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!

Homework Help: Geodesic equation in new coordinates question

  1. Apr 3, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose [tex]\bar{x}^{\mu}[/tex] is another set of coordinates with connection components [tex]\bar{\Gamma}^{\mu}_{\alpha\beta}[/tex]. Write down the geodesic equation in new coordinates.

    2. Relevant equations
    Using the geodesic equation: [tex]0 = \frac{d^{2}x^{\mu}}{ds^{2}} + \Gamma^{\mu}_{\alpha\beta}\frac{dx^{\alpha}}{ds}\frac{dx^{\beta}}{ds}[/tex]
    where [tex]s[/tex] represents the parameterised curve, i.e. [tex]x(s)[/tex]

    3. The attempt at a solution
    Now the question asks to move to a new coordinate system (i.e. bar over the terms). So, I began by identifying which terms will be effected by the transformation. Namely, the tangent vectors will transform according to
    [tex]\frac{dx^{\mu}}{ds} = v^{\mu} = \frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}}\frac{d \bar{x}^{\nu}}{ds}[/tex] ; [tex]\frac{dx^{\alpha}}{ds} = v^{\alpha} = \frac{\partial x^{\alpha}}{\partial \bar{x}^{\theta}}\frac{d \bar{x}^{\theta}}{ds}[/tex] ; [tex]\frac{dx^{\beta}}{ds} = v^{\beta} = \frac{\partial x^{\beta}}{\partial \bar{x}^{\phi}}\frac{d \bar{x}^{\phi}}{ds}[/tex]
    The metric connection will transform as
    [tex]\Gamma^{\mu}_{\alpha\beta} = \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}}[/tex][tex]\frac{\partial \bar{x}^\phi}{\partial x^{\beta}} \bar{\Gamma}^{P}_{\theta\phi} - \frac{\partial ^{2}x^{\mu}}{\partial \bar{x}^{\theta} \partial \bar{x}^{\phi}} \frac{\partial \bar{x}^{\theta}}{\partial x^{\alpha}} \frac{\partial \bar{x}^{\phi}}{\partial x^{\beta}}[/tex]
    I substituted these terms into the geodesic equation above, cancelled out partial derivatives on numerators and denominators to obtain:
    [tex]0 = \frac{d}{ds}\left(\frac{\partial x^{\mu}}{\partial \bar{x}^{\nu}} \frac{d \bar{x}^{\nu}}{ds}\right) + \frac{\partial x^{\mu}}{\partial \bar{x}^{P}}\frac{d \bar{x}^{\theta}}{ds}\frac{d \bar{x}^{\phi}}{ds} \bar{\Gamma}^{P}_{\theta\phi} - \frac{\partial ^{2}x^{\mu}}{\partial \bar{x}^{\theta} \partial \bar{x}^{\phi}} \frac{d \bar{x}^{\theta}}{ds} \frac{d \bar{x}^{\phi}}{ds}[/tex]

    I was just wondering if my final equation for the geodesic equation in new coordinates [tex]\bar{x}^{\mu}[/tex] was correct?

    I suspect a flaw in the mathematical logic when I cancel out the partial derivative in the numerators and denominators. Was this procedure allowed?
    Also, I think there may be a problem with my indices with respect to Einstein summation convention.
  2. jcsd
  3. Apr 4, 2010 #2


    User Avatar

    Sorry it took a while to get back to you. If you have it available, look at page 102 of Weinberg's "Gravitation and Cosmology. There he shows that the geodesic equation transforms like a vector.
  4. Apr 6, 2010 #3
    Cheers, AEM.

    I do have Weinberg's text, I will have a look.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook