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
[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!

 

[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 [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?

 

[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?