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Parallel propagator and covariant derivative of vector

  1. May 8, 2012 #1
    Hi all,

    I'm trying to figure out the link between the connection coefficients (Christoffel symbols), the propagator, and the coordinate description of the covariant derivative with the connection coefficients.

    As in http://en.wikipedia.org/wiki/Parall...ng_the_connection_from_the_parallel_transport one can write
    \nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - V_{\gamma(0)}}{h}= \frac{d}{dt}\left. \Gamma(\gamma)_t^0V_{\gamma(t)}\right|_{t=0}[/tex]
    However, we also know that
    \nabla_b V^a = \partial_b V^a + {\Gamma^a}_{cb}V^c.[/tex]
    I understand how, in some loose sense, one can think of the connection coefficients as the derivative of the parallel propagator:
    {\Gamma^a}_{cb} = \left.\frac{\partial}{\partial y^c}{[\Gamma(\gamma)^x_y]^a}_b\right|_{y \to x}.[/tex]
    However, I cannot see how to link the three together, and formalise things. Carrol's notes (http://preposterousuniverse.com/grnotes/grnotes-three.pdf [Broken]) give a hint, but without using the first sort of equation I gave--- he just jumps in with the second equation as an "assumption". What I'd ultimately be looking to do would be something like using the product rule:

    \nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - \Gamma(\gamma)^0_0V_{\gamma(0)}}{h} [/tex]
    \nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - \Gamma(\gamma)_0^0V_{\gamma(h)}+\Gamma(\gamma)_0^0V_{\gamma(h)}-\Gamma(\gamma)_0^0V_{\gamma(0)}}{h} [/tex]
    \nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - \Gamma(\gamma)_0^0V_{\gamma(h)}}{h}+\lim_{h\to 0}\frac{\Gamma(\gamma)_0^0V_{\gamma(h)}-\Gamma(\gamma)_0^0V_{\gamma(0)}}{h} [/tex]
    \nabla_X V =V_{\gamma(h)}\frac{d}{dt}\left.\Gamma(\gamma)^t_0\right|_{t=0} +\Gamma(\gamma)^0_0\frac{d}{dt}\left.V_{\gamma(t)}\right|_{h=0}
    So, that somehow
    \nabla_b V^a =V^c_{\gamma(0)}{\Gamma^a}_{cb} +\partial_b V^a_{\gamma(0)}
    However, this makes no sense, as it means you are subtracting vectors from different vector spaces (the whole reason the parallel propagator was introduced in the first equation I gave in this post). I've also jumped straight to coordinate components...

    Any bright ideas would be much appreciated.


    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. May 9, 2012 #2
    I just received an email that a post had been added, though it appears not to be here ???

    The post in the email read:

    As far as I am concerned, the propagator [itex]\Gamma(\gamma)_h^0[/itex] is a bitensor that takes a vector from [itex]T_{\gamma(h)}(M)[/itex] to [itex]T_{\gamma(0)}(M)[/itex]. I am not an expert by any means, but I guess they are linear in the sense that
    [tex]\Gamma(\gamma)_h^0 \left(V_{\gamma(h)} + W_{\gamma(h)}\right) = \Gamma(\gamma)_h^0\left(V_{\gamma(h)}\right)+ \Gamma(\gamma)_h^0\left(W_{\gamma(h)}\right)[/tex]

    Does that make sense?

  4. May 9, 2012 #3


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    I don't think the propagator can be considered a bitensor because it takes a vector and maps it to a vector in another space whereas a tensor "lives" in one space and so acts only one vectors and oneforms in that one space. The parallel propagator is simply a mapping between different vector spaces.

    I'm not entirely sure what you're going for.
  5. May 9, 2012 #4
    To quote Eric Poisson http://relativity.livingreviews.org/open?pubNo=lrr-2011-7&page=articlese15.html [Broken]

    I see your point, but is not a bitensor another way of describing a map from one vector space V1 to another V2? It "lives" in both spaces. For example you could write the one--form part of the bitensor as a one--form in V1*, and the vector part as a vector in V2...

    I need to read the above link in more detail (having just found it), but for reference, here are the notes I have been using: http://msor.victoria.ac.nz/twiki/pub/Courses/MATH465_2012T1/WebHome/notes-464-2011.pdf [Broken]

    See section 3.2 on the parallel propagator. (It is called a bitensor here.)

