Curved Space-time and Relative Velocity

[Anamitra]

I would request DaleSpam to address the following thread before the consideration of the chain of inertial states:https://www.physicsforums.com/showpost.php?p=2869524&postcount=128

The thread mentioned below has also not been addressed so far:
https://www.physicsforums.com/showpost.php?p=2862951&postcount=108

 

[Dale]

We consider a person at rest in four dimension.
Velocity Components
u(x)=0,u(y)=0 u(z)=0
[u(t)]^2[1-/2m/r]=1

If we are to transport this vector along a line of latitude or longitude the components do no change at any point.Attachment in thread #114
[https://www.physicsforums.com/showpost.php?p=2866003&postcount=114]

Given what we have worked out and given the definition of parallel transport, would you expect that this result is a general result for all paths in all curved spacetimes or a specific result for a particular class of paths in the Schwarzschild metric?

Hint: the non-zero Christoffel symbols for the t coordinate are $$\Gamma^t_{tr}$$ $$\Gamma^t_{rt}$$ and $$\Gamma^r_{tt}$$ so what kinds of paths might you expect to cause the timelike component to change?

 

[Anamitra]

It is true that certain specific curves were taken to illustrate the case of transport of the null vector. We could have other curves and other types of space-time surfaces.

Issues at hand:

1) The fundamental issue of the existence of relative motion:If I see a car passing by at a distance[of course in curved space-time],what do I understand,considering the fact that Dalespam considers relative motion an illogical issue? Can observation in reality be suppressed by the incapability of the mathematical apparatus,in case such an incapability exists for the sake of argument?

2)The chain of inertial states

3) The motion of a satellite is due to the curvature of space-time.Purely flat space-time cannot produce such motion.We have an estimate of such relative-motion both in theory and in experiment.Relative motion in curved space-time is an established fact.

4)Relative motion may be calculated in many instances as Dalespam has admitted in the previous thread . There are a great many cases when the null vector may be transported without any change.

 

[Dale]

It is true that certain specific curves were taken to illustrate the case of transport of the null vector. We could have other curves and other types of space-time surfaces.

Specifically for the Schwarzschild metric, if we look at the equation for parallel transport we find

$$\frac{dA^t}{d\tau}+\Gamma^t_{rt}\frac{dx^r}{d\tau}A^t+\Gamma^t_{tr}\frac{dx^t}{d\tau}A^r=0$$

So even if the only non-zero component of A is the time component we still have that A changes if the path goes in the radial direction. In our examples we fixed r and t so these terms dropped out.

In general spacetimes any and all of the Christoffel symbols may be non-zero.

2)The chain of inertial states

OK, let's deal with this one next.

In your previous work you said the following regarding your chain of inertial states idea:

2) A infinitesimal rotation which may be expressed by a matrix consisting of the Eulerian angles.
We leave the translation unchanged but we apply the inverse transformation to cancel the effect of rotation.

So if we consider two separate paths we will want to apply this process along each path and then compare the two resulting vectors to see if they are the same. In order to do so we will need to transform back into the initial coordinates, and to do that we will have to apply all of these little infinitesimal rotations.

Recall that rotations are not transitive, that is, applying the same rotations in a different order will give different results, let alone applying a different series of rotations. This should immediately cause you to suspect that the result of the above process will be different for the two different paths, but we would like to quantify that. In order to do this chain of rotations how do we determine the amount and direction to rotate at each point?

The proper way to do this is to start at your initial point and construct an orthonormal set of basis vectors. This gives us your locally inertial coordinates (Riemann normal coordinates) and in these coordinates at the initial point the metric is the Minkowski metric and the Christoffel symbols all vanish. Your starting vector is some unique linear combination of these basis vectors.

We then parallel transport the basis vectors along the two paths. At each point along both paths the basis vectors can be used to make a locally inertial coordinate system and, because parallel transport preserves the dot product, your starting vector is the same unique linear combination of these basis vectors. I.e. as you claimed the components of the starting vector do not change in the chain of inertial frames.

