0

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?

Qmechanic
  • 220,844

1 Answers1

3

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.

Zo the Relativist
  • 42,237