E. Noether contra B. Carter

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