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Checking derivation of the curvature tensor

  1. Aug 7, 2015 #1
    1. The problem statement, all variables and given/known data
    I am trying to derive the curvature tensor by finding the commutator of two covariant derivatives. I think I've got it, but my head is spinning with Nablas and indices. Would anyone be willing to check my work? Thanks

    2. Relevant equations
    I am trying to derive the curvature tensor by finding the commutator of two covariant derivatives: $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}.$$
    The covariant derivative of ##V_{a}## is given by$$\nabla_{b}V_{a}=\partial_{b}V_{a}-\Gamma_{ab}^{d}V_{d}.$$

    3. The attempt at a solution

    Start by taking another covariant derivative: $$\nabla_{c}\nabla_{b}V_{a}=\partial_{c}\left(\nabla_{b}V_{a}\right)-\Gamma_{ac}^{e}\nabla_{b}V_{e}-\Gamma_{bc}^{e}\nabla_{e}V_{a}.$$
    For the first right-hand side term $$\partial_{c}\left(\nabla_{b}V_{a}\right)=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}V_{d}\right).$$
    Then using the product rule I get $$\partial_{c}\left(\nabla_{b}V_{a}\right)=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}.$$
    For the second right-hand side term
    $$-\Gamma_{ac}^{e}\nabla_{b}V_{e}=-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right).$$
    And for the third right-hand side term $$-\Gamma_{bc}^{e}\nabla_{e}V_{a}=-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
    Put all these together to get $$\nabla_{c}\nabla_{b}V_{a}=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right)-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
    Interchanging ##b## and ##c##
    gives $$\nabla_{b}\nabla_{c}V_{a}=\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\left(\partial_{c}V_{e}-\Gamma_{ec}^{d}V_{d}\right)-\Gamma_{cb}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right).$$
    Subtract one from the other: $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\left(\partial_{b}V_{e}-\Gamma_{eb}^{d}V_{d}\right)-\Gamma_{bc}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right)-
    \left(\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\left(\partial_{c}V_{e}-\Gamma_{ec}^{d}V_{d}\right)-\Gamma_{cb}^{e}\left(\partial_{e}V_{a}-\Gamma_{ae}^{d}V_{d}\right)\right).$$
    Multiplying out the brackets in the second line gives $$\partial_{c}\partial_{b}V_{a}-\partial_{c}\left(\Gamma_{ab}^{d}\right)V_{d}-\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}-\Gamma_{ac}^{e}\partial_{b}V_{e}+\Gamma_{ac}^{e}\Gamma_{eb}^{d}V_{d}-\Gamma_{bc}^{e}\partial_{e}V_{a}+\Gamma_{bc}^{e}\Gamma_{ae}^{d}V_{d}-
    \left(\partial_{b}\partial_{c}V_{a}-\partial_{b}\left(\Gamma_{ac}^{d}\right)V_{d}-\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}-\Gamma_{ab}^{e}\partial_{c}V_{e}+\Gamma_{ab}^{e}\Gamma_{ec}^{d}V_{d}-\Gamma_{cb}^{e}\partial_{e}V_{a}+\Gamma_{cb}^{e}\Gamma_{ae}^{d}V_{d}\right).$$
    Note that ##\Gamma_{ac}^{e}\partial_{b}V_{e}=\Gamma_{ab}^{e}\partial_{c}V_{e}## and ##\partial_{c}\left(V_{d}\right)\Gamma_{ab}^{d}=\partial_{b}\left(V_{d}\right)\Gamma_{ac}^{d}##
    (because of the definition of ##\frac{\partial e_{x}}{\partial x^{z}}=\Gamma_{xz}^{y}e_{y}).##
    I then end up with $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=\left(\partial_{b}\left(\Gamma_{ac}^{d}\right)-\partial_{c}\left(\Gamma_{ab}^{d}\right)+\Gamma_{ac}^{e}\Gamma_{eb}^{d}-\Gamma_{ab}^{e}\Gamma_{ec}^{d}\right)V_{d}.$$
    The expression inside the brackets on the rhs is the Riemann tensor, meaning $$\nabla_{c}\nabla_{b}V_{a}-\nabla_{b}\nabla_{c}V_{a}=R_{\phantom{\mu}abc}^{d}V_{d}.$$
     
  2. jcsd
  3. Aug 7, 2015 #2

    DEvens

    User Avatar
    Education Advisor
    Gold Member

    It looks like more or less the right thing. I'm not going to go through the details. This is "textbook" stuff that you can find in lots of different differential geometry texts.
     
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