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Re: The Math is still Not Ready
Posted:
Jan 6, 2013 7:27 AM
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On Sunday, January 6, 2013 7:36:26 AM UTC+2, Koobee Wublee wrote:
> On Jan 5, 8:54 am, Tom Roberts wrote:
>
>
>
> > Here is General Relativity:
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> >
>
> > On a 4-d Lorentzian manifold M,
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> > G = T
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> >
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> > where G is the Einstein curvature tensor and T is the energy-momentum tensor.
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>
>
> Please allow Koobee Wublee reminds Tom where that overly simplified
>
> equation[s] above come from. Let?s follow Hilbert?s footsteps and
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> pull out the following so-called Lagrangian out of Hilbert?s ass.
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>
>
> ** L = (R / K + rho) sqrt(-det[g])
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>
>
> Where
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>
>
> ** L = Lagrangian
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> ** R = Ricci scalar
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> ** K = Constant
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> ** rho = Mass density
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> ** [g] = The metric (a 4x4 matrix)
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> ** det[] = Determinant of a matrix
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>
>
> For the language of convention in this case, [A] means a matrix with
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> elements [A]_ijk... or [A]^ijk...
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>
>
> The field equations can be derived in just one step by taking the
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> partial derivative of the Lagrangian above with respect to [g^-1]^ij
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> where [g^-1], a matrix, is the inverse of [g], another matrix, and
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> after setting each of the partial derivative to null, the result is
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> the following relationships of matrices.
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>
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> ** [R] ? R [g] / 2 = K rho [g] / 2
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>
>
> Where
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>
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> ** [R] = Ricci tensor (another 4x4 matrix)
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>
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> You would call the following.
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>
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> ** [G] = [R] ? R [g] / 2
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> ** [T] = K rho [g] / 2
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>
>
> Thus,
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>
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> ** [G] = [T]
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>
>
> Koobee Wublee would also like to remind Tom that the above equation
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> has never been tested with any experimentations, and the best Tom can
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> hope for is the following where the energy momentum tensor is null.
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>
>
> ** [G] = 0
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>
>
> Where
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>
>
> ** [T] = 0
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>
>
> Using only diagonal [g], the equation above simplifies into the
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> following where the effect of the ever so celebrated trace term is
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> nullified. The null Ricci tensor was basically Nordstrom?s work where
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> Schwarzschild had been working on the solution for years. That is why
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> within a couple months after Hilbert presented the field equations,
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> Schwarzschild published a solution.
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>
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> ** [R] = 0, first proposed by Nordstrom as the field equations
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>
>
> Where
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>
>
> ** R [g] / 2 = The trace term
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>
>
> <shrug>
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>
>
> > To get SR from GR:
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> >
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> > Riemann = 0
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> > Top(M) ~ R^4
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> >
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> > where Riemann is the Riemann curvature tensor on M, and Top(M) is the topology of M.
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>
>
> Nonsense, Tom. If the Riemann tensor is null, the Ricci tensor must
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> be null as well in which you end up with the null Ricci tensor above
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> where you can solve for the Schwarzschild metric and other equally
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> valid solutions that are able to degenerate into Newtonian law of
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> gravity at weak curvature in spacetime. <shrug>
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>
>
> The best way to get SR from GR is to set the gravitating mass, M, to
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> 0, duh! <shrug>
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>
>
> ** ds^2 = c^2 (1 ? 2 U) dt^2 ? dr^2 / (1 ? 2 U) ? r^2 dO^2
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>
>
> Where
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>
>
> ** U = G M / c^2 / r
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>
>
> > [Note that approximations are important in applying the theory,
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> > as this is PHYSICS, not math. Based on your writings around
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> > here, and your aversion to any intellectual effort, I estimate
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> > you will never understand this.]
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>
>
> Tom, in GR, physics = math, and math = physics. So, start
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> understanding the mathematics involved instead of wishing for what you
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> believe in. <shrug>
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>
>
> Faith should not come into any equations of science, no? <shrug>
IDIOT
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