Now, we have already established that, in general, parallel transport is path dependent. This applies for the basis vectors as well. So, the timelike basis vector transported along one path will be different from the timelike basis vector transported along the other path. Similarly with each of the three spacelike basis vectors. Therefore any linear combination of the basis vectors will also be different depending on which path was taken.

I know that is a lot to digest, and I skimmed rather rapidly, so feel free to ask for further details of any step that is unclear.

 

[Naty1]

Bravo! Dalespam for virtually infinite patience...just wanted to let you know I learned a LOT from your descriptions of curvature... and also the comments of Dr Greg much earlier and Ben crowell and George...I read all of both threads...except for the detailed math...and am glad I never attempted to learn all the detailed math on my own as time is limited...much better for my purposes to understand how experts interpret the math subtles...

While I think I understood the general concept of parallel transport of vectors and different paths making velocity comparisons in curved space at different postions ambiguous I had never seen so many examples mentioned in earlier these two thread discussions...those crystallized the concepts nicely...

Thanks for your efforts...

 

[Dale]

Thanks Naty1, a post like that is very encouraging for me, and a good reminder that such a thread may be useful to other people besides the main participants. I appreciate it a lot.

 

[Anamitra]

[Thanks for waiting.]

OK, let's deal with this one next.

In your previous work you said the following regarding your chain of inertial states idea:So if we consider two separate paths we will want to apply this process along each path and then compare the two resulting vectors to see if they are the same. In order to do so we will need to transform back into the initial coordinates, and to do that we will have to apply all of these little infinitesimal rotations.

Recall that rotations are not transitive, that is, applying the same rotations in a different order will give different results, let alone applying a different series of rotations. This should immediately cause you to suspect that the result of the above process will be different for the two different paths, but we would like to quantify that. In order to do this chain of rotations how do we determine the amount and direction to rotate at each point?

The proper way to do this is to start at your initial point and construct an orthonormal set of basis vectors. This gives us your locally inertial coordinates (Riemann normal coordinates) and in these coordinates at the initial point the metric is the Minkowski metric and the Christoffel symbols all vanish. Your starting vector is some unique linear combination of these basis vectors.

We then parallel transport the basis vectors along the two paths. At each point along both paths the basis vectors can be used to make a locally inertial coordinate system and, because parallel transport preserves the dot product, your starting vector is the same unique linear combination of these basis vectors. I.e. as you claimed the components of the starting vector do not change in the chain of inertial frames.

Now, we have already established that, in general, parallel transport is path dependent. This applies for the basis vectors as well. So, the timelike basis vector transported along one path will be different from the timelike basis vector transported along the other path. Similarly with each of the three spacelike basis vectors. Therefore any linear combination of the basis vectors will also be different depending on which path was taken.

I know that is a lot to digest, and I skimmed rather rapidly, so feel free to ask for further details of any step that is unclear.

It is clear that DaleSpam has entertained some serious misconceptions in his logic when he considers the "parallel transportation" of the basis vectors for the generation of the chain of inertial frames.
1)We consider as an instance the
$${e_{r}}{,}{e_{\theta}}{,}{e_{\phi}}$$
system as we move along a line of latitude,say for example the 45 degrees latitude.We do not parallel transport the basis vectors to obtain the chain of orthonormal reference frames as we move from point to point on the line of latitude.It is not at all necessary to do so in order to produce the chain of inertial reference frames.Such an action is not incumbent on us from any consideration whatsoever.
Movement from one frame to the next involves:
a)An infinitesimal translation.
b)Three infinitesimal rotations,involving the Eulerian angles.
We reverse the rotations keeping the translation intact.
2)Regarding Rotations:Finite rotations cannot be treated as vectors. But infinitesimally small rotations can be treated as vectors.

We may write:

$${d}{\theta}_{1}{+}{d}{\theta}_{2}{=}{d}{\theta}_{2}{+}{d}{\theta}_{1}$$
$${d}{\theta}$$ on either side is being treated as a vector.
No problem with that!

