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Covariant derivative of coordinates

  1. Jan 17, 2012 #1
    Hi,

    I am familiar with the covariant derivative of the tangent vector to a path, [itex]\nabla_{\alpha}u^{\beta}[/itex] and some interesting ways to use it. I am wondering about
    [tex]
    \nabla_{\alpha}x^{\beta}=\frac{\partial x^\beta}{\partial x^\alpha}+\Gamma^{\beta}_{\alpha\gamma}x^{\gamma}=\delta_{\alpha}^{\beta}+\Gamma^{\beta}_{\alpha \gamma}x^{\gamma}
    [/tex]
    Then if we let this equal some arbitrary (1,1)-tensor we can manipulate to get
    [tex]
    \frac{d\tau}{dx^{\alpha}}\frac{dx^{\beta}}{d\tau}+\Gamma^{\beta}_{\alpha\gamma}x^{\gamma}=\Omega^{ \beta}_{\alpha}
    [/tex]
    which can be rewritten as
    [tex]
    (\delta_{\alpha}^{\beta} +\Gamma^{\beta}_{\alpha\gamma}x^{\gamma}-\Omega_{ \alpha}^{\beta})u^{\alpha}=J_{ \alpha}^{\beta}u^{\alpha}=0
    [/tex]
    which looks like a classic homogeneous linear algebra problem (that's the simplification I made in the last equality, just aesthetic). Does this equation have a good physical meaning, or is this just non-sense?

    Thanks,
     
  2. jcsd
  3. Jan 17, 2012 #2

    Matterwave

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    Gold Member

    The covariant derivative applies to vector fields. As the coordinates do not form vector fields in a curved geometry, I don't believe taking their "covariant derivatives" will have any physical meaning. In the end, you don't have a (1,1) tensor since the x's are not vectors.

    At the very best, I guess if you consider the coordinates to be functions (i.e. f(x)=x), then you have 4 separate one forms at the end, where the beta index merely labels which one form you have. In this case, the "contraction" with the Christoffel symbol is not kosher as the covariant derivative of a function is equal to the regular partial derivative.
     
  4. Jan 17, 2012 #3
    okay, but I know one can take a parametrized covariant derivative, for quantities that evolve on a curve, [itex]x^{\alpha}(\tau)[/itex], of the form
    [tex]
    \frac{D}{D\tau}=\frac{d}{d\tau}+\Gamma^{\alpha}_{ \beta \gamma}u^{\gamma}
    [/tex]
    like
    [tex]
    \frac{DA^\alpha(\tau)}{D\tau}=\frac{dA^{\alpha}}{d\tau}+\Gamma^{\alpha}_{\beta\gamma}u^{\beta}A^{ \gamma}
    [/tex]
    These quantities are not vector fields but are only defined on curves, like the spin of a particle. I'm also aware that [itex]x^{\alpha}[/itex] does in general not transform covariantly, so I was worried about taking the derivative in the first place, which is why I asked the forum.

    Thanks for your answer.
     
  5. Jan 17, 2012 #4

    Matterwave

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    [itex]\frac{D}{D\tau}=\frac{d}{d\tau}+\Gamma^\alpha_{ \beta\gamma}u^\gamma[/itex] only works if what that operator is acting on is a vector. If that operator is acting on a tensor, then the form of that operator will change. If that operator is acting on a function (e.g. f(x)=x), then the Christoffel symbol term is not there.

    Essentially, [tex]\frac{D}{D\tau}=u^\gamma \nabla_\gamma[/tex]. Remember that for a rank 2 tensor, the covariant derivative is [tex]\nabla_\gamma T^{\mu\nu}=\frac{d}{dx^\gamma}T^{\mu\nu}+\Gamma^ \mu_{\gamma\rho}T^{\rho\nu}+\Gamma^\nu_{\gamma\rho}T^{\mu\rho}[/tex], and this form changes depending on the rank of the tensor. Since your x is not a vector (but can only at best be thought of as a set of 4 functions...which aren't even scalars), I don't think your expression holds any physical meaning.
     
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