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Spinning moving object

  1. Jan 3, 2014 #1
    If an "Object A" spins at nearly "c", and this object also is moving at any given posible speed. What happen with an "Object B" on the surface of "Object A" .I asume that there's a mechanism that "fix" this relation between angular and linear momentum to conserve the speed limit of light, but I can't realize what is the mechanism to get this explanation.

    Best regards big thinkers,

    AGZ
     
  2. jcsd
  3. Jan 3, 2014 #2

    HallsofIvy

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    Last edited: Jan 3, 2014
  4. Jan 3, 2014 #3
     
  5. Jan 3, 2014 #4
    In Relativity theory energy per unit mass [E/m], momentum per unit mass [p/m], and angular momentum per unit mass [L/m] are constants of the motion over an objects natural path [geodesic]. In the Kerr metric the angular momentum per unit mass is the rotation parameter [a]. In the appropriate metrics the angular momentum per unit mass and charge per unit mass sum to the Schwarzschild mass [what the mass would be if the object was spherically symmetric, non rotating, no charge]. When L/m=1, Q/m=1 [the extremal case] the mass of the system is 3 times the Schwarzschild mass.
     
    Last edited: Jan 3, 2014
  6. Jan 3, 2014 #5

    WannabeNewton

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    You are confusing a lot of different things here and this is not at all relevant to the OP's original question which has already been answered.
     
  7. Jan 3, 2014 #6
    Most likely true. I just wanted to add some information about how those quantities are conserved since he seemed confused about that in his original post. Maybe you learned something?
     
  8. Jan 3, 2014 #7

    WannabeNewton

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    The angular momentum parameter ##a## of the Kerr metric is the conserved angular momentum of the space-time, not the conserved angular momentum per unit mass along a geodesic in Kerr space-time. This is the main thing you confused in your post.
     
  9. Jan 3, 2014 #8
    I didn't confuse anything. To start with this is the constant of motion for all geodesics in the Kerr geometry.

    L/m=R^2 (dphi/dTau) - (2M^2/r)dt/dTau.

    There's only one. So I didn't confuse it with anything else. All I did was give a simple explanation for components in the metric. It's interesting and worth knowing when you're discussing a rotation parameter going to the limit 1. Whatever.
     
  10. Jan 3, 2014 #9

    WannabeNewton

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    Your exact statement was "In Relativity theory...angular momentum per unit mass [L/m] are constants of the motion over an objects natural path [geodesic]. In the Kerr metric the angular momentum per unit mass is the rotation parameter [a]." which is wrong for the reason stated in post #7. The rotation parameter ##a = \frac{J}{M}## of Kerr space-time is the reduced angular momentum of Kerr space-time obtained from ##J = \int _{S^{2}_{\infty}}\epsilon_{abcd}\nabla^c \psi^d## where ##\psi## is the axial killing field of Kerr space-time. The reduced angular momentum of an observer with 4-velocity ##u^{a}## on the other hand is gotten from ##L = u_{a}\psi^{a}##. You are confusing these two things, simple as that.

    Anyways like I noted this has absolutely nothing to do with the OP's original question.
     
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