Curved Space-time and Relative Velocity

[Dale]

Regarding #103: I have sufficient difficulty in accepting what DaleSpam has to say. I have no hesitation in working out any homework problem he suggests. I have done this before.Nevertheless I would like to clarify my stand on this issue once more.
A freely falling lift is an inertial frame in the gravitational field of the earth.Now we may assign different velocities to it without spoiling the inertial nature of the frame. This may be accepted in a general way. We consider two transformations from the same metric leading to the Minkowski matrix[1 -1 -1 -1] in a local way.From special relativity we know that they must be moving with uniform speed with respect to each other. Now we divide our path from A to B into small intervals (A ,A1),(A1,A2)...(A[n-1],B)
In each interval we choose a frame with the same velocity V. The intervals being very small we choose for every interval V=V+delta_V approximately.So we have several coordinate systems which are not in relative motion.We may view them as rectangular coordinate systems in consideration of the Minkowski matrix[1 -1-1-1]Then we move our vector through these intervals.It remains constant since dA(mu)/dx(i)=0 for each interval.The vector remains unchanged at the end point B. For any other path we repeat the same manoeuvre starting with the same velocity at A.
In case there is some mistake in my method it has to be pointed out in a specific way. Of course I am ready to work out any practice problem suggested.No harm in doing that.

OK, then using the Schwarzschild metric (in units where c=1, G=1, and M=1/2):
$$ds^{2} = -\left(1 - \frac{1}{r} \right) dt^2 + \frac{dr^2}{\displaystyle{1-\frac{1}{r}}} + r^2 \left(d\theta^2 + \sin^2\theta \, d\phi^2\right)$$

Start at $$t=0$$, $$r=2$$, $$\theta=45^{\circ}$$, $$\phi=0^{\circ}$$ and parallel transport an arbitrary vector to $$\theta=45^{\circ}$$, $$\phi=180^{\circ}$$ along:
a) the 45º colatitude line
b) the 0º and 180º longitude line

Work both cases using the standard parallel transport equation, and also try to work both using your "inertial frame" method.

 

[Anamitra]

Parallel Transport through Inertial Frames
We have the Schwarzschild metric for m=1/2

ds^2=[1-1/r]dt^2-[1-1/r]^(-1) dr^2 - r^2[d(theta)^2+[sin(theta)]^2 d(phi)^2]

In the above metric we hold r constant making r=R
We keep time variable . In the final stage we will consider dt->0 so that time becomes constant and we are back in a loop.
We have for the 45 degree latitude line:
ds^2=[1-1/R]dt^2 - R^2[d(theta)^2+1/2 d(phi)^2]
[Though theta is constant for a latitude we consider small variations in theta for better workability of the metric. Of course sin(theta) does not change appreciably. It may be taken as a constant]

We use the transformations:

T=root[(1-1/R)t]

x=R theta

y=R(1/root2)phi

The last two transformations apply locally.
Regarding time: It represents physical time [having the dimension of time]
Regarding x: The x-axis is parallel to the vector e(phi)
Regarding y: The y-axis is parallel to the vector e(theta)
[important to note that we have a z-axis along e(r) ]
Our metric now:
ds^2=dT^2 - dx^2-dy^2
Movement along the latitude:
We consider an infinitesimal movement of the coordinate frame along the line of latitude.In each new position the new [x,y,z] triad corresponds to the new position of [e(phi),e(theta),e(r)]There are two effects:
1) An infinitesimal translation that leaves the direction of the axes unchanged.
x=x+x'

y=y+y'

z=z+z'
The metric now:
ds^2=dT^2 - dx'^2-dy'^2-dz'^2
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.

Form of the metric:
ds^2=dT^2 - dx'^2-dy'^2-dz'^2

We carry on this process as we move along the line of latitude
It is to be noted that the axes do not change their orientation as we move from one frame to another. The frames are not in relative motion.

Parallel Transport Equation in each frame:

dA(mu)/dA(nu)=0

The components do not change as we pass in the same frame. They do not change as we pass from one frame to another.
The vector does not change its orientation at any instant.

Movement along the meridian: Similar arguments may be applied in the case of a meridian. We simply move the [x,y,z]triad along the meridian instead of the latitude. Each small movement can be decomposed into a translation and a rotation. We reverse the effect of rotation keeping the translation in tact.
Net Effect:
1) The metric has the diagonal form[1-1-1-1]
2) In each situation the axes remain parallel to the previous situation[respectively]
3) There is no relative motion between the frames.
4) In each frame we have
5) As we pass from frame to frame the components remain unchanged

 

[Anamitra]

A Suggested Experiment:

Two persons A and B are standing at two distinct points in curved spacetime where the metrics have different values[especially in relation to g(0,0)].B performs an experiment at his own location by sending a light ray(say gamma rays) through a cloud chamber and measures the length of the track and also records the time interval

Speed=[length of track]/time interval measured by B

This should be something less than "c" considering the the slowing down of the light ray by the medium

Let the observed speed be c/2

Observation of A:The clock of A runs at a different rate from that of B.Let us assume that the intervals measured by A are 100 times shorter than that of BSpeed measured by A=[Length of Track]/time interval of A
=[[Length of Track]/time interval of B] * 100
=50 * c
W are getting this result because A and B are non-local points.

