# Curved Space-time and Relative Velocity

1. Aug 20, 2010

### Anamitra

Is it meaningful to talk of relative velocity between two moving points at a distance in curved space-time? This interesting issue came up in the course of discussion of the thread "Curved Space-time and the Speed of Light".I remember Dalespam giving some formidable logic with a very good example[Thread :#25]The content of the thread #19 was also very interesting (and similar in some sense). I would like the to repeat the basic idea that could prevent relative velocity from becoming a valid concept in the framework of general relativity :
To calculate relative velocity we need to subtract one velocity vector from another at a distance.For this we have to bring the vectors to a common point by parallel transport. We could keep one vector fixed[let us call this the first vector]and parallel transport the other [the second vector]to the position of the first vector. Now parallel transport may be performed along several routes. If different routes lead to different directions of the second vector in the final position, relative velocity does not have a unique meaning and becomes mathematically unacceptable.Again I refer to the illustration in thread #25.

My queries: 1)If two points(or observers) are moving relative to each other in curved space-time how should the motion of one point appear to the other physically? Is such observation meaningless from the physical point of view if we are unable to interpret it in the existing framework of mathematics?
2)On Parallel transport: We start with a familiar example on parallel transport:
A vector e on a globe at point A on the equator is directed to the north along a line of longitude.We parallel transport the vector first along the line of longitude until we reach the north pole N and then (keeping it parallel to itself) drag it along another meridian to the equator.Then (keeping the direction there) subsequently transport it along the equator(moving the vector paralley) until we return to point A. Then we notice that the parallel-transported vector along a closed circuit does not return as the same vector; instead, it has another orientation. "

It is interesting to observe that the route followed in parallel transport [in the above example and the example cited by Dalespam in thread #25]involves sharp bends or joints where derivatives cannot be defined. Incidentally from the mathematical point of view the definition of parallel transport[Wald:page 34] involves derivatives[covariant derivatives to state accurately] and we know very well that derivatives do not exist at sharp junction.Can we entertain the above examples to refute the concept of relative velocity in curved space time, considering the fact that such examples use paths containing sharp junctions?
Interestingly one may perform the above examples by " rounding off" the north pole edge and the ambiguity will be removed.One may draw "smooth(closed) curves" on "curved surfaces" and carry out examples of parallel transport. The vectors will coincide in their initial and final positions![If one is to perform an experiment by drawing smooth curves on a basket ball he should take care to move the vector parallely without bothering about what angle it is making with the curve after the motion has been started.Angles should be noted only at the initial and the final points/stages.]

If the arguments in point (2) are correct I may conclude that
1)The concept of relative velocity is mathematically consistent in relation to the notion of curved space time.
2)The ideas portrayed in the thread "Curved Space-time and the Speed of Light" are correct.

2. Aug 20, 2010

### George Jones

Staff Emeritus
I'm not sure I follow you, so let's consider a simple example. Consider two static observers in Schwarzschild spacetime who both have the $\theta$ and $\phi$ values, but who hover at different values of $r$. Using the method of parallel transport, what is their relative velocity? I think that this is fairly straightforward to compute, at least for geodesic radial paths.

3. Aug 20, 2010

### bcrowell

Staff Emeritus
Last edited by a moderator: Apr 25, 2017
4. Aug 20, 2010

### bcrowell

Staff Emeritus
If I'm understanding you correctly, you're saying that this is an example that confirms, as claimed in DaleSpam's #25, that relative velocities of distant objects are not well defined. A free-falling observer moving along a radial geodesic passes static object A and records its velocity relative to him. He parallel-transports that velocity vector along with him as he continues on his geodesic, and when he passes static object B, he sees that B's velocity relative to him differs from A's velocity relative to him. This is different from the equality of velocities that most people would expect when initially presented with this example, and it's also different from the result you get with parallel transport along other paths. For example, if we go from A to B along an (approximately) elliptical orbit, we'll pick up a difference in direction due to the geodetic effect.

5. Aug 20, 2010

### bcrowell

Staff Emeritus
Right, the presence of kinks in the curves doesn't affect the argument.

6. Aug 20, 2010

### Anamitra

I have been ,for quite some time, trying to explore the possibility of breaking the speed barrier within the "confines of relativity"? Locally we cannot do it. The laws are very strong in this context.The only option would be to explore the matter in a "non-local" consideration. If I am standing at some point in curved space-time and a ray of light is coming from a distant point (close to some dense object)it would be my normal interest to know the speed of light at each and every point as it comes towards me[Of course I continue to stand at the same point]. With this idea in mind I wrote "Curved Space-time and the Speed of Light".

I am repeating the basic aspects of my considerations in the following calculations:
Let us consider two points A and B separated by a large distance with different values of the metric coefficients.Observers at A and B consider a light ray flashing past B.
Speed of light at B as observed from A= {Spatial separation at B}/{sqrt{g(00)} at A.dt}
Speed of light at B as observed from B ie, c ={Spatial separation at B}/{sqrt{g(00)} at B.dt}

[Noting that the speed of light is locally "c"]
Speed of light at B as observed from A= c sqrt{(g(00) at B)/(g(00) at A)}

The left side of the above relation exceeds the speed of light if

g(0,0) at B>g(0,0) at A
[spatial separation is the same for both the observers while the temporal separations[physical] are different---the clocks have different rates at the two points]

It is important to note that general relativity seems to avoid "non-local considerations" [Please do correct me if I am mistaken] and we can always take advantage of this fact to investigate the speed of light at one point as it is observed from another in case, we can find some interesting result.

