Dtyjlui

An observer in a fixed location relative to our coordinate system has a worldline with constant $r, theta, phi$, and thereofre has four velocity $U$ with only the first component non zero. Because $U^aU_a=1$ and $U^0 >0$, the four velocity components are

$$U^0=frac{1}{sqrt{1-2m/r}}, text{ } U^a=0 text{ for a=1,2,3}$$

How has this been derived mathematically?

.... As in special relativity, the acceleration felt by the observer is $sqrt{-alpha_aalpha^a}$

Where has the negative sign come from and why is it needed?

The relation
begin{equation}
U^0=frac{1}{sqrt{1-2m/r}}, U^a=0 text{ for a=1,2,3}
end{equation}
is obtained by solving the geodesic equation for the Schwarzschild metric
begin{equation}
- d tau^2=-(1-frac{2m}{r}) dt^2+(1-frac{2m}{r})^{-1} dr^2+r^2 d theta ^2+ r^2 sin^2(theta) d phi^2
end{equation}
The Schwarzschild metric is a smooth spherically symmetric metric interpreted as the spacetime modelling the exterior of a spherically symmetric body of relativistic mass m. As a matter of fact, the angular dependence of the Schwarzschild metric is precisely the same as that of a sphere. It is therefore almost always sufficient to consider the equatorial plane $θ = π/2$, so that $dθ = 0$. In this coordinate system the metric is invariant under time translations, and this allows us to define a class of stationary observers whose world lines are given by constant values $r$, $phi$ and $theta$. Consequently, the lapse of proper time τ between two events at a fixed spatial point in Schwarzschild space-time is
begin{equation}
ds^2 =-(1-frac{2m}{r}) dt^2= -g _{00}dt^2=-d tau ^2
end{equation}
The element of proper time $d tau$ is measured by a clock at the particular point, while the element of world time dt is fixed for the whole manifold. The 4-velocity of a stationary observer is $U = (U^0,0,0,0)$ with $U^0= dt/dtau$. The acceleration of the observer is
begin{equation}
alpha^a= U^b nabla_b U^a= U^b left(frac{partial U^a}{partial x^b}+ Gamma_{bc}^{a} U^c right)= U^0 left(frac{partial U^a}{partial x^0}+ Gamma_{00}^{a} U^0 right)=(U^0)^2 Gamma_{00}^{a}
end{equation}
Thus
begin{equation}
boldsymbol{alpha}=(0,m/r²,0,0)
end{equation}
The magnitude of the acceleration can be computed by using the dot product
begin{equation}
boldsymbol{alpha} cdot boldsymbol{alpha} = g_{ab} alpha^a alpha^b
end{equation}
and then computing the square root. By doing that we have
begin{equation}
|boldsymbol{alpha}| =left(1-frac{2m}{r} right)^{-1/2} frac{m}{r^2}
end{equation}
Since $g_{11}= -(1-frac{2m}{r})^{-1}$ when defining $sqrt{g_{ab} alpha^a alpha^b}=
sqrt{-g_{11} alpha^1 alpha^1}$
we have to use the sign minus to be consistent with definition of $g_{11}$.

For large values of $r$ this is essentially Newtonian, but as $r → 2m$, the acceleration becomes infinite.