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Curved Spacetime and Relative Velocity 
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#37
Aug2210, 05:43 PM

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"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 paralleltransported 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 


#38
Aug2210, 05:55 PM

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#39
Aug2210, 06:13 PM

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#40
Aug2210, 06:37 PM

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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 Spacetime and the Speed of Light" DaleSpam has tried to counter the concept of Relative Velocity in Curved Spacetime in by giving examples in relation to 4D 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 3velocities! 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. 


#41
Aug2210, 09:14 PM

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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 nongeodesic 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. 


#42
Aug2210, 09:55 PM

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The Sphere Again!
Let us consider the example of the spherical spacetime surface in a mathematical way: We calculate the distance[spacetime 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 4D considerations.Nevertheless I have a big interest in the 3D issues I have specified in Thread #40, the seriously heavy points. [NB: ds represents physical length on a spacetime surface] 


#43
Aug2210, 10:34 PM

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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.



#44
Aug2210, 11:24 PM

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I think you are asking about conjugate points. This is entertaining http://hawking.org.uk/oldsite/pdf/time.pdf



#45
Aug2310, 02:50 AM

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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 spacetime points. If they happen to be of unequal lengths is there going to be any problem, in the sense that spacetime 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. 


#46
Aug2310, 07:44 AM

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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. 


#47
Aug2310, 10:02 AM

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#48
Aug2310, 11:20 AM

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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. 


#49
Aug2310, 12:50 PM

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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 spacetime 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 spacetime separation between the poles? Do we need geodesics to calculate spacetime 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 zaxis 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 xt 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 timeaxis have to disappear to prevent infinitely fast motion. It seems ,that the sphere is in a certain amount of trouble] 


#50
Aug2310, 12:57 PM

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#51
Aug2310, 02:05 PM

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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 pathdependent. 


#52
Aug2310, 02:28 PM

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#53
Aug2310, 02:53 PM

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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 "spacetime separation" in GR.
But, as it relates to this thread, parallel transport need not be restricted to geodesics. 


#54
Aug2310, 03:37 PM

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