3 ms·
> If the interpretation were wrong, and that's not a force, then the amount of force in that equation would be 0. With no force, Gm₁m₂/r² = 0. Nonsense. The G
by pdonis 3y ago
> If the interpretation were wrong, and that's not a force, then the amount of force in that equation would be 0.
With no force, Gm₁m₂/r² = 0.
Nonsense. The GR interpretation is that G m1 m2 / r^2, when it is nonzero, is describing an effect of spacetime geometry, not a force. It can't be a force in GR because it isn't felt; an object moving solely under the influence of G m1 m2 / r^2 feels no weight--an accelerometer attached to it reads zero. GR does not change the numerical value of G m1 m2 / r^2 at all. It just reinterprets what the quantity represents.
- lazide 3y agoThis is the bookkeeping/accounting part of it, BTW. It clearly exists (as a thing). What we decide to call it, and how we account for it, while important, doesn't make it not exist in reality. Despite the motte and bailey argument to the contrary.