Acceleration of a static observer in Schwarzschild measured in locally inertial frame Consider a static observer at $r=R>2M$ with four velocity given by
$$U^\mu=\frac{dx^\mu}{d\tau}=(1-\frac{2M}{R})^{-1/2}\delta^\mu_0$$
in the Schwarzschild metric
$$ds^2=-(1-\frac{2M}{r})dt^2+(1-\frac{2M}{r})^{-1}dr^2+r^2d\Omega^2.$$
Question: What is the special relativistic acceleration as per Wikipedia
$$a^\hat{i}=\frac{d}{d\hat{t}}\frac{d}{d\tau}x^\hat{i}\,$$
of the static observer in locally inertial coordinates (hatted) which are momentarily at rest with respect the the static observer?
Presumably one could express the $a^\hat{i}$ in terms of $R$ and $M$ only using $U^\mu$ and the Jacobian of the coordinate transformation, even without explicitly finding Riemann normal coordinates, but I don't yet see how.
 A: For a static observer you typically would construct Fermi normal coordinates, not Riemann normal coordinates. The latter would only be valid around a single event on the observer's worldline, that at which we consider the local inertial frame momentarily comoving with the observer. Fermi normal coordinates on the other hand are valid all along the observer's worldline given a tetrad Fermi-Walker transported along the worldline. 
That being said, for what you wish to calculate there is no need to construct a Fermi normal coordinate system. This would only be required if we wanted to explicitly compute quantities that involved, say, the connection coefficients associated with the tetrad, or if we wanted to compute quantities evaluated some small distance away from the central worldline. 
We have simply $$ a^{\hat{\mu}}e_{\hat{\mu}} = a^{\mu}\partial_{\mu}$$ where $e_{\hat{\mu}}$ is the local inertial frame of the static observer and $\partial_{\mu}$ is a coordinate basis. All we have to do now is match the two sides. 
The RHS is easy to calculate and is simply $$ a^{\mu} = \frac{M}{r^2}\delta^{\mu}_r$$ 
and since $$e_{\hat{r}} = (1 - \frac{2M}{r})^{1/2}\partial_r $$ 
we find $$a^{\hat{r}} = (1 - \frac{2M}{r})^{-1/2}a^r = (1 - \frac{2M}{r})^{-1/2}\frac{M}{r^2}
$$ 
Note this blows up as we approach the event horizon, as we would expect. 
