Curved Space-time and Relative Velocity

[Anamitra]

A(r) could be zero only at the initial point but you can't hold it zero for all positions if you consider the parallel transport equations. There are three equations to be considered. [Attachment in Thread #118]
The equations are different from what you have got.

 

[George Jones]

No, you are confused. In the attachment in post #118, you have not parallel-transported a vector around a closed curved for "an ordinary sphere." In the attachment in post #118, you have parallel-transported a vector around a plane circle (with $\phi$ as parameter) in flat Euclidean 3-space! Of course this gives a null result!

 

[Dale]

Anamitra, I cannot read the pdf, but from George Jones' comments I understand the mistake. To verify, simply calculate the curvature tensor of the metric you used. You will find that it is all 0.

You are confusing the 3D flat space in spherical coordinates with the 2D curved space of (the surface of) a sphere.

 

[Anamitra]

How do we transport[I mean parallel transport] a three dimensional vector along a curve lying on a two dimensional surface? The vector cannot lose a component because we are moving it on a surface

[ If I move a four vector along a two dimensional surface it should no more be a four vector.It should immediately become a two vector!]

 

[Dale]

Do you know what an embedding space is? A lower dimensional curved space like the 2D surface of a sphere may be embedded in a higher dimensional flat space like ordinary 3D Euclidean space. Measures of the curvature which must be expressed in terms of the higher-dimensional flat embedding space are called extrinsic, and measures of curvature which can be expressed purely in terms of the lower-dimensional curved space are called intrinsic.

Parallel transport and all of the other machinery of GR deals with intrinsic curvature so the higher dimensional space is not necessary, but for pedagogical reasons it is often helpful to talk about familiar curved 2D spaces embedded in 3D.

 

[Dale]

Anamitra, are you now willing to admit that in curved spaces parallel transport is path dependent? We have shown this explicitly with three specific examples, anyone of which is conclusive.

1) Sphere - Transport 90 deg along equator v transport due north turn and due south.

2) Sphere - Transport around a lattitude loop v staying in place.

3) Schwarzschild - Transport along lattitude line v transport along longitude line.

If you are still insisting that parallel transport is not path dependent then you are clearly not a reasonable person who is willing to look at evidence or logic. This is not a matter of any possible doubt, it is a proven mathematical fact.

If you have any remaining doubts on this point then you tell me what it would take to convince you as I have already provided 3 clear examples. Once you accept this point we can move on to other points like the inertial frames and the relation to physics.

 

[Anamitra]

Regarding the lifting from the tangent plane, this is obvious and expected. In fact, if it didn't happen we would immediately know that there is a problem. Consider a vector pointing due north at the equator in the flat space limit of M=0, this vector is purely tangential at the equator and purely radial when transported to the poles.

[Thread #111]
You clearly subscribed to the fact that the parallel transported vector can rise out of the tangent plane.But now you seem to have changed your ideas in response to logical considerations.

But now let me come to something serious:I have a solid metal sphere in front of me and I try to understand its curvature by parallel transporting a vector round a line of latitude,say the 45 degree latitude.I don't have to think of embedding in the context of this issue.

If I take a three dimensional vector[even considering the initial value of A(r) =0] and take it in a round trip,it does not change its orientation.
[https://www.physicsforums.com/showpost.php?p=2867114&postcount=118]

If I take a "two dimensional" vector[that is I consider two components instead of three] it definitely changes its orientation on a round trip.

A four dimensional vector will reduce to a two dimensional one on embedding

I need to explain the whole situation to myself!

In the meantime I am taking a short leave from the audience so that I can take a correct decision. I am requesting the audience to go through the following threads during this time.
https://www.physicsforums.com/showpost.php?p=2862951&postcount=108
https://www.physicsforums.com/showpost.php?p=2861464&postcount=105
https://www.physicsforums.com/showpost.php?p=2862816&postcount=107

 

[Anamitra]

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]

For a two vector[corresponding to spatial rest] if we are to follow what George/DaleSpam are suggesting the vector does not change at any point of the path ,according to their own equations.

