Einsteins gravity theory, but for static objects

Question

https://www.youtube.com/watch?v=XRr1kaXKBsU this is a good video to explain Einsteins gravity.

The video claims that objects always move in a straight line, with an absence of a force of gravity, rather that space is curved, so that the straight line becomes curved.

It demonstrates it well, but there is one catch. If you were to place an object with 0 velocity, that means that it isn't moving, so therefore, it won't be moving through space, which means that, no matter how close that 0 velocity object is to another object, it won't budge.

However, if the aforementioned scenario were performed, the object would start moving, therefore demonstrating the existence of a force. That violates Einstein's theory.

Is there an explanation for this?

Answer 1

The video claims that objects always move in a straight line, with a absence of a force of gravity, rather that space is curved, so that straight line becomes curved.

Not quite. The video claims, spacetime is curved. And objects move through this $4$-dimensional curved spacetime in straight lines.

If you were to place a object with $0$ velocity, that means that it isn't moving, so therefore, it won't be moving through space,

When an object has $0$ velocity, this means it doesn't move through $3$-dimensional space. But it still moves in the $4$th dimension (i.e. in time direction). And if spacetime is curved, then this moving direction will change. The motion, which was previously only in time-direction, will (by following a straight line through curved spacetime) bend so that it gets some components in the space directions (i.e. in $x, y, z$ directions).

Answer 2

Let's break up your misconception in two parts. First, what some people mean when saying that "gravity is not a force it is merely the effect of curved spacetime" is that a broader definition of force can be made such that gravity ceases to be defined as one. Indeed every object possessing energy and or momentum will be affected by a "force of gravity", in the sense that its 3-dimensional momentum will change; that's the way we commonly define force, the rate of change of momentum. But that definition is somewhat challenged when we come to GR and have a 4-dimensional structure. In this setting, we can define the force (somewhat loosely) as the rate of change of 4-momentum, which is the generalization of 3-dimensional momentum to include the time coordinate (in fact, we define it as its covariant derivative). You can think of this covariant derivative as the projection of the derivative of the vector into the space tangent to the curved surface; so effectively you are measuring the "rate of change that actually lives in the surface". A geodesic is precisely defined as the curve for which the covariant derivative of its velocity (and hence momentum) is null. So, because gravitationally induced paths in spacetime are geodesics, it follows that such paths are under the influence of no force (using the latter covariant derivative definition). Now that that is out of the way, we come to your example of a still object in space. Like Andrea di Pinto stated in their comment, this object might have no 3 dimensional momentum but it has a 4 dimensional one (namely, the vector $(0,0,0,mc)$) where c is the speed of light. This object will trace out a straight line in the direction of time, which is equivalent to being still in space, if there are no sources of gravitation. When those are present, though, the curvature will change the straight line through time to whichever geodesic through spacetime is suitable to that curvature. So it becomes evident that GR predicts that all gravitational systems evolve in a geodesical way.

Answer 3

The geodesic equation of general relativity reduces to Newton's Law of Gravity, to a very high approximation, in such a case. Therefore, there is no contradiction. In the language of GR, the object is moving uniformly along a geodesic in spactime and, in the language of Newton, it is accelerating in space. It is simply two different ways of saying the same thing. The only way to really understand this is to follow the math and understand the equations for yourself. But I can't teach GR in real life or in a comment so I'd refer you to Exploring Black Holes by Wheeler.

https://www.eftaylor.com/exploringblackholes/

Answer 4

If you were to place an object with 0 velocity, that means that it isn't moving…

No, it doesn't, because that doesn't really mean anything. Velocity is a coordinate dependent quantity, so zero velocity (really zero 3-velocity) simply means that you've chosen a coordinate system in the 4-velocity is given by $(1,0,0,0)$.$^*$ Note that I said a coordinate system, rather than the, because while setting to velocity of a particular object uniquely specifies a global inertial coordinate system up to a choice of origin in special relativity, this is not the case in general relativity, due to the fact that a spacetime manifold need not have a global inertial coordinate system, so instead, you can only get a locally inertial coordinate system, which is not unique. The fact that it's not unique is important because, even if the 4-velocity remains as $(1,0,0,0)$ in one locally inertial coordinate system, there could be another locally inertial coordinate system in which the representation of the 4-velocity changes. In other words, it is not actually possible, in principle, to compare the velocity of the object at different times. This is a consequence of the fact that concepts like rest and motion can only be defined with respect to a local observer in curved spacetime, whereas these concepts can be defined with respect to an arbitrary observer in flat spacetime.

If you find all of this differential geometry stuff confusing, a more intuitive way summarize this is to say that Newton's First Law does not apply in general relativity, because "staying at rest" and "staying in motion" are not well defined in curved spacetime.

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$^*$using the $(+,-,-,-)$ metric signature and units with $c=1$