Re: Question on metric variation

On Mar 12, 7:31 am, Tom Roberts <tjroberts...@xxxxxxxxxxxxx> wrote:
babalu...@xxxxxxxxx wrote:
Carroll, Wald, and Schutz derive the geodesic equation using a taylor
expansion of the variation of the metric with respect to position in
space-time. In this expansion the partial derivative is used instead
of the covariant derivative. This seems to imply that the variation is
not a tensor. Is that true? And if so, this is confusing since the
vanishing (or non vanishing) of the variation and hence equilibrium,
is cooridnate dependent. Please help me out of this distressing trap.

First, remember that the covariant derivative of the metric is
identically zero. So it would be useless for this.

While I have not looked up those derivations, I believe they are
performed using s fixed but arbitrary coordinate system. So when one
derives the geodesic equation, the resulting geodesic path is an
extremum in ANY coordinate system. I see no requirement for the
"variation" to be a tensor, all that is required here is that the
geodesic path be an extremum in any coordinate system. That implies that
the vanishing of the variation on the path is NOT coordinate dependent,
as even though the variation is coordinate dependent, it is zero in
every coordinate system.

IOW: the path is independent of coordinates (as it must be). The fact
that the method of finding it used coordinates is irrelevant.

        Exercise for the reader: to demonstrate this, start with
        the geodesic equation in some coordinates, and transform
        the equation to a different set of coordinates. Verify that
        the result simplifies to the same equation in the new
        coordinates.

        [I believe that the original way the Christoffel symbols were
         defined was via a route similar to this exercise.]

Tom Roberts

This did help very much, Once I stopped thinking of the variation as a
tensor things became easier to understand. Thanks for the help again.
.

Relevant Pages

  • Re: Question on metric variation
    ... This seems to imply that the variation is ... While I have not looked up those derivations, I believe they are performed using s fixed but arbitrary coordinate system. ... So when one derives the geodesic equation, the resulting geodesic path is an extremum in ANY coordinate system. ... I see no requirement for the "variation" to be a tensor, all that is required here is that the geodesic path be an extremum in any coordinate system. ...
    (sci.physics.relativity)
  • Re: Question on metric variation
    ... In this expansion the partial derivative is used instead ... This seems to imply that the variation is ... not a tensor. ... vanishing of the variation and hence equilibrium, ...
    (sci.physics.relativity)
  • Question on metric variation
    ... space-time. ... In this expansion the partial derivative is used instead ... This seems to imply that the variation is ... vanishing of the variation and hence equilibrium, ...
    (sci.physics.relativity)
  • Re: Question on metric variation
    ... In this expansion the partial derivative is used instead ... This seems to imply that the variation is ... vanishing of the variation and hence equilibrium, ... I believe you are correct - the variation is not a tensor. ...
    (sci.physics.relativity)
  • Re: Question on metric variation
    ... In this expansion the partial derivative is used instead ... This seems to imply that the variation is ... vanishing of the variation and hence equilibrium, ... I believe you are correct - the variation is not a tensor. ...
    (sci.physics.relativity)
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