Date: Jan 6, 2013 12:36 AM
Author: Koobee Wublee
Subject: Re: The Math is still Not Ready

On Jan 5, 8:54 am, Tom Roberts wrote:

> Here is General Relativity:
> On a 4-d Lorentzian manifold M,
> G = T
> where G is the Einstein curvature tensor and T is the energy-momentum tensor.

Please allow Koobee Wublee reminds Tom where that overly simplified
equation[s] above come from. Let?s follow Hilbert?s footsteps and
pull out the following so-called Lagrangian out of Hilbert?s ass.

** L = (R / K + rho) sqrt(-det[g])


** L = Lagrangian
** R = Ricci scalar
** K = Constant
** rho = Mass density
** [g] = The metric (a 4x4 matrix)
** det[] = Determinant of a matrix

For the language of convention in this case, [A] means a matrix with
elements [A]_ijk... or [A]^ijk...

The field equations can be derived in just one step by taking the
partial derivative of the Lagrangian above with respect to [g^-1]^ij
where [g^-1], a matrix, is the inverse of [g], another matrix, and
after setting each of the partial derivative to null, the result is
the following relationships of matrices.

** [R] ? R [g] / 2 = K rho [g] / 2


** [R] = Ricci tensor (another 4x4 matrix)

You would call the following.

** [G] = [R] ? R [g] / 2
** [T] = K rho [g] / 2


** [G] = [T]

Koobee Wublee would also like to remind Tom that the above equation
has never been tested with any experimentations, and the best Tom can
hope for is the following where the energy momentum tensor is null.

** [G] = 0


** [T] = 0

Using only diagonal [g], the equation above simplifies into the
following where the effect of the ever so celebrated trace term is
nullified. The null Ricci tensor was basically Nordstrom?s work where
Schwarzschild had been working on the solution for years. That is why
within a couple months after Hilbert presented the field equations,
Schwarzschild published a solution.

** [R] = 0, first proposed by Nordstrom as the field equations


** R [g] / 2 = The trace term


> To get SR from GR:
> Riemann = 0
> Top(M) ~ R^4
> where Riemann is the Riemann curvature tensor on M, and Top(M) is the topology of M.

Nonsense, Tom. If the Riemann tensor is null, the Ricci tensor must
be null as well in which you end up with the null Ricci tensor above
where you can solve for the Schwarzschild metric and other equally
valid solutions that are able to degenerate into Newtonian law of
gravity at weak curvature in spacetime. <shrug>

The best way to get SR from GR is to set the gravitating mass, M, to
0, duh! <shrug>

** ds^2 = c^2 (1 ? 2 U) dt^2 ? dr^2 / (1 ? 2 U) ? r^2 dO^2


** U = G M / c^2 / r

> [Note that approximations are important in applying the theory,
> as this is PHYSICS, not math. Based on your writings around
> here, and your aversion to any intellectual effort, I estimate
> you will never understand this.]

Tom, in GR, physics = math, and math = physics. So, start
understanding the mathematics involved instead of wishing for what you
believe in. <shrug>

Faith should not come into any equations of science, no? <shrug>