# Curved Space-time and Relative Velocity

by Anamitra
Tags: curved, relative, spacetime, velocity
 P: 621 "Again, this is demonstrably false; DrGreg provided a very detailed counter example. http://www.physicsforums.com/showpos...2&postcount=22 Starting from the same point the components of the final parallel-transported vector are in fact changed depending on the path. This is fundamental to understanding the very basic concept of curvature, and instead of trying to learn it you are just going around in circles making the same false assertion over and over." Dalespam, Thread 35 This "demonstrably false" notion arises out of the fact that the singularities north and south poles have been chosen simultaneously.I have demonstrated my reason as to why they should not be chosen simultaneously in thread #15 ,http://physicsforums.com/showpost.ph...8&postcount=15 If you chop off either the north or the south pole from the sphere the demonstration provided by Dr Greg fails
P: 621
 Quote by DrGreg Remember geodesics in spacetime do not connect points in space, they connect events in spacetime. The decision as to which spacetime geodesic the particle will follow is determined entirely by the particle's velocity (or to be more precise and coordinate independent, by its 4-velocity tangent 4-vector). The point is, given an event, there are lots of particles that can pass through that event on a geodesic, and no way to single out one of those geodesics as being the "correct" one (unless you believe in aether).
Dr Greg has considered here several geodesics emanating from the same point. This is possible. If all these geodesics terminated on the same point we encounter a serious discrepency as shown in--->http://physicsforums.com/showpost.ph...8&postcount=15

 Quote by DaleSpam Thanks for the great explanations DrGreg The Schwarzschild solution has all of the symmetry of a 2-sphere plus a lot of symmetries that the sphere does not have. These types of problems are actually more of an issue in 4D, not less.
In his example Dr Greg has used spatial geodesics!
P: 8,430
 Quote by Anamitra Dr Greg has considered here several geodesics emanating from the same point. This is possible. If all these geodesics terminated on the same point we encounter a serious discrepency as shown in--->http://physicsforums.com/showpost.ph...8&postcount=15
Anamitra, look at page 118 from Relativity on Closed Manifolds--the diagram clearly shows two geodesics which intersect at two different points, and the text even gives a name for this phenomenon:
 When two neighboring geodesics intersect twice, the points of intersection are termed conjugate
Similar diagrams and discussions can be found in this book and this one.
 P: 621 A "Seriously Heavy Point" It is very important to consider the addition/subtraction of three velocities. If I am standing at Point A and I see a light ray flashing past past another B I would be interested in the three velocity of the light ray[my own three velocity being the null vector]. This is relevant to the issue in thread#6 --->http://physicsforums.com/showpost.ph...10&postcount=6 In threads #19 and #25 [http://physicsforums.com/showpost.ph...7&postcount=19 , http://physicsforums.com/showpost.ph...ostcount=25]of "Curved Space-time and the Speed of Light" DaleSpam has tried to counter the concept of Relative Velocity in Curved Space-time in by giving examples in relation to 4-D space,I mean by referring to four vectors.[ Of course these examples have failed in their mission] He has kept silent on the issue of the addition/subtraction of 3-velocities! The issue of subtraction of three velocities at a distance,especially when one is null, is extremely relevant to the discussion in thread #6 I am saying all this not to give any extra fortification to what ever I have said in relation to 4D considerations but because these points are seriously heavy. My assertions in relation to 4D concepts are strong enough to stand on their own feet.
Mentor
P: 15,586
 Quote by Anamitra This "demonstrably false" notion arises out of the fact that the singularities north and south poles have been chosen simultaneously.
There are no singularities on a sphere, the curvature is everywhere finite.

 Quote by Anamitra If you chop off either the north or the south pole from the sphere the demonstration provided by Dr Greg fails
No. Because of the symmetry you can do this from any point on the sphere, it is just easier to describe verbally from the poles.

 Quote by Anamitra In his example Dr Greg has used spatial geodesics!
So what? Spacelike paths are perfectly acceptable paths for parallel transport and need not even be geodesic.

Anamitra, you don't seem to understand the very basics of parallel transport and intrinsic curvature. The most important example of parallel transport is to transport a vector around a closed loop back to its original position (NB a loop is generally a non-geodesic path). In a curved space the parallel transported vector will be rotated from the original vector by an amount which depends on the area enclosed by the loop as well as the direction of the loop. The Riemann curvature tensor describes exactly this property of curved space in the limit of infinitesimal loops.

In parallel transport the covariant derivative which is zero is the covariant derivative of the transported vector, the path along which it is transported need not have a zero covariant derivative, and need not even be smooth.

