zoltrix said:
I supposed it was something like that
however I definitely understand why relativity of simultaneity can explain schrink of body in motion of special relativity
I think also to grasp the reason why the acceleration of a body in presence of a mass is caused by the space time curvature
is the body at rest with respect to a certain observer ?
which one ?
may you expand on that ?
Are you familiar with the concept of geodesics - and geodesic motion?
I would suggest re-interpreting your question in terms of "geodesic motion", mainly because it's rather unclear exactly and precisely what you mean by "a body at rest".
Let's take an example. An airplane is flying a parabolic trajectory which makes it's passengers weightless. We consider only the part of the airplane's trajectory in which the passengers on the airplane are "weightless".
Is the airplane "at rest"? I would say that would be an unusual statement. But the airplane is undergoing geodesic motion.
Now consider someone standing on the surface of the Earth. Said person is not undergoing geodesic motion. But you'd probably consider them to be "at rest", at least I'd assume you might. It's hard to be sure.
Now, consider a person on a train moving at a constant groundspeed around the equator. Is the train "at rest?" It would be possibly a bit unusual to say that the train was at rest, but not unheard of.
In terms of whether the motion is geodesic or not, the example of the train moving at a constant groundspeed along the equator may or may not be undergoing geodesic motion. If the train's velocity relative to the ground is such that the passengers on it are weightless, it is following a geodesic path. This requires one specific velocity of the train. If the train does not have that exact velocity, then it is not undergoing geodesic motion.
Finally, let's consider someone on Einstein's elevator. They are not in the presence of any significant mass - the elevator is assumed to have negligible mass. Is the person standing on the elevator "at rest?" It's unclear. But they're not undergoing geodesic motion.
Having talked about "bodies at rest" and the more precise concept of "bodies undergoing geodesic motion"geodesic motion, we can now talk about gravity and curved space-time.
In the flat space-time of special relativity, two bodies undergoing geodesic motion have zero relative acceleration. We'd say they have a constant relative velocity.
In curved space-time, in the presence of mass, two bodies undergoing geodesic motion will not, in general, have zero relative acceleration.
And that's mostly all there is to gravity, the fact that two bodies undergoing geodesic motion have a non-zero relative acceleration. This is formalized by the "geodesic deviation equation" of General relativity. See for instance the wiki article
https://en.wikipedia.org/wiki/Geodesic_deviation, though I'm not sure how clear it will be. The relative acceleration betwen two initially parallel geodesics can be used to define the Riemann curvature tensor, called the Riemann tensor and usually denoted by R, which appears on the right hand side of the geodesic deviation equation:
$$D^2 \frac{x^a}{d\tau^2} = R^a{}_{bcd} u^b u^c \delta(x^c} $$.
The Riemann tensor here is the mathematical entity R.
There's one final clarification to be made. The notion of "geodesic motion" actually applies precisely only to a point particle, not an extended body. If the body is small enough, the distinction isn't too important. But for a large enough body, it becomes significant.
Consider again the example of the train running around the equator such that the passengers are weightless. If you look at the problem more precisely, the notion that I'm calling weightless, and also calling "geodesic motion", applies exactly only at a point. (In terms of space-time, it applies to a worldline.) For instance, a point on the floor of the train might be undergoing geodesic motion, while a point on the ceiling would then have the wrong velocity (the train is assumed to be rigid), and the point on the ceiling would not be precisely following a geodesic. But if the train isn't too tall, the "weight" (aka proper acceleration) of a point (worldline) on the ceiling would be small enough to be ignored.