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If gravity is just space distortion, which affects trajectories of moving objects, then a static object (not moving, thus no trajectory) will not suffer any type of accelerating force from gravity?

If so, this would also mean that we are always moving, even galaxies. The universe as a whole is moving and the definition of "static" has no physical manifestation.

Moreover if this holds true, it means that gravity just "bends" the kinetic energy (momentum?) that objects already contain for being in constant motion, which explains even further why gravity is not free energy. It just changes the direction of already existing kinetic energy.

So we are attracted to earth because the earth's mass bends space and we are moving in that space, so our trajectory gets bent down to earth..!

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closed as unclear what you're asking by John Rennie, Pranav Hosangadi, ACuriousMind, bobie, Kyle Kanos Jan 2 '15 at 15:02

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The curvature of spacetime is independent of whether a test mass moves or not. The concept of a "static object" is already not defined in physics. Whether an object moves or not depends on the relative velocity between object and observer. Now, in the reference frame of the falling observer spacetime appears to be maximally flat except for a second (and higher) order term that describes tidal forces. So even the extended free falling observer can feel that spacetime around them is not flat. In the extreme case of an observer falling into a black hole this leads to spaghettification. $\endgroup$ – CuriousOne Jan 2 '15 at 11:48
  • $\begingroup$ I don't understand why bent space attracts objects. I don't like the analogy of the trampoline, I still can't get it. I can understand that if a object is moving in a straight line, if space is bent then the line will be bent (to an external observer only I guess), however that object will feel the accelerating force bending its trajectory right? Now, if the object is not moving, there is no trajectory to bend, so no gravity effects should be felt by that non moving object. Or else... This is so confusing and existing illustrations are too simplistic, they fail to explain these corner cases. $\endgroup$ – PedroD Jan 2 '15 at 11:59
  • $\begingroup$ The curvature of spacetime doesn't attract objects, it's just a mathematical description of how objects move under the influence of gravity. Gravity is not a force but an acceleration. The acceleration is independent of how fast a test particle moves, but it does, of course change the trajectories of fast moving objects less than it does with slow moving ones (relative to e.g. the center of mass of a gravitating system). This is no different than in the Newtonian picture. For weak gravity there is basically no noticeable difference. Yep, pictures don't work... you have to learn the math. $\endgroup$ – CuriousOne Jan 2 '15 at 12:07
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    $\begingroup$ Gravity is not just space distortion. It's distortion of spacetime. That's why static objects do accelerate - the time dimension is also distorted. $\endgroup$ – Ruslan Jan 2 '15 at 12:41
  • $\begingroup$ But by distorting time to me was like making objects age faster, instead of moving towards the earth. This is utterly counter intuitive. $\endgroup$ – PedroD Jan 2 '15 at 13:06
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Newton's second law tells us that the acceleration of an object is related to the force acting on it by the equation:

$$ \frac{d^2x}{dt^2} = \frac{F}{m} \tag{1} $$

Note that if the force, $F$, is zero then equation (1) reduces to:

$$ \frac{d^2x}{dt^2} = 0 $$

i.e. the acceleration is zero so the velocity of the object doesn't change. If the force is zero a stationary object will remain stationary.

The corresponding equation in general relativity is the geodesic equation:

$$ {d^2 x^\mu \over d\tau^2} = \Gamma^\mu_{\alpha\beta} {dx^\alpha \over d\tau} {dx^\beta \over d\tau} \tag{2} $$

This is a lot more complicated that equation (1), but the left side is basically just an acceleration and the right hand side can be thought of as the gravitational force. The quantity $\Gamma^\mu_{\alpha\beta}$ describes the spacetime curvature in a complicated way that only us nerds need to worry about. What's interesting about this equation is that the right hand side contains the terms $dx^\alpha/d\tau$ and this is sort of a velocity. So if all the $dx^\alpha/d\tau$ terms, i.e. all the velocities, were zero equation (2) would simplify to:

$$ {d^2 x^\mu \over d\tau^2} = 0 $$

and just as for our original Newton equation this would tell us that the velocity of the object is constant i.e. there is no gravitational force - a stationary object wouldn't feel a force, just as you say.

The trouble is that in GR we consider motion in spacetime not just space, and while you can stop moving in space you can't stop moving in time. All of us move through time (normally at one second per second) no matter how much we might wish it otherwise. That means there is no such think as a stationary object, and no way to escape gravity.

You are right to be cautious about the rubber sheet analogies for spacetime curvature. While these have various shortcomings a major one is that they don't portray the curvature in the time dimension.

Response to comment:

Using the geodesic equation to show how movement in time causes an object to move in a gravitational field is straightforward, but I doubt the maths would be very illuminating. So let me attempt an intuitive explanation of why it happens. Nota bene that like any intuitive explanation this will be simplified and potentially misleading if you pursue it too far. Still, let's give it a go.

Let's put you at some distance from a black hole and let go. And just to make it more fun we'll put an opaque shell around you so you can't see out. The key point is that you will feel no force (you'll be in free fall just like the astronauts in the International Space Station) so you're going to assume space is flat. You would draw a spacetime diagram to describe your motion that looks like this:

Flat space

We're approximating you as a green dot, and in your coordinates you stay still in space but move in time, so you just move up the time axis.

Now suppose I'm watching you from well away from the black hole. Because spacetime is curved, my $r$ and $t$ axes won't match yours. I've drawn my $r$ and $t$ axes as curves, but don't take the shape of the curve too literally as I have just sketched any old curve:

Curved space

Now let's superimpose my axes and your axes:

Superimposed

Because of the curvature your time axis and my time axis won't match. That means if you're moving along your time axis you're not moving along my time axis. From my perspective you've moved off the time axis by a distance shown by the red arrow, and that means you must have moved in space.

So even though, as far as you're concerned you're just moving in time, as far as I'm concerned you've moved in space as well. And that's why a stationary object falls.

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  • $\begingroup$ Ok, but how can that time motion accelerate a static object? This is so tricky to ask because in the end we cannot have static objects, and if we don't have two objects we cannot tell that one has momentum or not, just as everything is relative. However I cannot understand how does the passing of time result in accelerating an object. By the way can't we make time slower if we approach the light speed, thus making time not passing at one second per second, for outside observers of such object? $\endgroup$ – PedroD Jan 2 '15 at 13:24
  • $\begingroup$ You voted to close this question as unclear, despite your very high quality answer? $\endgroup$ – user259412 Jan 2 '15 at 16:27
  • $\begingroup$ @PeterHorvath: yes, I voted to close then changed my mind and decided something could be made of it but forgot to retract my vote. Now it's too late! $\endgroup$ – John Rennie Jan 2 '15 at 16:29

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