Interesting Point to Note:

We may write:

$${d}{\theta}_{1}{+}{d}{\theta}_{2}{+}{...}{d}{\theta}_{n}{=}{d}{\theta}_{n}{+}{...}{+}{d}{\theta}_{2}{+}{d}{\theta}_{1}$$

We have considered each quantity as a vector on either side.

But we should never write:
$${d}{\theta}_{1}{+}{d}{\theta}_{2}{+}{...}{d}{\theta}_{n}{=}{d}{[}{\theta}_{1}{+}{\theta}_{2}{+}{...}{\theta}_{n}{]}$$

considering the vector nature of the individual infinitesimals on the left side.
For the same reason,
$${\theta}{=}{d}{\theta}_{1}{+}{d}{\theta}_{2}{+}{...}{d}{\theta}_{n}$$
also would be an incorrect expression.
[On the left hand side of the last equation we have a scalar while on the right hand side we have a sum of vectors]

 

[Dale]

It is clear that DaleSpam has entertained some serious misconceptions in his logic

We do not parallel transport the basis vectors to obtain the chain of orthonormal reference frames as we move from point to point on the line of latitude.It is not at all necessary to do so in order to produce the chain of inertial reference frames.Such an action is not incumbent on us from any consideration whatsoever.

Yes, you do parallel transport the basis vectors. In fact, it is completely implied by your idea. The coordinate basis of an inertial reference frame (one where the metric is the Minkowski metric and the Christoffel symbols all vanish) is always orthonormal. Therefore, as soon as you have specified an inertial frame you have implicitly specified an orthonormal basis.

If a vector is parallel transported through a succession of inertial frames then at each point the vector has some coordinates in the local inertial frame and therefore the vector is some linear combination of the orthonormal coordinate basis vectors at that point. In your case, you are further requiring that the coordinates of the vector be the same in all of the inertial frames. This in turn implies that the dot products of the vector with the respective coordinate bases is the same in each inertial frame. Since parallel transport preserves the dot product the coordinate basis must also have been parallel transported.

Movement from one frame to the next involves:
...
b)Three infinitesimal rotations,involving the Eulerian angles.
We reverse the rotations keeping the translation intact.

If not by parallel transport then how do you propose to determine these angles given the metric and the path? It is not sufficient to merely say "reverse the rotations", you must provide a procedure to determine the rotations that need to be reversed. What is that procedure if not parallel transport?

 

[Anamitra]

Let us try to figure out the problem in this way:

From points A to B we take two paths[curves] L1 and L1. A is our initial point and B our final point.

1)We take a chain of inertial frames from A to B along L1 wrt orthogonal bases whose axes are parallel to each other individually. This may be achieved by parallel transport of the axes.The basis at B is slided down to A by parallel transport to A along L2.
i) When a vector T[$$T^{0},T^{1}T^{2}T^{3}$$ ] moves from A to B along L1 the components do not change.
ii) The two bases at A ate not identical. The same vector will have different components in the two bases at A.let A1 be the basis wrt L1 and A2 the basis wrt L2.
2) We choose a point P on L2.Between A and P along L2 we create non inertial states by some suitable transformation so that the vector T when parallel transported from A to P along L2 has the components $$T^{0},T^{1}T^{2}T^{3}$$ at P.This has to be done without changing the bases. Again Between P and B we consider a chain of inertial states with parallel transported axes.
It is to be noted that for the same base we may use several transformations to our advantage. These may lead to inertial or non-inertial states.

4)Let the components of the tensor T at A be $$T^{0},T^{1}T^{2}T^{3}$$ in the basis A1 and $$T^{'0},T^{'1}T^{'2}T^{'3}$$ in the other[A2].WE choose a transformation in such a way that when we move from A to B along L2 the tensor at P has the components $$T^{0},T^{1}T^{2}T^{3}$$ at P

