Curved Space-time and Relative Velocity

In summary, the conversation discusses the concept of relative velocity between two moving points in curved space-time. The argument is that in order to calculate relative velocity, we need to subtract one velocity vector from another at a distance and bring them to a common point through parallel transport. However, the use of different routes in parallel transport can result in different directions of the second vector at the final position, making the concept of relative velocity mathematically unacceptable. The discussion also includes examples of parallel transport on curved surfaces and the potential impact of sharp bends on the calculation of relative velocity. One example involves two static observers in Schwarzschild spacetime, where their relative velocity is found to be different when calculated using parallel transport along different paths. The conclusion is
  • #316
JDoolin said:
It is my "opinion" that the easiest way to determine the observations of Barbara, at any given time, is to apply the Lorentz Transformations until we are looking at the reference frame in which Barbara is at currently at rest. (Then to do further calculation to account for the finite speed of light) It is my opinion that Tom Fontenot's procedure to use the Momentarily Comoving Inertial Reference Frame (MCIRF) to calculate the Current Age of Distant Objects" (CADO), is a better method for describing simultaneity for accelerating observers than the Einstein Convention.
Is it your opinion that that is the only correct way to determine the observations of Barbara at any given time?

Assuming that your answer to the above is that it is not your opinion that it is the only way, then the only potential disagreement we have is here:
JDoolin said:
Another fact is that the Lorentz Transformation already provides a one-to-one mapping of the events.
This may only require clarification and we may actually agree. Are you referring here to a single inertial reference frame, or are you referring to a non-inertial reference frame formed by stitching together Barbara's MCIF's?
 
Last edited:
Physics news on Phys.org
  • #317
DaleSpam said:
Is it your opinion that that is the only correct way to determine the observations of Barbara at any given time?

Let me be explicit.

Pick an arbitrary event on Barbara's worldline. Construct an inverted light-cone down from that event into the past. This lightcone represents the locus of events that Barbara will see at that instant. This lightcone will correctly tell you what events Barbara sees at that instant, but will not tell you the correct distances.

If you want to know the correct distances, the procedure I would follow would be to perform the Lorentz Transformation so that the tangent of Barbara's world-line at that instant is "vertical" and her lines of simultaneity are perpendicular to the tangent world-line. Then the x, y, and z coordinates of the events in the inverted light cone will be the correct distances that Barbara sees.

There may be mathematical methods to produce the same results, so I can't say this is the only way, but if you find another way to determine the observations, it should produce the same results.

Assuming that your answer to the above is that it is not your opinion that it is the only way, then the only potential disagreement we have is here:This may only require clarification and we may actually agree. Are you referring here to a single inertial reference frame, or are you referring to a non-inertial reference frame formed by stitching together Barbara's MCIF's?

Actually, I think possibly, we've been meaning different things by "one-to-one"

When I'm saying the Lorentz Transformation produces a one-to-one mapping of events, I mean for every event in Barbara's current comoving inertial reference frame, the same event happens in every other inertial reference frame.

I think, possibly what you have been calling the one-to-one mapping is: for every event that happens to Barbara (a continuous locus of events forming her curving worldline) there is an event that happens to Alex (a continuous locus of events forming his straight worldline). So by using the radar time, the Einstein convention, you can construct a one-to-one mapping between the time on Barbara's clock and the time on Alex's clock.

Is your main concern, then, mapping time to time, or is your main concern mapping event to event?

Edit: Just to be clear, The MCIRF and CADO (Current Age of Distant Objects) definitely do not create a 1-to-1 mapping of tBarbara to tAlex.
 
Last edited:
  • #318
PAllen said:
Two way speed of light can be objectively defined for one observer. One way speed of light being constant (even for an inertial observer) is an additional assumption that cannot be directly verified, and is one of many equally possible conventions (read all the papers you listed above carefully). Assuming one way speed of light is constant is exactly Einstein's convention. There is no way for one observer to measure one way speed of light. You need two separated observers, with distant clock synchronization established.

Parallax requires distant simultaneity between two separate observers (your eyes or even opposite ends of the Earth are no good for astronomic events). So you are back to a conventions about distant simultaneity to actually measure parallax - which all the papers under discussion here agree is impossible to define objectively.

Trying to be explicit about the operational definitions behind real measurements is exactly what leads to relativity, and to clarifying which parts of it are fundamental features of the universe and which parts are possibly useful conventions. This is also part of what leads to the quantum revolution.

I'm not sure if I'm entirely following your argument.

Yes, I mean to make the assumption that the one-way speed of light is the same as the two-way speed of light. I believe that assumption is implicit in the Lorentz Transformations.

I agree with your point that we do not have the resources to put two comoving cameras in positions far away from each other so that they can get parallax for astronomical distances. But in your earlier post you proposed a 200 million mile meter stick. Instead I would prefer a pair of eyes with perfect (infinite density) resolution. Then I would have no trouble perceiving the parallax of Jupiter.

Obviously in real life, I can't have the infinite resolution pair of eyes. But I can easily put my brain in the center of those eyes, and have the two images processed simultaneously. I have an operational definition of simultaneous; that the events that reached my brain simultaneously are simultaneous* in my brain's frame of reference.

