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Parallel transport approximation

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

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

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

    \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}
    = \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}

    and then substitute these into

    \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}),

    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.
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