Observation in the physical world cannot be suppressed by by the incapability of the mathematical apparatus to cope up with the situation[in case such an incapability exists, for the sake of argument]

[Instead of a light ray we could use a fast moving electron (or an alpha particle)also,say one that moves at 0.99c]

 

[Dale]

Please finish working the problems. Then we can discuss the results which should help you understand the next parts of the discussion.

 

[Anamitra]

Regarding DaleSpam's Problems:
Four equations for parallel transport have been considered for each case,the latitude and the longitude.
1)Changes of A(theta) A(phi) A(r) and A(t) for changes in phi have been considered for theta=const and r=constant
2)Changes of A(theta) A(phi) A(r) and A(t)[for changes theta)have been considered for phi=const and r=constant

The initial values of A(r) and A(t) are zero. But even with that the equations yield non-zero values of A(r) in general. A(t) is not showing any problem. The vector seems to lift out of the tangent plane to satisfy the equations. That needs an explanation.

[The work has been shown in the attachment]

 

[Dale]

Hi Anamitra,

I have looked at the first part, transport along the 45º lattitude line. My results are pretty close to yours, with I suspect only a single small mistake that had minimal impact.

In the third differential equation you are missing a sin² term:
$$\frac{dA^r}{d\phi}-(r-2M) A^{\phi} sin^2(\theta)=0$$

This then leads to a couple of factors of 2 (at 45º) in the solution:
$$A^r=\frac{r-2M}{2k}\left( C_1 sin(k\phi) - C_2 cos(k\phi) \right)-2 r C_3 cos^2(\theta)$$

The errors don't affect the form of the solution, just a couple of small details, so I don't think they are a big deal.

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.

I will get to the second half in a bit.

 

[Dale]

On the second half I have a different expression fro the second equation. Specifically, I have:

$$\frac{dA^{\phi}}{d\theta}+A^{\phi}cot(\theta)=0$$

You have an extra term in there. Unfortunately, this one makes a big difference. I don't get even close to your solutions.

 

[Anamitra]

Yes ,I made a mistake with a Christoffel symbol.And I am repeating the thing.In any case cot(theta) causes some problem for theta=0 [and theta=pi]. How to get over it?I mean to say, one must write a separate equation for the neighborhood of theta=0[and pi].The values are enormously large here.
.

 

[Anamitra]

The revised calculations have been uploaded.The effect of the infinite discontinuity reflects itself in the problem.A(phi) is undefined at theta=0 and theta=pi.It tends to an unbounded value as theta tends to 0 or pi.

 

[Dale]

Your revised section is correct. Also, the cot and csc functions are correct. These don't have anything particularly to do with curved spacetime or parallel transport, but simply are features of spherical coordinates where near the poles a small change in $\theta$ leads to a large change in $A^{\phi}$. You can set M=0 and work in a flat spacetime as we did before to see that this is correct.

So, let us now review the recent results. Using the definition of parallel transport you have parallel transported a vector from one location to another location in curved spacetime using two different paths and obtained two different results.

Do you now agree with that and understand that parallel transport is indeed path dependent in a curved spacetime?

If you agree with that then we can investigate in a little more detail the "inertial frames" approach.

 

[Anamitra]

$$\frac{dA^{\phi}}{d\theta}+A^{\phi}cot(\theta)=0$$

If you the solve the equation in thread#112[ https://www.physicsforums.com/showpost.php?p=2865831&postcount=112] you get:

A(phi)=c* Cosec (theta)

Now this blows up in the neighborhood of theta =0 or theta =pi.

A small change in theta produces a large change in A(phi)----that part is ,perhaps,OK. But the value of A(phi) getting infinitely large at the poles[in the actual sense of its value] cannot be entertained.The concept of parallel transport is simply undefined at the poles.

 

[Dale]

If you the solve the equation in thread#112[ https://www.physicsforums.com/showpost.php?p=2865831&postcount=112] you get:

A(phi)=c* Cosec (theta)

Now this blows up in the neighborhood of theta =0 or theta =pi.

Yes, as it should in spherical coordinates.

A small change in theta produces a large change in A(phi)----that part is OK. But the value of A(phi) getting infinitely large at the poles[in the actual sense of its value] cannot be entertained.The concept of parallel transport is simply undefined at the poles.

Parallel transport is fine. It is just spherical coordinates where phi is undefined at the poles. This has nothing to do with parallel transport or curved spacetime, this is just "business as usual" in spherical coordinates. Your objections in both .pdf files are not about parallel transport, but spherical coordinates. I get the impression that you have not worked in spherical coordinates very much, or you would have seen this type of behavior previously.

