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Misner: exercise 6.8

  1. Nov 6, 2007 #1
    This is a multi part question so I'll just ask one part.

    I never understood where the equation for the Fermi Walker Transport came from and I'd really like to understand this because I think it would be really good for pedagogical value and for better understanding parallel transport, maybe I'm wrong.

    In this question I'll be using c=1

    For those of you who don't know the equation is
    [tex]
    \frac{{dv^\mu }}
    {{d\tau }} = - \Omega ^{\mu \tau } v_\tau
    [/tex]
    where

    [tex]
    \Omega ^{\mu \tau } = a^\mu u^\tau - a^\tau u^\mu + u_\alpha \theta _\beta \varepsilon ^{\alpha \beta \mu \tau }
    [/tex]

    Where [tex]{\mathbf{a}}[/tex] and [tex]{\mathbf{u}}[/tex] are the proper acceleration and velocity respectively,of your frame of reference, [tex]\theta[/tex] is an angle of rotation and [tex]\varepsilon[/tex] is the levi civita pseudo tensor.

    Here's where I'm at

    ignoring electromagnetism or anything like that particles follow geodesics given by the equation

    [tex]
    \frac{{dv^\alpha }}
    {{d\tau }} + \Gamma _{\beta \gamma }^\alpha v^\beta v^\gamma = 0
    [/tex]

    where [tex]{\mathbf{v}}[/tex] is the velocity of the particle being observed and [tex]
    \Gamma [/tex] is the affine connection:

    [tex]
    \left\langle {{\nabla _\gamma {\mathbf{e}}_\beta }}
    \mathrel{\left | {\vphantom {{\nabla _\gamma {\mathbf{e}}_\beta } {{\mathbf{\omega }}^\alpha }}}
    \right. \kern-\nulldelimiterspace}
    {{{\mathbf{\omega }}^\alpha }} \right\rangle = \Gamma _{\beta \gamma }^\alpha = \left\{ {\begin{array}{*{20}c}
    \alpha \\
    {\beta \gamma } \\

    \end{array} } \right\} + \frac{1}
    {2}\left( {c_{\beta \gamma } ^\alpha + c_\beta ^\alpha _\gamma + c_\gamma ^\alpha _\beta } \right)
    [/tex]

    Where

    [tex]
    \left\{ {\begin{array}{*{20}c}
    \alpha \\
    {\beta \gamma } \\

    \end{array} } \right\}
    [/tex] is the christoffel symbol of the second kind


    [tex]
    \left\{ {\begin{array}{*{20}c}
    \alpha \\
    {\beta \gamma } \\

    \end{array} } \right\} = \frac{1}
    {2}g^{\alpha \tau } [\beta \gamma ,\tau ] = \frac{1}
    {2}g^{\alpha \tau } \left( { - g_{\beta \gamma } ,_\tau + g_{\beta \tau } ,_\gamma + g_{\gamma \tau } ,_\beta } \right)
    [/tex]

    and [tex]c_{\beta \gamma } ^\alpha [/tex] are your structure coefficients

    [tex]
    [{\mathbf{e}}_\beta ,{\mathbf{e}}_\gamma ] = \nabla _\beta {\mathbf{e}}_\gamma - \nabla _\gamma {\mathbf{e}}_\beta = c_{\beta \gamma } ^\alpha {\mathbf{e}}_\alpha
    [/tex]

    Lets consider a tetrad formulation for our locally at rest coordinate system

    [tex]
    g_{\alpha \beta } = \eta _{\mu \tau } e^\mu _\alpha e^\tau _\beta
    [/tex]

    pick your tetrad to always orthonormal, [tex]
    e^\mu _\alpha = \delta ^\mu _\alpha [/tex], in which case it can be shown that

    [tex]
    {\mathbf{\omega }}^\alpha _\beta = \Gamma _{\beta \gamma }^\alpha {\mathbf{\omega }}^\gamma = \frac{1}
    {2}\left( {c_{\beta \gamma } ^\alpha + c_\beta ^\alpha _\gamma + c_\gamma ^\alpha _\beta } \right){\mathbf{\omega }}^\gamma
    [/tex]

    where [tex]{\mathbf{\omega }}^\alpha _\beta[/tex] is the spin connection and in these orthonormal coordinates it has the property [tex]
    {\mathbf{\omega }}_{\beta \alpha } = - {\mathbf{\omega }}_{\alpha \beta }
    [/tex]

