e^a = I^a(globally flat LNIF inertial force observer effect) + A^a(intrinsically curved geodesic deviation)
e^a = e^audx^u
these 4 tetrad fields are GCT group local frame invariant scalars, but they are Lorentz group O(1,3) 4-vectors.
e^au = I^au(globally flat LNIF inertial force observer effect) + A^au(intrinsically curved geodesic deviation)
The total O(3,1)xGCT scalar invariant e = Identity on Tangent Bundle is like
1 = Sum |i>
i.e. e = identity on tangent bundle is an invariant constraint in all local frames.
e = e(observer frame effect) + e(intrinsic curved geometry)
The two terms on RHS are not separately identity actions on the tangent bundle only their sum is. They are separately local frame invariants however.
On Aug 21, 2006, at 11:18 AM, Jack Sarfatti wrote:
Yes thanks for reminding me of course I should have explicitly mentioned that, but I split
e = I(LNIF observer) + A(intrinsic curved geometry)
where
A has the curvature information.
A ~ dTheta/\dPhi
e = I is simply the globally flat Minkowski spacetime trivial tetrad since
guv(curvilinear) equivalent to globally flat Minkowski metric in that case.
i.e.
e = Trivial Globally Flat Tetrad (LNIF Non-Geodesic Observer Inertial Force Contingencies) + Non-Trivial Geodesic Deviation Tetrad from modulation of the Goldstone phases of the vacuum ODLRO "inflation" Higgs field.
The electroweak Higgs and the inflation field may be closely related in my program for a theory.
All curvilinear effects when A = 0 are then simply non-geodesic LNIF observer effect inertial forces from non-gravity forces on the local detectors. There is no geodesic deviation when A = 0. That's the key idea that
A is analogous to v = (h/m)dTheta in superfluid helium
The formal issue is can we have things like
(p/q)/\(r/s) = (-1)^|pr/qs|(r/s)/\(p/q)
p,q,r,s all integers
If so, that begins to suggest anyons with fractional quantum statistics and fractional charges according to Frank Wilzcek.
On Aug 21, 2006, at 9:24 AM, Waldyr A. Rodrigues Jr. wrote:
Dear Jack,
Objects like (1/2) = (1/2)u^v^wdx^u/\&xv/\&xw have nothing to do with fractals. However they have a nice algebra. I call them Clifford valued-differential forms and use them in the attached paper which has been published in 2004 in the Int. J. Mod. Phys. D. You also can find it in the arXiv.
By the way, your e = eu^adx^u&a, is nothing more than the identity operator acting on the tangent bundle (or in the cotangent bundle). . . I already explained that, but it seems that you forgot.
Best regards,
Waldyr
-----Mensagem original-----
De: Jack Sarfatti [mailto:sarfatti@pacbell.net]
Enviada em: segunda-feira, 21 de agosto de 2006 12:16
Para: Sarfatti_Physics_Seminars
Assunto: What is a fractal Cartan form anyway?
Next question since the idea only popped into my mind for the first
time late last night.
For example what would a 1/2 form be?
Maybe
?
Where &xv is a dual basis to dx^u?
Then a (p,q) form has p covariant & q contravariant indices.
On Aug 20, 2006, at 9:49 PM, Jack Sarfatti wrote:
>
>>
>> e = eu^adx^u&a
>>
>> is contracted over both the GCT indices u and the tangent space
>> indices a.
>> Therefore it's locally frame invariant both under GCT Diff(4) for
>> COINCIDENT non-geodesic LNIFs at fixed physical event E as well as
>> for O(1,3) COINCIDENT LIF transformations at same E.
>>
>> Cartan's whole idea of differential forms is that they are local
>> frame invariants - local coordinate independent.
>>
>> A general 1 form is
>>
>> 1 = 1udx^u
>>
>> that's scalar invariant
>>
>> 1 = 1udx^u = 1u'dx^u'
>>
>> A general 2-form is
>>
>> 2 = 2uvdx^u/\dx^v
>>
>> 2uv = - 2vu
>>
>> etc.
>>
>> d(p/\q) = dp/\q + (-1)^|p|p/\dq
>>
>> d^2 = 0
>>
>> If p is a zero form scalar O
>>
>> d(O/\q) = dO/\g + O/\dq is a q + 1 form
>>
>> If p is a 1-form
>>
>> d(1/\q) = d1/\q - 1/\dq
>>
>> etc.
>>
>> p/\q = (-1)^|pq|q/\p
>>
>> If p & q are both 1-forms then they anti-commute sort of like
>> fermion operators.
>>
>> ckck' + ck'ck = 0
>>
>> when k = k' that's the Pauli exclusion principle.
>>
>> If p & q are both 0-forms then they commute like bosons.
>>
>> bkbk' - bk'bk = 0
>>
>> You get Heisenberg uncertainty principle by taking canonical
>> conjugates
>>
>> e.g. c*k = d/dck is conjugate to ck, one must assume c*K is also a
>> 1-form?
>>
>> c*kck + ckc*k = 1
>>
>> So now let p & q be rational numbers, i.e. fractal forms.
>>
>> This gives fractional quantum statistics & fractional charges!
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