Time Dilation: Is it Counterintuitive?

[yuiop]

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?

Hi again George,

I reread your last post and note you have already answered my question about the effect of the density distribution in the negative.

If we assume the perfect required density distribution and a careful setup such that clocks A,B and E all return to the location of surface clock C at the same time, then my best guess for the elapsed proper time intervals of the clocks is:

$$t_A > t_C > (t_B = t_E ) > 2t_D$$


I am assuming here that the geodesic of clock B is another mimimal proper time extremum (along with clock A) relative to location C and at the same time a maximum proper time extremum relative to location D. Clock D would be appear to be its own miminimal proper time extremum for location D. Does that sound reasonable?

 

[DrGreg]

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.

When you are specifying a geodesic, the zeroth derivative tells you where it is, and the first derivative tells you which direction it is pointing. You are free to specify whatever you like for these quantities -- they are the "initial conditions". Once you have done so, the geodesic is completely determined. The second derivative is forced upon you.

In flat space, the second derivative must be zero. In curved space, there is a difference between "coordinate derivative" and "covariant derivative". The coordinate derivative includes within it the curvature of spacetime but the covariant derivative (along a curve) cancels out the spacetime curvature leaving only the curvature of the curve itself "relative to spacetime" so to speak, using the "parallel transport" concept. It is the second covariant derivative that has to be zero.

It's not all that helpful to think of a geodesic as being the shortest or longest route between two points. Better to think of it as being a "straight" line, that is, the closest to straight you can get in a curved spacetime. (Indeed you can take geodesics as defining what "straight" means.) When you zoom in and look at a small section where spacetime curvature is negligible, it will indeed be the shortest distance/longest time between any two close points.

(WARNING: I'm a novice in this subject so I can't guarantee I'm always correct.)

 

[MeJennifer]

When you zoom in and look at a small section where spacetime curvature is negligible, it will indeed be the shortest distance/longest time between any two close points.

You must mean the longest distance/longest time right?

 

[George Jones]

If we assume the perfect required density distribution and a careful setup such that clocks A,B and E all return to the location of surface clock C at the same time, then my best guess for the elapsed proper time intervals of the clocks is:

$$t_A > t_C > (t_B = t_E ) > 2t_D$$

My guess is that, unlike for Newtonian gravity, $t_B \ne \t_E$, but I would have to do the calculation to be sure.

I am assuming here that the geodesic of clock B is another mimimal proper time extremum (along with clock A)

Minimum over what set? Certainly not over timelike worldlines, as the elapsed time can be made arbitrarily small.

Over timelike geodesics? Without thinking some more, I'm not sure how to approach proving this.

 

[Fredrik]

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?

Yes, that's what I meant.

That would make a geodesic not even a local minimum.

Hmm...that's weird. I don't even think about that, but it looks like you're right. Anyone else have a comment about this?

It's interesting that if we stop before we get all the way to the other side, e.g. if we go from (0,1,0) along the shortest geodesic towards (0,-1,0) but choose the endpoint of the curve to be, say at (0,0,1), we get a curve that locally minimizes the length between those two points. If we go in the other direction (passing through (0,0,-1) to the same endpoint, we get a curve between the same two points, which I think is a local minimum when the ellipsoid is shaped like this, but would have been a local maximum if the ellipsoid hadn't been elongated enough in the x direction. I wonder if there's a shape of the ellipsoid that makes the path neither a local minimum nor a local maximum.

We would get the result I described above even if we had started out from (0,1,0) in a slightly different direction. I think the reason why the "slightly longer" geodesics you considered are neither local maxima nor local minima is that you picked the endpoints so that one of the endpoints is a conjugate point to the other. (See the post George linked to for a short introduction to conjugate points).

 

[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?)

Wait, I just thought of a name: the "distance" between the two points.

 

[Mentz114]

Yes, that's what I meant.

Hmm...that's weird. I don't even think about that, but it looks like you're right. Anyone else have a comment about this?