We write the parallel transport equation for the curve L2:

$$\frac {dT^{\mu}}{d\tau}{+}{\Gamma^{\mu}}_{\nu\lambda}\frac{dx^{\lambda}}{d\tau}A^{\nu}{=}{0}$$
Let the solution of this equation be:

$$T^{\mu}{=}T^{\mu}{[}x^{0}{,}x^{1}{,}x^{2}{,}{x^{3}}{]}$$

Using the values of T at the point P[ $$T^{0}{,}T^{1}{,}T^{2}{,}T^{3}$$] we solve these four equations to get the coordinates of P.If these coordinates lie between the points A and B on the curve L2 there should be no problem.Otherwise we change the non-inertial transformation for the portion between A and P to get P in the portion between A and B

In fact we can always have a huge number of orthogonal transformations. In fact if one system is orthogonal any linear transformation should give us another orthogonal system.With this enormous choice we should have no problem in achieving our goal.

When we parallel transport the tensor T from A to B along L1 the components do not change.

When we parallel transport T from A to P along L2 the components of our tensor change to the value at A in basis A1 when we arrive at P.Henceforth there is no change in the components.At B we have the same basis with respect to the two paths and the components of the two vectors remain unchanged.

[For creating the chain of inertial states we use a separate transformation for each and every point ]

 

[Anamitra]

I have a four vector in spacetime which is described some metric,say the Schwarzschild metric.To investigate parallel transport wrt to the Schwarzschild metric we move it [by parallel transport]along a curve described on a two dimensional surface.The surface may be described by conditions like t=const and r= const.The vector should not become two dimensional in such a case,I believe.

If I am to parallel transport a four vector along curve on a 2-Dimensional surface we must consider all the four equations of parallel transport.

In case my assertion is correct then the parallel transported four vector may rise out of the tangent surface [described by $${e^{\theta}{e^{\phi}$$]when it is being transported over a sphere.
If the above surface is treated as a 4D surface with t=const and r=const then of course the vector does not rise out of the tangent plane.

I am requesting George [and of course DaleSpam]to comment on this issue .

 

[Dale]

Let's stick to inertial frames. If you are going to be using non-inertial frames then the usual process that we already covered in great depth applies. If you use non-inertial frames then you are simply doing standard parallel transport and there is no conceptual difference between one non-inertial coordinate system and another.

1)We take a chain of inertial frames from A to B along L1 wrt orthogonal bases whose axes are parallel to each other individually. This may be achieved by parallel transport of the axes.The basis at B is slided down to A by parallel transport to A along L2.
i) When a vector T[$$T^{0},T^{1}T^{2}T^{3}$$ ] moves from A to B along L1 the components do not change.
ii) The two bases at A ate not identical. The same vector will have different components in the two bases at A.let A1 be the basis wrt L1 and A2 the basis wrt L2.

Yes, exactly.

4)Let the components of the tensor T at A be $$T^{0},T^{1}T^{2}T^{3}$$ in the basis A1 and $$T^{'0},T^{'1}T^{'2}T^{'3}$$ in the other[A2].WE choose a transformation in such a way that when we move from A to B along L2 the tensor at P has the components $$T^{0},T^{1}T^{2}T^{3}$$ at P

I may be misunderstanding what you are saying, but from how I read this statement that is already the case with the inertial coordinate system approach, by definition. Let's make this a little concrete. Suppose the vector at A is a timelike vector of magnitude 5, then we can choose an orthonormal basis at A such that T=(5,0,0,0). If we then parallel transport the bases and T from A to B along L1 and L2 we are guaranteed that at each point along the path (including B) the coordinates of T=(5,0,0,0) in the respective bases. But because the bases themselves are different, having the same coordinates does not imply that the vectors are the same.

 

[Dale]

I have a four vector in spacetime ...

I am requesting George [and of course DaleSpam]to comment on this issue .

I am glad to address this as the next topic, but I want to finish one topic before moving to the next.

Are we done with the chain of inertial frames? Do you now understand how the fact that the coordinates do not change along each path does not imply that the vectors are the same at the final point?

 

[Anamitra]

Let us assume that we cannot transport a vector from one point to another over a finite distance since it ends up with different directions as we proceed along different paths.This simply favors the concept underlined in "Curved spacetime and the Speed of Light" or in the thread: https://www.physicsforums.com/showpost.php?p=2846710&postcount=6
If we could calculate the relative velocity at a point by the parallel transport method we would have two local vectors and the speed of light could never be exceeded, considering the fact that parallel transport never changes the norm of a vector.An alternative evaluation of relative motion may favor my considerations.