This exercise does not contradict any law in Special or General Relativity.

Last edited: Aug 20, 2010
7. Aug 20, 2010

### Anamitra

George's reply seems to favor me if I am not incorrect.And I have liked his signature very much.

A relevant point:
If I am standing still at one point V=0. What do I get if I transport a null vector? The nature of the curve will not be a big factor.

8. Aug 20, 2010

### Anamitra

Geodetic effect is a physical effect . It should not be confused with the concept of parallel transport which is an imaginary procedure related to the geometry of the transport.

9. Aug 20, 2010

### bcrowell

Staff Emeritus
The way you calculate the geodetic effect is by using parallel transport. See
http://www.lightandmatter.com/html_books/genrel/ch06/ch06.html#Section6.2 [Broken] , subsection 6.2.5.

Last edited by a moderator: May 4, 2017
10. Aug 20, 2010

### George Jones

Staff Emeritus
Careful; think some more about this.

11. Aug 21, 2010

### Anamitra

An interesting line to quote from the above site:

"The definition of a geodesic is that it parallel-transports its own tangent vector, so the
velocity vector has to stay constant."

If a particle moves solely under the influence of gravity it should follow a geodesic and the above definition(which is common to all texts) settles the issue.

Now we may think of forced motions where forces other than gravity are operating and the bodies are constrained to move along lines that are not geodesics. [We may think of an aircraft following a line of latitude instead of a great circle and another one which moves along a path which is neither a line of latitude or longitude or a great circle]In such cases one may follow the logic suggested in thread#1 ,Point 2, to make relative velocity unique.

Last edited by a moderator: May 4, 2017
12. Aug 21, 2010

### Anamitra

It is important to note that the space-time around the earth is approximately flat and the nature of this space-time should not be confused with the example of the two aircraft I have given or with the spherical shape of the earth. The aircraft example have been given to emphasize that we may have motion along a geodesic (when gravity is the only agent) and we may have a non-geodesic motion if some other force,I mean some inertial force, is operating. If the effect of the inertial force is taken to be similar/equivalent to gravity we may think of adjusting the original metric to take care of the inertial force. At this point I may refer to Thread #8 of the posting "On the Speed of Light Again!" where the equivalence/similarity of gravity and acceleration has been highlighted with reference to papers in the archives of the "Scientific American" and the "Physical Review" as cited by Robert Resnick.
[If the original metric is altered/adjusted the lines of geodesic should change ]

13. Aug 21, 2010

### Staff: Mentor

Even if you restrict parallel transport to only be defined along geodesics it is still not unique. Consider parallel transporting a vector from the north pole to the south pole. Each longitude line is a geodesic and each one will result in a different vector.

Many issues in curved geometry can be worked around or defined away, but the non-uniqueness of parallel transport is something we just have to live with. Distant vectors are in different tangent spaces and cannot be compared.

PS when refering to posts in other threads a link would be helpful.

14. Aug 21, 2010

### Anamitra

Let us consider the definition of parallel transport as we know in general relativity:

A vector is parallely propagated along a curve if its covariant derivative along the curve vanishes at each point.
So if the velocity vector is parallely propagated along a curve --"IT CANNOT CHANGE". Simple as that.

Conclusions:
1) The notion of Relative velocity is consistent with the mathematics of curved spacetime.
2)My assertions in the posting "Curved Space-time and the Speed of Light" are correct.

15. Aug 21, 2010

### Anamitra

Can a space-time surface be exactly spherical?
Let us see.
A particle at the south pole sees several geodesics connecting it to the north pole. Which direction is to follow? It will be in a state of indecision.[We are assuming the presence of gravity only]

We may have several geodesics emerging from the same point.But they should not terminate on the same and the identical point on the other side.

16. Aug 21, 2010

### yuiop

I guess it makes some sort of sense if we define two particles on opposite sides of the "ball" to be infinitely far apart. When two particles are opposite each other like this, there is no preferred direction to move in, so they do not move. This is what we would intuitively expect if the consider the "force of gravity" loosely speaking to be proportional to the inverse of the distance squared, so that the "force" tends to zero when the particles are infinitely far apart. I might well be wrong. I am only just starting to "study" non Euclidean geometry and embedded surfaces.

17. Aug 21, 2010

### Anamitra

Well, the space-time structure referred to in thread #15 was never created by the particles themselves,I mean the particles between whom the relative velocity is to be calculated. The mechanism of creation of the spacetime surface has not been described[and possibly cannot be described] . I have simply assumed its existence to prove that it cannot exist. The particles have been used as "test particles" whose fields are of negligible strength.They should not disrupt the existing field or interact between themselves but they should respond to the existing gravitational field created by some "other means".

