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 $\sqrt{-\alpha_a\alpha^a}$
Where has the negative sign come from and why is it needed?
Best Answer
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 $U^0= dt/d\tau$. 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.