Curved Space-time and Relative Velocity

[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?

 

[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.

We may have an infinite number of inertial frames at each point and therefore we may have several such chains connecting a pair of initial and final point on the space time surface. For each such chain dA(mu)/d(xi)=0.For a "particular inertial frame" at the initial point we may choose several inertial paths connecting the initial point to the final point.We end up with the same tensor finally.If we change the inertial frame at the initial point we adjust the "inertial paths". There should be no problem in such a procedure.

Of course parallel transport is causing a lot of problem in defining Relative velocity if we leave aside the aspect of the chain of inertial frames.
Now let us come to a relevant issue.If somebody sees a moving car at some distance in front of him in curved space-time he should have some idea/report of the motion.If there are problems in defining relative motion in curved spacetime[I am assuming this for argument's sake] it does not mean that relative motion is meaningless.I tried to highlight this in Query 1 of thread #1.The incapability of the mathematical apparatus in defining a physical quantity[in case such an incapability exists] does not imply the non-existence of the physical quantity itself.

 

[Anamitra]

An Important Point

For flat space-time the equation for parallel transport is given by:

dA(mu)/dx(i)=0 as we move along the path[The christoffel tensors are equal to zero]

So the components should not change individually .But this not true for spherical or polar coordinates. Though the vector in the final state remains parallel to its initial position the components do change for a path that is not closed.
"Gravity" by James Hatley,page 456,Figure 20.3
[Chapter 20,Section 20.4:The covariant derivative]

 

[Dale]

We may have an infinite number of inertial frames at each point and therefore we may have several such chains connecting a pair of initial and final point on the space time surface. For each such chain dA(mu)/d(xi)=0.For a "particular inertial frame" at the initial point we may choose several inertial paths connecting the initial point to the final point.We end up with the same tensor finally.

Not in general, no. If you cannot accept this then you need to do some homework problems from your favorite GR textbook. Wald's problems may not be of the "practical" sort that you need.

I don't see any reason to move to other points when you are still stuck on the basics. If we cannot agree on the simple and obvious mathematical fact of the non-uniqueness of parallel transport in curved spaces then any other discussion will be pointless.

 

[Dale]

An Important Point

For flat space-time the equation for parallel transport is given by:

dA(mu)/dx(i)=0 as we move along the path[The christoffel tensors are equal to zero]

No, the equation for parallel transport remains the same in flat spacetime. Some coordinate systems in flat spacetime have non-zero Christoffel symbols as you point out, therefore you cannot simply drop them.

However, in flat spacetime it is always possible to perform a global coordinate transform to a standard Minkowski inertial frame where the Christoffel symbols are all 0 and the parallel transport equation simplifies as you propose. In this simplification you immediately see that the parallel transport equation becomes independent of the path in a Minkowski inertial frame, and therefore (since the covariant derivative is a tensor operation) it is true in any coordinate system in flat spacetime.

[nitpick]The Christoffel symbols are not tensors. The correct terminology is "Christoffel symbols".[/nitpick]

 

[Anamitra]

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.

On Parallel transport:For parallel transport we have simply come to the conclusion that for a round trip the orientation of the vector may change or it may not change even for a curved surface.So the relationship between parallel transport and the curvature is not as strong or conclusive as we might be tempted to think of.

A fundamental issue has been raised in thread #101 in the second paragraph. It has not been addressed.

Regarding Thread #102: Yes ,there has been a mistake