Even if my eyes can't make out parallax at any great distance, the mathematical form for calculating parallax still gives a finite value; a value smaller than the resolution power of my rods and cones, of course, but it is still an unambiguous determination of the distance.

*Edit: (reached my eyes simultaneously)
 
Last edited:
  • #319
JDoolin said:
There may be mathematical methods to produce the same results, so I can't say this is the only way, but if you find another way to determine the observations, it should produce the same results.
OK, we agree here then.

JDoolin said:
Actually, I think possibly, we've been meaning different things by "one-to-one"

When I'm saying the Lorentz Transformation produces a one-to-one mapping of events, I mean for every event in Barbara's current comoving inertial reference frame, the same event happens in every other inertial reference frame.
Yes, we have been meaning different things, thanks for the clarification.

The mapping I am referring to is the mapping from points in the manifold (events in spacetime) to coordinates (ordered 4-tuples of real numbers). For any mathematically valid coordinate system this mapping must be 1-to-1 otherwise the mapping will not be invertible and you will not be able to transform coordinate systems.

When we speak of an observer's "perspective" in relativity we are referring to this mapping between points and numbers. For an inertial observer there is a standard mapping given by the Einstein/radar convention, and it is 1-to-1. For a non-inertial observer there is no standard mapping, and we have to be a little more explicit in defining it. The radar approach gives a strange mapping (bent lines of simultaneity) that is 1-to-1, and the many MCIF's approach gives a less strange (straight lines of simultaneity) mapping that is not 1-to-1.
 
  • #320
DaleSpam said:
OK, we agree here then.

Yes, we have been meaning different things, thanks for the clarification.

The mapping I am referring to is the mapping from points in the manifold (events in spacetime) to coordinates (ordered 4-tuples of real numbers). For any mathematically valid coordinate system this mapping must be 1-to-1 otherwise the mapping will not be invertible and you will not be able to transform coordinate systems.

When we speak of an observer's "perspective" in relativity we are referring to this mapping between points and numbers. For an inertial observer there is a standard mapping given by the Einstein/radar convention, and it is 1-to-1. For a non-inertial observer there is no standard mapping, and we have to be a little more explicit in defining it. The radar approach gives a strange mapping (bent lines of simultaneity) that is 1-to-1, and the many MCIF's approach gives a less strange (straight lines of simultaneity) mapping that is not 1-to-1.

I'm not sure if I fully grasp the idea of a manifold, but are there two other possible one-to-one mappings that we are neglecting? Does a manifold have to be about distant simultaneity, or could it be a manifold mapping mapping causality? In determining "radar time" you use a signal from Barbara to Alice, and a signal back from Alice to Barbara. These two signals both represent unambiguous one-to-one mappings between Alice's world-line and Barbara's world-line. If we extended these out to inverted lightcones from each observer's positions, and allowed the lightcones to stack as the observers progressed forward through time, would these stacks of lightcones form manifolds?

Is there a way we could have a separate manifold for each observer representing distant "observations" instead of distant "simultaneity?" ...or am I misunderstanding the manifol idea?
 
  • #321
Causality; the idea that events from the future cannot affect the past.

Radar simultaneity relies on the mathematical tautology that two one-to-one functions added together will also be one-to-one. This method does not rely in any way on the results of Special Relativity. It only relies only on causality; that the future cannot affect the past.

The MCIRF/CADO method relies on the assumption that the results of special relativity are valid; that the one-way and two-way speeds of light are the same; that the Lorentz Transformations map between intertial reference frames. Within this framework "simultaneity" is already defined as a function of the observer's velocity; a line of constant t, or a line of constant t' in spacetime represents a line of simultaneity.

The two methods have an equal number of assumptions, I think.

(1) They both require the causality requirement to be met.

(2a) Radar time also requires that the definition of simultaneity be one-to-one.

(2b) MCIRF/CADO requires that the Lorentz Transformations apply and uses lines of constant t to define simultaneity.
 
  • #322
JDoolin said:
I'm not sure if I fully grasp the idea of a manifold, but are there two other possible one-to-one mappings that we are neglecting? Does a manifold have to be about distant simultaneity, or could it be a manifold mapping mapping causality? In determining "radar time" you use a signal from Barbara to Alice, and a signal back from Alice to Barbara. These two signals both represent unambiguous one-to-one mappings between Alice's world-line and Barbara's world-line. If we extended these out to inverted lightcones from each observer's positions, and allowed the lightcones to stack as the observers progressed forward through time, would these stacks of lightcones form manifolds?

Is there a way we could have a separate manifold for each observer representing distant "observations" instead of distant "simultaneity?" ...or am I misunderstanding the manifol idea?
A manifold is a very fundamental topological space. A plane, the surface of a sphere, and the surface of a torus are all examples of 2D manifolds. Manifolds, by themselves, don't have any notion of angles, relative velocities, durations, distances, or causality. They are topological spaces rather than geometrical spaces.