Interstingly relative motion is a physical concept which we cannot override by mathematical failures.The incapability of the mathematical apparatus does not imply that the physical effect "Relative Motion" does not exist.If I am standing in curved spacetime and I observe a car in motion at a certain distance what do I conclude about its motion? If I cannot calculate/measure the distance from New York to Boston should it mean that the concept of distance is meaningless?

I would request the audience to consider the following thread with full consideration of the aforesaid ideas:

https://www.physicsforums.com/showpost.php?p=2862951&postcount=108

 

[Dale]

So again. Are we done with the inertial states? If yes, then which topic would you like to pursue next?

1) embedded spaces (vectors lifting out of tangent planes or not)
2) observation and math

I am glad to pursue either one next (I would recommend number 1), but this "shotgun" form of conversation where you repeatedly bring up multiple topics is counterproductive. Because experience has shown that it can take more than 100 posts to explain a single point, I will not work on a new point until we have finished the current one.

So again. Have we finished the inertial frames topic? Do you now understand how the fact that the coordinates do not change along each path does not imply that the vectors are the same at the final point? If yes, then which topic would you like next? Pick only one for now, and we will get to the rest later.

 

[Anamitra]

I am requesting DaleSpam to address the issues in thread #153

 

[Dale]

And I am requesting Anamitra to address the questions in post 152 first.

Are we done with the inertial frames? Do you now understand how the fact that the coordinates do not change along each path does not imply that the vectors are the same at the final point?

 

[Anamitra]

We simply cannot transport a vector from one point to another along different paths without changing its orientation[whether we choose inertial or non inertial paths]. This renders the calculation of relative velocity impossible by standard methods.

Now DaleSpam should have no difficulty in addressing thread #153,now that we have finished with the inertial states.

We can definitely pass to the issues of 1)Embedded spaces 2)Observation and Math at a later stage.

 

[Dale]

Thanks for the answer, I will address the post 153 topic next.

 

[Dale]

Let us assume that we cannot transport a vector from one point to another over a finite distance since it ends up with different directions as we proceed along different paths.This simply favors the concept underlined in "Curved spacetime and the Speed of Light" or in the thread: https://www.physicsforums.com/showpost.php?p=2846710&postcount=6
If we could calculate the relative velocity at a point by the parallel transport method we would have two local vectors and the speed of light could never be exceeded, considering the fact that parallel transport never changes the norm of a vector.An alternative evaluation of relative motion may favor my considerations.

I think that your statements here are essentially correct. Parallel transport does not change the norm of a vector, so a timelike vector will never be parallel transported into a spacelike vector, meaning that you never get v>c. Parallel transport also results in a well-behaved coordinate-independent tensor. However it is not unique.

On the other hand, it is possible to simply define some number by some other formula and just call that number "relative velocity". While you can avoid the uniqueness issue that way you often introduce v>c problems, and you also will wind up with a coordinate-dependent number rather than a good tensor. This is the case with your example in a static spacetime, it is also the case with the redshift in the FLRW spacetime which is probably the most commonly used measure of relative velocity in any curved spacetime. Also, such measures don't usually generalize very well to other spacetimes.

Interstingly relative motion is a physical concept which we cannot override by mathematical failures.The incapability of the mathematical apparatus does not imply that the physical effect "Relative Motion" does not exist.If I am standing in curved spacetime and I observe a car in motion at a certain distance what do I conclude about its motion?

This depends in large measure on your definition of "physical concept". I honestly have no strong opinion on the matter, so let me ask a few questions of you about your personal definition of "physical effect":

1) Do physical effects need to be measurable in principle?
2) Can a physical effect depend on the coordinate system used or should it be coordinate independent?
3) Are physical effects tied to any particular theories of physics?
4) Are concepts in other theories excluded?
5) Anything else that can help me understand your idea of a physical effect?