If I am at rest and I want to calculate the relative velocity at some distant point there should be no problem at all!

For two two objects individually in motion,say cars A and B,[of course we are in curved space time] the driver in one car should have some idea of the motion of the other car.

If DaleSpam[and of course George] cannot calculate the relative velocity ,Relative velocity should not exist.
If I cannot measure the distance from Boston to New York,distance is obviously a meaningless concept-----That is exacly what Dalespam and George are suggesting
While I do some more thinking on the problem of embedding ,George and Dalespam should have no problem in addressing these issues.

 

[Dale]

Anamitra, I am still on a mobile device. It would really help if you would simply put the LaTeX equations directly into a post instead of in a pdf file.

Also, I wish you would stop referring to "thread N". This whole conversation is a thread and each numbered reply is a post, not a thread.

 

[Dale]

[Thread #111]
You clearly subscribed to the fact that the parallel transported vector can rise out of the tangent plane.But now you seem to have changed your ideas in response to logical considerations.

My position has not changed a bit. As I said just a few posts ago you are confusing higher dimensional flat spaces with lower dimensional curved spaces. In the quoted reference I very carefully and explicitly mentioned that I was referring to the flat Schwarzschild metric with M=0. That is a 4D flat space with spherical coordinates, not an embedded 2D curved space. As I mentioned before, you can verify the flatness by calculating the curvature tensor, which you obviously have not done or you would not be trying to use a flat space to prove something about curved spaces.

Please answer the following question which I have asked repeatedly:

Given the overwhelming evidence, including no less than two problems which you worked yourself, do you now understand that parallel transport is path dependent in curved spaces?

If after more than 100 posts of proof you are still unable to grasp such a basic mathematical fact then the other topics are pointless to discuss. So answer the question, you have had proof enough.

PS If the three problems we have worked are not sufficient proof then I would turn your attention to the definition of parallel transport. The path is in the second term, and therefore the equation is only path independent if this second term drops out for all possible paths. This, in turn, only occurs for spaces where there exists a global transformation to a metric with no non-zero Christoffel symbols. Such spaces are caled flat. For all other spaces there is no coordinate transformation which can remove all of the Christoffel symbols everywhere, and therefore the second term is not everywhere nonzero and the result depends on the path. The path dependence is absolutely obvious from the definition, and there is no avoiding that simple mathematical fact.

 

[Anamitra]

I have tried to calculate the components of the Riemannian-Christoffel tensor in three dimensions. That is ,the rank-four tensor has been evaluated in three dimensions with reference to the spherical coordinates.

$${R^{\alpha}}_{\beta\gamma\delta}$$=$$\frac{d{{\Gamma}^{\alpha}}_{\beta\delta}}{dx^{\gamma}}$$-$$\frac{d{{\Gamma}^{\alpha}}_{\beta\gamma}}{dx^{\delta}}$$+$${{\Gamma}^{\alpha}}_{\gamma\epsilon}{{\Gamma}^{\epsilon}}_{\beta\delta}$$-$${{\Gamma}^{\alpha}}_{\delta\epsilon}{{\Gamma}^{\epsilon}}_{\beta\gamma}$$

Each index runs over three values.

For
$${\gamma}$$=$${\delta}$$

$${R^{\alpha}}_{\beta\gamma\delta}$$=0
$${R^{\alpha}}_{\beta\gamma\gamma}$$=0

Therefore,
$${R_{\alpha\beta\gamma\gamma}$$=0

$${R_{\gamma\gamma\alpha\beta}$$=$${R_{\alpha\beta\gamma\gamma}$$=0

The above results hold for any three dimensional system.

Now for spherical coordinates:
$${R_{\alpha\gamma\beta\gamma}$$=0
$${R_{\gamma\alpha\gamma\beta}$$=0

Therefore in three dimensional spherical coordinates the Riemannian-Christoffel tensor is always zero!How do we understand the curvature of a sphere by using it.

