1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Parallel transport approximation

  1. Mar 2, 2008 #1
    The parallel transport equation is

    [tex]
    \frac{d\lambda^{\mu}}{d\tau} = -\Gamma^{\mu}_{\sigma\rho} \frac{dx^{\sigma}}{d\tau} \lambda^{\rho}
    [/tex]

    If I take the derivative of this with respect to tau, and get

    [tex]
    \frac{d^2\lambda^{\mu}}{d\tau^2} = -\partial_{\nu}\Gamma^{\mu}_{\sigma\rho}\frac{dx^{\nu}}{d\tau}\frac{dx^{\sigma}}{d\tau} \lambda^{\rho} \;- \;\Gamma^{\mu}_{\sigma\rho}\frac{d^2x^{\sigma}}{d\tau^2}\lambda^{\rho} \;-\; \Gamma^{\mu}_{\sigma\rho}\frac{dx^{\sigma}}{d\tau}\frac{d\lambda^{\rho}}{d\tau}
    [/tex]
    [tex]
    = \Big(-\partial_{\nu}\Gamma^{\mu}_{\sigma\rho} \frac{dx^{\nu}}{d\tau}\frac{dx^{\sigma}}{d\tau} \; - \;\Gamma^{\mu}_{\sigma\rho}\frac{d^2x^{\sigma}}{d\tau^2}\;+\; \Gamma^{\mu}_{\sigma\nu}\Gamma^{\nu}_{\alpha\rho} \frac{dx^{\sigma}}{d\tau}\frac{dx^{\alpha}}{d\tau}\Big)\lambda^{\rho}
    [/tex]

    and then substitute these into

    [tex]
    \lambda^{\mu}(\tau) = \lambda^{\mu}(0) \;+\; \tau\frac{d\lambda^{\mu}(0)}{d\tau} \;+\; \frac{1}{2}\tau^2\frac{d^2\lambda^{\mu}(0)}{d\tau^2} \;+\; O(\tau^3}),
    [/tex]

    have I already done something wrong, or is this a valid way to do parallel transport small distances?
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2
    I was getting wrong answers in one exercise, and started doubting this, but I located the mistake (or a mistake), and right now I have no particular reason to believe that there was anything wrong in this.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Parallel transport approximation
  1. Parallel Transport (Replies: 20)

  2. Parallel transport (Replies: 14)

Loading...