# Series expansion tetrad Fermi coordinates

1. Jul 26, 2014

### WannabeNewton

Hi all. I'm working on a project that requires me to perform calculations in Fermi normal coordinates to certain orders, mostly 2nd order in the distance along the central worldline orthogonal space-like geodesics. In particular I need a rotating tetrad propagated along the central worldline obeying an arbitrary transport law which is parallel transported along the space-like geodesics so as to generate a tetrad field at each point in the coordinates to 2nd order. I found a very handy paper http://arxiv.org/pdf/gr-qc/0010096.pdf which computes the series expansion I need to arbitrary order but I'm not able to reproduce the calculation fully.

Appendix B of the paper is the relevant section for my question. In (66) the tetrad is expanded in a power series about the central worldline $\gamma$ in terms of the distance $u$ along the space-like geodesics orthogonal to and emanating from the worldline. In (67), $\xi^{\mu}$ is the unit tangent field to each space-like geodesic. The first thing to note is that $\xi^{\mu}(e_0)_{\mu}|_{u = 0} = 0$ by construction hence $\xi^{\mu}(e_0)_{\mu} = 0$ all along the space-like geodesic since both $\xi^{\mu}$ and $e^{\mu}_0$ are parallel transported along the geodesic. Now $e^{\mu}_0|_{u = 0} = \delta^{\mu}_{0}$ so $\xi^{\mu}(e_0)_{\mu}|_{u = 0} = 0 \Rightarrow \xi^{0}|_{u = 0} = 0$ but this does not mean that $\xi^0 = 0$ for all $u$ because $\xi^{\mu}(e_0)_{\mu} = 0$ does not imply $\xi^{\mu}\delta^0_{\mu} = 0$ as is clear both intuitively and from (65). So I will write $\xi^{i}$ in place of $\xi^{\mu}$ below because the final expression involves only $\xi^{\mu}$ and in the end we evaluate at $u = 0$ so we only need to consider the $\xi^i$ terms but you should keep the above in mind.

My first problem then is with (68). We differentiate (67) once with respect to $\frac{d}{du}$ to get $$\frac{d^2}{du^2}e^{(\nu)}_{\mu} - \Gamma^{\sigma}_{\mu i, l} e^{(\nu)}_{\sigma} \xi^l \xi^i - \Gamma^{\sigma}_{\mu i}\xi^i \frac{d}{du}e^{(\nu)}_{\sigma} - \Gamma^{\sigma}_{\mu i}e^{(\nu)}_{\sigma} \frac{d \xi^i}{du} \\= \frac{d^2}{du^2}e^{(\nu)}_{\mu} - \Gamma^{\sigma}_{\mu i, l} e^{(\nu)}_{\sigma} \xi^l \xi^i - \Gamma^{\sigma}_{\mu i}\Gamma^{\lambda}_{\sigma l}\xi^i \xi^l e^{(\nu)}_{\lambda} - \Gamma^{\sigma}_{\mu i}e^{(\nu)}_{\sigma} \frac{d \xi^i}{du}\\= \frac{d^2}{du^2}e^{(\nu)}_{\mu} - \Gamma^{\sigma}_{\mu i, l} e^{(\nu)}_{\sigma} \xi^l \xi^i - \Gamma^{\sigma}_{\mu i}\Gamma^{\lambda}_{\sigma l}\xi^i \xi^l e^{(\nu)}_{\lambda} + \Gamma^{\sigma}_{\mu i}e^{(\nu)}_{\sigma} \Gamma^{i}_{l m}\xi^l \xi^m \\ = \frac{d^2}{du^2}e^{(\nu)}_{\mu} - \Gamma^{\nu}_{\mu i, l} \xi^l \xi^i - \Gamma^{\sigma}_{\mu i}\Gamma^{\nu}_{\sigma l}\xi^i \xi^l + \Gamma^{\nu}_{\mu i}\Gamma^{i}_{l m}\xi^l \xi^m = 0$$
where I've used $\frac{d \xi^i}{du} = -\Gamma^{i}_{lm}\xi^l \xi^m$ from the geodesic equation. We thus have $$\frac{1}{2!}\frac{d^2 e^{(\nu)}_{\mu}}{du^2}|_{u = 0}u^2 = \frac{1}{2!}(\overset{0}{\Gamma^{\nu}_{\mu i, l}} + \overset{0}{\Gamma^{\sigma}_{\mu i} }\overset{0}{\Gamma^{\nu}_{\sigma l}} - \overset{0}{\Gamma^{\nu}_{\mu m}}\overset{0}{\Gamma^{m}_{i l}})X^i X^l$$
since $X^i = u \xi^i|_{u = 0}$. As you can see this is clearly not what the paper has in (68). It doesn't have the extra $- \overset{0}{\Gamma^{\nu}_{\mu m}}\overset{0}{\Gamma^{m}_{i l}}$ term. I however do not see how this term necessarily vanishes. Could anyone help me out with this? Why does the aforementioned term vanish in (68)? Thanks in advance.

2. Jul 26, 2014

### George Jones

Staff Emeritus
Look at (2).

3. Jul 26, 2014

### WannabeNewton

Haha yes of course, thanks George! I completely skipped over the section on Riemann coordinates and went straight to the section on Fermi coordinates so that was entirely my fault.

To be sure though, (2) is as given valid in Riemann normal coordinates, not necessarily Fermi normal coordinates. But it should hold in Fermi coordinates as well for exactly the same reason, as explained on p.331 of MTW, yes?

4. Jul 28, 2014

### George Jones

Staff Emeritus
Didn't have MTW at home on the weekend.

Yes, it looks like it is given by something like the equation before (13.69b), i.e., by using the geodesic equation along the spacelike geodesic for the coordinates $X^i = u \xi^i|_{u = 0}$.

Last edited: Jul 28, 2014