Curved Space-time and Relative Velocity

[Dale]

May I refer to Wald: page 30[3.2 Curvature]
The diagram given[fig 3.3] we have a curve with sharp edges.The proof seems to be concerned with four separate parallel transports and not with a single transport at a stretch

I don't know how you reach that conclusion when the figure caption clearly reads "The parallel transport of a vector v around a small closed loop".

If you have this book then please examine carefully equation 3.1.19. Note that the tangent vector to the path appears in this equation, but not any derivatives of the tangent vector. So a sharp bend in the path does not cause any trouble. Note also the second sentence of section 3.2 where he explicitly states that the result is path-dependent.

 

[JesseM]

If you mean by 'space-time separation' the metric distance then only a geodesic represents that.

What do you mean? You can certainly calculate the integral of ds along a non-geodesic worldline, for a timelike path this would just be the proper time along that worldline (or i*c times the proper time, if you're using a definition of ds that is real-valued for spacelike paths). In GR I don't think physicists talk about the "space-time separation" between two events since there can be multiple geodesics between the same pair of events, the metric is understood to give you a notion of "distance" along particular worldlines.

 

[Dale]

Hi JesseM, I think I agree with Passionflower on this point. In flat spacetime the invariant interval ("spacetime separation") between two events is the integral of ds along a straight line from one event to the other. So in curved spacetime it should be the integral of ds along a geodesic. The invariant interval then may become ambiguous if there are multiple geodesics connecting the events, which is probably why, as you note, physicists do not speak of "space-time separation" in GR.

But, as it relates to this thread, parallel transport need not be restricted to geodesics.

 

[JesseM]

Hi JesseM, I think I agree with Passionflower on this point. In flat spacetime the invariant interval ("spacetime separation") between two events is the integral of ds along a straight line from one event to the other. So in curved spacetime it should be the integral of ds along a geodesic. The invariant interval then may become ambiguous if there are multiple geodesics connecting the events, which is probably why, as you note, physicists do not speak of "space-time separation" in GR.

Well, the fact that physicists don't really speak of "space-time separation" between events in GR was basically the point I was making to PassionFlower, with the additional point that it is still physically meaningful to integrate ds along non-geodesic wordlines.

 

[Dale]

I agree with both those points.

 

[Anamitra]

We consider a metric of the type shown below:

ds^2=g(00) dt^2-g(1,1) dx^2 - g(2,2) dy^2 - g(3,3) dz^2

ds^2= dT^2- dL^2 [dT--->Physical time,ds----> physical distance]

ds^2/dT^2 = 1- [dL/dT]^2

[ds/dT]^2 = 1-v^2

ds/dT = sqrt[1-v^2]

dT = gamma ds

[ c=1 here and ds is analogous to proper time]

This seems to be a counterpart of special relativity with a variable gamma . Here "v" corresponds to the notion of the three velocity.

Integrating we have,
T2-T1=integral[ s1 to s2 along some path] gamma ds

Since the left side is path independent the right side is also path independent. While in special relativity we can take gamma outside the integral,here we cannot perform such an action.

 

[Anamitra]

This is in relation to what has been said in the Threads #40[https://www.physicsforums.com/showpost.php?p=2849247&postcount=40] and #46[https://www.physicsforums.com/showpost.php?p=2849745&postcount=46]


The space-time we live in is "nearly flat". The slight amount of curvature it has, is extremely important. For instance it keeps satellites/planets in their orbits. We do calculate the three velocities of these satellites from the ground---that is we calculate/assess experimentally the speed of objects at distance. And all this is done in disregard to whatever objections Dalespam has raised against the concept of relative speed in curved space-time.That should settle the issue.

 

[Dale]

Certainly. I never said otherwise.

In a flat space parallel transport is unique and path independent, so we can consider all vectors at any point in the manifold to be part of the same vector space and thus we can compare velocities of distant objects. In a "nearly flat" space, by definition, we can ignore the curvature and treat it as flat.

