Misner Ex 6.8: Understanding Fermi-Walker Transport Equation

[Jim Kata]

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
<br /> \frac{{dv^\mu }}<br /> {{d\tau }} = - \Omega ^{\mu \tau } v_\tau <br />
where

<br /> \Omega ^{\mu \tau } = a^\mu u^\tau - a^\tau u^\mu + u_\alpha \theta _\beta \varepsilon ^{\alpha \beta \mu \tau } <br />

Where {\mathbf{a}} and {\mathbf{u}} are the proper acceleration and velocity respectively,of your frame of reference, \theta is an angle of rotation and \varepsilon 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

<br /> \frac{{dv^\alpha }}<br /> {{d\tau }} + \Gamma _{\beta \gamma }^\alpha v^\beta v^\gamma = 0<br />

where {\mathbf{v}} is the velocity of the particle being observed and <br /> \Gamma is the affine connection:

<br /> \left\langle {{\nabla _\gamma {\mathbf{e}}_\beta }}<br /> \mathrel{\left | {\vphantom {{\nabla _\gamma {\mathbf{e}}_\beta } {{\mathbf{\omega }}^\alpha }}}<br /> \right. \kern-\nulldelimiterspace}<br /> {{{\mathbf{\omega }}^\alpha }} \right\rangle = \Gamma _{\beta \gamma }^\alpha = \left\{ {\begin{array}{*{20}c}<br /> \alpha \\<br /> {\beta \gamma } \\<br /> <br /> \end{array} } \right\} + \frac{1}<br /> {2}\left( {c_{\beta \gamma } ^\alpha + c_\beta ^\alpha _\gamma + c_\gamma ^\alpha _\beta } \right)<br />

Where

<br /> \left\{ {\begin{array}{*{20}c}<br /> \alpha \\<br /> {\beta \gamma } \\<br /> <br /> \end{array} } \right\}<br /> is the christoffel symbol of the second kind


<br /> \left\{ {\begin{array}{*{20}c}<br /> \alpha \\<br /> {\beta \gamma } \\<br /> <br /> \end{array} } \right\} = \frac{1}<br /> {2}g^{\alpha \tau } [\beta \gamma ,\tau ] = \frac{1}<br /> {2}g^{\alpha \tau } \left( { - g_{\beta \gamma } ,_\tau + g_{\beta \tau } ,_\gamma + g_{\gamma \tau } ,_\beta } \right)<br />

and c_{\beta \gamma } ^\alpha are your structure coefficients

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

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

<br /> g_{\alpha \beta } = \eta _{\mu \tau } e^\mu _\alpha e^\tau _\beta <br />

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

<br /> {\mathbf{\omega }}^\alpha _\beta = \Gamma _{\beta \gamma }^\alpha {\mathbf{\omega }}^\gamma = \frac{1}<br /> {2}\left( {c_{\beta \gamma } ^\alpha + c_\beta ^\alpha _\gamma + c_\gamma ^\alpha _\beta } \right){\mathbf{\omega }}^\gamma<br />

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

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

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

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

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

The particle is at rest in our coordinate system so

<br /> {\mathbf{u}} = {\mathbf{e}}_0 <br />

So its acceleration is

<br /> {\mathbf{a}} = \frac{{d{\mathbf{u}}}}<br /> {{d\tau }} = \frac{{d{\mathbf{e}}_0 }}<br /> {{d\tau }} = {\mathbf{e}}_i \Gamma _{00}^i <br />

where <br /> \Gamma _{00}^0 = 0 since {\mathbf{a}} \cdot {\mathbf{u}} = 0

and a^i = \Gamma _{00}^i since x^0 = \tau

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

The answer in Jackson is

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

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
<br /> \Omega ^{\mu \tau } = a^\mu u^\tau - a^\tau u^\mu + u_\alpha \theta _\beta \varepsilon ^{\alpha \beta \mu \tau } <br />

I don't see it

I mean I understand

<br /> \Lambda = \exp \left( {\theta \cdot {\mathbf{S}}} \right)\exp \left( { - \zeta \cdot {\mathbf{K}}} \right)<br />

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

Help!

 

[Jim Kata]

Sorry to be a nag about this, but Help!

I know someone has derived the fermi walker transport before

 

[Jim Kata]

I am so confused. Let's assume in the rest frame

you have <br /> \left\langle {{\omega ^i }}<br /> \mathrel{\left | {\vphantom {{\omega ^i } {\nabla _{\mathbf{u}} {\mathbf{e}}_0 }}}<br /> \right. \kern-\nulldelimiterspace}<br /> {{\nabla _{\mathbf{u}} {\mathbf{e}}_0 }} \right\rangle =\Gamma ^i _{00} = a^i <br /> 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

<br /> \bar \Gamma ^\alpha _{\beta \gamma } = - \frac{{\partial ^2 \bar x^\alpha }}<br /> {{\partial x^\mu \partial x^\tau }}\frac{{\partial x^\mu }}<br /> {{\partial \bar x^\beta }}\frac{{\partial x^\tau }}<br /> {{\partial \bar x^\gamma }} + \frac{{\partial \bar x^\alpha }}<br /> {{\partial x^\kappa }}\Gamma ^\kappa _{\mu \tau } \frac{{\partial x^\mu }}<br /> {{\partial \bar x^\beta }}\frac{{\partial x^\tau }}<br /> {{\partial \bar x^\gamma }}<br />

in the rest frame
\frac{{dx^\alpha }}<br /> {{d\tau }}{\mathbf{e}}_\alpha = {\mathbf{e}}_0 <br />

<br /> u^\alpha \equiv \frac{{d\bar x^\alpha }}<br /> {{d\tau }} = \frac{{\partial \bar x^\alpha }}<br /> {{\partial x^\beta }}\frac{{dx^\beta }}<br /> {{d\tau}} = \frac{{\partial \bar x^\alpha }}<br /> {{\partial x^0 }}<br />

so

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

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

The answer is supposed to be

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

The last part of my answer checks but what about

<br /> - \frac{\partial }<br /> {{\partial \bar x^\beta }}\left( {u^\alpha } \right)T^\beta <br />

That isn't equal to u^\alpha a_\beta T^\beta

is it?

 

[George Jones]

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

You might find https://www.physicsforums.com/showthread.php?p=680081" interesting and/or useful.

 

[Jim Kata]

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?