Series expansion tetrad Fermi coordinates

[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.

 

[George Jones]

Look at (2).

 

[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?

 

[George Jones]

for exactly the same reason, as explained on p.331 of MTW, yes?

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 $X^i = u \xi^i|_{u = 0}$.

 

[sardar]

Hello there,

As a fellow scientist, I can understand the struggle of trying to reproduce calculations from a paper. After carefully examining the equations and your reasoning, I believe I have found the answer to your question.

In your derivation, you have correctly identified the terms that contribute to the second derivative of the tetrad, but there is one term that you have overlooked. In the geodesic equation, the term $\Gamma^{i}_{lm}\xi^l \xi^m$ is only valid for the tangent vector $\xi^i$, but in the expression for the second derivative of the tetrad, we are also differentiating with respect to the parameter $u$. This means that the term $\Gamma^{i}_{lm}\xi^l \xi^m$ needs to be multiplied by $\frac{d\xi^i}{du}$, which is equal to $-\Gamma^{i}_{lm}\xi^l \xi^m$. This results in the term $\Gamma^{i}_{lm}\xi^l \xi^m \frac{d\xi^i}{du}$, which cancels out the extra term in your calculation and gives the correct result as shown in (68) of the paper.

I hope this helps clarify your doubts. Keep up the good work on your project!

Best regards,