In what way is Idea 1 simpler than what Newtonian gravity gives you? It’s (roughly) an integral of a Fourier transform of a convolution (three nested integrals) versus two nested integrals for Newton. What’s simpler about the stuff inside those integrals than the stuff inside Newton’s?
Well, I think we can simplify the calculations for your Idea 1. Let’s have a go. Your proposed potential is
∫|F{ρ∗f}|2∥ω∥2 which we can write as ∫|F{ρ∗f}|2(ω2x+ω2y+ω2z) where subscripts indicate components.
This equals
∫|F{ρ∗f}ωx|2+⋯ and this is (up to some multiplicative constant whose value will depend on exactly what conventions you’re using for Fourier transforms) the same thing as ∫|F{∂(ρ∗f)/∂x}|2+⋯ which provided everything is reasonably smooth and well-behaved at infinity equals ∫|F{(∂ρ/∂x)∗f}|2+⋯ which by Parseval’s theorem equals (again maybe up to a multiplicative constant) ∫|(∂ρ/∂x)∗f|2+⋯ which if we don’t mind convolving a vector function with a scalar one, with the obvious meaning, equals ∫∥∇ρ∗f∥2.
So, how could this give us “interaction terms” resembling Newtonian gravity? It seems like we need f to look like 1/r (at least away from the origin), so that you get something that looks like (local+remote/r)2 which has a local . remote . 1/r in it. Maybe if you do it right this somehow gives you exactly Newtonian gravity or something … but, also, this is beginning to look rather similar to the original Newtonian integral, and I don’t see what about it is simpler. (And it seems to me we got here by starting with your proposal and simplifying it, so I don’t see how the original proposal can be simpler than Newton.)
I’m not sure I understand Idea 2, so I won’t try to comment on it :-).
In what way is Idea 1 simpler than what Newtonian gravity gives you? It’s (roughly) an integral of a Fourier transform of a convolution (three nested integrals) versus two nested integrals for Newton. What’s simpler about the stuff inside those integrals than the stuff inside Newton’s?
Well, I think we can simplify the calculations for your Idea 1. Let’s have a go. Your proposed potential is
∫|F{ρ∗f}|2∥ω∥2 which we can write as ∫|F{ρ∗f}|2(ω2x+ω2y+ω2z) where subscripts indicate components.
This equals
∫|F{ρ∗f}ωx|2+⋯ and this is (up to some multiplicative constant whose value will depend on exactly what conventions you’re using for Fourier transforms) the same thing as ∫|F{∂(ρ∗f)/∂x}|2+⋯ which provided everything is reasonably smooth and well-behaved at infinity equals ∫|F{(∂ρ/∂x)∗f}|2+⋯ which by Parseval’s theorem equals (again maybe up to a multiplicative constant) ∫|(∂ρ/∂x)∗f|2+⋯ which if we don’t mind convolving a vector function with a scalar one, with the obvious meaning, equals ∫∥∇ρ∗f∥2.
So, how could this give us “interaction terms” resembling Newtonian gravity? It seems like we need f to look like 1/r (at least away from the origin), so that you get something that looks like (local+remote/r)2 which has a local . remote . 1/r in it. Maybe if you do it right this somehow gives you exactly Newtonian gravity or something … but, also, this is beginning to look rather similar to the original Newtonian integral, and I don’t see what about it is simpler. (And it seems to me we got here by starting with your proposal and simplifying it, so I don’t see how the original proposal can be simpler than Newton.)
I’m not sure I understand Idea 2, so I won’t try to comment on it :-).