Quantum Gravity Via Expansion-Contraction 6.0: Riccati 2 Version Using Special Relativity Beta Squared

From Osher Doctorow mdoctorow@xxxxxxxxxxx

The Riccati-2 Differential equation version of the Quantum Gravity
equation is obtained by a method which also sheds light on how we can
determine whether various models and theories relate causally to each
other. Recall that the square root of the quantity:

1) 1 - v^2/c^2

or its inverse is the Special Relativity Beta or Gamma factor that
yields the Lorentz or Lorentz-Fitzgerald contraction. Let's set y of
the Riccati Differential equation dy/dt = A(t) + B(t)y + C(t)y^2 equal
to 1 - v^2/c^2, which yields:

2) -2va/c^2 = A(t) + B(t)(1 - v^2/c^2) + C(t)(1 - v^2/c^2)^2

with acceleration a = dv/dt. Expanding, we obtain:

3) -2va/c^2 = A(t) + B(t) - B(t)v^2/c^2 + C(t)(1 - 2v^2/c^2 + v^4/c^4)

and expanding further yields:

4) -2va/c^2 = A(t) + B(t) - B(t)v^2/c^2 + C(t) - 2C(t)v^2/c^2 +
C(t)v^4/c^4

Now factor similar terms on the right hand side of (4):

5) -2va/c^2 = [A(t) + B(t) + C(t)] - (v^2/c^2)[B(t) + 2C(t)] +
C(t)v^4/c^4

Let's write D(t) for A(t) + B(t) + C(t) for brevity, and then divide
both sides of (5) by v/c^2 or in other words multiply by c^2/v:

6) -2a = D(t)c^2/v - v[B(t) + 2C(t)] + C(t)v^3/c^2

Multiply both sides of (6) by m (assumed constant) to obtain:

7) -2F = mc^2 D(t)/v - p[B(t) + 2C(t)] + C(t)Kv/c^2, K = kinetic
energy, p = momentum

Use E = mc^2 for energy E in (7) to obtain:

8) -2F = ED(t)/v - p[B(t) + 2C(t)] + C(t)Kv/c^2

Again force relates to energy and momentum as in the previous Sections
of this thread, but this time there is no separate mass term and there
are separate energy and kinetic energy terms and there are some hints
of a sign relationship between F and p which I'll try to explore later.

Osher Doctorow

.

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