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Co-ordinate vs. true acceleration

  1. Dec 7, 2006 #1
    In G.R. we do physics in any kind of frame, inertial or not. To compensate for that, we get nice little Christoffel symbols in all our derivatives. And straight lines become geodesics.

    But is there truly no way to distinguish between co-ordinate acceleration and "real" acceleration? Or is there no such thing as real acceleration, since we can always transform it away? Or is is that in a reference frame where we did transform the acceleration of a particle away, geodesics would no longer be straight lines? But if the curvature tensor's components all vanished, then we know that the space is Minkowski, and hence we can distinguish between these two types of acceleration.

    Hmm, I'm not sure if I've made sense here. Is every question [posed above] answerable by a resounding "yes" or have I made a mistake somewhere?
     
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  3. Dec 7, 2006 #2

    Garth

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    You have answered your own question I think:
    It's all in the Riemannian.

    Garth
     
  4. Dec 7, 2006 #3

    George Jones

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    No coordinate transformation will tranform away the real 4-acceleration of an observer. Let [itex]\left\{ x^\mu \left( \tau \right) \right\}[/itex] be the coordinate description of an observer's worldline parametrized by proper time. Then the observer's 4-velocity is

    [tex]u = \frac{dx^\mu}{d\tau} \frac{\partial}{\partial x^\mu},[/tex]

    and 4-acceleration is

    [tex]
    \begin{equation*}
    \begin{split}
    a &= \nabla_u u \\
    &= u^\mu \nabla_\mu \left( u^\nu \frac{\partial}{\partial x^\nu} \right) \\
    &= u^\mu \left[ \left( \nabla_\mu u^\nu \right) \frac{\partial}{\partial x^\nu} + u^\nu \nabla_\mu \frac{\partial}{\partial x^\nu} \right] \\
    &= u^\mu \left[ \frac{\partial u^\nu}{\partial x^\mu} \frac{\partial}{\partial x^\nu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \frac{\partial}{\partial x^\alpha} \right] \\
    &= u^\mu \left[ \frac{\partial u^\alpha}{\partial x^\mu} + u^\nu \Gamma^{\alpha}_{\nu \mu} \right] \frac{\partial}{\partial x^\alpha}.
    \end{split}
    \end{equation*}
    [/tex]

    If an observer's 4-acceleration is zero in one coordinate system (freely falling obsrever), it is zero in all coordinate systems; if an observer's 4-acceleration is non-zero in one coordinate system, it is non-zero in all coordinate systems.
     
    Last edited: Dec 7, 2006
  5. Dec 7, 2006 #4

    Garth

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    The operative word here of course is "4-acceleration".

    In the OP it sounded as if 3-acceleration was being talked about.

    Was that the cause of your confusion masdur?

    Garth
     
  6. Dec 7, 2006 #5

    George Jones

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    But, [itex]1 = g \left( u , u \right)[/itex] gives that

    [tex]0 = \nabla_u g \left( u , u \right) = g \left( \nabla_u u , u \right) + g \left( u , \nabla_u u \right),[/tex]

    so

    [tex]0 = g \left( a , u \right).[/tex]

    Thus, since the 4-velocity [itex]u[/itex] is timelike, the 4-acceleration [itex]a[/itex] is spacelike. Therefore, if the 4-acceleration is non-zero, the three "spatial" (assuming that three of the coordinates are spacelike and one is timelike) coordinates of the 4-acceleration cannot be simultaneously transformed away.
     
    Last edited: Dec 7, 2006
  7. Dec 8, 2006 #6
    Thanks for your timely responses.

    This probably demonstrates my lack of experience in thinking in GR terms, but I was just concerned that there should be something physically different in the situation of using spherical polars and detecting acceleration, and using Cartesians and detecting acceleration. But in G.R. neither system has an advantage: both types of acceleration should be deemed equivalent. This seems especially troublesome to me, especially when considering non-gravitational motion, i.e. in a flat space.

    Again this probably demonstrates my lack of knowledge, but I fail to see how your conclusion follows directly from your expression for the 4-acceleration.
     
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