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### Author Topic: Introduction into the Final Equation  (Read 1781 times)

#### Mr. Scientist

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##### Introduction into the Final Equation
« on: 17/11/2009 14:58:40 »
The equations for Verns Gravito-electromagnetic charge curvature hypothesis is given, which holds quite a few differential relationships, resulting in a geometry which tends to mark that as inertial energy. I think the explanations to the equations should follow last - just to make references on certain mathematical properties.

Theorem

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0(M²ψ-M²ψ]+½[ξε_g(M²ψ*-M²ψ*]) (8) - Charged equation in curved spacetime

A=e^i ∫d^4 x(½[ξε_0(M²ψ-M²ψ]+½[ξε_g(M²ψ*-M²ψ*]) (9)

(vortex) k=Dφ*,Dφ* (9)

and ∫k=D_{φ*,φ*}(x,1... x,2)... D_φ*(x_n) which is also equal to =

∏ D_{φ*,φ*}(x_i) (6)
n=i

so the four-vortex is given as:

k²=Dφ*Dφ*DφDφ (7)

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i∫d^4 x(½[ξε_0(-∂t(ψ)+▼²ψ-(-∂t(ψ)+▼²ψ)]+½[ξε_g(-∂t(ψ)+▼²ψ*-(-∂t(ψ)+▼²ψ)]) (8) - plane wave solutions

ξ1,ξ1_{ε_g,ε_0} (8)

Substution

μ1,μ2_{g,e,0} (9)

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i∫d^4 x(½[μ_0(-∂t(ψ)+▼²ψ)-(-∂t(ψ)+▼²ψ)]+½[μ_g(-∂t(ψ)+▼²ψ*-(-∂t(ψ)+▼²ψ)]) (10) - electromagnetic and graviatational charge

Epsilon ε in the plane wave solution is itself a tensor so has a factor of √g' which is respectively negative. In mathematical terms, its the determinent of the metric tensor.

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξ√g'_0(M²ψ-M²ψ)]+½[ξ√g'_g(M²ψ*-M²ψ*)]) (11)

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξ√g'_0((∂²/∂t²-Δ + M²)ψ-(∂²/∂t²-Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂t²-Δ + M²)ψ*-(∂²/∂t²-Δ + M²)ψ*]) (12)

Where (∂²/∂t²-Δ + M²)ψ = 0 (13)

→ = g_μν∂^ν∂^μ (14)

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 δ(x-x')(½[ξ√g'_0((∂²/∂δt²-Δ + M²)ψ-(∂²/∂δt²-Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂δt²-Δ + M²)ψ*-(∂²/∂δt²-Δ + M²)ψ*]) (14)

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0(M²ψ-M²ψ]+½[ξε_g(M²ψ*-M²ψ*]) (15)

is balanced, because it takes into respect the electromagnetic permittivity added with that of the gravitational permittivity with a Langrangian term for M². More interestingly enough, M²ψ is similar to the Klein-Gorden relationship. Here are some interesting reationships:

M²ψ=-∂t(ψ)+ ▼²ψ (16)

which results in plane wave solutions. By substitution, you can reconfigurate eq.(1) into:

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0(-∂t(ψ)+ ▼²ψ)-(-∂t(ψ)+ ▼²ψ)]+½[ξε_g(-∂t(ψ)+ ▼²ψ*)-(-∂t(ψ)+ ▼²ψ)]) (17)

Which is very attractive as a wave equation.

We could manipulate the equation even more to have nuetral components after taking ino account, from a Klein-Gorden relationship, where for manipulative convenience we can rewrite the plane wave solutions in  quantized form as:

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 x(½[ξε_0((∂²-M²)ψ*-(∂²-M²)]+½[ξε_g((∂²-M²)ψ*-(∂²-M²)ψ*]) [18]

This is I suppose the mechanism which would cancel them out, or at least, this is my interpretation of the equation.

