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

<br /> t = 4 \int^0_R \frac{dr}{v},<br />

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 &gt; t_C &gt; t_B.

Since A and B are freely falling and C is accelerated, it might be expected that t_A &gt; t_C and t_B &gt; t_C, so t_C &gt; 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 &gt; 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 &gt; 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 &gt; t_C &gt; 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 &gt; t_C &gt; t_E is correct.

 

[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 &gt; t_C &gt; (t_B = t_E ) &gt; 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 &gt; t_C &gt; (t_B = t_E ) &gt; 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.