 

[JesseM]

We consider a metric of the type shown below:

ds^2=g(00) dt^2-g(1,1) dx^2 - g(2,2) dy^2 - g(3,3) dz^2

ds^2= dT^2- dL^2 [dT--->Physical time,ds----> physical distance]

Anamitra, can you respond to my question here about whether your notion of "physical time" and "physical distance" matches DrGreg's interpretation in post #18 of that thread? Do you indeed define them using two metrics which are different from the usual spacetime metric? If so, note that unlike the integral of ds which is always the same regardless of your choice of coordinate system, the integral of dT and dL along a particular path will be coordinate-dependent, since terms like g(00) will have different equations in different coordinate systems.

ds^2/dT^2 = 1- [dL/dT]^2

[ds/dT]^2 = 1-v^2

ds/dT = sqrt[1-v^2]

dT = gamma ds

Note that since you are defining gamma as a function of dL/dT, here gamma would be coordinate-dependent as well.

Integrating we have,
T2-T1=integral[ s1 to s2 along some path] gamma ds

Since the left side is path independent the right side is also path independent.

Why do you think the left side is path independent? If we consider a pair of events with two different worldlines that pass through the pair, the integral of dT (i.e. the integral of sqrt(g(00)) dt) on the first worldline may be different than the integral of dT along the second.

 

[Anamitra]

Certainly. I never said otherwise.

In a flat space parallel transport is unique and path independent, so we can consider all vectors at any point in the manifold to be part of the same vector space and thus we can compare velocities of distant objects. In a "nearly flat" space, by definition, we can ignore the curvature and treat it as flat.

We are not treating space-time as flat when we are considering the motion of planets or satellites.We are considering "with seriousness" the curvature of space-time---- the deviations from the flatness.

 

[Dale]

And when the deviations from flatness are significant then, by definition, you cannot compare distant velocities.

 

[TrickyDicky]

And when the deviations from flatness are significant then, by definition, you cannot compare distant velocities.

Would this be due to the fact that in GR there isn't even an unambiguous way to compare in general the ticking "rates" of inertial clocks, which leads us to not being able to compare velocities since for doing that you need an agreement about the proper time of the objects whose velocity you want to compare?
In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?
I'm just starting with the study of GR and wanted to check if I understand anything at all

 

[TrickyDicky]

Would this be due to the fact that in GR there isn't even an unambiguous way to compare in general the ticking "rates" of inertial clocks, which leads us to not being able to compare velocities since for doing that you need an agreement about the proper time of the objects whose velocity you want to compare?
In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?
I'm just starting with the study of GR and wanted to check if I understand anything at all

Anyone agrees?

 

[Passionflower]

Would this be due to the fact that in GR there isn't even an unambiguous way to compare in general the ticking "rates" of inertial clocks, which leads us to not being able to compare velocities since for doing that you need an agreement about the proper time of the objects whose velocity you want to compare?
In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?
I'm just starting with the study of GR and wanted to check if I understand anything at all

Take a simple example of a 1 meter rod free falling radially in the length in a Schwarzschild spacetime. Gravitation will cause an inertial acceleration between the front and the back of this rod (and of course everything in between). If it gets stretched do you consider the rod to be still one meter long? If the rod is very strong and accelerates everything inwards (proper acceleration!) to maintain its structure is the distance between the ends still one meter? How do you even want to define a meter in this situation?

You see that even in a simple scenario we could already get issues with distance.

 

[TrickyDicky]

Take a simple example of a 1 meter rod free falling radially in the length in a Schwarzschild spacetime. Gravitation will cause an inertial acceleration between the front and the back of this rod (and of course everything in between). If it gets stretched do you consider the rod to be still one meter long? If the rod is very strong and accelerates everything inwards (proper acceleration!) to maintain its structure is the distance between the ends still one meter? How do you even want to define a meter in this situation?

