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More questions on parallel transport

  1. Sep 9, 2014 #1
    In my recent studies of curvature, I worked with the Riemann tensor and the equation:

    ([itex]\delta[/itex]V)a= A[itex]\mu[/itex]B[itex]\nu[/itex]Rab[itex]\mu[/itex][itex]\nu[/itex]Vb

    Now previously, I worked in 2D with the 2 sphere. While doing so, I learned that if I set my x1 coordinate to be θ and my x2 coordinate to be ø, then the vectors that serve to be the legs of the loop that I am transporting around would be as follows:

    A[itex]\mu[/itex] = <θ, 0>

    B[itex]\nu[/itex] = <0, ø>

    and then of course the vector that I parallel transport would be as follows:

    Vb= <θ, ø>

    Now this may work for 2 dimensions, but what if I have 3 or more dimensions? With only 2 vectors being the legs of the loop, there wouldn't be enough vectors for me to give each individual coordinate its own leg with every other component being 0 (as shown above with A[itex]\mu[/itex] and B[itex]\nu[/itex]).

    How do I deal with this? Is it even a requirement for every coordinate to have its own leg that is reminiscent of a unit vector? Page 5 on the following PDF gave me the impression that it is a requirement:


    Is it possible for one of the legs of the loop to have more than one type of coordinate in it (like A[itex]\mu[/itex] = <r , θ, 0>) ?
  2. jcsd
  3. Sep 9, 2014 #2


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    Staff: Mentor

    Yes, and indeed that's usually the case; the Riemann tensor tells you how a vector is affected (usually differently) by [parallel transport along a closed curve with any orientation. If you have more than one non-zero component in the A aor B vectors, then you just end up with additional terms when you do the implied summation on ##\mu## and ##\nu##.
    Last edited: Sep 9, 2014
  4. Sep 9, 2014 #3


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    Staff Emeritus
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    You have several options when you want to evaluate a loop that doesn't point in the direction of the coordinate basis vectors.

    The easiest way is probably to take advantage of the linearity property of tensors.

    Suppose ##A^\nu = C^\nu + D^\nu## and ##B^\mu = D^\mu + E^\mu##, where we make C,D,E,and F some multiples of the coordinate basis vectors

    A^\nu B^\mu R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} = [(C^\nu + D^\nu)(E^\mu + F^\mu)]R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} =

    [C^\nu E^\mu + C^\nu F^\mu + D^\nu E^\mu + D^\nu F^\mu]R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} =
    C^\nu E^\mu R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} + C^\nu F^\mu R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} + D^\nu E^\mu R^{\rho}{}_{\sigma \mu \nu} V^{\sigma} + D^\nu F^\mu R^{\rho}{}_{\sigma \mu \nu} V^{\sigma}

    Then you just need to evalutate the for terms for CE, CF, DE, and DF which you can do by inspection since they align with the coordinate axis.

    Other methods include keeping tract of all the components, or transforming the basis vectors of the tensor by the tensor transformation rules.
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