Re: [Vo]: Re: Einstein's Elevator Le Sage's Gravity Theory
Stephen A. Lawrence wrote: Harry Veeder wrote: Consider the situation far from any planets or stars. If the ball-bearing is initially at the centre of the shell it will remain there. If it is initially off centre, the ball bearing and the shell will move so as to minimize the distance between the point on the shell that was initially closest to the ball bearing. This is my prediction. It does not violate conservation of momentum, but it is not based on Newton's or Einstein's conception of gravity. I haven't worked it out, but I think a force going as 1/r^3 would have that effect. Actually any rate of falloff faster than 1/r^2 should do that, I think. I wouldn't try to work it out. The theory which gives rise to this prediction is in the very early stages of gestation. But ... the fact that Mercury's orbit precesses as it does is evidence that real gravity around a spherically symmetric object doesn't actually fall off as 1/r^2 (falls off a bit faster, IIRC), which makes me wonder whether real gravity would also show that effect, albeit weakly (my proof using the Ricci tensor notwithstanding ... among other things I assumed a massless ball bearing, which is a little wrong). In Newton's mathematical model of gravity an equivalency exists between power-of-inertia and power-of-attraction. The properties of inertia (a.k.a inertial mass-points) are transferred to points-of-attraction (a.k.a. gravitational mass points) through the equivalency. Also since the power-of-inertia is communicated by contact the transference leads to the paradox of action at a distance. One way to address the paradox is to extend inertia into space and time, which is what Einstein did. In this way we are never out of touch with inertia. My way is to begin by giving gravity a distinct power from that of inertia. Harry
[Vo]: Re: Einstein's Elevator Le Sage's Gravity Theory
Will a sphere within a sphere (a ball-bearing in a transparent hollow sphere) due to the gravitational attraction between them, center itself during free fall? http://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation http://en.wikipedia.org/wiki/Introduction_to_general_relativity Or? Fred
Re: [Vo]: Re: Einstein's Elevator Le Sage's Gravity Theory
Frederick Sparber wrote: Will a sphere within a sphere (a ball-bearing in a transparent hollow sphere) due to the gravitational attraction between them, center itself during free fall? First, in Newtonian gravitation: Inside a uniform spherical shell there's no gravitational field (no field due to the shell, that is), so the ball bearing will ignore the shell and fall normally. It won't move to the center. The force of gravity exerted by the inner ball on the entire surface of the shell will cancel, and the shell won't accelerate as a result of the ball bearing's field -- so it, too, will therefore fall normally, and won't center itself around the ball bearing. You can check this by integrating the field of a point particle over an offset spherical shell (and noting that the ball bearing consists of a big blob of point particles), or you can just use conservation of momentum to argue that since the ball bearing doesn't feel a force from the shell, the shell must not feel a (net) force from the ball bearing either. Now, in relativity: It's a whole lot harder, but I think the answer's the same, based on this very sloppy argument: Within the chamber in the big sphere, the stress/energy tensor is zero (assuming the lights are off, and ignoring the contribution of the ball bearing). If the stress tensor is zero, then the Ricci tensor must be zero too, by Einstein's field equation. And if the Ricci tensor is zero, then a small ball of initially comoving particles won't change in volume as time goes by (though it _may_ deform), since that's what the Ricci tensor measures. What that says is that there aren't any points of attraction or repulsion in empty space, but there may be tidal effects. Tidal effects = a saddle point in the field. Inside the chamber, the field due to the shell must be spherically symmetric, so we _can't_ have a saddle point in the center of the sphere. Therefore, particles can't be attracted to the center of the sphere (nor repelled from it). Since momentum is still conserved in GR, I _think_ we can again argue that the big sphere can't be pulled to the center by the small sphere, either, in that case. *http://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation* _*http://en.wikipedia.org/wiki/Introduction_to_general_relativity*_ Or? Fred http://en.wikipedia.org/wiki/Introduction_to_general_relativity
[Vo]: was-Einstein's Elevator Le Sage's Gravity Theory
From the link Frederick Sparber provided on LeSage's theory of gravity: http://en.wikipedia.org/wiki/Le_Sage%27s_theory_of_gravitation In the context of mainstream science (albeit not as an explanation of gravitation), the Lesage mechanism has been identified as a significant factor in the behavior of dusty plasma. A.M. Ignatov [95] has shown that an attractive force arises between two dust grains suspended in an isotropic collisionless plasma due to inelastic collisions between ions of the plasma and the grains of dust. This attractive force is inversely proportional to the square of the distance between dust grains, and can counterbalance the Coulomb repulsion between dust grains. I thought this might have some bearing on CF. Harry
Re: [Vo]: Re: Einstein's Elevator Le Sage's Gravity Theory
Consider the situation far from any planets or stars. If the ball-bearing is initially at the centre of the shell it will remain there. If it is initially off centre, the ball bearing and the shell will move so as to minimize the distance between the point on the shell that was initially closest to the ball bearing. This is my prediction. It does not violate conservation of momentum, but it is not based on Newton's or Einstein's conception of gravity. Harry