    I claim that the spin connection is basically the same as [tex]\Omega ^{\mu \tau }[/tex]

    Using the geodesic equation and the fact that we are using orthonormal tetrads we basically have it, but we have to work out [tex]\Omega ^{\mu \tau }[/tex]

    Now, since we are doing coordinate changes from orthonormal coordinate system to another orthonormal coordinate system we have [tex]
    \eta _{\alpha \beta } = \eta _{\mu \tau } \Lambda ^\mu _\alpha \Lambda ^\tau _\beta [/tex]

    Now picking our coordinate systems to be right handed and assuming they're orthochronous too we have that [tex]
    \Lambda \varepsilon SO(3,1)
    [/tex]

    The particle is at rest in our coordinate system so

    [tex]
    {\mathbf{u}} = {\mathbf{e}}_0
    [/tex]

    So its acceleration is

    [tex]
    {\mathbf{a}} = \frac{{d{\mathbf{u}}}}
    {{d\tau }} = \frac{{d{\mathbf{e}}_0 }}
    {{d\tau }} = {\mathbf{e}}_i \Gamma _{00}^i
    [/tex]

    where [tex]
    \Gamma _{00}^0 = 0[/tex] since [tex]{\mathbf{a}} \cdot {\mathbf{u}} = 0[/tex]

    and [tex]a^i = \Gamma _{00}^i[/tex] since [tex]x^0 = \tau[/tex]

    Now this problem is a lot like problem 11.12 in Jackson which I got right

    The answer in Jackson is

    [tex]
    A_T = I - \left( {\gamma ^2 \delta {\mathbf{v}}_\parallel + \gamma \delta {\mathbf{v}}_ \bot } \right) \cdot {\mathbf{K}} - \frac{{\gamma ^{\mathbf{2}} }}
    {{\gamma + 1}}\left( {{\mathbf{v}} \times \delta {\mathbf{v}}_ \bot } \right) \cdot {\mathbf{S}}
    [/tex]

    but there's factors of gamma that do not match the fermi walker equation so what's the difference between these two equations

    basically how do I get
    [tex]
    \Omega ^{\mu \tau } = a^\mu u^\tau - a^\tau u^\mu + u_\alpha \theta _\beta \varepsilon ^{\alpha \beta \mu \tau }
    [/tex]

    I don't see it

    I mean I understand

    [tex]
    \Lambda = \exp \left( {\theta \cdot {\mathbf{S}}} \right)\exp \left( { - \zeta \cdot {\mathbf{K}}} \right)
    [/tex]

    where [tex]
    {\mathbf{\zeta }} = {\mathbf{\hat v}}\tanh ^{ - 1} v
    [/tex] but I can't seem to make the Jackson equation agree with the Misner equation.

    Help!!!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 8, 2007 #2
    Sorry to be a nag about this, but Help!!!!!!!

    I know someone has derived the fermi walker transport before
     
  4. Dec 17, 2007 #3
    I am so confused. Let's assume in the rest frame

    you have [tex]
    \left\langle {{\omega ^i }}
    \mathrel{\left | {\vphantom {{\omega ^i } {\nabla _{\mathbf{u}} {\mathbf{e}}_0 }}}
    \right. \kern-\nulldelimiterspace}
    {{\nabla _{\mathbf{u}} {\mathbf{e}}_0 }} \right\rangle =\Gamma ^i _{00} = a^i
    [/tex] and that all the other parts of the connection are zero

    Now say your in another coordinate system just watching this what would the connection be in this coordinate system?

