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Parallel transport and geodesics 
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#1
Oct1609, 09:34 AM

P: 16

A vector field is parallel transported along a curve if and only if the the corariant derivative of the vector field along the path is 0. That is
[tex]\frac{d}{d\lambda} V^\mu + \Gamma^\mu_{\sigma \rho} \frac{dx^\sigma}{d\lambda} V^\rho = 0[/tex] This is basically what every book says. But what exactly does it mean? 1) if you have a vector (field), there is a specific set of paths, that  when you transport the vector along them  will leave the vector 'unchanged'. That means that there are certain paths that change the vector i.e. stretches it or turns it or whatever. 2) the equation describes how the vector (field) should change when you move it along a certain path (any path you choose) in order to keep it constant with respect to the connection. I can see arguments for both interpretations and it's driving me crazy. ;) In flat space space you can parallel transport a vector in any direction you want  there are no right or wrong paths. This speaks for 2). A geodesic is a path that parallel transports a vector that is the tangent vector to the path itself. This speaks for 1) (i.e. for a vector there is a specific parallel transport path, for the tangent vector to the path this is the path itself) If you have a sphere and parallel transport the vector [tex]V = (1,0)[/tex] around a circle of constant [tex]\theta[/tex] (altitude) the vector changes like [tex]V(\theta,\phi) = \left(\cos(\phi \cos\theta), \frac{1}{\sin \theta}\sin(\phi \cos\theta)\right)[/tex] This speaks for 2) as you can clearly choose any path (any altitude (except the north/south pole of course)). It also shows the vector is not constant with respect to the path (the path parameter being [tex]$\lambda = \phi$[/tex]) as it changes when you go round. So...which one is it? What point am I missing? ;) 


#2
Oct1609, 10:18 AM

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Vectors can be parallel transported in any direction. There are no preferred directions for parallel transport.



#3
Oct1609, 10:29 AM

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In (1), you seem to be implying that somehow it is that path that changes the vector. The vector is defined in the tangent space at a point p. If you have a path through p, this vector may be parallel transported along it to another point. You also refer to vector fields instead of vectors, so again I don't quite understand what you're implying.



#4
Oct1609, 03:55 PM

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P: 7,659

Parallel transport and geodesics
The formal definitions of parallel transport are very good for calculation, but for understanding the idea I prefer geometrical constructs, like Schild's ladder.
http://en.wikipedia.org/wiki/Schild%27s_ladder Consider a parallelogram  if opposing sides are of equal lengths, the opposing sides are parallel. And we know how to calculate lengths if we have a metric, so there isn't any problem in the notion of "sides of equal length". So if we have a metric, the idea that the sides of a parallelogram "should be" parallel if the parallelogram is small enough gives a unique idea of parallel transport. As long as you don't mind representing vectors as line segments, that is..... 


#5
Oct1709, 08:08 AM

P: 707

If you parallel transport a vector around a closed curve then the vector may come back to a different vector even in a flat space. Also the change in the vector when it returns to the starting point may depend on the curve, even in a flat space. 


#6
Oct1709, 11:39 AM

P: 3,967

It might help to think of a imaginary parallel transport vehicle. This vehicle is designed so that whenever the vehicle turns x degrees clockwise, the vector it is transporting turns x degrees anticlockwise (or vice versa) so that on a two dimensional plane the transported vector always remains parallel to its starting position. Now imagine the following journey. The vehicle starts at the equator and heads to the North pole. At the North pole it makes a 90 degree turn clockwise and heads South back towards the equator. At the equator it makes another 90 degree turn clockwise and heads back to its starting point and makes a final 90 degree turn so that the vehicle is aligned with its original heading. The vector it was transporting is now rotated 270 degrees anticlockwise relative to its starting position. The internal angles of the triangle formed by this journey add up to 270 degrees rather than the usual 180 degrees we would normally expect for a triangle in flat space. If this vehicle has its wheels locked in the straight ahead position it will be forced to follow a great circle or geodesic on the globe and its transported vector will not precess from its original direction when it returns to its starting point. Now if the vehicle follows a line of latitude it has to turn its front wheels to stay on the line of latitude because if it tries to keep pointing straight ahead it will be forced to deviate from the line of latitude. This turning of the front wheels of the vehicle to stay on the line of latitude, causes the vector it is transporting to precess. In a nutshell, deviating from a geodesic path causes precession of the parallel transported vector. I think the source of your confusion is the use of the term "parallel transport" for something that is not necessarily parallel transported in the casual sense. Page 26 of this document http://www.shef.ac.uk/physics/teachi..._lecture_8.pdf shows a nice trick with a cone on sphere that may help visualise the situation. 


#7
Oct1709, 01:37 PM

P: 16

The change of coordinates that they calculate, describe how the vector will look, when it has been transported to a new point, from the new point's view, right? Also, the figure at page 24 is sort of how I thought a tangent vector to the equator would be parallel transported around the equator  i.e. along its own geodesic. I thought that transport around a circle in flat space (which is sort of like the top part of the sphere) would keep the vector pointing 'up' (whith up I here mean at 12 o'clock in the drawing). 


#8
Oct1709, 03:16 PM

P: 3,967

The equation that they give for the precession of a vector transported along a line of latitude is: [tex]\alpha = \phi*\cos(\theta)[/tex] If the vector is transported half a circle around a line of latitude the equation becomes: [tex]\alpha = \pi*\cos(\theta)[/tex] At the equator [itex]\theta[/itex] is 2*pi so [itex]\alpha = 0[/itex] so there is no precession and the vector is pointing in the same direction as the vehicle. Very near the North pole [itex]\theta[/itex] is close to zero (say 0.1 radians) so for a half circle path [itex]\alpha = \pi*\cos(0.1) = 0.995*\pi [/itex] radians, so by the time the vehicle has travelled to the 9 o'clock position in the diagram the vector will have been rotated nearly half a turn clockwise relative to the vehicle (and in the plane tangential to the surface the vehicle is on) and still pointing roughly in the 12 o'clock direction. (The vector is now pointing roughly to the rear of our imaginary vector transportation vehicle.) 


#9
Oct1909, 02:45 PM

P: 16

Thank you for your answers  I think my understanding has improved. :)



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