You see that even in a simple scenario we could already get issues with distance.

Yes, there is ambiguity also from the lengths part,as well as with the times (I assume what I say in my quote is basically correct, then), in the end, the final consequence is that as has been said here distant velocities can't be compared as opposed to what the OP was asking.

 

[Dale]

Sorry I missed this earlier. Essentially, yes. There is just no way to do this and arrive at a single unique answer. Meters, seconds, and directions are all affected by curvature.

 

[Anamitra]

"Additional Information"

In Thread #1, I have remarked that parallel transport of a vector in a round trip along a closed smooth on some surface (which in general may be curved)should not produce any change in the orientation of the vector. Let me investigate this in relation to an arbitrary vector without any reference to an inertial frame:

Covariant derivative along the curve=0 [the curve is parametrized by the variable t]

=> dA(mu)/dt + Appropriate Christoffel symbol * dA(lambda)/dt * A(nu) = 0 ---- (1)

[The second term on the left side is a summation term. In the aforesaid Christoffel tensor "mu" is the superscript while "nu" and "lambda" are subscripts]

When the vector comes back to the original location ,after the round trip,the value represented by the Christoffel tensor does not change. We have four identical differential equations at the beginning and the end of our journey as we perform the parallel transport.The equations should produce the same results, that is, the tensor components do not change------the vector does not change its orientation. But if we have sharp bends on our journey of parallel transport, derivatives become undefined on these "sharp bends" and the orientation of the vector is not expected to remain the same when it returns to its original location.
[Boundary conditions change at each sharp bend since the differential equations themselves become undefined on these sharp bends]
For a "smooth curve" which is "not closed" the vector remains parallel if we refer to a chain of inertial states .I have tried to show in Thread #33.

 

[Dale]

Anamitra, please post your equations (particularly tensor equations) using LaTeX. Also, please define your terms. What is t and what is A.

Hopefully when you do so you will note that the above equation is missing an impoortant variable.

 

[Anamitra]

I have uploaded the file which contains the formula in Latex.The formula may be referenced from "Gravitation and Cosmology" by Steven Weinberg, Chapter 4,Tensor Analysis,Section 9[Covariant Differentiation along a curve]

We must always keep in mind the boundary conditions when we are trying /considering to solve the four equations shown in the attachment[The equations have been represented by a single one]. The values of A(mu)/dt contribute to the boundary conditions. At the sharp bends we consider their limiting values. These values are unequal on either sides of any bend

 

[Dale]

OK, so looking at the equation you should be able to see that the curve along which you are parallel transporting is x, and A is the vector which is being parallel transported along x. Also, note that only first derivatives of x (the tangent vector) appear, so a sharp bend in x causes only a finite discontinuity in the tangent vector.

Note further that the equation is a first-order ordinary differential equation in A, and as such it has an analytical solution which involves integration. The integration is unaffected by any finite number of finite discontinuities in the tangent vector. This is why sharp bends are perfectly acceptable in parallel transport.

Furthermore, even if you restrict yourself to smooth curves you will find that your claim is completely false. When a vector comes back to its original location after a round trip (even with smooth curves) it will not match the original vector. This is essentially the DEFINITION of curvature. You are trying to flatten something that cannot be flattened, you need to let go of your Euclidean preconceptions.

I challenge you to take the formula that you have posted, use the metric on a sphere, and parallel transport a vector around the 45º lattitude. Work the problem out for yourself and see if the rest of us are wrong or not.

 

[Anamitra]

Your example was a good one DaleSpam. Thank you!

