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Weird version of covariant derivative on wikipedia

  1. Jul 12, 2012 #1

    At the bottom of that page, the author provides the generalization of four force in general relativity, where the partial derivative is replaced with the covariant derivative.

    However if you notice on the second term in the third equality, there is a four vector summed on mu with the gamma symbols, is that there to select the 0th component of the gamma symbol? Since in a MCRF frame, the spatial components of U are all 0.
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  3. Jul 12, 2012 #2


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    That's because it's not really the covariant derivative. It's the absolute derivative, which is the covariant derivative taken along the particle's world line. That is,

    Fλ = DPλ/Dτ = PλUμ
  4. Jul 12, 2012 #3

    George Jones

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    In special relativity, the chain rule gives

    [tex]\frac{d}{d\tau} = \frac{dx^\alpha}{d\tau} \frac{\partial}{\partial x^\alpha} = U^\alpha \frac{\partial}{\partial x^\alpha}.[/tex]

    In general relativity, the partial derivative have to be replace by covariant derivatives, so

    [tex]U^\alpha \nabla_{\partial_\alpha},[/tex]


    [tex]U^\alpha \nabla_{\partial_\alpha} = \nabla_{U^\alpha \partial_\alpha} = \nabla_{U}[/tex]

    Differentiating with respect to proper time is replaced by a covariant directional derivative in the direction of the 4-velocity.

    [edit]didn't see Bill_K's post[/edit]
  5. Jul 12, 2012 #4
    Um because proper time is the parameter of the world line that the four velocity is tangent to?
  6. Jul 13, 2012 #5


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    Just a comment: the distinction isn't Special Relativity vs General Relativity, it's inertial, cartesian coordinates vs. noninertial, curvilinear coordinates.

    The covariant derivative must be used whenever the basis vectors vary from place to place. That can be due to spacetime curvature, but it can also be due to using curvilinear coordinates.

    For example, let's do ordinary Newtonian physics in 2D space. We have a particle whose position as a function of time is given by [itex]x=A[/itex], [itex]y=vt[/itex] where A and v are constants. So this is just constant-velocity motion. The velocity vector [itex]U[/itex] is given by [itex]U^x = 0[/itex], [itex]U^y = v[/itex].

    Now, in terms of polar coordinates [itex]r[/itex] and [itex]\theta[/itex], the same motion looks like:

    [itex]U^r = \dfrac{dr}{dt} = \dfrac{v^2 t}{r}[/itex]
    [itex]U^\theta = \dfrac{d \theta}{dt} = \dfrac{A\ v}{r^2}[/itex]

    That doesn't look like "constant velocity" at all:
    [itex]\dfrac{dU^r}{dt} = \dfrac{v^2 (r^2 - (v t)^2)}{r^3} = \dfrac{v^2 A^2}{r^3}[/itex]
    [itex]\dfrac{dU^\theta}{dt} = \dfrac{- A v^3 t}{r^4}[/itex]

    But it is constant, if you use covariant derivatives, with
    [itex]\Gamma^r_{r r} = 0[/itex]
    [itex]\Gamma^r_{r \theta} = 0[/itex]
    [itex]\Gamma^r_{\theta r} = 0[/itex]
    [itex]\Gamma^r_{\theta \theta} = -r[/itex]
    [itex]\Gamma^\theta_{r r} = 0[/itex]
    [itex]\Gamma^\theta_{r \theta} = \dfrac{1}{r}[/itex]
    [itex]\Gamma^\theta_{\theta r} = \dfrac{1}{r}[/itex]
    [itex]\Gamma^\theta_{\theta \theta} = 0[/itex]
  7. Jul 13, 2012 #6

    George Jones

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    Yes, good point.
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