It's interesting that if we stop before we get all the way to the other side, e.g. if we go from (0,1,0) along the shortest geodesic towards (0,-1,0) but choose the endpoint of the curve to be, say at (0,0,1), we get a curve that locally minimizes the length between those two points. If we go in the other direction (passing through (0,0,-1) to the same endpoint, we get a curve between the same two points, which I think is a local minimum when the ellipsoid is shaped like this, but would have been a local maximum if the ellipsoid hadn't been elongated enough in the x direction. I wonder if there's a shape of the ellipsoid that makes the path neither a local minimum nor a local maximum.

We would get the result I described above even if we had started out from (0,1,0) in a slightly different direction. I think the reason why the "slightly longer" geodesics you considered are neither local maxima nor local minima is that you picked the endpoints so that one of the endpoints is a conjugate point to the other. (See the post George linked to for a short introduction to conjugate points).

Are you working in a Euclidean space ? You will get unique geodesics if you work in 4D in a pseudo-Riemannian space ( i.e. signature -+++ or +--- ). In this space you can't choose your endpoints without considering the light-speed limitation.

 

[MeJennifer]

You will get unique geodesics if you work in 4D in a pseudo-Riemannian space ( i.e. signature -+++ or +--- ). In this space you can't choose your endpoints without considering the light-speed limitation.

That is not true as spacetime has timelike, null, and spacelike geodesics. But obviously no particle can travel on a spacelike geodesic.

 

[Mentz114]

MeJennifer,

That is not true as spacetime has timelike, null, and spacelike geodesics. But obviously no particle can travel on a spacelike geodesic.

I don't understand your objection. Isn't that what I said with 'In this space you can't choose your endpoints without considering the light-speed limitation.' ?

M

 

[yuiop]

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.

Hi Snoopies,

I am not sure what you are getting here, but the "adjacent" geodesic that starts out at slightly different angle from (0,1,0) does not arrive at (0,-1,0) in one revolution but spirals outwards around the x-axis and then returns sometimes along a very much longer spatial path. See the attached image below that was generated by this nice http://www.vismath.de/vgp/content/curve/PaSurfCurve.html" that plots manifolds and geodesics in 3D that can be rotated etc in real time.

If we take a cross section of the ellipsoid that is slightly offset from the circular cross section in the x=0 plane, the path that connects (0,1,0) and (0,-1,0) is an ellipse, but it is not a geodesic (apparently it is a "shadowline"). Here I am using the definition by DrGreg and others that a geodesic is the straightest line locally and can be generating by parallel transport of a tangent vector (that you mentioned earlier) or by the practical method of pulling a string taut between the two end points on a convex surface or running some sticky tape along the surface without creasing it. None of these methods produce a path with an elliptical cross section that connect (0,1,0) with (0,-1,0) with the exception of the non local path in the z=0 plane that goes via (2,0,0) or (-2,0,0). It is possible to connect the two furthest points on the y-axis with a spiralling geodesic that is much longer than the elliptical path but maybe that does not count as a geodesic due to conjugate points?

I think part of the problem in this thread is that we talk about "distance" without defining it in the context, sometime meaning distance in 3 space and sometimes in 4 space or in temporal terms of proper time, local time or even sometimes in coordinate time. Local time varies with the location of the observer and can be different for different end points of the geodesic. The manifolds also differ. Some like the sphere are presumably a surface of constant gravitational potential, while others that look like embedding diagrams have varying gravitational potential. In the latter case presumably the speed or energy of the particle is constant? Spatial distance also varies depending upon whether it measured by local non inertial observers and were they are located or if it is measured by the observer traveling between the two points. Ruler distances and radar distances differ too. So what does "distance" mean? Maybe the spacetime interval would be the most consistent definition of distance but that has its problems too. What is the distance between two points located on the photon sphere of a black hole? The spacetime interval of a photon traveling between those two points is always zero for a photon whichever route it takes.

Say a clock is dropped from a given height above a planet and we plot its local velocity against displacement from the centre of the planet and then rotate the plotted curve to produce something that looks like an embedding diagram. What sort of manifold would that be? Neither the local velocity or the gravitational potential would be constant in this case. (What, if anything, would be constant on such a surface?)

