I am an newbie general relativistic learner and I learnt that gravity is bending of space-time and since objects move in straight-lines but since its curved they follow curved movement through space thus creating the effect we know as gravity.

That said, what if an object has velocity of 0 and since its not moving at all, (except in Time) why does the object fall or move in curved space-time (geodesics)? Is there external force that also pulls the object if so why and how does it work?

If the object moves in an curved space-time, why does an object just move in an normal space-time (not curved) rather than stay at velocity 0?

Is there any reason why this happens?

  • $\begingroup$ In general relativity objects do not move on straight lines, but on geodesic lines. The geodesics of test masses that start in different points "at infinity" meet at the center of the gravitating body. Just like in Newtonian mechanics, where the motion of a body depends on its initial conditions, the geodesic that a test mass in general relativity will take also depends on the initial conditions. What geodesics have in common with the "straight lines" of free Newtonian particles is that they, too, minimize the distance between two points, but not in space, but in spacetime. $\endgroup$
    – CuriousOne
    Commented Sep 12, 2014 at 19:28
  • $\begingroup$ As the ensuing comment discussion has moved to chat I've removed it from here. $\endgroup$
    – David Z
    Commented Sep 14, 2014 at 1:57

1 Answer 1


You're thinking of space with an external time, not spacetime.

In spacetime, all objects move, because they trace out timelike curves through spacetime. The statement that an object is "stationary" in a coordinate system adapted to a reference frame is merely a statement that it's 4-velocity is of the form $(a,0,0,0)$, where $|a|^{2} = \frac{1}{|g_{tt}|}$. And note, that this initial 4-velocity might not be sufficient to keep an object stationary, since for any one of the spatial coordinates (let's say $x$), you might have a nonzero Christoffel symbol with $tt$ in the lower two indices, and then, you'll have${}^{1}$:

$${\ddot x} + \Gamma^{x}{}_{tt}{\dot t}{\dot t} = 0,$$

which means that for $t > 0$, $\dot x\neq 0$.

${}^{1}$Note that this fact is why old-style GR texts refer to $g_{tt}$ as one minus twice the classical potential, since, for a diagonal metric, you have:

$$\Gamma^{i}{tt} = \frac{1}{2}g^{i\mu}\left(g_{t\mu,t} + g_{t\mu,t} - g_{tt,\mu}\right) = -\frac{1}{2}g^{ii}g_{tt,i}.$$

Which, if you map to three dimensions, and call $g_{tt} = -\left(1-2\phi\right)$, is equal to ${\vec \nabla}\phi$, and the above equation reduces to ${\ddot {\vec r}} + {\vec \nabla} \phi = 0$, which is the Newtonian version of gravitational kinematics.

  • $\begingroup$ And as a note, I should say that it is NOT fair to say that $g_{tt}$ can always be interpreted in this way. In particular, this is only true for a particular low-velocity limit of relativity, where all of the velocities are suppressed by factors of $c$. $\endgroup$ Commented Sep 13, 2014 at 20:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.