aug 13, 2008

### Fredrik

Staff Emeritus
That's what I had in mind when I described clocks A and B in #4, but I realized that it may not be possible to get the clocks to meet twice unless you adjust the size of the "planet" carefully. It seemed easier to adjust the height from which we drop clock A. (This should be sufficient to guarantee that they meet again, but I said we would adjust the height of the orbit of B as well, just in case).

Why didn't I think of that? That's a great addition to the scenario I suggested. Now the paths of A and D are two geodesics connecting the same two events at the center, and the paths of A and B are two geodesics connecting the same two events at (or above) the surface. I'd like to know which clock measures the longest proper time in both cases.

13. Aug 13, 2008

### George Jones

Staff Emeritus
In order to do a calculation, the metric in the interior needs to be known. I assumed a constant density spherical body, the metric for which Schwarzschild gave at the same as he gave his vacuum metric.

Between meetings, more time elapses for the oscillating clock A than elapses for the central clock D.

I'll try try and outline the computation later today or tomorrow. This involves the the numerical calculation of a definite improper integral. Even thought the integrand diverges at one of the endpoints of the interval of integration, the integral itself is still finite.

Very roughly, $v = dr/dt$ gives

$$t = 4 \int^0_R \frac{dr}{v},$$

where $R$ is the radius of the spherical body, and clock A has $v = 0$ at $r = R$.

An appropriate change of variable removes the singularity, and the numerical integration can be calculated.
I'm scrambing to find my hardcopy of a very relevant 40-year-old reference, about which I found out (accidently) after I had already done my calculation.

14. Aug 13, 2008

### Fredrik

Staff Emeritus
Thanks. Don't work too hard though. I'm just curious. It's not like I have to know the answer.

15. Aug 13, 2008

### snoopies622

So between any two points in spacetime, there might be not just one but many possible geodesics, each with a unique length? If this is the case, is there a name for the longest one? (Likewise for the shortest one in space?)

16. Aug 13, 2008

### George Jones

Staff Emeritus
Right, where length could mean elapsed proper time.

If there are such names, I don't know them.

17. Aug 13, 2008

### DrGreg

Geodesics

I had to think about it for a while, but George's example does seem to prove that, even with a positive-definite metric (i.e. space-only), a geodesic can be longer than a neighbouring curve. To make it even clearer, instead of the North Pole and Greenwich, consider two points just one inch apart, with the North Pole halfway between them. There are two geodesics, one an inch long via the North Pole and the other going via the South Pole.

It is pretty clear that if you replace the Great Circle through the South Pole by a slightly-less-than-great circle through the two points that misses the South Pole by an inch, you will get a slightly smaller circle. So the Great Circle is longer than some neighbouring curves. But you could also choose neighbouring curves that are longer e.g. one that zig-zags around the Great Circle. So the Great Circle neither maximises nor minimises the lengths of neighbouring curves between the same endpoints. But I'm guessing the length is "stationary", in the calculus sense?

Of course, the property of "being geodesic" is not a global property of a whole curve, it is a local property that is valid at every point along the curve. You can chop a geodesic into lots of bits and each bit must be a geodesic in its own right.

18. Aug 13, 2008

### yuiop

In Newtonian physics it is known that the two times are the same, but it would be interesting to now how much they differ in GR terms.

19. Aug 13, 2008

### snoopies622

I was wondering this because for years I thought the definition of "geodesic" was "the shortest path between two points", which of course implies uniqueness. Since then I have learned a different definition -- "a path which parallel transports its tangent vector" -- but I assumed the two were synonymous, with the going-around-the-world-in-the-opposite-direction as a kind of freak case.

I don't have a question here, I just wanted to comment that this non-uniqueness of geodesics (and of their corresponding lengths) is surprising to me and a bit vexing.

20. Aug 13, 2008

### MeJennifer

That is simply not true.

For instance think about how to go from here to the other side of the Earth taking the shortest path (assuming the Earth is a perfect sphere), there are many, in fact an infinite, number of such paths.

