Time Dilation: Is it Counterintuitive?

[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?

 

[Fredrik]

Yes, it's correct.

 

[George Jones]

Yes, it's correct.

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!

 

[Fredrik]

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?

 

[George Jones]

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?

Yes, see

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

 

[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?

 

[Fredrik]

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?

That's what my memory says too.

 

[A.T.]

I have a question on that scenario too:

Consider this scenario: Suppose that we drill a hole along the rotational axis of a solid spherical planet without atmosphere. Now we drop clock A down the hole from some altitude, so that it oscillates back and forth along the rotational axis.

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?

 

[George Jones]

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?

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

 

[Ich]

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

 

[A.T.]

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

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?

 

[Fredrik]

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

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).

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?

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.

 

[George Jones]

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?

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'd like to know which clock measures the longest proper time in both cases.

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.

 

[Fredrik]

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.

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

 

[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?)

 

[George Jones]

So between any two points in spacetime, there might be not just one but many possible geodesics, each with a unique length?

Right, where length could mean elapsed proper time.

If this is the case, is there a name for the longest one? (Likewise for the shortest one in space?)

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

 

[DrGreg]

Geodesics

Yes, see

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

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?

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.

 

[yuiop]

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

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.

 

[snoopies622]

So between any two points in spacetime, there might be not just one but many possible geodesics, each with a unique length?

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.

 

[MeJennifer]

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.

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.

 

[yuiop]

Clock tossing solution.

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.

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.

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

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 traveling 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.

 

[yuiop]

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.

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.

 

[Garth]

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

Garth

 

[snoopies622]

..think about how to go from here to the other side of the Earth..

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 .

 

[Fredrik]

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 .

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.

 

[A.T.]

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

Yes, I think this is the key element. Talking about the shortest possible path in regards to geodesics only confuses people. It's better to talk about a locally straight path. Like adhesive tape sticked to a curved surface without any folds on the tape's edges, The initial direction of taping fully defines the further path.

 

[A.T.]

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.

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

I'm also just curious. But I've been thinking about it(clock in center vs. oscillating clock) for a while already and would be thankful for any references.

 

[atyy]

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.

I think you are remembering a correct statement incompletely. The shortest path on a curved surface in Euclidean space doesn't imply uniqueness (I think). However, there is a unique geodesic between infinitesimally separated points on a curved surface in Euclidean space (looks like this can be seen from the tangent vector definition), but not points that are separated by some finite distance. The Riemannian metric of the surface describes the behaviour of a taut, non-stretchable string on a hard, smooth surface. If I visualize the string, the difference between infinitesimally and finitely separated points seems reasonable.

 

[snoopies622]

..there is a unique geodesic between infinitesimally separated points on a curved surface in Euclidean space..

On an infinitesimal scale the space is flat and we're talking about a straight line, correct?

 

[snoopies622]

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.

Are you saying that if we take the ellipsoid
$$(\frac {x^2}{4})+y^2+z^2=1$$
and want to travel along its surface from (0,1,0) to (0,-1,0), then not only is one of the semi-circles contained in the plane x=0 a geodesic, but so are the (slightly longer) paths adjacent to it? That would make a geodesic not even a local minimum.

 

[Dale]

Talking about the shortest possible path in regards to geodesics only confuses people. It's better to talk about a locally straight path.

IIRC the two are mathematically equivalent (e.g. "the shortest distance between two points is a straight line" axiom from Euclidean geometry).

I think that if you have a starting event and velocity then the "locally straight path" approach is most natural, but if you have a starting and ending event then the "locally maximal interval" is most natural. The concept of a local maximum already implies all of the potential non-uniqueness issues brought up here.

 

[MeJennifer]

Note that a geodesic in spacetime is not the shortest but generally the longest or, more accurately, the extremum, path between two points in spacetime.

 

[atyy]

On an infinitesimal scale the space is flat and we're talking about a straight line, correct?

I don't know the answer. I am going to guess. Someone should help us with the technical bits.

On an infinitesimal-1 scale the space is flat. But the unique geodesic between infinitesimal points is also true for a slightly larger infinitesimal-2 distance over which you can sense curvature (otherwise it would be a pretty useless idea).