Anyway, in Euclidean geometry the shortest distance between two points is a straight line and the reverse is true. If you can show that a path between two end points is a straight line then that path is shortest possible route between those two points. It seems the same is not true for geodesics. The shortest coordinate spatial distance between two points is always a geodesic but the reverse is not true. If the path between two end points is a geodesic it is not necessarily the shortest coordinate spatial distance between those two end points.

Jennifer has defined a geodesic as the longest distance between two points. I assume she is talking about the longest proper distance as measured by the traveling particle? A non inertial traveller could measure the proper distance to be shorter, but then again a non inertial traveller going slowly via a meandering route could (I think) measure the proper distance to be longer than that measured by the inertial traveller too? The slow meandering traveller could also measure a longer proper time than the inertial traveller that takes the shortest possible geodesic. I guess it only makes sense if we compare ratios of time, in other words the "shortest" path always has the greatest ratio of proper time to coordinate time and that path would always be a geodesic.

 

[DrGreg]

You must mean the longest distance/longest time right?

For a timelike geodesic, the longest proper time.

For a spacelike geodesic, the shortest proper distance.

For a null geodesic, both proper time and proper distance are zero; I'm not clear what happens in this case.

I am not sure what you are getting here, but the "adjacent" geodesic that starts out at slightly different angle from (0,1,0) does not arrive at (0,-1,0) in one revolution but spirals outwards around the x-axis and then returns sometimes along a very much longer spatial path.

But we're not considering adjacent geodesics, we're comparing our geodesic against nearby non-geodesic curves.

So what does "distance" mean? Maybe the spacetime interval would be the most consistent definition of distance but that has its problems too. What is the distance between two points located on the photon sphere of a black hole? The spacetime interval of a photon traveling between those two points is always zero for a photon whichever route it takes.

A geodesic always (locally) minimises/maximises the spacetime interval relative to other non-geodesic curves between the same nearby endpoints.

__________________

I think (but I'm not sure) I may have come up with an example in 2D space embedded in 3D Euclidean space (the only kind of curved space that's easy to picture) where the shortest distance is not a geodesic.

Consider a flat horizontal 2D plane with a hole in it. I don't mean a hole that has been cut out and removed, I mean a depression, or "well" if you like, where there is a still a continuous surface but it drops below ground level. Let's assume the hole has a circular horizontal cross section at all levels, so it will have a sort of smooth U-shaped vertical cross-section.

Now consider two points each on the flat part of the surface on diametrically opposite sides of the hole. It ought to be clear that the line in the surface straight through the middle of the hole is a geodesic, by symmetry. (This line would look like a straight line looking down on it vertically from above.) However, providing the hole is a lot deeper than it is wide, there will be a shorter, but non-geodesic, route within the flat part taking a detour around the hole.

There might, conceivably be another shorter route that dropped partly down one side of the hole (a bit like a light ray refracting through a lens), but I think we could arrange the geometry (maybe distort it from being circular?) such that this other route was still not a geodesic. However I'm not 100% convinced by this last paragraph yet, so I might be wrong.

 

[MeJennifer]

For a spacelike geodesic, the shortest proper distance.

Ok, as long as everybody understand it is the largest positive number for a timelike geodesic and the largest negative number for a spacelike geodesic.

 

[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.

If we take a cross section of the ellipsoid that is slightly offset from the circular cross section in the x=0 plane, the path that connects (0,1,0) and (0,-1,0) is an ellipse, but it is not a geodesic...

That's what I was confused about. I took "interval of path lengths" to mean that they were all geodesics. At this point I think I'm best off forgetting about any reliable connection between geodesics and extremes of path length, except on the very small scale where one assumes the manifold is flat and the problem becomes trivial.

 

[Fredrik]

But we're not considering adjacent geodesics, we're comparing our geodesic against nearby non-geodesic curves.