If I cannot calculate/measure the distance from New York to Boston should it mean that the concept of distance is meaningless?

The modern view is that yes, if something cannot be measured then it is not physically meaningful. That is essentially the objection to Lorentz's Aether Theory and the reason that Einstein's SR was adopted instead.

 

[Dale]

This depends in large measure on your definition of "physical concept". I honestly have no strong opinion on the matter, so let me ask a few questions of you about your personal definition of "physical effect":

1) Do physical effects need to be measurable in principle?
2) Can a physical effect depend on the coordinate system used or should it be coordinate independent?
3) Are physical effects tied to any particular theories of physics?
4) Are concepts in other theories excluded?
5) Anything else that can help me understand your idea of a physical effect?

After thinking about it for a while I realized that it might be helpful if I posted the way that I would answer these questions. As I mentioned before, I don't have any strong opinions on this so if you can clearly articulate a different definition then I will be glad to use it instead.

I would describe a physical effect or a physical concept as a measurable quantity in a specified theory of physics. So I wouldn't make statements like "X is a physical effect" but rather "X is a physical effect in theory Y".

1) They do need to be measurable (or calculated from measurements) in principle.
2) They can depend on the coordinate system.
3) They are tied to a theory of physics, just specify it.
4) Physical effects/concepts in one theory may not exist in another.

 

[Dale]

So Anamitra, do you want to move on to embedded spaces and tangent vectors now, or are you clear about everything?

 

[Anamitra]

Extremely sorry for the delay. I was busy with some other work.But I am very much interested in the issues like embedding etc and most certainly I would like DaleSpam and others to continue in their participation in the thread.

 

[Anamitra]

Regarding the previous two threads of DaleSpam[#159 and #160]:They contain important material and I would try to give full consideration to them.

Regarding the concept of definition I have the following ideas:

1)A definition should be self consistent and also consistent with other concepts in physics.
2) It should help in understanding some reality or fact in physics.It should be connected with something that can be observed. This is again related to measurements being performed.

Parallel transport fails to define relative velocity.Does it mean that relative velocity cannot be measured or defined especially if some alternative procedure is adopted?Relative velocity is something very much "tangible" to us from the perspective of daily considerations. Should it become something "un-viewable" in curved spacetime?
If something is seen moving towards a black hole or a neutron star what could this observation mean if relative velocity is a meaningless concept?[We assume that the observation has been made from a space ship]

I believe that there many issues have to be considered.

I would try to reflect on these ideas in a serious manner. I will come back to them at a later stage in these threads.

Presently it would be interesting to hear from DaleSpam regarding embedding and other issues he has referred to.

 

[Anamitra]

An important issue to consider is: Is it essential to define relative velocity in curved space-time?

I would like to suggest this idea in view of what one may observe in the vicinity of curved spacetime for example particles/antiparticles flying outwards from a black hole[from beyond the Schwarzschild radius]For an observer at a distance the relative speed should be of greater concern than the local speed!

But I have already invited DaleSpam to talk on the issue of embedding. More important I would like to have some more time to reflect on the concept of relative speed in relation to curved spacetime .

 

[pervect]

You can define relative velocity if two objects are at the same point in space time (colliding with each other, or passing close-by). It's just an angle on the space-time diagram. You can also define relative velocity if you have a static space-time - which covers the black hole case, at least if it's not rotating. (It can probably be extended to cover a rotating black hole too, but I'd have to think about that case some more to nail down exactly how you'd do it.). This is just the velocity of the object relative to some static observer hovering above the black hole at constant coordinates. While you *can* define velocity this way, you'd be well advised to be careful to explain your definition - if you just talk about "the velocity relative to the black hole" without any more explanation, it won't really be clear what you meant.

 

[Dale]

1)A definition should be self consistent and also consistent with other concepts in physics.

I agree with this one.

2) It should help in understanding some reality or fact in physics.It should be connected with something that can be observed. This is again related to measurements being performed.

I don't really like this one. The word "real" is notoriously difficult to define and such discussions always seem to degenerate.