But how infinitesimal is infinitesimal? Technically, derivatives of any order can exist at a point, and there are no such things as infinitesimals. But, heuristically, a zeroth order derivative is just the value at the point. A first order derivative compares the difference in values at two points, so it sees infinitesimally further than the zeroth order derivative. A second order derivative compares the difference in the difference between two pairs of points, so it involves at least 3 points (one point shared by both pairs), and sees infinitesimally even further than a first order derivative.

A curved surface are infinitesimally-1 flat in the sense the first order derivatives of distance are the same as a flat surface. But a curved surface is infinitesimally-2 flat in the sense the that second order derivatives of the distance are different from a flat space.

I think this is true because curvature at a point is defined using second derivatives, but can also be thought of it terms of how nearby geodesics deviate.

 

[George Jones]

In this post, I will summarize the results, and the I will gives an explanation of the results in another post.

Consider a spherical planet of uniform density and five clocks (changing notation slightly):

clock A is thrown straight up from the surface and returns to the surface;
clock B is dropped from rest through a tunnel that goes through the centre of the planet;
Clock C remains on the surface;
clock D remains at the centre of the planet;
clock E orbits the body right at the surface.

Assume that A is thrown at the same time that B is dropped, and that the initial velocity of A is such that A and B arrive simultaneously back at the starting point. The times elapsed on the clocks A, B, and C between when they are all are together at the start and when they are all together at the end satisfy $t_A > t_C > t_B$.

Since A and B are freely falling and C is accelerated, it might be expected that $t_A > t_C$ and $t_B > t_C$, so $t_C > t_B$ seems strange.

Assume that clock E is coincident with clocks A, B, and C when A and B start out. As Fredrik has noted, unless the density of the planet has a specific value, E will not be coincident with with A, B, and C when A and B arrive back, but E will be coincident again with C at some other event. The elapsed times between coincidence events of E and C satisfy $T_C > T_E$. Again, since E is freely falling and C is accelerated, this seems strange.

If B is allowed to oscillate, B and D continually meet at the centre of the planet, and, for B, the time between meetings is $t_B/2$, where $t_B$ is as above. In this case, $t_B/2 > t_D$.

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.

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.

Since A, B, and D all follow geodesics in spacetime, it might be thought that $t_D = t_B/2 = t_A/2$, but there is actually no reason to expect this. Elapsed proper time is determined by integrating the metric along worldlines. Since the metric is a tensor field that varies from event to event, there is no reason to expect elapsed proper times to be the same between coincidence events for geosecics that pass through different events between the coincidence events.

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?

That's what my memory says too.

Because of the the possibility of conjugate points, this isn't always true (as DrGreg has demonstrated for a positive-definite spatial case), and I hope to say more about this in another post.

 

[yuiop]

Consider a spherical planet of uniform density and five clocks (changing notation slightly):

clock A is thrown straight up from the surface and returns to the surface;
clock B is dropped from rest through a tunnel that goes through the centre of the planet;
Clock C remains on the surface;
clock D remains at the centre of the planet;
clock E orbits the body right at the surface.
.
.
.
Since A, B, and D all follow geodesics in spacetime, it might be thought that $t_D = t_B/2 = t_A/2$, but there is actually no reason to expect this.

Hi George,

An easy demonstration that the geodesics of particles departing a given point do not have equal proper times when they all rejoin is this:

Arrange the experiment so that A is thrown upwards with an initial velocity such that it returns to the ground at exactly the time clock E completes one orbit. When the proper times of the clocks are compared to clock C that remains on the surface the results are

$$t_A > t_C > t_E$$,

Inertial clocks A and E follow geodesics so they are extremums where the proper time of A is a maximum and proper time of E is a minimum.

I am not sure where that leaves clock B that drops through the tunnel. By definition it is following a geodesic and should be a maximum or a minimum. Maybe the experiment can not be arranged so that B returns at the same time as A and E. I know tunnel clock B returns at the same time as orbiting clock E in Newtonian physics but is that true in GR for any internal density distribution of the planet? It may require use of the interior Schwarzschild solution to get a definite solution and people are not used to or comfortable working with that. Is there a simple axiomatic solution to what happen to tunnel clock B?

I did the math for clocks A, C, and E so I am fairly sure $t_A > t_C > t_E$ is correct.