Actually we were talking about adjacent geodesics. I introduced some confusion into this thread by incorrectly assuming that in this specific case, the adjacent (in the sense that they pass through the same point, but in a slightly different direction) geodesics would pass through the antipodal point on the ellipsoid. If that had been true, we would have had a whole bunch of geodesics connecting the same two points, but with very different lengths. It seemed weird to me when I suggested it, but I didn't immediately see what was wrong with it, so I waited for someone else to find the flaw, and after a while kev did just that. (Thanks kev).

 

[Dale]

When you talk about a function value being an extremum (minimum or maximum) you don't talk about nearby or adjacent extrema, you talk about nearby points, which differ from the point of interested by some infinitesimal amount. If the function has a slope of 0 at that point then the nearby point may also be an extremum, but in general the most "nearby" extremum may be separated by a finite distance.

Similarly when you are talking about functionals (such as path length) and functions (the path) that extremize the functional. A nearby path is one that differs from the path under consideration by an infinitesimal amount. If two such nearby paths both extremize the functional then they must have the same value for the length. But if two extreme paths are not nearby then they may have different lengths, corresponding to different local extrema.

I am no GR expert, but my understanding is that geodesics are extrema in this sense.

 

[Happy Jack]

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

What about gravity, as the stationary clock D has no gravity, where the osccilating clock ranges from 1g to 0g to 1g? In orbit, the clocks speed up (relative to the surface) as gravity drops, and slow down as the speed increases.

Has this been tested, the effect of decread gravity on a clock in a deep hole/mine?

 

[yuiop]

...

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 A is thrown up as E passes C so that A returns to C at the same time as E returns.

These are the results of some calculations I have done for the proper times of the clocks where R is the Radius of the planet and H is the Height that clock A reaches at its apogee when thrown upwards. F is the coordinate time according to a clock at infinity. I have not calculated the proper time for clock B oscillating inside the tunnel as that will require a real expert to figure out ;)


$$\begin{tabular}{l l l l l l l} R : 2.13m & 2.80m & 3.21m & 4.93m & 45.23m & 448.85m & 44,847.79m\\ H : 3m & 5m & 6m &10m & 100m & 1,000m & 100,000m\\ \hline A : 7.52 & 19.32 & 25.95 & 57.79 & 1882.18 & 59658.94 & 59673959.04 \\ C : 4.81 & 15.71 & 22.11 & 53.07 & 1868.45 & 59615.90 & 59673528.99 \\ D : -2.52 & 8.87 & 12.82 & 45.19 & 1847.07 & 59549.27 & 59672863.68 \\ E : (12.48) & (7.90) & 9.13 & 43.08 & 1846.72 & 59549.16 & 59672863.67 \\ F : 19.52 & 29.41 & 36.06 & 68.83 & 1911.19 & 59749.16 & 59674859.61\\ \end{tabular}$$

Note that tossed clock A consistently records the longest proper time period than any of the other clocks, (except for the coordinate clock at infinity of course.) Orbiting clock E consistently records the least proper time. (The figures in brackets for clock E should not be compared as they are virtual periods because there are no real circular orbits at radii less than the photon orbit radius at 3m.). Centre clock D always records less proper time than stationary clock C at the surface as would be expected but it is somewhat surprising to me that also generally records more proper time than orbiting clock E. The proper time for clock D is calculated using the interior Schwarzschild solution. Generally the interior solution requires knowledge of the density distribution of the mass of the planet but in the case of a stationary clock exactly at the centre, the enclosed mass is zero and the density distribution becomes irrelevant.

The next table compares the coordinate spatial distances for the paths of the various clocks. Clocks C, D and F have not been included as they are always stationary.

$$\begin{tabular}{l l l l l l l} R : 2.13m & 2.80m & 3.21m & 4.93m & 45.23m & 448.85m & 44,847.79m\\ H : 3m & 5m & 6m &10m & 100m & 1,000m & 100,000m\\ \hline A : 1.74 & 4.40 & 5.59 & 10.14 & 109.54 & 1102.30 & 110304.42\\ B : 8.52 &11.19 & 12.82 & 19.73 & 180.92 & 1795.40 & 179391.15\\ E : 13.38 & 17.58 & 20.14 & 30.99 & 284.18 & 2820.21 & 281786.96\\ \end{tabular}$$

Now it can be seen that clock A not only consistently has the greatest proper time but it also has the shortest spatial path distance and orbiting clock E that always has the least proper time also always has the longest path distance.

