A side note which is sort of interesting but rather useless: Carter
constant *does* follow from
Noether's theorem after all. OK, let's rewind a bit. Let gamma(s) be
any geodesic in
the Kerr geometry.
There are 3 "obvious" constants of the motion in that geometry:
(1) q = <gamma', gamma'> ("mass"). This constant is a freebe because in
*any* geometry
*any* geodesic has this property,
(2) E = -<gamma', @/@t> ("energy"), the minus sign is traditional in
order to make
the number E positive for timelike future-pointing geodesics outside
the outer
horizon. This is a constant because the metric does not depend on t (translational
symmetry in the t direction),
(3) L = <gamma', @/@phi> ("angular momentum"), again constant because
g_ab does not
depend on phi (rotational symmetry).
There does not seem to exist any other symmetry of the Kerr geometry to
yield another
constant. Having four motion constants would be great because it would
reduce the
geodesic equations to first-order equations.
As is well-known, such fourth constant K does exist, it was found as a separation constant of
a certain PDE, so not related to any symmetry. It can be written e.g. in
this form
(a is the rotational parameter of the Kerr metric):
K = rho^4 theta'^2 - q a^2 cos^2(theta) - [L - a E sin^2(theta)]^2/sin^2(theta) (*)
..where rho^2 = r^2 + a^2 cos^2(theta), (a shortcut notation) and the components of gamma are: (t, r, theta, phi).
Looks unpleasant but it's extremely useful as it allows one to avoid
dealing with
the geodesic equation.
But once the formula is known, one can try to reverse engineer it by
other methods.
One can play a game with Noether's theorem. I'll skip the details but
the (rather useless) result surprised me, I haven't seen it mentioned anywhere:
assume gamma(s) is a geodesic, with the usual components (t, r, theta,
phi)
(four functions of s). Consider the following, surprisingly simple
*and* totally
un-illuminating variation involving *just* the theta coordinate:
t_epsilon = t
r_epsilon = r
theta_epsilon = theta - epsilon * 2 rho^2 * theta'
phi_epsilon = phi
Although the corresponding variated Lagrangian does NOT have the zero epsilon-derivative
(i.e., the above variation is not a symmetry), it's equal to a total
(d/ds) derivative of...
something. Denoting the epsilon-derivative at epsilon=0 by delta (the usual
variation procedure), we get:
delta(Lagrangian) = d/ds{ -rho^4 theta'^2 - q a^2 cos^2(theta) -
- [L - a E sin^2(theta)]^2/sin^2(theta) }
OTOH by Noether's theorem:
delta(Lagrangian) = d/ds{ (@L/@theta') delta(theta) }
(since delta(t) = delta(r) = delta(phi) = 0 by the above variation definition).
Hence:
-(@L/@theta') delta(theta) - rho^4 theta'^2 - q a^2 cos^2(theta) -
- [L - a E sin^2(theta)]^2/sin^2(theta) = const.
(**)
But:
(@L/@theta') delta(theta) = g_{theta,theta} * theta' * (-2 rho^2
theta')
so simplifying (**) we have (recall g_{theta,theta} = rho^2):
rho^4 theta'^2 - q a^2 cos^2(theta) - [L - a E
sin^2(theta)]^2/sin^2(theta) = const.
This is the Carter constant.
Similar result can be obtained by varying the r-coordinate instead:
r_epsilon = r + epsilon * 2 rho^2 r'
OK, so it's totally uninspiring because the Lagrangian variation is
merely a
total derivative, not zero. So the geometric meaning remains as murky
as ever.
Probably the cleanest geometric meaning is this one:
K = < r^2 gamma' - rho^2 P(gamma') , gamma' >
..where P denotes the projection onto the principal plane. Still completely
unexpected.
--
Jan
On Thu, 19 Sep 2024 7:31:58 +0000, The Starmaker wrote:
JanPB wrote:[...]
Your math is correct but doesn't seem to have any meaning.
No geometric meaning (which I was hoping for).
It's like a Kamala Number Salad...
My previous post has disappeared mysteriously(?) so I'll repeat,
more or less: Kamala is like Basil Fawlty at the White House.
It's amazing it's real life.
We'll see if this post survives.
--
Jan
JanPB wrote:
On Thu, 19 Sep 2024 7:31:58 +0000, The Starmaker wrote:
JanPB wrote:[...]
Your math is correct but doesn't seem to have any meaning.
No geometric meaning (which I was hoping for).
i see, you have your own approach to doings things no one else does..
On Sat, 21 Sep 2024 6:01:55 +0000, The Starmaker wrote:
The Starmaker wrote:
JanPB wrote:
On Thu, 19 Sep 2024 7:31:58 +0000, The Starmaker wrote:
JanPB wrote:[...]
Your math is correct but doesn't seem to have any meaning.
No geometric meaning (which I was hoping for).
i see, you have your own approach to doings things no one else does..
Also the P isn't standard and needs more clarification
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
Typo:
"a principal" --> "a principal geodesic"
--
Jan
JanPB wrote:
Typo:
"a principal" --> "a principal geodesic"
--
Jan
"Typo:"???? Don't you mean...error? a mistake??..error error error
My post! My post is melting! Melting! Oh, what a world, what a world!
On Sat, 21 Sep 2024 6:01:55 +0000, The Starmaker wrote:
The Starmaker wrote:
JanPB wrote:
On Thu, 19 Sep 2024 7:31:58 +0000, The Starmaker wrote:
JanPB wrote:[...]
Your math is correct but doesn't seem to have any meaning.
No geometric meaning (which I was hoping for).
i see, you have your own approach to doings things no one else does..
Also the P isn't standard and needs more clarification
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
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