# Misner: exercise 6.8

1. Nov 6, 2007

### 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
$$\frac{{dv^\mu }} {{d\tau }} = - \Omega ^{\mu \tau } v_\tau$$
where

$$\Omega ^{\mu \tau } = a^\mu u^\tau - a^\tau u^\mu + u_\alpha \theta _\beta \varepsilon ^{\alpha \beta \mu \tau }$$

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

$$\frac{{dv^\alpha }} {{d\tau }} + \Gamma _{\beta \gamma }^\alpha v^\beta v^\gamma = 0$$

where $${\mathbf{v}}$$ is the velocity of the particle being observed and $$\Gamma$$ is the affine connection:

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

Where

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

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

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

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

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

$$g_{\alpha \beta } = \eta _{\mu \tau } e^\mu _\alpha e^\tau _\beta$$

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

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

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

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 $$\eta _{\alpha \beta } = \eta _{\mu \tau } \Lambda ^\mu _\alpha \Lambda ^\tau _\beta$$

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

The particle is at rest in our coordinate system so

$${\mathbf{u}} = {\mathbf{e}}_0$$

So its acceleration is

$${\mathbf{a}} = \frac{{d{\mathbf{u}}}} {{d\tau }} = \frac{{d{\mathbf{e}}_0 }} {{d\tau }} = {\mathbf{e}}_i \Gamma _{00}^i$$

where $$\Gamma _{00}^0 = 0$$ 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

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

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
$$\Omega ^{\mu \tau } = a^\mu u^\tau - a^\tau u^\mu + u_\alpha \theta _\beta \varepsilon ^{\alpha \beta \mu \tau }$$

I don't see it

I mean I understand

$$\Lambda = \exp \left( {\theta \cdot {\mathbf{S}}} \right)\exp \left( { - \zeta \cdot {\mathbf{K}}} \right)$$

where $${\mathbf{\zeta }} = {\mathbf{\hat v}}\tanh ^{ - 1} v$$ 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. Nov 8, 2007

### Jim Kata

I know someone has derived the fermi walker transport before

3. Dec 17, 2007

### Jim Kata

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

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

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

in the rest frame
$$\frac{{dx^\alpha }} {{d\tau }}{\mathbf{e}}_\alpha = {\mathbf{e}}_0$$

$$u^\alpha \equiv \frac{{d\bar x^\alpha }} {{d\tau }} = \frac{{\partial \bar x^\alpha }} {{\partial x^\beta }}\frac{{dx^\beta }} {{d\tau}} = \frac{{\partial \bar x^\alpha }} {{\partial x^0 }}$$

so

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

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

The answer is supposed to be

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

$$- \frac{\partial } {{\partial \bar x^\beta }}\left( {u^\alpha } \right)T^\beta$$

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

is it?

4. Dec 18, 2007

### George Jones

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