I have tried to reason out the problem in this way.As we move along the curve the Christoffel tensor changes according to the changing values of the metric coefficients. If we divide the curve into small segments the differential equations themselves are different over each segment(due to the changing values of the Christoffel Tensor). So the boundary conditions go on changing as we move along it and so on the last lap of the journey we may end up with different boundary conditions. This can of course change the orientation of the vector. Perhaps this is the reason .I thank you for the example.
[But then again if you refer to a chain of inertial states there is no problem at all]. I have a very strong point here.[Please refer to the attachment]
Now my further confusions[I am simply trying to get my own ideas clear]:

1)I would like to refer to three vectors now. We represent a four dimensional space-time surface by f(x,y,z,t)=0 --------------- (1)
We may split the four dimensional space into a product of two spaces--a three dimensional space consisting of the variables x,y and z and a time space.The three dimensional space cannot have a surface restriction. Say it had some restriction like z=f1(x,y) .Then we could have substituted the value of z by f1(x,y) and reduced the number of variables in equation (1). It is a well known fact that a three dimensional region becomes a surface in the four dimensional space.What I would like to emphasize is that there is no surface restriction on the three dimensional region and parallel transportation of a three dimension vector should be quite "classical" in its mode.There should be no problem at all in adding or subtracting vectors at a distance.I would like you to consider the threads #40 ,#57 and #60
Links: https://www.physicsforums.com/showpost.php?p=2849247&postcount=40
https://www.physicsforums.com/showpost.php?p=2850358&postcount=57
https://www.physicsforums.com/showpost.php?p=2850394&postcount=60
2) The highlighted sentence towards the end of the first paragraph is a serious one in relation to four dimensional space-time . It stands out as a very strong argument.
3) In relation to what I have said in the first paragraph, if I distort the space-time surface slightly here and there but not at the starting point we may make the boundary conditions the same at the segment in the initial and final positions.
Obviously we end up with the same vector[with the same orientation,on the new surface].

Does it relate to curvature now?
Of course the matter may be defended by considering a small curve. Suppose we have a non-zero value of the Ricci tensor at some point on a small closed curve. We may still think of distorting the surface again to get the same orientation of the transported vector in the initial and final positions. If we try to make the surface too small we are in effect considering flat space-time.So what is the Ricci tensor measuring in reality? Probably this due to the fact that higher order terms are usually excluded from theory.Though it is inconvenient to include them,such exclusions may sometimes lead to problems[perhaps].

 

[Anamitra]

The Dot Product and Parallel Transport

We consider the parallel transport of a pair of distinct vectors a(mu) and b(nu) from the same initial point in a round trip.We assume,their directions do not coincide initially. As the pair moves along the curve the dot product is always preserved as a result of the the definition of parallel transport.When the pair returns to the initial position, not only does the dot product remain same but the values of the metric coefficients get restored.

We have,
g(mu,nu)a(mu)b(nu)=g(mu,nu)a'(mu)b'(nu)

=> a(mu)b(mu)=a'(mu)b'(mu) for all mu [Please see the attachment in thread #76]
For norm [a(mu)]^2=[a'(mu)]^2

Now the norm of each vector[which is the modulus of the of product with itself] remains constant as it moves along the curve. This is an important fact.Not only the angle between the vectors is preserved but their components seem to be preserved individually.[Thread #76]The components,it a seems,should not change.Did we parallel transport the parallel vector along the 45 degree latitude with sufficient accuracy? Or was there some human mistake somewhere?
Perhaps I am wrong. I am not claiming any thing with confidence. I am waiting for your approval or rejection.
Please go to the attachment in thread #76 for proof and clarification

 

[Passionflower]

This forum has a great feature that allows you to use Latex in a posting, why not use it instead of taking resources by attaching pdf files just for formulas? Personally I feel if someone has a question that requires extended formulas he should at least take the trouble to use Latex, it is very easy.

 

[Anamitra]

I have been having a big problem with the Latex toolbar provided by the forum. Every time I am writing something I am getting a different preview. But I am not having any such problem with the one I use on the MS word editor.I really don't know what to do and I am ready to take any advice.
Thank you.

 

[Passionflower]

I have been having a big problem with the Latex toolbar provided by the forum. Every time I am writing something I am getting a different preview. But I am not having any such problem with the one I use on the MS word editor.I really don't know what to do and I am ready to take any advice.

