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: Christoffel Symbol / Covariant derivative

  1. May 1, 2012 #1
    1. The problem statement, all variables and given/known data
    My teacher solved this in class but I'm not understanding some parts of tis solution.

    Show that [itex]\nabla_i V^i[/itex] is scalar.

    2. Relevant equations
    [itex]\nabla_i V^i = \frac{\partial V^{i}}{\partial q^{i}} + \Gamma^{i}_{ik} V^{k}[/itex]

    3. The attempt at a solution
    To start this, I'll solve this [itex]\Gamma^{i}_{ik}[/itex] first.

    [itex]\Gamma^{i}_{ik} = \frac{1}{2} g^{il} (\frac{\partial g_{lk}}{\partial q^{i}} + \frac{\partial g_{il}}{\partial q^{k}} - \frac{\partial g_{ki}}{\partial q^{l}}[/itex]

    [itex]\Gamma^{i}_{ik} = \frac{1}{2} g^{il} \frac{\partial g_{il}}{\partial q^{k}} = \frac{1}{2g} \frac{\partial g}{\partial q^{k}}[/itex]

    [itex]\Gamma^{i}_{ik} = \frac{1}{\sqrt g}\frac{\partial \sqrt{g}}{\partial q^{k}}[/itex]

    (THIS PART: how this [itex]\sqrt{g}[/itex] appeared??)


    [itex]\nabla_i V^i = \frac{\partial V^{i}}{\partial q^{i}} + \frac{V^{k}}{\sqrt{g}} \frac{\partial \sqrt{g}}{\partial q^{k}}[/itex]

    [itex]\nabla_i V^i = \frac{1}{\sqrt g} \partial_i (\sqrt g V^{i})[/itex]

    And this last part... What happened to [itex]\partial q^{i}[/itex] and [itex]V^{k}[/itex]?
  2. jcsd
  3. May 1, 2012 #2


    User Avatar
    Homework Helper

    take this with a grain of salt as I haven't played with tensors in a while... anyway here I go... I think both these are tricks that can be understood by working back

    I think the first part is just normal chain rule differentiation, what is the standard derivative of sqrt(g)?

    then try expanding the very last expression using the standard product rule, what do you end up with?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook