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Kerr-Newman charged geodesics

  1. Aug 15, 2014 #1
    I've found the equations of motion for a charged test particle in the Kerr-Newman geometry from a number of sources. However, they aren't very reliable and disagree on small details, so I'm trying to derive it myself. I'm completely stuck at the last step though, where you "use" the separability of the Hamilton-Jacobi equation to get the EOM of [itex]\dot{r}[/itex] and [itex]\dot{\theta}[/itex] without their second derivatives. Here is what I've got so far:

    Metric + Potential (only part not derived/checked by me):
    [tex]
    ds^2\equiv-\dfrac{\rho}{\Delta}dr^2-\rho^2d\theta^2+\dfrac{\Delta}{\rho^2}\left(dt-asin^2\theta d\phi\right)^2-\dfrac{sin^2\theta}{\rho^2}\left((r^2+a^2)d\phi-adt\right)^2 \\
    A_\mu\equiv \dfrac{Qr}{\rho^2}\left(\delta^t_\mu-asin^2\theta\delta^\phi_\mu\right) \\
    \Delta\equiv r^2+a^2-2Mr+Q^2 \\
    \rho^2\equiv r^2+a^2cos\theta
    [/tex]
    for a black hole with parameters M,a,Q

    Geodesic Equation+Hamiltonian:
    [tex]
    \dfrac{d}{ds}\left(g_{\mu\nu}\dot{x}^\nu -\dfrac{q}{m}A_\mu\right) = \dfrac{1}{2}\partial_\mu g_{\sigma\rho}\dot{x}^\sigma\dot{x}^\rho - \dfrac{q}{m}\dot{x}^\sigma \partial_\mu A_\sigma \\
    H=\dfrac{1}{2}mg_{\mu\nu}\dot{x}^\mu\dot{x}^\nu=\dfrac{\kappa}{2}m=(2m)^{-1}g^{\mu\nu}\left(p_\mu+qA_\mu\right)\left(p_\nu+qA_\nu\right)
    [/tex]
    for a test particle with mass m and charge q, with [itex]\kappa\equiv sgn(m^2)[/itex]. I've also scaled everything by m just to keep it consistent with usual notation in the non-gravitational limit.

    The time and rotational symmetries immediately give the first two equations of motion:
    [tex]
    E\equiv mg_{tt}\dot{t}+mg_{\phi t}\dot{\phi}-qA_t \\
    L\equiv qA_\phi - mg_{\phi\phi}\dot{\phi}-mg_{\phi t}\dot{t}
    [/tex]

    The 2 other solutions to the EL equations give second derivatives of r and θ, so I'm following the steps of the sources I found (all originally from Carter,68). They all claim they are "using" the Hamilton-Jacobi equation to derive the next set of equations. However, as far as I can tell all they really do is assume that [itex]p_r[/itex] is a function of only r and [itex]p_\theta[/itex] is a function of only theta, and then set the two forms of H equal. The Hamilton-Jacobi equations and the S function never even need to be mentioned, but for some reason every one of them does?

    Anyway, when I do this step I find:
    [tex]
    2mH = |m|^2 = g^{tt}\left(E+qA_t\right)^2 + g^{\phi\phi}\left(L-qA_\phi\right)^2-2g^{t\phi}\left(E+qA_t\right)\left(L-qA_\phi\right) - \dfrac{\Delta}{\rho^2} p_r^2-\dfrac{1}{\rho^2}p_\theta^2
    [/tex]

    I calculated the inverse metric elements to be:
    [tex]
    g^{\phi\phi}=\dfrac{1}{\rho^2}\left(\dfrac{a^2}{\Delta}-\dfrac{1}{sin^2\theta}\right) \\
    g^{t\phi}=\dfrac{a}{\Delta\rho^2}\left(2Mr-Q^2\right) \\
    g^{tt}=\dfrac{1}{\rho^2}\left(\dfrac{(r^2+a^2)^2}{\Delta}-a^2sin^2\theta\right)
    [/tex]
    which can be plugged in to the Hamiltonian expression. Multiplying both sides by [itex]\rho^2[/itex] gives you an easily separable equation of r and θ IF q=0. In this case, you can continue by setting the separated equations equal to a constant K and finding the 2 remaining EOM. However, if q≠0, the A terms add a very non-trivial dependence of both r and θ. As far as I can tell this general formula is NOT separable. This doesn't really make sense though because even Carter's original derivation made the general assumption of q≠0! Anybody know what I'm doing wrong?

    *edit*
    The paper I mentioned can be found at http://journals.aps.org/pr/abstract/10.1103/PhysRev.174.1559

    Also, I just realized that by including the m factor everywhere I invalidated these equations for anything massless. So just assume we're dealing with a timelike massive test particle and it all works
     
    Last edited: Aug 15, 2014
  2. jcsd
  3. Aug 25, 2014 #2
    ^bump, Nobodies got any insight on this?
     
  4. Sep 4, 2014 #3
    Comon... 400 views and not one response? The Q=0 case is interesting, but I'm looking for the MOST general geodesic formula for a test particle near a black hole
     
  5. Sep 18, 2014 #4
    Last shot at this and I'm giving up
     
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