Just refresh as the preview is showing cached images.

 

[Anamitra]

In the uploaded attachment I have tried to prove that the vector components remain unchanged after a round trip over a smooth closed curve with the help of the dot product. This is relevant to thread #72.

[The summation in the attachment makes an interesting application of the fact that the metric coefficients vanish for unequal indices.]

 

[Passionflower]

I will start using the forum Latex tool bar at my earliest.

If you are not comfortable with the refresh as an alternative use a latex editor (such as Led) veryy that your equations are right and them simply cut and paste. There is actually no need to use the toolbar it is only there for convenience.

 

[Dale]

I thank you for the example.
[But then again if you refer to a chain of inertial states there is no problem at all]. I have a very strong point here.[Please refer to the attachment]

A local inertial frame is not unique. There are an infinite number of such frames. If you go on different chains of inertial states you may still wind up with different final inertial frames and different final vectors. A chain of locally inertial frames is not sufficient to establish uniqueness. The argument is not as strong as you seem to believe.

Now my further confusions[I am simply trying to get my own ideas clear]:

1)I would like to refer to three vectors now. We represent a four dimensional space-time surface by f(x,y,z,t)=0 --------------- (1)
We may split the four dimensional space into a product of two spaces--a three dimensional space consisting of the variables x,y and z and a time space.The three dimensional space cannot have a surface restriction. Say it had some restriction like z=f1(x,y) .Then we could have substituted the value of z by f1(x,y) and reduced the number of variables in equation (1). It is a well known fact that a three dimensional region becomes a surface in the four dimensional space.What I would like to emphasize is that there is no surface restriction on the three dimensional region and parallel transportation of a three dimension vector should be quite "classical" in its mode.There should be no problem at all in adding or subtracting vectors at a distance.I would like you to consider the threads #40 ,#57 and #60
Links: https://www.physicsforums.com/showpost.php?p=2849247&postcount=40
https://www.physicsforums.com/showpost.php?p=2850358&postcount=57
https://www.physicsforums.com/showpost.php?p=2850394&postcount=60

You can test this out for yourself. Simply take the Schwarzschild metric and do a similar exercise to what I suggested earlier. Set t and r to some constant values (r outside the event horizon) and calculate the parallel transport of a vector around some lattitude line, eg 45º.

Please work this problem through and post your work. After you have done that then we can discuss the dot product.

 

[Anamitra]

A local inertial frame is not unique. There are an infinite number of such frames. If you go on different chains of inertial states you may still wind up with different final inertial frames and different final vectors. A chain of locally inertial frames is not sufficient to establish uniqueness. The argument is not as strong as you seem to believe.

A local inertial frame is characterized by a unique diagonal matrix [1,-1,-1,-1]. But you can have many of them if they are moving with uniform velocity with respect to each other.But then again you can have a chain of inertial states.You can always select one set and get the same conclusion for each set

Now in thread #76, I have a small attachment where I have made some calculations based on the invariance of the dot product. The components don't seem to change.

One may think that if the norm of each vector remains constant and the whole system rotates preserving the angle between the vectors. But in my calculations in thread #76 the components themselves don't seem to change.. I would request the audience to look into the calculation..I have simply used the fact that the metric coefficients become zero if the indices are different ,that is,

g(mu,nu)=0 if mu not equal to nu

 

[Dale]

Please work the problem I posed.

 

[Anamitra]

Please goto thread number #84

 

[Dale]

I claim with full confidence ...

Then confidently work the problem. If you really are interested in learning then you will find it a valuable exercise. If you just have some sort of weird agenda to promote then I won't waste my time any longer.

 

[Anamitra]

I have made important modifications to thread #81 in response to what DaleSpam has said in #82

 

[Anamitra]

For parallel Transport along a curve:

dA(mu)/dt + Appropriate Christoffel symbol * dA(lambda)/dt * A(nu) = 0 ---- (1)

Solving these equations we have:

A(mu)=A(mu)(coordinates)

A(mu) is a function of coordinates.

