# Fermi normal coordinates

1. Jul 15, 2005

### mikeu

I'm following the derivation of Fermi coordinates in MTW, section 13.6. Equation 13.60 states
$$\mathbf{\nabla_ue}_{\hat{\alpha}} = -\mathbf{\Omega\cdot e}_{\hat{\alpha}}$$
where $\Omega^{\mu\nu}$ is antisymmetric (and $\hat{\alpha}$ is the tetrad label). My question is, over which index is the contraction of Omega with e performed? We have
$$\Omega^{\mu\nu}e_\mu = - \Omega^{\nu\mu}e_\mu$$
so the result changes sign depending on the index chosen, but I can't find the authors' definition of the dot product between anything besides two vectors. Also, just want to check that my interpretation of the rest of the equation is correct... It's equivalent to
$$u^\beta\left(\partial_\beta e^\mu_{\hat{\alpha}} - \Gamma^\mu_{\beta\gamma}e^\gamma_\hat{\alpha}\right) = -g_{\beta\gamma}\Omega^{\mu\beta}e^\gamma_\hat{\alpha}$$
maybe with the mu and beta swapped on the Omega, right?

2. Jul 16, 2005

### George Jones

Staff Emeritus
I don't have MTW handy, so I'm just going to do some calculations and see what happens, and I'll attempt answers to your 2 questions in 2 separate posts.

First question. You're right about the ambiguity in the definition of contraction, so I'll start by defining

$$\begin{equation*} \begin{split} \mathbf{\Omega\cdot v} &= \Omega^{\gamma\beta} v_{\gamma} \mathbf{e}_{\beta}\\ &= \Omega_{\gamma}{}^{\beta} v^{\gamma} \mathbf{e}_{\beta}, \end{split} \end{equation*}$$

giving

$$\begin{equation*} \begin{split} \mathbf{\Omega\cdot e}_{\hat{\alpha}} &= \Omega_{\hat{\gamma}}{}^{\hat{\beta}} \delta^{\hat{\gamma}}_{\hat{\alpha}} \mathbf{e}_{\hat{\beta}}\\ &= \Omega_{\hat{\alpha}}{}^{\hat{\beta}} \mathbf{e}_{\hat{\beta}}. \end{split} \end{equation*}$$

MTW's antisymmetric $\mathbf{\Omega}$ is just the connection, which has a pair of components that are antisymmetric with respect to a tetrad. To see this, take the covariant derivative of (the constants) $\eta_{\hat{\mu}\hat{\nu}} = \mathbf{g}\left( \mathbf{e}_{\hat{\mu}}, \mathbf{e}_{\hat{\nu}} \right)$, remembering that the connection is metric compatible, i.e., that $\mathbf{\nabla g} = 0$. This gives

$$\begin{equation*} \begin{split} 0 &= \mathbf{g}\left( \mathbf{\nabla}_{\hat{\alpha}}\mathbf{e}_{\hat{\mu}}, \mathbf{e}_{\hat{\nu}} \right) + \mathbf{g}\left( \mathbf{e}_{\hat{\mu}}, \mathbf{\nabla}_{\hat{\alpha}}\mathbf{e}_{\hat{\nu}} \right)\\ &= \mathbf{g}\left( \Gamma^{\hat{\beta}}{}_{\hat{\mu}\hat{\alpha}} \mathbf{e}_{\hat{\beta}}, \mathbf{e}_{\hat{\nu}} \right) + \mathbf{g}\left( \mathbf{e}_{\hat{\mu}}, \Gamma^{\hat{\beta}}{}_{\hat{\nu}\hat{\alpha}} \mathbf{e}_{\hat{\beta}} \right)\\ &= \Gamma^{\hat{\beta}}{}_{\hat{\mu}\hat{\alpha}} \eta_{\hat{\beta}\hat{\nu}} + \Gamma^{\hat{\beta}}{}_{\hat{\nu}\hat{\alpha}} \eta_{\hat{\mu}\hat{\beta}}\\ &= \Gamma_{\hat{\nu}\hat{\mu}\hat{\alpha}} +\Gamma_{\hat{\mu}\hat{\nu}\hat{\alpha}} \end{split} \end{equation*}$$

I hope that I have used MTW's convention for the connection.