However, if you want to modify the path so that it avoids the poles then feel free to do so. I am glad to work with any alternative path you choose (other than the latitude line since we already did that one). Do you have a suggestion?

 

[Anamitra]

For the Swarzschild Sphere parallel transport is working well with the exclusion of the poles which admit themselves to multiple values for phi.

Now let us look into the fact that an ordinary sphere is commonly used to illustrate the concept of parallel transport in the texts. You did the same thing the following thread:
https://www.physicsforums.com/showpost.php?p=2758350&postcount=25

If you write the three equations [for A(r),A(phi) and A(theta)] for a line of latitude and solve them you get functions like Sin(phi) and cos(phi) in the solutions which are periodic functions of 2pi .

So the vector returns to its original orientation in this case!
[File has been uploaded for consideration]

 

[Dale]

I will be on a mobile device for the next couple of days. Please use LaTeX directly in the forum. There is no need to keep posting .pdf files.

For the Swarzschild Sphere parallel transport is working well with the exclusion of the poles which admit themselves to multiple values for phi.

What do you mean by "working well"? Do you mean that you finally understand parallel transport and how it is path dependent?

If so then we can begin with the other issues.

 

[George Jones]

Now let us look into the fact that an ordinary sphere is commonly used to illustrate the concept of parallel transport in the texts. You did the same thing the following thread:
https://www.physicsforums.com/showpost.php?p=2758350&postcount=25

If you write the three equations [for A(r),A(phi) and A(theta)] for a line of latitude and solve them you get functions like Sin(phi) and cos(phi) in the solutions which are periodic functions of 2pi .

So the vector returns to its original orientation in this case!

Does it?

My daughter was constantly pestering me while I was working on this, so what follows could be full of mistakes.

Tangent spaces of an "ordinary sphere'' are 2-dimensional, and vectors in these tangent spaces don't have $r$ components. Therefore, set $r=1$ and $A^{r}=0$. Your three equations then become two equations,

$$\begin{equation*} \begin{split} \frac{dA^{\phi }}{d\phi }+\cot \theta A^{\theta } &=0 \\ \frac{dA^{\theta }}{d\phi }-\sin \theta \cos \theta A^{\phi } &=0. \end{split} \end{equation*}$$

This set of equations can be solved elegantly using 2x2 matrices, but I'll solve it using a different method. Differentiating the first equation gives

$$\frac{d^{2}A^{\phi }}{d\phi ^{2}}+\cot \theta \frac{dA^{\theta }}{d\phi }=0,$$

and using the second equation in this gives

$$\frac{d^{2}A^{\phi }}{d\phi ^{2}}+\cos ^{2}\theta A^{\phi }=0.$$

Since $\theta$ is constant along a line of longitude, this is just a harmonic oscillator equation with solution

$$A^{\phi }\left( \phi \right) =c_{1}\sin \left( \phi \cos \theta \right) +c_{2}\cos \left( \phi \cos \theta \right)$$

Then,

$$\begin{equation*} \begin{split} \frac{dA^{\phi }}{d\phi }\left( \phi \right) &= \cos \theta \left[ c_{1}\cos \left( \phi \cos \theta \right) -c_{2}\sin \left( \phi \cos \theta \right) \right] \\ &=-\cot \theta A^{\phi } \end{split} \end{equation*}$$

and, from the first equation,

$$A^{\theta }\left( \phi \right) =\sin \theta \left[ c_{1}\cos \left( \phi \cos \theta \right) -c_{2}\sin \left( \phi \cos \theta \right) \right].$$

Consequently,

$$\begin{equation*} \begin{split} A^{\phi }\left( 0\right) &= c_{2}\cos \left( \phi \cos \theta \right) \\ A^{\theta }\left( 0\right) &= c_{1}\sin \theta \cos \left( \phi \cos \theta \right) \end{split} \end{equation*}$$

and

$$\begin{equation*} \begin{split} A^{\phi }\left( 2\pi \right) &= c_{1}\sin \left( 2\pi \cos \theta \right) +c_{2}\cos \left( 2\pi \cos \theta \right) \\ A^{\theta }\left( 2\pi \right) &= \sin \theta \left[ c_{1}\cos \left( 2\pi \cos \theta \right) -c_{2}\sin \left( 2\pi \cos \theta \right) \right]. \end{split} \end{equation*}$$

As a simple example, take $c_{1}=1/\sin \theta$ and $c_{2}=0$, so that

$$\begin{equation*} \begin{split} A^{\phi }\left( 0\right) &= 0 \\ A^{\theta }\left( 0\right) &= 1 \end{split} \end{equation*}$$

and

$$\begin{equation*} \begin{split} A^{\phi }\left( 2\pi \right) &= \sin \left( 2\pi \cos \theta \right) /\sin \theta \\ A^{\theta }\left( 2\pi \right) &= \cos \left( 2\pi \cos \theta \right). \end{split} \end{equation*}$$

What does this mean?

 

[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.