Coordinates at the initial and final points are the same.
A(mu) should not change for parallel transport in a round trip

How is DaleSpam getting a different result for a smooth continuous curve?

In finding the constants of integration we have to use the boundary condition at the initial point.Only if sharp bends are involved new boundary conditions would come into the picture giving different solutions for different segments.Only in such cases the vector would reorient itself for a round trip.

 

[Anamitra]

FINAL CONCLUSIONS

1) Parallel transport of a vector along a smooth continuous closed curve should not change its orientation.
https://www.physicsforums.com/showpost.php?p=2857409&postcount=84
https://www.physicsforums.com/showpost.php?p=2855092&postcount=67

There could be the following exception to this rule say for the case of a sphere. . The same point(initial and final ) can be characterized by several values of the same coordinate.For instance on a line of latitude we may have phi=0,phi=360,phi=720 etc. at the same time. If the values of the component tensors solve out to be functions that change with these different values for the same point,it will produce tensors of different orientations after a round trip.[We may try to eliminate these terms with suitable choice of initial values]But this will never happen to surfaces that do not allow multiple values of the coordinates at each point or some point or for solutions that do remain the same for periodic changes of the coordinates
2) Parallel Transport of a vector along a closed curve with sharp corners may change its orientation
https://www.physicsforums.com/showpost.php?p=2855134&postcount=69
3)For curves which are not closed we can always devise a chain of inertial frames between the initial and final points. A vector referred to these frames does not change its orientation when transported parallely.[See first attachment]
4)For three dimensional vectors, we can add or subtract them ,even when they are at a distance.
https://www.physicsforums.com/showpost.php?p=2850394&postcount=60
https://www.physicsforums.com/showpost.php?p=2850765&postcount=62

My assertions in the article "Curved Space-time and the Speed of Light" are Valid
[I had inadvertently admitted to DaleSpams considerations. Now it is quite clear that I am on the Right track]

 

[Anamitra]

We consider the transport of two vectors A(mu) and B(nu) along a line of latitude say the 45 degree latitude. We start from the point P on the latitude. The first vector is moved along the line of latitude parallel to itself. The other vector is moved by parallel transport in a direction perpendicular to the line of latitude that is as we move it, the vector always points towards the north pole.
Both the vectors return to their original position without any change of orientation.
We now consider a third vector C(lambda) with initial position at P. We move the three vectors together by parallel transport the first two moving by the method described in the first paragraph. We observe the following points:
1)The norm of each vector remains the same, since dot product is preserved
2) The angle between each pair is preserved since the norm of each vector is preserved and the dot product between each pair is also preserved
3)If C(lambda) vector is constrained to lie on the surface of the sphere there is only one position for C(lambda).Its angle should not change either wrt to A (mu) or B(nu).

 

[Dale]

Both the vectors return to their original position without any change of orientation.

Show your work. The last three posts are just so much static noise until you do.

 

[Dale]

Regarding your rather premature "final conclusions" you have not made one single argument to support your point, nor have you shown any work to support it. You have made 3 irrelevant points that do not in any way establish the uniqueness of parallel transport:

1) You can define locally inertial frames along any path:
Not true in general - there are no locally inertial frames along a spacelike path
Insufficient - locally inertial frames are not unique so they do not establish the uniqueness of parallel transport

2) Parallel transport preserves the dot product:
Insufficient - rotation is an operation which preserves dot products but is not the identity operation, therefore merely preserving the dot product does not establish the uniqueness of an operation like parallel transport

3) You think parallel transport requires smooth curves:
Wrong - a sharp bend introduces a finite discontinuity which is integrated out as you can see in equation 3.1.19 in Wald, also you even cited figure 3.3 in Wald where he clearly does parallel transport around a curve with sharp bends

You have not even addressed most of these objections, nor have you overcome any of them. You have not demonstrated that you are correct in even a simple situation of either a sphere or a Schwarzschild metric. I have not seen any indication that you are actually interested in learning anything, but merely interested in repeating your completely wrong and unsubstantiated assertions.