Relative velocity is something very much "tangible" to us from the perspective of daily considerations.

Sure, so is simultaneity. That is a deficiency of relying on our daily experience.

If something is seen moving towards a black hole or a neutron star what could this observation mean if relative velocity is a meaningless concept?

There are some spacetimes (eg toroids) where it is generally not even clear if two objects are moving towards or away from each other.

Presently it would be interesting to hear from DaleSpam regarding embedding and other issues he has referred to.

Do you understand the difference between a manifold and a tangent space? Specifically, do you understand why a tangent space is a vector space and a manifold is not?

 

[Anamitra]

Regarding Simultaneity:Simultaneity is a well defined concept.The only point is that a pair of events which are simultaneous in a particular frame of reference may not be simultaneous in another.

Regarding manifolds and tangent spaces:One may think of establishing one to one correspondence between the points of a sphere and the $${R^{2}}$$. But this is not possible. But we may break up the surface of a sphere into small parts and establish one to one correspondence between the small surfaces and $${R^{2}}$$.
The normal practice is to set up local charts at each point of the curved surface[the sphere in this case] . For each point on an infinitesimally small piece of surface[which we obtain by subdividing the original curved surface] we associate a point of $${R^{2}}$$ by some mapping satisfying conditions of continuity,differentiability etc.Now at each point we have a tangent space [ $${R^{2}$$] and here we can define our vectors.

The entire union of the small curved surfaces[and against each such surface we have a tangent space] is the manifold. Incidentally the manifold and the tangent spaces are of the same dimension.

The idea may be extended to higher dimensions

 

[Anamitra]

Regarding the"real" and the "unreal":Common sense perceptions may lead us to believe something which is "not real " to be "real".This has happened in realm of physics on several occasions. But on each such occasion the matter was clearly tested out both from theoretical and experimental considerations.
The failure of a definition may not be a sufficient ground for dismissing a commonsense point of view.One has to be very careful !
Very often the commonsense point of view has undergone a modification instead of a complete dismissal. The rule of velocity addition has changed on the passage from the classical to the Special Relativity concepts--but the concept of velocity addition/subtraction is still there very much in conformity with what is observed.Does this concept of velocity addition/subtraction need to disappear totally in curved spacetime?

 

[Dale]

One may think of establishing one to one correspondence between the points of a sphere and the $${R^{2}}$$. But this is not possible. But we may break up the surface of a sphere into small parts and establish one to one correspondence between the small surfaces and $${R^{2}}$$.
The normal practice is to set up local charts at each point of the curved surface[the sphere in this case] . For each point on an infinitesimally small piece of surface[which we obtain by subdividing the original curved surface] we associate a point of $${R^{2}}$$ by some mapping satisfying conditions of continuity,differentiability etc.

This is correct except for the very minor point that the charts need not be infinitesimal. In fact, you can easily map the entire sphere minus a single point onto $${R^{2}}$$.

Because you can map the sphere (minus a point) onto $${R^{2}}$$ and because $${R^{2}}$$ is a vector space you might be tempted to think that the sphere is a vector space, but it doesn't have the vector operations which define a vector space. Ie, you don't add scalar multiples of two points to get a new point.

Now at each point we have a tangent space [ $${R^{2}$$] and here we can define our vectors.

Yes. At each point we can take the derivatives of our coordinates to get the point's tangent space. This is a vector space since it does have a vector addition and scalar multiplication.

Are you comfortable with that?

 

[Anamitra]

Addition of Tensors/Vectors

Tensors admit themselves to a simple law of addition:if $${a^{\alpha}}$$ and $${b^{\alpha}}$$ are tensors then $${a^{\alpha}}{+}{b^{\alpha}}$$ is a tensor.