Nuetral Charge in Linear Metric Space

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i∫d^4 x(½[μ_0(-∂t(ψ)+▼²ψ)-(-∂t(ψ)+▼²ψ]+½[μ_g(-∂t(ψ)+▼²ψ*-(-∂t(ψ)+▼²ψ]) [19]

Assuming not only a small box space d^3, we are using the notion of d^4 which takes into respect an equally tiny amount of time. This will be given the form as:

|(∫F_g vt)²_<A_k²>|=∫-▼²φ²(ћ(c/G))_g β²t²(e^i ∫d^4 δ(x-x')(½[ξ√g'_0((∂²/∂δt²-Δ + M²)ψ-(∂²/∂δt²-Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂δt²-Δ + M²)ψ*-(∂²/∂δt²-Δ + M²)ψ*]) (20)

Any actions that can come from this are restricted by the presence of d^4 δ(x-x'). Finding this in the equation however has itself a bit of interesting qualities. The term d^4(x) in the equation above Some has the products of a 0th, 1st, 2nd and 3rd component, or dx^0dx^1dx^2dx^3  which inexorably makes it a zeroeth rank tensor. However, integral to it is in fact displacement from one position x to another x'. This immediately makes this part a vector quantity, and thus, the entire configuaration d^4 δ(x-x') as rank 1 tensor.

There is also a rank 3 tensor in the equation

eq.1

.........∞
E²/c²- ∏_(3)|p²|(x_i)=U(xn)
........n=i

Where U is a scalar quantity, definiting presicely the rest mass on the left hand side under Lorentz Covariant-relations. Since the expression U(x) describes the scalar mass quantity at some point, then the contraction of the tensor is calculated as:

eq.2

.........i
U(x)=∑ T
......i..i

Using Einstein notation. The right hand side of equation 1 can now be equated to exactly the value of equation 2, with the interesting geometrical formation of the left hand side of equation 1.

The rank 3 tensor can be found when scaling out a matrix for the three-momentum as a non-covariant vector:

(p,x)
(p,y)=η_νμ U^μ
(p,z)

The final equation involving the substitutions of equations 1. and eq 2 we can derive a much more simpler form of:

= ∫-▼²φ²(ћ(c/G))_g β²t²(e^i U(x_{1,2})A

Where A is simply

A=(½[ξ√g'_0((∂²/∂δt²-Δ + M²)ψ-(∂²/∂δt²-Δ + M²)ψ)]+½[ξ√g'_g((∂²/∂δt²-Δ + M²)ψ*-(∂²/∂δt²-Δ + M²)ψ*])

|(∫F_g vt)²_<A_k²>|= ∫-▼²φ²(ћ(c/G))_g β²t²(e^i U(x_{1,2})A

Decriptions of the intermediate coefficients and the lot

This was hard to do, because with the standard amount of equations - and to their lengthly detail, this was not a matter of a day or so job. It's taken me at least 2 years to create these, and a few latter equations where inspired by someone you know; Vern.

Implicit things to know

F_g=-▼φMg (1)

Describes a particle moving in a gravitational field, sometimes interpreted as the gravitational moving charge

Multiply (vt)^2 on both sides → (F_gvt)² (2) - which is obviously a rest-energy-gravitational-force relation.

φ' - is the energy scalar field

Vφ'- is the potential of this energy scalar field

we define wave functions explicitely under retarded and advanced waves, which are concurrently hot in the Transactional Interpretation. Incoming fields will be represented as Ψ' where outgoing simply Ψ.

Obviously under amplitude theory the chances of finding an eigenstate of either wave functions is given as ∫|ψ|²=1. This will become important for the function of k represented in the expression |(∫F_g vt)²_<A_k²>| because it involves highly complicated vortex's - complicated to keep up with i mean.

In the expression given as (F_gvt)², it has dimensions (M_Gc²)², AND WE will soon find out how tensor contraction is related to this.

Other things needed to be known are the following derivations made by Planck himself:

M²=ћ(c/G) (the famous mass-sqaured term)

ћc=GM²

and that β=v/c (a trig-function)

and supercomplex no.'s denoted as ξ, and this beside any imaginary number to the power of two which usually yields a negative value does not. It remains perfectly positive.

Vortex Information

First for the record:

∫φ(x)=D(ψ,ψ)(x,1... ...x,2) ... dφ(x_n)

=πdψdψ(x_i) (A)
i

has a conjugate form

∫φ(x)=D(ψ*,ψ*)(x,1... ...x,2)

=πdψ*dψ*(x_i)
i

The first vortex of [A] is

k=Dψ*Dψ*DψDψ

A new vortex is added into the mix as to assertain the energy level of the gravitational potential Vφ so it acts as a gradient of the gravitational energy; it too has its own vortex made of incoming and outgoing signalling.

