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Exercise [14.05]b http://www.roadtoreality.info/viewtopic.php?f=19&t=1590 
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Author:  Roberto [ 22 Jun 2010, 09:08 ]  
Post subject:  Exercise [14.05]b  
Attached my solution, with an approach different from Robin's one.

Author:  DimBulb [ 19 Dec 2010, 23:07 ]  
Post subject:  Re: Exercise [14.05]b  
I am confused. See attached

Author:  Roberto [ 25 Dec 2010, 20:38 ]  
Post subject:  Re: Exercise [14.05]b  
I tried to answer in attachment.

Author:  DimBulb [ 29 Dec 2010, 22:12 ] 
Post subject:  Re: Exercise [14.05]b 
Thanks for answering. I am thinking that the difference between the two covariant derivatives of a vector field along a path that starts out in direction w would be due to the different curvatures of the paths. The two paths may have the same direction at the point where the covariant derivatives are being taken, and so have the same w, but if the curvature of the paths are different, you will get different "Gamma" factors. Taking the example of the surface of a sphere, if the first path is a geodesic or along a great circle of the sphere, the path is "straight" and then the "Gamma" factor would be 0. If however the second path were along a (nongreat) circle tangent to the first path at the point where the covariant derivatives are being taken, the w direction would be the same as the straight path, but the second path would have a curve to it, which would cause the "Gamma" factor to be nonzero. Does this make any sense? 
Author:  Roberto [ 05 Jan 2011, 10:52 ] 
Post subject:  Re: Exercise [14.05]b 
Quote: I am thinking that the difference between the two covariant derivatives of a vector field along a path that starts out in direction w would be due to the different curvatures of the paths. In my opinion, you are not quite right. The connection coefficients Gamma are a function of the point only, not of the curvature of the path. The reason for having more covariant derivatives is simply that there is some freedom in defining how "parallel transport" can be made and therefore what "covariant derivative" exactly is. Quote: Taking the example of the surface of a sphere, if the first path is a geodesic ...Does this make any sense? In order to be able to speak of a geodesic, you must have defined a metric. But a metric defines uniquely the connection (i.e. the covariant derivative), see chapter 14.7 and exercise 14.28. So after having defined a metric you have lost the freedom I mentioned above and in your example there is just one covariant derivative, defined by the metrics. And the Gamma are nonzero because there is a surface curvature, independently from the path curvature. By the way, probably we should have this discussion in the Exercise Discussion section of the forum, not in the Solutions section. 
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