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Is acceleration due to gravity a 4-vector?

  1. Oct 18, 2013 #1
    We can write the geodesic condition as

    [tex]\frac{d^2 x^\alpha}{d\tau^2}={-\Gamma^\alpha}_{\mu\nu}\frac{dx^\mu}{d\tau}\frac{dx^\nu}{d\tau}[/tex]

    On the RHS we have two contravariant vectors ( or (0,1) tensors ) contracted with the connection. But the connection is NOT a tensor. So is the LHS a vector? And if so, how does that work?
     
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  3. Oct 18, 2013 #2

    WannabeNewton

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    ##a^{\mu} = \ddot{x}^{\mu} + \Gamma^{\mu}_{\alpha\beta}\dot{x}^{\alpha}\dot{x}^{\beta}## is a 4-vector but the two individual terms on the RHS are not 4-vectors. It's very easy to show that they both fail to transform as 4-vectors but that their sum does transform as a 4-vector.
     
  4. Oct 18, 2013 #3
    Thanks. Regarding [itex]\ddot{x}^\mu[/itex] ... where do I go wrong in the following?

    Suppose we have a change of coordinates [itex]x^\mu={\Lambda^\mu}_\alpha x'^\alpha[/itex]. Elements of the tangent space are vectors and likewise transform according to [itex]V^\mu={\Lambda^\mu}_\alpha V'^\alpha[/itex]. In particular the tangent to a curve [itex]x^\mu(\tau)[/itex] is [itex]dx^\mu/d\tau[/itex] which as a tangent vector transforms as

    [tex]\frac{dx^\mu}{d\tau} ={\Lambda^\mu}_\alpha \frac{dx'^\alpha}{d\tau}[/tex]

    but if we take the derivative of both sides of

    [itex]x^\mu={\Lambda^\mu}_\alpha x'^\alpha[/itex]

    we get

    [itex]\frac{dx^\mu}{d\tau}=\frac{d}{d\tau}\left({\Lambda^\mu}_\alpha x'^\alpha\right)[/itex]

    This is equivalent to the previous expression only if [itex]\frac{d}{d\tau}{\Lambda^\mu}_\alpha =0[/itex]. In this case if we take the second derivative, we get

    [tex]\frac{d^2 x^\mu}{d\tau^2} ={\Lambda^\mu}_\alpha \frac{d^2 x'^\alpha}{d\tau^2}
    [/tex]

    and so [itex]\ddot{x}^\mu[/itex] transforms as a vector.

    ?????
     
  5. Oct 19, 2013 #4

    WannabeNewton

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    ##x^{\mu}## are not position vectors. The transformation ##x^{\mu} \rightarrow x^{\mu'}(x^{\nu})## will not be the same as the transformation matrix ##V^{\mu} \rightarrow V^{\mu'}## which is where your error is; in fact if we write the former transformation as ##x^{\mu'} = x^{\mu'}(x^{\nu})## then ##V^{\mu'} = \frac{\partial x^{\mu'}(x^{\nu})}{\partial x^{\mu}}V^{\mu}## so clearly we are talking about different transformations for the two.
     
  6. Oct 19, 2013 #5

    stevendaryl

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    To simplify the notation, I'm going to drop the indices. It should be clear how to put them back.

    In general, if you converting from a coordinate [itex]x'[/itex] to a coordinate [itex]x[/itex], then it's a nonlinear, continuous, invertible function:

    [itex]x = F(x')[/itex]

    So,

    [itex]\frac{dx}{d \tau} = \frac{\partial F}{\partial x'} \frac{dx'}{d\tau}[/itex]

    So even if [itex]F(x')[/itex] is nonlinear, the relationship between velocities is linear:

    [itex]\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}[/itex]

    where [itex]\Lambda = \frac{\partial F}{\partial x'}[/itex]

    But note: the transformation between coordinates is not the same as the transformation between velocities:

    [itex]\frac{dx}{d \tau} =\Lambda \frac{dx'}{d\tau}[/itex] versus [itex]x = F(x')[/itex]

    The only time that you can use the same matrix [itex]\Lambda[/itex] to transform coordinates or velocities is when:

    [itex]F(x') = \Lambda x'[/itex]

    which is only possible when [itex]\frac{\partial \Lambda}{\partial x'} = 0[/itex]
     
  7. Oct 19, 2013 #6
    I got this one straight now, guys. Thanks!
     
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