• Re: E. Noether contra B. Carter

    From The Starmaker@21:1/5 to JanPB on Thu Sep 19 00:31:58 2024
    JanPB wrote:

    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

    Your math is correct but doesn't seem to have any meaning.


    It's like a Kamala Number Salad...




    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
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  • From The Starmaker@21:1/5 to JanPB on Thu Sep 19 21:14:31 2024
    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..



    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

    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
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  • From The Starmaker@21:1/5 to The Starmaker on Fri Sep 20 23:01:55 2024
    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' >




    further more, there are too many errors for me to list them all...


    you finally reached ...kooksville.



    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
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  • From The Starmaker@21:1/5 to JanPB on Sun Sep 22 11:25:53 2024
    JanPB wrote:

    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


    I'ts not clear what "projection onto the principal plane" means in this context.


    "orthogonal projection"???? Now you sound like Einstein throwing in
    more words to hide his mistakes!


    Einstein said: "Nothing travels faster than light." ...then he throws
    in..."in a vacuum".


    How about outside the vacuum?


    Einstein sez: "Well, no, I mean.. I don't think so...but vacuum sounds
    soooo cool!"



    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
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  • From The Starmaker@21:1/5 to JanPB on Sun Sep 22 11:33:25 2024
    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!




    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From The Starmaker@21:1/5 to The Starmaker on Mon Sep 23 11:39:03 2024
    The Starmaker wrote:

    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!





    In America we say "I made a typo error."

    i don't know how dey say it in London town, but might be "I say o'l
    chap, I mauk a typo!"


    with a spoon in the mouth...


    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)
  • From The Starmaker@21:1/5 to JanPB on Mon Sep 23 20:21:02 2024
    JanPB wrote:

    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.

    "Talk is cheap."???? Do you think I have time to do ALL your homework???!


    Let me give you another example of the error of your ways...


    You wrote: "There does not seem to exist any other symmetry of the Kerr geometry to
    yield another constant."

    but, but..the hidden symmetry leading to the Carter constant is associated with the Killing tensor, which you didn't mention...
    i cannot complete your homework for you.

    there are too many errors for me to list them all...


    I would suggest you take up another hobby like ...playing the piano but. that doesn't seem to
    work for you either.

    i throw up my hands.

    I'm failing for this class. I give you a B minus.




    and PLEASE, don't mix Carter with Noether...you're making a mess!



    and don't play the violin!



    buy ten cats.










    --
    Jan

    --
    The Starmaker -- To question the unquestionable, ask the unaskable,
    to think the unthinkable, mention the unmentionable, say the unsayable,
    and challenge the unchallengeable.

    --- SoupGate-Win32 v1.05
    * Origin: fsxNet Usenet Gateway (21:1/5)