Gravitational force equation derived from GR

In summary: Newton's 3rd law is not valid in GR, except if you use the usual "Newtonian" definition of force, but that's not very helpful...).The geodesic equation is a second-order ODE, so to solve it you need two initial conditions on the particle's position and velocity. In addition you need to know the geometry of the spacetime, which determines the connection coefficients needed to compute the derivatives of the particle's position. The geodesic equation is also covariant, meaning that it is valid in any (sufficiently smooth) coordinate system. This means that there is no unique way to break the second-order ODE into two first order O
  • #36
sha1000 said:
PAllen wrote: " As for relating force needed to maintain hovering to proper acceleration of a hovering body, note that in a momentarily comoving local inertial frame (in this case, a momentarily stationary free fall frame coinciding with some hovering body), you have that coordinate acceleration equals proper acceleration..."

I'm really not an expert. Does this mean that we can define a co-moving local inertial frame (stationary free fall frame coinciding with hovering body)? From this point would it be possible to apply F = γ3ma?
Since we are dealing with co-moving local inertial frame?
NO, you would just apply F=ma in that frame. ##\gamma## would be 1 by definition in such a frame.
 
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  • #37
sha1000 said:
Ok I got it. Thanks

Just last question if you don't mind. Can we apply the same reasoning to the object in circular oribit around massive body. Is it possible to use naively the same proper acceleration equation obtained for a hovering object?
No. Circular orbits can be powered or unpowered, and the necessary proper acceleration can be anywhere between ##\pm\infty##. However, the typical use of the term "circular orbit" refers to an unpowered orbit. This is zero proper acceleration.
 

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