1. Limited time only! Sign up for a free 30min personal 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!

Coordinate-independence of equation for the parallel transport

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Please show that the defining equation for the parallel transport of a contravariant vector along a curve [itex]\dot{\lambda}^a+\Gamma^a_{bc}\lambda^b\dot{x}^c=0[/itex] is coordinate-independent, given that the transformation formula for the christoffel symbol being ##\Gamma^{a'}_{b'c'}=(\Gamma^{d}_{ef}X^{a'}_d-X^{a'}_{ef})X^{e}_{b'}X^{f}_{c'}##.

    3. The attempt at a solution
    I have stuck by the following derivation ##X^{a'}_{ab}\dot x^b\lambda^a+X^{a'}_a\dot\lambda^a+(\Gamma^{d}_{ef}X^{a'}_d-X^{a'}_{ef})X^{e}_{b'}X^{f}_{c'}(X^{c'}_{ef}\dot x^fx^e+X^{c'}_e\dot x^e)##, where I can't simplify it to the unprimed equation [itex]\dot{\lambda}^a+\Gamma^a_{bc}\lambda^b\dot{x}^c=0[/itex].
    Any advice will be appreciated!
     
  2. jcsd
  3. Jun 17, 2012 #2
    Hello!
    After think twice, I met some mistakes in the derivation.
    First, dummy indexes were repeated in the same term.
    Second, tangent vectors are transformed by ##\dot{x} ^{a'}=X^{a'}_a\dot x^a##

    Now, I describe my derivation as follows.

    The parallel transport equation for contravariant vector ##\lambda ^{a'}## along a curve ##\gamma## parametrized by t
    ##\dot\lambda^{a'}+\Gamma^{a'}_{b'c'}\lambda^{b'} \dot{x}^{c'}=0##
    was transformed as
    ##\frac{dX^{a'}_a\lambda^a}{dt}+(\Gamma^{d}_{ef}X^{a'}_dX^{e}_{b'}X^f_{c'}-X^e_{b'}X^f_{c'}X^{a'}_{ef})(X^{b'}_b\lambda^bX^{c'}_c\dot x^c)##
    ##=X^{a'}_{ab}\lambda^a\dot x^b+X^{a'}_a\dot\lambda^a+\Gamma^d_{ef}X^{a'}_d \lambda ^e \dot x^f-\lambda^e \dot x^fX^{a'}_{ef}##
    ##=X^{a'}_a(\dot\lambda^a+\Gamma^a_{bc}\lambda^b \dot x^b)=0##
    Since ##X^{a'}_a ## are arbitrary coefficients of transformation, ##\dot\lambda^a+\Gamma^a_{bc}\lambda^b \dot x^c)=0## in ##x^a## coordinate frame, as desired.

    If I have fault, please kindly inform me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Coordinate-independence of equation for the parallel transport
  1. Parallel Transport (Replies: 3)

Loading...