Now, with $\mathbf{u} = \mathbf{e}_{\hat{0}}$,

$$\begin{equation*} \begin{split} \mathbf{\nabla}_{\hat{0}} \mathbf{e}_{\hat{\alpha}} &= \Gamma^{\hat{\beta}}{}_{\hat{\alpha}\hat{0}} \mathbf{e}_{\hat{\beta}}\\ &= -\Gamma_{\hat{\alpha}}{}^{\hat{\beta}}{}_{\hat{0}} \mathbf{e}_{\hat{\beta}}\\ &= - \Omega_{\hat{\alpha}}{}^{\hat{\beta}} \mathbf{e}_{\hat{\beta}}\\ &= -\mathbf{\Omega\cdot e}_{\hat{\alpha}}, \end{split} \end{equation*}$$

where $\Omega_{\hat{\alpha}\hat{\beta}} := \Gamma_{\hat{\alpha}\hat{\beta}\hat{0}}$.

3. Jul 16, 2005

### George Jones

Staff Emeritus
Second question. Let $\left\{ \mathbf{e}_{\mu} \right\}$ be a coordinate basis and let $\left\{ \mathbf{e}_{\hat{\alpha}} \right\}$ be an orthonormal tetrad along the observer's worldline. Since both are bases, along the wordline there is a transformation that relates them:

$$\mathbf{e}_{\hat{\alpha}} = e^{\mu}_{\hat{\alpha}} \mathbf{e}_{\mu}.$$

Then,

$$\begin{equation*} \begin{split} \mathbf{\nabla_{u} e}_{\hat{\alpha}} &= u^{\beta} \mathbf{\nabla}_{\beta} \left( e^{\gamma}_{\hat{\alpha}} \mathbf{e}_{\gamma} \right)\\ &= u^{\beta} \left[ \left( \partial_{\beta} e^{\gamma}_{\hat{\alpha}} \right) \mathbf{e}_{\gamma} + e^{\gamma}_{\hat{\alpha}} \mathbf{\nabla}_{\beta} \mathbf{e}_{\gamma} \right]\\ &= u^{\beta} \left[ \left( \partial_{\beta} e^{\mu}_{\hat{\alpha}} \right) \mathbf{e}_{\mu} + e^{\gamma}_{\hat{\alpha}} \Gamma^{\mu}{}_{\gamma\beta} \mathbf{e}_{\mu} \right]\\ &= u^{\beta} \left[ \left( \partial_{\beta} e^{\mu}_{\hat{\alpha}} \right) + e^{\gamma}_{\hat{\alpha}} \Gamma^{\mu}{}_{\gamma\beta} \right] \mathbf{e}_{\mu}. \end{split} \end{equation*}$$

From my previous post,

$$\begin{equation*} \begin{split} \mathbf{\nabla_{u} e}_{\hat{\alpha}} &= -\Omega_{\hat{\alpha}}{}^{\hat{\nu}} \mathbf{e}_{\hat{\nu}}\\ &= -\Omega_{\hat{\alpha}}{}^{\hat{\nu}} e^{\mu}_{\hat{\nu}} \mathbf{e}_{\mu}\\ &= -\eta_{\hat{\alpha}\hat{\delta}} \Omega^{\hat{\delta}\hat{\nu}} e^{\mu}_{\hat{\nu}} \mathbf{e}_{\mu}\\ &= -e^{\gamma}_{\hat{\alpha}} e^{\beta}_{\hat{\delta}} g_{\gamma \beta} \Omega^{\hat{\delta}\hat{\nu}} e^{\mu}_{\hat{\nu}} \mathbf{e}_{\mu}\\ &= -e^{\gamma}_{\hat{\alpha}} g_{\gamma\beta} \Omega^{\beta\mu} \mathbf{e}_{\mu} \end{split} \end{equation*}$$

I'm often quite careless when I do calculations, so there could well be mistakes in these posts.

Regards,
George

4. Jul 16, 2005

### pervect

Staff Emeritus
I think that there should be a plus sign inside of the parenthisis on the left hand side, similar to George's remarks.

I find that Wald is a lot clearer on the definition of the derivative operator $\nabla_u$. In any coordinate basis we have

$$\nabla_u t^a = \partial_u t^a + \Gamma^a{}_{ub} t^b$$

Note that this does require that we be in a coordinate basis.

This is the same as your LHS with the minus sign replaced by a plus sign, except that you have additionally expanded $\partial_u$ by the chain rule.

I don't see any problems with your right hand side, except for the possible sign issue that you've already noted (which index you contract with).

It's not needed for this problem, but it's probably good to know that

$$\nabla_u t_a = \partial_u t_a - \Gamma^b{}_{ua} t^b$$

You might also find it handy to note that you can make the spacing in the Christoffel symbols look right by the following latex

\Gamma^a{}_{bc}

the empty pair of braces after the 'a' makes the vertical position of {bc} line up correctly.

5. Jul 16, 2005