21. Aug 13, 2008

### yuiop

Clock tossing solution.

Hi Snoopies,

I think I have found the mathematical solution to the question you have posed.

The total proper time of the tossed clock to go up and come back down again is:

(Eq 1) $$\tau = R_{H}(a +sin(a))\sqrt{\frac{R_{H}}{2m}}$$

where 2m = the Schwarzschild radius and

$$a = cos^{-1} \left(2 \frac{ R_{L}}{R_{H}}-1 \right)$$

R(L) = The low radius that the clock is tossed from.

R(H) = The maximum height the clock gets to (apogee).

The total coordinate time is:

(Eq 2) $$t = (R_{H} + 4m)Qa + 4m LN\left(\frac{Q+tan(a/2)}{Q-tan(a/2)}\right) + R_{H}Q sin(a)$$

where LN is the natural logarithm and

$$Q = \sqrt{(R_{H}/2) -1 }$$

The total local time on the clock that remained at R(L) is simply (Eq 2) adjusted by the gravitational time dilation facter to give:

(Eq 3) $$t = \left((R_{H} + 4m)Qa + 4m LN\left(\frac{Q+tan(a/2)}{Q-tan(a/2)}\right) + R_{H} Q sin(a) \right)\sqrt{1-\frac{2m}{R_{O}}}$$

where R(o) is the radial location of the stationary observer, which in this case is R{L).

It turns out that the proper time that elapses on the clock that stays on the ground according to (Eq 3) is indeed less than the proper time of the tossed clock as per (Eq 1), which agrees with your assumption in the opening post of this thread.

However, if the location R(o) of the stationary observer is higher than the apogee R(H) of the tossed clock, the time according to that observer is greater than the proper time of the tossed clock.

The above equations are my interpretation of the ones given here
Ref: http://www.mathpages.com/rr/s6-04/6-04.htm and I have doubled them to allow for the return journey.

First you have to define "distance". Usually in common terms I think of least distance as the route that takes the least proper time. As DrGreg pointed out there can be many geodesics between two points A and B but it useful to think there is always at least one geodesic that is the shortest possible route (in terms of proper time) between two points for a given initial velocity. In 3 space the route may appear curved but in fact it takes longer to travel the Euclidean "straight" line between A and B.

The above equations show that when the inertial clock travelling on the geodesic returns to the stationary non inertial clock, the free falling inertial clock shows the greater elapsed proper time. I agree that in some ways this is counter intuitive because in Special relativity the clock moving relative to your reference frame experiences less proper time than the clocks that are stationary relatived to you. To me, my intuitive in GR is that the clock that spends the most time in the "fast time zone" higher up experiences the most proper time even when time dilation due to velocity is taken into account, but I am not sure how that would be expressed formally or even if it is strictly true.

Last edited: Aug 13, 2008
22. Aug 13, 2008

### yuiop

What if we take the idea that the initial velocity as a vector uniquely defines the geodisec?

For example if a particle has the required orbital tangential velocity at the equator and going due East then the only geodesic for that particle is along the great circle going East along the Equator. I guess the exception would be if it was going to a point on the opposite side of the world and chose not to stop and go around two or more times.

23. Aug 13, 2008

### Garth

One example of a "tossing a clock" experiment was the 1976 Gravity Probe A rocket borne experiment.

Garth

24. Aug 13, 2008

### snoopies622

Yes of course if there is enough symmetry there can be more than one geodesic, but even in that case (opposite sides of a sphere) the length of each path is the same. In this thread I'm reading that between two points there may be not only more than one geodesic, but each with its own length. And for now that thought makes my head hurt, but I suppose I'll get used to it .

25. Aug 13, 2008

### Fredrik

Staff Emeritus
It's not that strange really. Imagine an ellipsoid instead of a sphere. Suppose that you are at one of the points on the ellipsoid where the curvature is the minimum and you want to follow a geodesic to the opposite side. Now there's a whole interval of possible path lengths. Of course, when I try to picture a time dimension my head starts to hurt too.

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