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| Ken S. Tucker |
 Posted: Tue Feb 15, 2005 2:03 am |
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Guest
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Bjoern Feuerbacher wrote:
[quote:fc7b5c75e2]Ken S. Tucker wrote:
Bjoern Feuerbacher wrote:
Ken S. Tucker wrote:
Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,
It even makes perfect sense when the curvature tensure
has any other form. It's just not an invariant then.
Can you prove that?
Which part of what I said? The first or the second?
The latter, it need not be rigorous a hint
on how I could do it would helpful, then I'll
try to work it out myself.
Sorry. My original statement was not right.
Please rephrase it as:
"x_u dx^u makes perfects sense for any form of the
curvature tensor. It's just not invariant under general
coordinate transformations."
[/quote:fc7b5c75e2]
I take it you presume "general coordinate transforms",
are able to exclude a "curvature tensor" in some cases?
Ken |
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| Bjoern Feuerbacher |
| Posted: Tue Feb 15, 2005 5:10 am |
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Ken S. Tucker wrote:
[quote:eab54f0ae6]Bjoern Feuerbacher wrote:
Ken S. Tucker wrote:
Bjoern Feuerbacher wrote:
Ken S. Tucker wrote:
Semantics vary, x_u dx^u makes perfect sense when
the metric satisfies R^a_bcd=0,
It even makes perfect sense when the curvature tensure
has any other form. It's just not an invariant then.
Can you prove that?
Which part of what I said? The first or the second?
The latter, it need not be rigorous a hint
on how I could do it would helpful, then I'll
try to work it out myself.
Sorry. My original statement was not right.
Please rephrase it as:
"x_u dx^u makes perfects sense for any form of the
curvature tensor. It's just not invariant under general
coordinate transformations."
I take it you presume "general coordinate transforms",
are able to exclude a "curvature tensor" in some cases?
[/quote:eab54f0ae6]
Sorry, I have no clue what you mean with "exclude a curvature tensor"
here.
Coordinate transformations obviously act on *all* tensors.
Bye,
Bjoern |
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