In order to add notions of angles, relative velocities, durations, distances, and causality a manifold can be equipped with a metric. The metric defines these coordinate-independent geometrical relationships in the manifold. The purpose of expressing the laws of physics in terms of tensors is to identify these underlying geometrical relationships that do not depend on the coordinate system.

If we wish to use notions like simultaneity or co-location then we need to equip the manifold with a coordinate system. A coordinate system is not necessary for these geometric relationships, but it makes working with the math a lot easier. When you express the metric in terms of your coordinates then you provide the key connection beteween your coordinate system and the underlying geometry.

I hopes this helps clarify the differences between manifolds, metrics, and coordinates.
 
  • #323
(2a) Radar time also requires that the definition of simultaneity be one-to-one.
This isn't a requirement of radar time. It is a requirement of any coordinate system on any manifold. If a mapping isn't 1-to-1 then it isn't a coordinate system by definition.
 
  • #325
  • #326
DaleSpam said:
Of course, if you smooth out the corners then you are taking a different path so you will, in general, get a different result. However, if the path is only changed very slightly and the Christoffel symbols vary only slightly over that change then the final result will differ only slightly. Since a sphere is so symmetric I wouldn't expect a large difference without a large change in the path, but I would have to work it out for myself to be sure.

On rounding off the edges the direction of the parallel transported vector does not change or it changes by a very small amount/insignificant amount.If the corners are sharp there is a significant change in the orientation of the vector notwithstanding the fact that the nature of the surface included does not change in a significant manner.This seems quite peculiar and I would request some clarification from the side of the audience.In case Dalespam has done some calculations he is requested to present them. But this is not mandatory

[This is in relation to the second link in Thread #325]
 
  • #327
Anamitra said:
On rounding off the edges the direction of the parallel transported vector does not change or it changes by a very small amount/insignificant amount.If the corners are sharp there is a significant change in the orientation of the vector notwithstanding the fact that the nature of the surface included does not change in a significant manner.
Do you have the metric for the two spaces in question? I do not.
 
  • #328
The metric relating to the surface does not change at all,so far its form is concerned. We are simply changing the curves --that too slightly --to investigate how the large change in the direction of the parallel-transported vector corresponds to the fact the of surface enclosed has changed by a small amount.The metric does not change in form.
 
  • #329
I don't know the form of the metric for the space you are talking about here. Do you?
 
Last edited:
  • #330
Just think of an ordinary sphere--you don't need to consider a space-time sphere to under the particular aspect of the problem being considered.

Metric:
[tex]{ds}^{2}{=}{r}^{2}{(}{{d}{\theta}}^{2}{+}{{sin}{\theta}}^{2}{{d}{\phi}}^{2}{)}[/tex]

We simply do not have two spaces here as referred to in thread #327
 
  • #331
OK, now I am completely lost. Would you stop referring to posts which refer to other posts and simply post your question in one complete post. In post 325 you referred vaguely back to post 137 where you described two spaces:

Anamitra said:
1)I consider a "Semi-hemispherical spherical" bowl with a flat lower surface[I can have it by slicing a sphere at the 45 degree latitude].A vector is parallel transported along the circular boundary a little above the flat surface[or along the boundary of the flat surface as a second example] . The extent of reorientation of the vector seems to attribute similar characteristics of the surfaces on either side of the curve.How do we explain this?

2) We come to the typical example of moving a vector tangentially from along a meridian,from the equator to the north pole and then bringing it back to the equator along another meridian, by parallel transport and then back to the old point by parallel transporting the vector along the equator. It changes its direction . Now if we make the corners "smooth" it seems intuitively that the vector is not changing its orientation. Even if it changes its orientation it is not going to be by any large amount while the curvature of the included surface remains virtually the same. How does this happen?
Your semi-hemispherical bowl with a flat lower surface and the same space but with smooth corners.

If you are not talking about those two spaces then just be explicit with your complete question in one self-contained post where you describe the issue in detail without referring back to any previous posts.
 
  • #332
Anamitra said:
2) We come to the typical example of moving a vector tangentially from along a meridian,from the equator to the north pole and then bringing it back to the equator along another meridian, by parallel transport and then back to the old point by parallel transporting the vector along the equator. It changes its direction . Now if we make the corners "smooth" it seems intuitively that the vector is not changing its orientation. Even if it changes its orientation it is not going to be by any large amount while the curvature of the included surface remains virtually the same. How does this happen?
Thread #327 clearly refers to the above quoted problem. My subsequent postings/replies are related to the above example.
 

Similar threads

  • Special and General Relativity
Replies
26
Views
853
  • Special and General Relativity
Replies
9
Views
2K
  • Special and General Relativity
2
Replies
35
Views
3K
  • Special and General Relativity
2
Replies
63
Views
3K
  • Special and General Relativity
Replies
3
Views
830
  • Special and General Relativity
Replies
3
Views
768
  • Special and General Relativity
Replies
9
Views
1K
  • Special and General Relativity
Replies
14
Views
1K
  • Special and General Relativity
Replies
10
Views
2K
  • Special and General Relativity
Replies
27
Views
4K
Back
Top