Now let us consider a pair of gravitating particles separated by a large distance. The lines of force between them are basically parallel lines and the spacetime structure is "not a sphere". The particles if released from a large distance will move along a straight line. The geodesic is simply a straight line and the space you have described is flat space-time .[the shortest distance being a straight line]. If you kept the two initial particles fixed and released smaller particles midway between them ,they should be moving along straight lines.Of course, the "smaller particles" must be "small enough" not to disrupt the existing field.This in fact would be a better interpretation of the situation.

Now in the first paragraph I have used the term "test particles". It is important that you understand them in relation to the study of gravitational fields/spacetime structure.

I would refer to the book "Gravity" by James B. Hartley , Chapter 8,"Geodesics" in understanding the concept of "Test Particles"

18. Aug 21, 2010

### Staff: Mentor

Correct. The parallel transport of a vector along a curve is unique, but there is not a single unique curve connecting two events.

Neither of these conclusions is correct.

19. Aug 21, 2010

### Anamitra

Let's see:

We consider two points A and B being connected by a pair of curves L1 and L2 lying on the curved space-time surface.We start with a vector V from A and propagate it parallely to B along L1.
V remains constant because its covariant derivative vanishes at each point during the parallel propagation.

Again we start from A with the same vector V and propagate it parallely along L2 to B. V does not change because the covariant derivative of V(as we move along L2)= always zero

In each case we start with the same vector and finish with the same vector.

It is to be seriously noted that constant vectors are not defined by constant components except in rectangular Cartesian systems. Rather the constancy of vectors is defined by the concept of parallel transport[the mathematical concept of parallel transport to state accurately].

Q)If I transport a vector from one point on the curved space time surface to another point,how do I make sure that it has remained constant?

Ans)By the concept of parallel transport. A parallel transport of the vector has to be executed from the first point to the second point.

20. Aug 21, 2010

### Staff: Mentor

I already gave counterexamples. It is useless to make assertions to which there are known counterexamples. Regardless of how compelling you find the chain of logic you know that there is a flaw since the conclusion is demonstrably false.

Can you find the flaw? Specifically, what is the geometric meaning of your phrase "remains constant"?

21. Aug 21, 2010

### Anamitra

The parallel transport of a vector along a curve is unique, but there is not a single unique curve connecting two events.
DaleSpam in thread #18
This statement is indeed correct

It does not in any way contradict the assertions in thread #19

Specifically, what is the geometric meaning of your phrase "remains constant"?
Dalespam in Thread #20

You will find the answer in thread #19
Nevertheless I am repeating:
Q)If I transport a vector from one point on the curved space time surface to another point,how do I make sure that it has remained constant?

Ans)By the concept of parallel transport. A parallel transport of the vector [as defined mathematically]has to be executed from the first point to the second point.

You may perform this action along several curves connecting them.I have talked of the curves L1 and L2 connecting the points A and B in thread #19[but these two curves should not be geodesics at the same time connecting the same pair of points,A and B]

One should not confuse the concept of geodesics and the parallel transport of a vector along a curve[or along number of curves connecting a pair of points]. All good texts give us a clear concept of these two important(but distinct concepts) -------- that is, "geodesics" and "parallel transport".

Conclusions:
1) The notion of Relative velocity is consistent with the mathematics of curved spacetime.
2)My assertions in the attached article in "Curved Space-time and the Speed of Light"[the essence of which may be viewed in thread #6] are correct.

Last edited: Aug 21, 2010
22. Aug 21, 2010

### DrGreg

Anamitra, you don't seem to have understood the following example:
In this case the two dimensional surface of the Earth is a Riemannian manifold. Consider a vector at the North Pole pointing along the 0° meridian.

If you parallel transport it along the 0° meridian (a geodesic) it will at all times make an angle of 0° with the meridian, pointing south, and when you get to the South Pole it will be pointing along the 180° meridian.

If you parallel transport it along the 90°W meridian (another geodesic) it will at all times make an angle of 90° with the meridian. When you get to the Equator it will be pointing east and when you get to the South Pole it will be pointing along the 0° meridian.

This is an example in a 2D Riemannian manifold (space); something similar could happen in a 4D pseudo-Riemannian manifold (spacetime).

23. Aug 21, 2010

### Anamitra

You are getting these results because you are considering a spherical space-time surface which should not exist in practice. I have tried to explain this in thread #15

24. Aug 21, 2010

### DrGreg

The spherical space-only manifold is just an easy-to-understand example. It can happen in other manifolds too.

In spacetime, the geodesic that a particle follows depends on its velocity. There are an infinite number of different geodesics through any single event, corresponding to particles with different velocities. In flat spacetime there is only one geodesic joining a given pair of events, but in curved spacetime there can be more than one. In fact, a good example is lots of different satellites all at the same height following geodesics orbiting a planet. There are lots of ways of orbiting a planet to reach an antipodal point.

25. Aug 21, 2010

### Staff: Mentor

Thanks for the great explanations DrGreg

The Schwarzschild solution has all of the symmetry of a 2-sphere plus a lot of symmetries that the sphere does not have. These types of problems are actually more of an issue in 4D, not less.

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