3
$\begingroup$

I understand how an object moving through space would be affected if gravity were a curvature of space as opposed to being a force. However, I do not understand why this creates a "pull" on an object.

If, for example, an object were to be placed on the edge of the "wall" created by gravity, would it not simply stay there? What would cause it to accelerate "downwards" towards the planet, if it had no initial velocity?

$\endgroup$
  • 1
    $\begingroup$ Suppose you put an object (such as a ball) on a slope. Does it only roll downwards, if you add an initial velocity? $\endgroup$ – Semoi Jan 26 at 19:54
  • $\begingroup$ Here is a decent video from PBS youtube.com/watch?v=AwhKZ3fd9JA $\endgroup$ – Adrian Howard Jan 26 at 20:24
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/102910/2451 , physics.stackexchange.com/q/3009/2451 and links therein. $\endgroup$ – Qmechanic Jan 26 at 21:42
  • 1
    $\begingroup$ What wall created by gravity are you talking about? Did you mean well? $\endgroup$ – G. Smith Jan 26 at 21:50
  • 1
    $\begingroup$ In curved spacetime, objects don’t accelerate under gravity. The acceleration in Newtonian gravity is an artifact of the coordinate systems we use for Newtonian mechanics. $\endgroup$ – G. Smith Jan 26 at 21:57
2
$\begingroup$

The answer to your question is the four vector. You just have to accept it that the universe is set up so and the four vector is set up so that its magnitude must always be c.

in short, the magnitude of the four-velocity for any object is always a fixed constant: $ \| \mathbf{U} \|^2 = c^2 \,$

https://en.wikipedia.org/wiki/Four-vector

Now if as you say you put a stationary (in space) object close to Earth, why will it start moving towards the center of gravity?

The answer is that the object is stationary in space, but it is moving in time at the speed of light.

It is in a gravitational field, and this slows down the object (relatively) in the temporal dimension, this is GR time dilation. But the magnitude of the four vector must stay c, so if the speed of the object decreases in the temporal dimension, its speed in the spatial dimensions must increase, thus the object will start moving in space towards the center of gravity.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ +1 You just have to accept it that the universe is set up so - try to read this though: physics.stackexchange.com/a/133821/248459 $\endgroup$ – Pagoda Jan 26 at 21:55
  • $\begingroup$ @Pagoda thank you so much $\endgroup$ – Árpád Szendrei Jan 26 at 22:29
  • $\begingroup$ This answer is mind opening for me. So if I use the Schwarzchild metric it will give me the relative coefficients of the 4-vector as a function of distance r from the center of the mass? $\endgroup$ – dllahr Jan 27 at 1:16
2
$\begingroup$

In curved spacetime, objects don’t accelerate under gravity. The 3D acceleration in Newtonian gravity is merely an artifact of the coordinate systems we use for Newtonian physics. Einstein taught us that objects moving under gravity are simply moving with constant (i.e., parallel-transported) four-velocity through curved spacetime.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The idea of relativity is that the laws of nature are independent of the reference frame.

And the basic for gravity in GR is the equivalent principle. That means: you can think as the surface of the planet was accelerating (upwards) all the time. That is (almost) all. GR doesn't try to explain why that happens.

But there is a little more than that principle. Planets are spherical, and gravity is stronger close the surface than at a mountain. The direction of that acceleration also changes for different points of the sphere.

So, the vector field of accelerations is not constant. That is the meaning of the space-time curvature.

If a spaceship is drifting in the space, and see a planet orbiting it, the crew (even while floating around the ship) knows that they are in a gravitational field, because otherwise all external objects should follow a straight line. If it is not the case there is space-time curvature.

So, space-time curvature is not a cause for gravity, but a way that floating crew in the orbiting space station agree with fixed crew in the base station that both are under Earth gravitational field.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.