If you are so stuck on your preconcieved notions that you are not willing to learn these basic and fundamental geometric concepts then you may as well just stop even attempting to learn general relativity as it will be completely futile. I would recommend that you view Leonard Susskind's lectures on General Relativity which are available on YouTube.
 P: 621 The Sphere Again! Let us consider the example of the spherical space-time surface in a mathematical way: We calculate the distance[space-time separation] between the north pole and the south poles along the meridians and of course for a sphere we get the same value.One should use the relation ,space -time separation=integral ds along any meridian.Now if by some suitable trans formation we change the sphere to some other surface.The "meridians" will have different "lengths". The same pair of events[4D events] will have different separations. Since the sphere is full of antipodal points,better to leave aside the example of the sphere. But these were 4-D considerations.Nevertheless I have a big interest in the 3-D issues I have specified in Thread #40, the seriously heavy points. [NB: ds represents physical length on a space-time surface]
 PF Patron Sci Advisor Emeritus P: 5,309 Anamitra, three people have been spending a lot of time trying to help you. In my opinion, all three know general relativity pretty well. You would be well advised to get out of this mode where you feel you have to defend a position you've staked out. It's not going to serve you well in learning general relativity.
 P: 621 I am very much interested in receiving replies in regard to Thread#40 Regarding Thread#42: It is true that in many standard texts we have examples of several geodesics connecting a pair of space-time points. If they happen to be of unequal lengths is there going to be any problem, in the sense that space-time separation for the pair is no more unique?I am keen on receiving some answer to this issue so that I can improve my knowledge. This is just a request.
 Mentor P: 15,586 Anamitra, regarding post 40: You cannot compare vectors unless they are in the same vector space. The tangent space at each point in a manifold is a different vector space, and it is only once you have mapped the vector in one tangent space to a vector in the other tangent space that you can make any comparisons. The process for doing this is called parallel transport. Parallel transport must come first, before any other vector operation is possible. Now, please address post 41.
Mentor
P: 15,586
 Quote by Anamitra We calculate the distance[space-time separation] between the north pole and the south poles along the meridians and of course for a sphere we get the same value.One should use the relation ,space -time separation=integral ds along any meridian.Now if by some suitable trans formation we change the sphere to some other surface.The "meridians" will have different "lengths".
This is not true. ds is an invariant quantity (a tensor of rank 0), so it remains unchanged under any coordinate transformation.
P: 621
 Quote by DaleSpam Anamitra, you don't seem to understand the very basics of parallel transport and intrinsic curvature. The most important example of parallel transport is to transport a vector around a closed loop back to its original position (NB a loop is generally a non-geodesic path). In a curved space the parallel transported vector will be rotated from the original vector by an amount which depends on the area enclosed by the loop as well as the direction of the loop. The Riemann curvature tensor describes exactly this property of curved space in the limit of infinitesimal loops. In parallel transport the covariant derivative which is zero is the covariant derivative of the transported vector, the path along which it is transported need not have a zero covariant derivative, and need not even be smooth.
We consider the definition of the covariant derivative:

covariant derivative= dA(mu)/dx(i) + affine connection part
Now How does one calculate dA(mu)/dx(i) on a sharp bend? I am quite confused

Now on to the aspect of the Rimannian curvature tensor.

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
If such a procedure defines the curvature of a surface in a proper manner it really does not contradict any thing.

But if one is interested in the parallel transporting a vector at a stretch along a curve it should not be one with sharp bends. In such an instance it cannot be called parallel transport in the totality of the operation.
P: 621
 Quote by DaleSpam This is not true. ds is an invariant quantity (a tensor of rank 0), so it remains unchanged under any coordinate transformation.

This is correct and I do not have any means to reject it.

But I will place certain questions to clarify my own concepts and not to contradict any body.

Well the length of any line connecting a pair of points and lying on the space time surface seems to be the space-time separation between them[s = integral ds along the said line]

Now if I take a line from the south to the north pole winding it several times on the body of the sphere is its length going to represent the space-time separation between the poles? Do we need geodesics to calculate space-time separations?

A subsidiary issue:

We may represent the space time sphere by the equation

x^2+y^2+t^2=a^2
I have taken the z-axis to be the time axis(t). Any motion perpendicular to the time axis represents infinitely fast motion forbidden by relativity.So the meridian perpendicular the time axis goes off.Keeping the axes fixed we may rotate the aforesaid plane[perp. to the time axis and going through the origin] about the x or y axis and remove a huge number of meridians. Of course a huge number of meridians do . One may think of chopping off certain parts of the sphere using the light cone .Fact remains,I am confused.One may consider the meridian in the x-t plane.A particle moving round and round along it has "oscillatory time". It it were a human being his age would undergo periodical movement in the forward and backward directions of time.If the particle stays quiet at one point time would not flow. I am again confused.

[Lines of latitude perpendicular to the time-axis have to disappear to prevent infinitely fast motion. It seems ,that the sphere is in a certain amount of trouble]
P: 1,555
 Quote by Anamitra Well the length of any line connecting a pair of points and lying on the space time surface seems to be the space-time separation between them[s = integral ds along the said line] Now if I take a line from the south to the north pole winding it several times on the body of the sphere is its length going to represent the space-time separation between the poles?
If you mean by 'space-time separation' the metric distance then only a geodesic represents that.
Mentor
P: 15,586
 Quote by Anamitra 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.