 

[bcrowell]

In the case of "nearly flat" scenarios with no significant curvature where you have the objects paths sharing two events, you can compare their velocities, right?

You'll probably get better replies to this kind of thing if you start a separate thread, since this one degenerated a long time ago into a situation where knowledgeable, helpful people were failing to get through to Anamitra. But anyway, yes, if there's no curvature then you're dealing with SR, and in SR you can find relative velocities of distant objects.

 

[Anamitra]

I have uploaded the calculations DaleSpam asked for. I am getting exactly what I am supposed to get. Please note that in the calculations dA(theta)/d(phi) and dA(phi)/d(phi) at the initial point have been taken to be zero.

[ Actually the formulas/equations have been written on MS Word using a latex toolbar]

 

[Dale]

I am on a mobile device and cannot open the attachment. Can you put it in LaTeX? If not I will look at it in a couple of days when I return.

 

[Anamitra]

In my last [Thread #90]I got solutions of the type:
T(theta)=A cos[phi/root2] + B sin[phi/root2] ------------------------- (1)
And
T(phi)= C cos[phi/root2] + D sin [phi/root2] ---------------- (2)
[I have used T for the tensor instead of A]

To keep the norm constant we should have:
R^2 [ T(theta)^ 2+ [sin(theta)]^2 T(phi)^2]=const
We denote sin (theta) by K
R^2[T(theta)^2+k^T(phi)^2]= const
=>[T(theta)^2+k^T(phi)^2]= const
We may make the norm constant by the following choice
A^2+K^2 C^2 = B^2+K^2 D^2
K^2 CD=-AB
This may be achieved by choosing
A=B
C=-D
K^2 C^2=A^2
We have,
T(theta)/T(phi)=A[Cos (phi/root2) +Sin (phi/root2)]/C[Cos(phi/root2)-Sin (phi/root2)/]
For theta=0 , we have,
T(theta)/T(phi)=A/B
For theta =360 degrees
T(theta)/T(phi) obtains a different value
If norm is kept constant the vector is changing its orientation as it comes back to its original location.
Otherwise of course we may have variations, if the norm is allowed to change. We may write B=-A and D=-C to relate the constants. This gives the same value of T(theta)/T(phi) for theta=0 and theta=360 . At the same time the sign of each individual expression for T(theta) and T(phi) remains unaltered.

 

[Dale]

Hi Anamitra, I was able to look at your work this evening. The first page is correct. Starting on the second page you had a few mistakes.

First, you have a system of two first order differential equations, so there are only two unspecified constants of integration, not four. The correct form for the solution of this system is:
A^\theta=C_1 \; cos\left(\frac{\phi}{\sqrt{2}}\right)+\frac{C_2}{\sqrt{2}} \; sin\left(\frac{\phi}{\sqrt{2}}\right)
A^\phi=C_2 \; cos\left(\frac{\phi}{\sqrt{2}}\right)-\sqrt{2} \, C_1 \; sin\left(\frac{\phi}{\sqrt{2}}\right)

Then in the next section you got a little confused. When we set \phi=0 we get:
A^\theta(0)=C_1
A^\phi(0)=C_2

Plugging that back into the above we clearly see that
A^\theta(2 \pi) \ne A^\theta(0)
A^\phi(2 \pi) \ne A^\phi(0)
therefore the vector does not return to itself when parallel transported around a smooth curve, contrary to your above assertions.

Overall, you did fine setting up the equations, you just solved them wrong.

 

[Anamitra]

Yes you have got the constants so well by keeping the norm constant. In fact I have just now tested the norm by using the relation:

norm=R^2[A(theta)^2 + sin(theta)^2 A(phi)^square]
For your equations[solutions] the norm happens to be constant.In fact I have said this in my previous thread -that the vector changes its orientation if the the norm is kept constant. But I chose a different relation between the constants.


