Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [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]
    However, we also know that
    [tex]
    \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:
    [tex]
    {\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:

    [tex]
    \nabla_X V = \lim_{h\to 0}\frac{\Gamma(\gamma)_h^0V_{\gamma(h)} - \Gamma(\gamma)^0_0V_{\gamma(0)}}{h} [/tex]
    [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]
    [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]
    [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}
    [/tex]
    So, that somehow
    [tex]
    \nabla_b V^a =V^c_{\gamma(0)}{\Gamma^a}_{cb} +\partial_b V^a_{\gamma(0)}
    [/tex]
    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.

    Cheers,

    Ianhoolihan.
     
    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?

    Ianhoolihan
     
  4. May 9, 2012 #3

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    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]

    and

    [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].

    Hence

    [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

    [tex]
    \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]
    and
    [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]

    Therefore,

    [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}
    \right]_{t=t_0}e_{a_{t_0}}[/tex]

    Given that

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

    and defining

    [tex]
    {\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]

    then

    [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}} +
    {\Gamma^{a_{t_0}}}_{b_{t_0}c_{t_0}}V^{b_{t_0}}
    \right]e_{a_{t_0}}[/tex]

    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 +
    {\Gamma^a}_{bc}V^b
    \right]e_a
    =
    X^c \nabla_c V^a e_a.
    [/tex]

    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.

    Thoughts?

    Ianhoolihan
     
  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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Parallel propagator and covariant derivative of vector
  1. Covariant derivative (Replies: 8)

Loading...