Using the Schwarzschild metric and the mostly minus signature, how can I calculate the $9.81$ m$/$s$^2$ of acceleration on Earth using general relativity and the geodesic equation?
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In general relativity, gravity is not a force, and hence it causes no acceleration. This is clearly different from the perspective in Newtonian physics, in which we have an acceleration due to gravity.
Nevertheless, it turns out that a stationary orbit in general relativity does have an acceleration. In order to stay at the surface of the Earth, you must accelerate upwards, and this acceleration can be easily computed. Namely, it is given by $a^a = u^b \nabla_b u^a$, where $u^a$ is the four-velocity for the stationary observer. Using this result, you can conclude that the acceleration for a freely falling observer relative to a stationary observer must have the same numerical value (up to a sign).
Níck Aguiar Alves
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You can write the geodesic equation for $r$ and suppose $\theta = 90^{\circ}$ for simplicity. With a great degree of approximation for a planet with mass and radius as the Earth and velocities much smaller than light speed, the equation becomes near the Earth surface: $$\frac{d^2r}{d{\tau}^2} + \frac{GM}{R^2} + r\omega^2 = 0$$
$\frac{GM}{R^2} = g$, and if there is only radial movement, the radial acceleration is $g$. If the radial movement is zero (circular orbit), the centripetal acceleration is $g$.
By the way, we can not suppose that the body is at rest on the Earth surface, because it is not a solution for the geodesic equation.
Claudio Saspinski
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