    I used the affine connection

    [tex]
    \bar \Gamma ^\alpha _{\beta \gamma } = - \frac{{\partial ^2 \bar x^\alpha }}
    {{\partial x^\mu \partial x^\tau }}\frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}\frac{{\partial x^\tau }}
    {{\partial \bar x^\gamma }} + \frac{{\partial \bar x^\alpha }}
    {{\partial x^\kappa }}\Gamma ^\kappa _{\mu \tau } \frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}\frac{{\partial x^\tau }}
    {{\partial \bar x^\gamma }}
    [/tex]

    in the rest frame
    [tex]\frac{{dx^\alpha }}
    {{d\tau }}{\mathbf{e}}_\alpha = {\mathbf{e}}_0
    [/tex]

    [tex]
    u^\alpha \equiv \frac{{d\bar x^\alpha }}
    {{d\tau }} = \frac{{\partial \bar x^\alpha }}
    {{\partial x^\beta }}\frac{{dx^\beta }}
    {{d\tau}} = \frac{{\partial \bar x^\alpha }}
    {{\partial x^0 }}
    [/tex]

    so

    [tex]
    \begin{gathered}
    \bar \Gamma ^\alpha _{\beta \gamma } T^\beta \frac{{d\bar x^\gamma }}
    {{d\tau }} = - \frac{{\partial ^2 \bar x^\alpha }}
    {{\partial x^\mu \partial x^\tau }}\frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}\frac{{\partial x^\tau }}
    {{\partial \bar x^\gamma }}T^\beta \frac{{d\bar x^\gamma }}
    {{d\tau }} + \frac{{\partial \bar x^\alpha }}
    {{\partial x^\kappa }}\Gamma ^\kappa _{\mu \tau } \frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}\frac{{\partial x^\tau }}
    {{\partial \bar x^\gamma }}T^\beta \frac{{d\bar x^\gamma }}
    {{d\tau }} \hfill \\
    = - \frac{{\partial ^2 \bar x^\alpha }}
    {{\partial x^\mu \partial x^\tau }}\frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}T^\beta \frac{{dx^\tau }}
    {{d\tau }} + \frac{{\partial \bar x^\alpha }}
    {{\partial x^\kappa }}\Gamma ^\kappa _{\mu \tau } \frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}T^\beta \frac{{dx^\tau }}
    {{d\tau }} \hfill \\
    = - \frac{{\partial ^2 \bar x^\alpha }}
    {{\partial x^\mu \partial x^0 }}\frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}T^\beta + \frac{{\partial \bar x^\alpha }}
    {{\partial x^i }}\Gamma ^i _{00} \frac{{\partial x^0 }}
    {{\partial \bar x^\beta }}T^\beta \hfill \\
    = - \frac{\partial }
    {{\partial x^\mu }}\left( {u^\alpha } \right)\frac{{\partial x^\mu }}
    {{\partial \bar x^\beta }}T^\beta + \frac{{\partial \bar x^\alpha }}
    {{\partial x^i }}a^i \frac{{\partial x^0 }}
    {{\partial \bar x^\beta }}T^\beta \hfill \\
    = - \frac{\partial }
    {{\partial \bar x^\beta }}\left( {u^\alpha } \right)T^\beta - a^\alpha u_\beta T^\beta \hfill \\
    \end{gathered}
    [/tex]

    where [tex]
    a^\alpha = \frac{{\partial \bar x^\alpha }}
    {{\partial x^i }}a^i
    [/tex] sorry for the poor notation and [tex] - u_\beta = \frac{{\partial x^0 }}
    {{\partial \bar x^\beta }}[/tex] since [tex]u^\alpha = \frac{{\partial \bar x^\alpha }}
    {{\partial x^0 }}[/tex]

    The answer is supposed to be

    [tex]
    \bar \Gamma ^\alpha _{\beta \gamma } T^\beta \frac{{d\bar x^\gamma }}
    {{d\tau }} = \left( {u^\alpha a_\beta - a^\alpha u_\beta } \right)T^\beta
    [/tex]

    The last part of my answer checks but what about

    [tex]
    - \frac{\partial }
    {{\partial \bar x^\beta }}\left( {u^\alpha } \right)T^\beta
    [/tex]

    That isn't equal to [tex]u^\alpha a_\beta T^\beta[/tex]

    is it?
     
  5. Dec 18, 2007 #4

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    I am too busy at work right now to think about this, but I might get back to it.

    You might find this thread and this thread interesting and/or useful.
     
    Last edited: Dec 18, 2007
  6. Feb 6, 2008 #5
    Sorry to keep bringing this up, but I still don't get it. Can anyone give a fairly rigorous derivation of the fermi walker transport, and a better way of understanding it?
     
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