The geodesics of clocks A and E are the only ones that are directly comparable as they are the only geodesics that pass directly through the same point C. The clock at C is not on a geodesic as it is not inertial. The path of clock B is a geodesic that passes through point C but it remains to be determined what its proper time period is and whether there is a density distribution that allows it to return to point C at an interval that coincides with clocks A and E. An experiment could be set up to send clock A up to an altitude such that it returns to point C at the same time as clock B and clock E could be sent on an approximately elliptical orbit that has its short axis of the same length of the tunnel, but the maths would be very complicated and the fact that orbit of clock E would be precessing would add further complication.

Note: The negative proper time for clock D when R=2.13 is not a typo. The proper time of a clock at the centre of a gravitational mass is negative always negative for radii less than 2.25m or less than $$\frac{9}{8} R_{s}$$

 

[Stoonroon]

A reference that may clarify and tie loose ends in this thread is a 1961 paper by N. W. Taylor, 'Note on the Harmonic Oscillator in General Relativity,' in the Journal of the Australian Mathematical Society, vol. 2.

Google: journals.cambridge.org/abstract_S1446788700026677

Taylor's calculation refers to the difference in elapsed time between a clock at the center (clock D, in the above discussion) and a radially oscillating clock (clock B, in the above discussion). [These designations correspond to Dr. Jones' "revised" notation used in the latter half of this thread.]

The calculation is based on an approximation of the Schwarzschild interior solution for a uniformly dense sphere. Except for a difference of order v^4/c^4, the result (which corroborates an earlier analysis by O. Bergmann) is that clocks B and D show the same elapsed times when clock B is at the center and at the limits of its path.

It is not clear to me how this difference would be affected by a calculation based on the exact interior solution. In any case, it shows that Epstein's intuitively deduced result is at least very nearly correct. It would be interesting to hear back from Dr. Jones as to the magnitude of the time difference that his numerical integration yielded for this problem.

On the basis of general relativity, the agreement between clocks D and E – the latter being in a grazing circular orbit -- can be easily seen for the following reason. The frequency of a clock at the center (D) is supposed to be slower than a clock at infinity by the square root of the temporal coefficient (1 – 3GM/rc^2) from the Schwarzschild interior solution. And the frequency of the orbiting clock is slower by the square root of (1 – 2GM/rc^2 – GM/rc^2). These coefficients are obviously equal. In the latter coefficient the middle term represents the effect of gravity, while the right hand term represents the effect of the orbiting velocity.

Interesting as this all may be theoretically, gaps in the empirical evidence must be pointed out. We have ample evidence regarding clocks E (orbiting) and C (on the surface). But we have no evidence for clocks A, B, and D. Garth has mentioned Gravity-Probe A (the Vessot-Levine experiment) which surely involved a tossed clock. But the elapsed time on this clock was never measured; only its apparent frequency was deduced from a distance. So we do not have direct empirical evidence bearing on the time given by clock A (tossed straight up).

Furthermore, nobody knows for certain that a clock at the center ticks at the predicted rate. The times that would be given by clocks B and D are quite unknown. We cannot even claim to know for certain whether a clock dropped into the hole oscillates according to the textbook prediction. As scientists, we cannot presume to know that the book answer agrees with Nature's answer until we check.

It would be possible to find out with a laboratory experiment (using, e.g., a modified Cavendish balance).

http://www.ptep-online.com/index_files/2011/PP-24-03.PDF

If we would at least confirm that the oscillation prediction is correct, then this could be adduced as indirect evidence of the clock rate prediction. As it presently stands, the interior solution predictions regarding clock rate and simple harmonic motion have no empirical support; they are only mathematical extrapolations.