I tried to express the importance of

φ' - is the energy scalar field

Vφ'- is the potential of this energy scalar field

these very early on, because their roles are very complicated. It's geometrical in a sense, as it makes up in total an eight-way vortex when both vortexes acts on the final equation.

Since this new vortex needs a new description, it is given as:

∫(φ(x))²=D(φ',φ')(x,1... ... x,2)D(φ,φ')(x,1... ... x,2)

So equally one can assume for the gravitational gradient of the gravitational field to be:

∫(φ'(x))'=D(φ*,φ*)(x,1... ... x,2)D(φ*,φ*)(x,1... ... x,2)

And are subsequent in the same probability density solutions. The two fields of D compactified gives an inter-related vortex as:

D(K²(k²))=(Dφ*Dφ*DφDφ)Dφ'Dφ'Dφ'Dφ'Dφ'

Mind to express k as as k sqaured, as it deal with the incoming waves and the outgoing.

This is the first of probably four posts which will explain the entire equationa. I cannot continue for now but i will return later in admission to complete the thing, as requested.

« Last Edit: 17/11/2009 15:07:12 by Mr. Scientist »

#### Mr. Scientist

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##### Introduction into the Final Equation
« Reply #1 on: 17/11/2009 19:22:18 »
Excerpt

The GEM equations you remember are closely related to these expressions which seems to cancel out the magnetic charge:

▼•B_g + ∫E_gΦ e^i^∫d^4δ(x'-x)(η²_iν²_f(Pμ,ν) -(η²_iν²_i))<A>_∫∫|φ'_i_{ij}φ'_i_{ji}

This part decribes mass dispersion. The dispersion of what mass is questionable, (η²_iν²_f(Pμ,ν) -(η²_iν²_i). And ∫∫|φ'_i_{ij}φ'_i_{ji} represents the complex vortex interactions. One vortex operates the mechanics (a rank 3 tensor) inside the equation and another vortex connects as a

.........i
U(x)=∑ T
......i..i

which reveals a tensor - invariant from the last equjations.

|(∫F_g vt)²_<A_k²>|= ∫-▼²φ²(ћ(c/G))_g β²t²(e^i U(x_{1,2})A

Concerning the final equation above |(∫F_g vt)²_<A_k²>|

To remove the magnetic interaction, this first expression negatates its use ▼•B_g + ∫E_gΦ e^i^∫d^4δ(x'-x)(η²_iν²_f(Pμ,ν) -(η²_iν²_i))<A>_∫∫|φ'_i_{ij}φ'_i_{ji} since ▼•B_g = 0, only for a static field. So the magnetic field should become static if we are to believe an electron can have an electric charge only when passing round warped spacetime.

#### Mr. Scientist

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##### Introduction into the Final Equation
« Reply #2 on: 17/11/2009 19:38:43 »
Just a quick proof

If ▼•B_g = 0 is true, then so it can be for ▼•E_g = 0.

I form:

∫E_gΦ(x) = E_gΦ(x,1... x,2)

so equally

∫B_gΦ(x) = B_gΦ(x,1... x,2)

For a Gravito-electric equation ignoring the magnetic influence of the field, we have:

E_g=μ_gv X G' - ∂A*/∂t - ▼Φ

Where

E_g -energy due to gravity

μ_g - gravitational permitivvity

G' - Gravitatizing Constant

A* - Electric Potential

Which leads to the dispersion formula and its reults in treating the magnetic charge to dissipate under a gravitational stressed region.

#### Mr. Scientist

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##### Introduction into the Final Equation
« Reply #3 on: 17/11/2009 19:47:39 »
Excerpt 2:

A New Lorentz Force Law Under a Gravitational Form

v²ћ(c/G)=μ_gν X G'(∂A*/∂t)−F_g

where v<c

#### The Naked Scientists Forum

##### Introduction into the Final Equation
« Reply #3 on: 17/11/2009 19:47:39 »