Sujet : Re: E. Noether contra B. Carter
De : film.art (at) *nospam* gmail.com (JanPB)
Groupes : sci.physics.relativityDate : 22. Sep 2024, 17:18:11
Autres entêtes
Organisation : novaBBS
Message-ID : <a0471517e47f3b15422996f86e638a0f@www.novabbs.com>
References : 1 2 3 4 5
User-Agent : Rocksolid Light
On Sat, 21 Sep 2024 6:01:55 +0000, The Starmaker wrote:
The Starmaker wrote:
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JanPB wrote:
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On Thu, 19 Sep 2024 7:31:58 +0000, The Starmaker wrote:
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JanPB wrote:
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[...]
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Your math is correct but doesn't seem to have any meaning.
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No geometric meaning (which I was hoping for).
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i see, you have your own approach to doings things no one else does..
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Also the P isn't standard and needs more clarification
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K = < r^2 gamma' - rho^2 P(gamma') , gamma' >
OK, fair point. Surprisingly perhaps, the Kerr spacetime has
an orthogonal moving frame almost everywhere (away from the
usual suspects like the singular set and the horizons):
e0 = (r^2 + a^2)@/@t + a @/@phi
e1 = @/@r
e2 = @/@theta
e3 = a sin^2(theta) @/@t + @/@phi
e2 and e3 are always spacelike while e0, e1 are always of
the opposite causal character, so e0 and e1 span a
Minkowski signature plane, called the principal plane.
The P I used above denotes the orthogonal projection onto
that plane.
BTW, a geodesic is called principal if its tangent vector
lies in the principal plane.
EXERCISE. Let gamma be a timelike geodesic. Then:
K = 0 if and only if gamma is a principal in
the equatorial plane (theta = pi/2).
further more, there are too many errors for me to list them all...
There are no errors.
you finally reached ...kooksville.
Talk is cheap.
-- Jan