 

[George Jones]

Euclidean 3-space is (locally) flat, so, with respect to any coordinate system (including spherical coordinates), the components of the Riemann curvature tensor will all be zero. In spherical coordinates, The metric tensor for Euclidean 3-space is given by

$$g = dr^2 + r^2 \left( d\theta^2 + sin^2 \theta d \phi^2 \right).$$

On a 2-sphere, $r$ is constant, so, without loss of generality, take $r =1$. Consequently, on the 2-spehere, $dr = 0$, and the metric for the 2-sphere is given by

$$g = d\theta^2 + sin^2 \theta d \phi^2.$$

Calculate the curvature tensor for the metric of a 2-spehere (the second metric).

 

[Dale]

$${R^{\alpha}}_{\beta\gamma\delta}$$=$$\frac{d{{\Gamma}^{\alpha}}_{\beta\delta}}{dx^{\gamma}}$$-$$\frac{d{{\Gamma}^{\alpha}}_{\beta\gamma}}{dx^{\delta}}$$+$${{\Gamma}^{\alpha}}_{\gamma\epsilon}{{\Gamma}^{\epsilon}}_{\beta\delta}$$-$${{\Gamma}^{\alpha}}_{\delta\epsilon}{{\Gamma}^{\epsilon}}_{\beta\gamma}$$

Hi Anamitra, I second what George Jones said, calculate the curvature of the embedded 2D spherical metric that he gave and you will find non-zero terms. This will show how a lower-dimensional curved space may be embedded in a higher dimensional flat space.

But I just wanted to add a congratulations for figuring out the LaTeX directly within a post. It is really a better approach than posting .pdf files since .pdf files are the most common vector for malware, as a security-minded friend often reminds me.

 

[Anamitra]

Of Course George is right!

 

[Dale]

So then, do you now understand that parallel transport is path dependent in curved spaces?

 

[Anamitra]

Its Ok.

I have tried to interpret George in the following way.

Reimannian tensor is a property of the metric and the metric itself gets transformed by the constraint r= const.
I was trying to work out the problem for r=K Cos$${\theta}$$. Obviously the christoffel symbols themselves will change because r cannot be held constant wrt $${\theta}$$ .We have different metric coefficients now.It is highly probable that the Riemannian curvature will not be zero.
Before I move out of this parallel transport issue [to other ones]I have some last questions to ask.
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?

[I am asking these questions not to contradict parallel transport but to understand it]

 

[Dale]

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?

Parallel transport depends on the metric and its derivatives only along the path. So if the metric and derivatives along the path are the same as the spherical metric then along that path it will give exactly the same result as the spherical case. However, I am not exactly certain what the metric would be for the space you described in general and along the path you describe in particular, so I cannot be more specific.

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?

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.

 

[Dale]

Hi Anamitra,

If you are comfortable with the path-dependence of parallel transport then I think we should next look into the "chain of inertial frames" approach, which I believe will have some pedagogical value.

 

[Anamitra]

Of course DaleSpam

But before that I would like to say something:

Now the geodesic is a curve along which the tangent vector gets parallel transported. We write the equation for a geodesic.

$$\frac{d^{2}x^{\alpha}}{d{\tau}^{2}}{= }{-}{{\Gamma}^{\alpha}}_{\beta\gamma}{\frac{dx^{\beta}}{d\tau}}{\frac{dx^{\gamma}}{d\tau}}$$

Now,$$\frac{dx^{\beta}}{d\tau}$$ may be regarded as velocity referred to proper time and if we are to parallel transport it we can do so only along a geodesic!.There is no problem at all with the velocity vector. We may transport it along a unique path except for conjugate points. We may try similar methods with the momentum vector etc.

[The chain of inertial points is there as a reserve consideration]

 

[Dale]

Now, frac{dx^{\beta}}{d\tau} may be regarded as velocity referred to proper time and if we are to parallel transport it we can do so only along a geodesic!

We have been over this before. There is nothing in the definitions of parallel transport or tangents that would suggest this restriction. In fact, this restriction would make the definition of parallel transport circular.

 

[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 .