    OK, I guess I could have been clearer. I have two expressions for the covariant derivative:

    [tex]\nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - V_{\gamma(0)}}{h}= \frac{d}{dt}\left. \Gamma(\gamma)_t^0V_{\gamma(t)}\right|_{t=0}[/tex]


    [tex]\nabla_b V^a = \partial_b V^a + {\Gamma^a}_{cb}V^c.[/tex]

    I simply want to get from the first to the second.
    Last edited by a moderator: May 6, 2017
  6. May 9, 2012 #5
    Right, let's see if this works. Firstly, since we're working with bitensors, I'll expand on what I hinted at in the last post. To make things easier notationally, let the propagator be [itex]\Gamma(t \to t_0;\gamma)[/itex], which takes [itex]T_{\gamma(t)}(M)[/itex] to [itex]T_{\gamma(t_0)}(M)[/itex]. Then we can write it as

    [tex]\Gamma(t \to t_0;\gamma) = \Gamma^{a_{t_0}}_{b_t} e_{a_{t_0}} \otimes \omega^{b_t}[/tex]

    with the notation on the indices to indicate whether they are at [itex]\gamma(t_0)[/itex] or [itex]\gamma(t)[/itex]. That is [itex]e_{a_{t_0}}[/itex] are the basis vectors at [itex]\gamma(t_0)[/itex] and [itex]\omega^{b_t}[/itex] are the dual basis vectors at [itex]\gamma(t)[/itex].


    [tex]\Gamma(t \to t_0;\gamma)V_{\gamma(t)} = \Gamma^{a_{t_0}}_{b_t} e_{a_{t_0}} \otimes \omega^{b_t}\left(V^{c_t}e_{c_t}\right) = V^{b_t}\Gamma^{a_{t_0}}_{b_t}e_{a_{t_0}}[/tex]

    for some vector [itex]V_{\gamma(t)}[/itex]. Therefore

    \left.\frac{d}{dt}\Gamma(t \to t_0;\gamma)V_t \right|_{t=t_0}=\left[\frac{d}{dt}\left(V^{b_t}\right)\Gamma^{a_{t_0}}_{b_t} + V^{b_t} \frac{d}{dt} \left(\Gamma^{a_{t_0}}_{b_t}\right)\right]_{t=t_0}e_{a_{t_0}}[/tex]

    Letting [itex]x=x(t)[/itex] be the local coordinates, then
    [tex]\frac{d}{dt}\left(V^{b_t}\right) = \frac{dx^{c_{t_0}} }{dt} \partial_{c_{t_0}} V^{b_t}[/tex]
    [tex]\frac{d}{dt}\left( \Gamma^{a_{t_0}}_{b_t} \right) = \frac{dx^{c_{t_0}} }{dt} \partial_{c_{t_0}} \Gamma^{a_{t_0}}_{b_t}.[/tex]


    \left.\frac{d}{dt}\Gamma(t \to t_0;\gamma)V_t \right|_{t=t_0}=\frac{dx^{c_{t_0}} }{dt}\left[
    \partial_{c_{t_0}} V^{b_t}\Gamma^{a_{t_0}}_{b_t} +
    V^{b_t} \partial_{c_{t_0}} \Gamma^{a_{t_0}}_{b_t}

    Given that

    \left.\Gamma^{a_{t_0}}_{b_t}\right|_{t=t_0} = \delta ^{a_{t_0}}_{b_{t_0}}[/tex]

    and defining

    {\Gamma^{a_{t_0}}}_{b_{t_0}c_{t_0}}\equiv\left. \partial_{c_{t_0}} \Gamma^{a_{t_0}}_{b_t}\right|_{t \to t_0}[/tex]


    \left.\frac{d}{dt}\Gamma(t \to t_0;\gamma)V_t \right|_{t=t_0}=X^{c_{t_0}} \left[
    \partial_{c_{t_0}} V^{a_{t_0}} +

    or, dropping all the [itex]t_0[/itex]

    [tex]\nabla_X V = \frac{d}{dt}\left. \Gamma(t\to t_0;\gamma)V_{\gamma(t)}\right|_{t=0}
    =X^c \left[
    \partial_c V^a +
    X^c \nabla_c V^a e_a.

    That seems alright for me, but I understand I've skipped over a few of the nuances along the way. However, I suspect if I do things in the coincidence limit of the bitensors (or something like that) this will turn out to be correct.


  7. May 10, 2012 #6
    Yes, I deleted it, I thought you were referring to the QFT propagator. When I read your link I realized it was a different thing.
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