But in the case of light, if it is treated as a four vector how do we calculate its time component?[ I am asking this for my own understanding]

A Subsidiary Point

If along the line of latitude we make small adjustments of curvature to change the values of the metric coeffocients at some point or some lenghths (which could be small enough )we could generate a parallel transport that does not change the orientation of the vector after a round trip.In such a case the differential equations would change in different parts of the trip.The idea is to produce solutions which are periodic functions of 2*pi.But the inner part of the surface remains the same.If we make the enclosed surface very small to prevent this manoevre we are coming very close to flat space-time

 

[Dale]

Yes you have got the constants so well by keeping the norm constant. In fact I have just now tested the norm by using the relation:

norm=R^2[A(theta)^2 + sin(theta)^2 A(phi)^square]
For your equations[solutions] the norm happens to be constant.In fact I have said this in my previous thread -that the vector changes its orientation if the the norm is kept constant. But I chose a different relation between the constants.

Two quick points:

1) It is not simply a matter of choosing a different relation between the constants, you had too many constants. Any system of two first order differential equations has two constants, not four.

2) You are not free to choose a "different relation between the constants". If the norm does not remain constant then the dot product is not preserved and therefore it is not parallel transport.

With this counter-example using a smooth curve, do you now understand and agree that parallel transport is path-dependent? Do you have a better feeling for the geometric idea of curvature and the effect it has on parallel transport?

 

[Anamitra]

It is true that a relation between the constants is not sufficient--we need to have only two constants and not four.What physical condition did you use for that?We can assign arbitrary values to A(theta) and A(phi). But it is difficult to predict the initial value of the derivatives.Did you find the constants by the invariance of the dot product or you have you used some other physical condition? I am very much interested in knowing that for my own understanding.

Another point to be addressed is my Subsidiary Point in thread #94.

 

[Dale]

It is true that a relation between the constants is not sufficient--we need to have only two constants and not four.What physical condition did you use for that?We can assign arbitrary values to A(theta) and A(phi). But it is difficult to predict the initial value of the derivatives.Did you find the constants by the invariance of the dot product or you have you used some other physical condition?

The value of the two-dimensional vector A at anyone point along the curve (e.g. phi=0) provides the physical condition needed to determine the two constants of integration. There is no need to "predict the initial value of the derivatives". The invariance of the dot product falls out naturally from the definition of parallel transport and does not need to be added in by hand later, i.e. it is already in the equations that you correctly set up on the first page.

Regarding your subsidiary point: Even if you find an example of a closed path that maps the vectors back onto themselves that still does not make the result of parallel transport independent of the path in general.

 

[Anamitra]

Regarding your subsidiary point: Even if you find an example of a closed path that maps the vectors back onto themselves that still does not make the result of parallel transport independent of the path in general.

So parallel transport does not give us a report of curvature for every instance!

Next issue: How do we calculate the time component of the velocity of light if it is to be treated as a four vector?

 

[nismaratwork]

So parallel transport does not give us a report of curvature for every instance!

Next issue: How do we calculate the time component of the velocity of light if it is to be treated as a four vector?

The time component of light is 0, or the equal of its space components, but with an opposite sign to cancel. Whatever gets you a null-geodesic works in my view, but that's just my opinion.

 

[Dale]

Anamitra, I don't think we are ready to pursue any next issues yet. We still need to come to a resolution on the many previous issues:

Do you now understand how parallel transport is path dependent in curved spaces? Essentially, do you understand that your "final conclusions" post was wrong and (more importantly) why it was wrong?

Do you understand how different chains of inertial frames could lead to different results? Do you understand how the dot product could be preserved even though parallel transport is path-dependent?

On a less important note, are you still stuck on smooth paths or do you understand how a finite number of sharp bends is acceptable?