Gravitational collapse and making a Higgs

We need some way to permanently make the c thing a relative sample rate. What is it really? The number of bubbles in a Higgs squared, no? Gravity here seems to be another Avogadro thing, upside down.  But it has force in it, what is Force? Bubble gradient, or delta bubble/bubble. Is this really all about fitting an Avogadro into a sphere?

G also has a (m/kg)^2 in it. Are they both exponents here?  A Newton is kg*m/sec^2, all the fake variables. Gravity is N* (m*kg)^2. So the top is what? (m*kg)^3/sec^2. The seconds cancel, A m^2 cancels, so we get a (Kg^3)*m on top, and a radius to the left. But m is a kg on top, making a (kg^4).

 We get one of these when it happens. where M is one half of that that thing above, so the twos cancel.

OK, I can tell you that the singular solution, the formula just above, is the SHannon orthogonality point.  That is the point where you get this great Avogadro 2^79-1, otherwise known as the Higgs density of a radius. So this is really all about Avogadro and the sphere. Take the G, invert it, square it, and it puts you just below a Higgs wave. But it needs to be scaled, but it is an Avogadro, I am sure.

So what Einstein had discovered was a Higgs, and M, after being scaled to normal, is the exponent 79, or there abouts. He wanted to know when we quantized a humongous ball of gravity.

Einstein's general relativity was about when the speed of light found a point of symmetry with respect to other space dimensions. In sampling theory, that is all about when the sample space is not Nyquist with respect to the band width of the signals. So he will get forms that look like 2^n -1 = SNR, and so will Shannon. So just convert to sample space and use the sample rate of light.

However, gravitational collapse will not work unless protons migrate from the center to the outside, making gravity steeper without making heavy Lagrange points.