If we treat $${b^{\alpha}}$$ and $${b^{\alpha}}$$ as velocity vectors their norms are always "c"[individually]

Referred to orthogonal coordinates,where $${g_{\mu\nu}}{=}{0}$$ if $${\mu}{\neq}{\nu}$$, we have:

$${g_{\alpha\alpha}}{{(}{a^{\alpha}}{)}}^{2}}{=}{c^{2}}$$
$${g_{\alpha\alpha}}{{(}{b^{\alpha}}{)}}^{2}}{=}{c^{2}}$$
Again:
$${g_{\alpha\alpha}}{{{(}{a^{\alpha}{+}{b^{\alpha}}{)}^{2}}{=}{c^{2}}$$

We have $${4c^{2}}{=}{c^{2}}$$ if $${a^{\alpha}}{=}{b^{\alpha}}$$ ----------------?

Can we treat the velocity four vector as a tensor of rank one, in view of the constancy of the norm and as such and apply the parallel transport equation to it?

 

[Dale]

You are correct. The set of all velocity four-vectors is not a vector space, for the same reason that the set of all unit vectors is not a vector space. A four-velocity is a unit tangent vector, and is a member of the tangent space which is a vector space, but it includes un-normalized vectors which are not four-velocities.

 

[Anamitra]

Velocity as a Four Vector:

Let us consider the flat spacetime

Four velocity: $${(}{c \frac{dt}{{d}{\tau}}}{,}{\frac{dx_{1}}{dt} \frac{dt} {{d}{\tau}}}{,}{\frac{dx_{2}}{dt} \frac{ dt}{{d}{\tau}}}{,}{\frac{dx_{3}}{dt} \frac{dt}{{d}{\tau}}}{)}$$
Or,
Four Velocity: $${(}{c \frac{1}{\sqrt{1-{v^{2}/{c^{2}}}}}{,}{\frac{dx_{1}}{dt} \frac{1}{\sqrt{1-{v^{2}/{c^{2}}}}}{,}{\frac{dx_{2}}{dt}\frac{1}{\sqrt{1-{v^{2}/{c^{2}}}}}{,}{\frac{dx_{3}}{dt} \frac{1}{\sqrt{1-{v^{2}/{c^{2}}}}}{)}$$

We consider the following case:
$$dx_{1}/{dt}{=}{0.9c}$$
$$dx_{2}/{dt}{=}{0}$$
$$dx_{1}/{dt}{=}{0}$$
=>v=0.9c
$${\gamma}{=}{2.29415}$$

Four velocity=(2.29415c,2.06474c,0,0)

[$${2.29415^{2}}{-}{2.06474^{2}}{=}{1}$$ ----------------approximately,because of decimal approximations]

But the spatial part of the four velocity[speed referred to proper time is exceeding the value of "c"].

I am requesting Dalespam[and of course others] to comment on the issue.

 

[Dale]

Yes. That is correct.

 

[Anamitra]

Considering the four velocity of light we have four infinitely large components but the norm is finite.How do you interpret this?The whole thing appears to be indeterminate.Of course the spatial part of the four velocity[celerity] is remotely connected to the concept of three velocity[quite distinct from it to be accurate] as we understand in the physical world.This becomes more conspicuous in the high speed regions.

 

[JDoolin]

Is it meaningful to talk of relative velocity between two moving points at a distance in curved space-time?

My own impression is that you revoke almost all meaning once you invoke curved spacetime.

Relative speed ,it appears from your condiderans,is a troublesome concept.The concept of parallel transport adds a serious attribute of non-uniqueness to the whole idea.

Good point.

Even then if I am standing at some point in curved space-time and an object flies past some other point(we consider somthing which is not a light ray), I should have some observation of relative motion physically.And I hope that from the physical point of view this observation should be of a unique natue!Is there any procedure in General Relativity that allows me to make such predictions ? In case there is some procedure/method can we apply it to a light ray?

In General Relativity, there is no way to tell, via direct observation, how much of the velocity that you perceive is due to actual motion through space, and how much is due to the motion of space.

So, no.

The Standard Model in General Relativity assumes first that the matter-density of the universe is uniform in space and time, then wherever our perceptions disagree with our assumptions, one simply applies a coordinate transformation, or "metric." The science of GR is to find the metric that will fulfill this function precisely.

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.

The Schwartzchild metric is of another category of GR than the Standard Model. The fact that it predates Eddington is a strong point in its favor.