      # Wormhole Metric Continued.

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### #1 VictorMedvil

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Posted 11 April 2019 - 11:09 PM

I have decided to continue my Quaternionic Equation from the original Wormhole Metric Thread = https://www.sciencef...his-screwed-up/

This is mainly to check all the variables in the differential Equation to make sure that they all solve correctly and to make sure the Quaternion is anomaly free and solve the equation for (x,y,z,t,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β)

∇'(x,y,z,t,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (d2/((ħ /(2Erest/C2)) 3a =1 (d2/d((C2/Erest)Ni = 1 MiRi)2) + (1/2)3a,β = 1  μ(PΠa)(Pβ - Πβ) + U - (ħ2/2)3N-6s=1(d2/dq2) + V)((|(Log(DgDaDψDφ-W)(((2ħGC2))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμ(((Lghost QE  - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)ψi +(ψ-bar)iLVijφψjr + (aji) - (μ2((φ-Dagger)φ) + λ((φ-Dagger)φ)2)/-(((Lghost QE   - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)2)|)-e2S(r,t)/h)) - ((Erest/C2)ωs((((8πGTab/C4) + Λgab  - Rab) * gab-1))1/2 + (S/ (((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp ))))Rs2/2))) / (ħ2/2(Erest/C2))))1/2(((1-(((2(Erest/C2)G / Rs) - (Isωs((((8πGTab/C4) + Λgab  - Rab) * gab-1))1/2 + (S/(((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp )))))/2(Erest/C2))+ (((8πG/3)((g/(2π)3)∫(((Erelativistic2 - Erest2 / C2((Ar(X) + (ENucleon binding SNF εμ/mu) - Ar(XZ±)/Z) / mu)2)(1/2)(1/e((ERelativistic  - μchemical)/TMatter)±1)(ħω + ħωs) - ((ksC2)/ Rs2) + (((8πGTab/C4) + Λgab  - Rab) * gab-1))1/2(ΔxKiloparsec)))2/(C2)))1/2)

(d2/∇') - (Ctp) = ds2

https://www.wolframalpha.com/input/?i=(d%5E2+%2F+%E2%88%87%27)+-+(C+t)%5E2

(Universe Volumetric Planck State @ size of universe in radius) =(4/3)π((RUniverse/(tpC))Luniverse

https://www.wolframa... ((R/(t C)) )^3

https://www.wolframa... π ((R/(t C))^3

Luniverse = (Charge,∇Color,∇flavour,∇gravity  - ∇Dark Energy)

https://www.wolframalpha.com/input/?i=%E2%88%87+d

https://www.wolframa...i=(∇ g) - (∇ d)

Charge possible states per point (1,2/3, 1/3, 0,-1/3,-2/3,-1)

Color Possible states per point(R,B,G,0,antiG,antiB,antiR)

Flavour possible states per point (I,II,III,0,darkIII,darkII,darkI)

Gravity/Dark Energy possible states per point of space (Energy,Mass,Spin,0,-spin,-mass,-Energy)

Atleast the graphing equation and Equivalence principal are in working order having A.I. do the work.

Edited by VictorMedvil, 14 April 2019 - 03:02 AM.

### #2 VictorMedvil

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Posted 14 April 2019 - 04:08 AM

I have decided to use this equation for a proton instead of the entire universe as it would be too much data to ever complete.

(Universe Volumetric Planck State @ size of universe in radius) =(4/3)π((RUniverse/(tpC)) (Charge,∇Color,∇flavour,∇gravity  - ∇Dark Energy)

RUniverse = RProton = 10-15 meters

The Equation Yields a Planck State of 9.9023511969154288921026543960449 * 1059 (Charge,∇Color,∇flavour,∇gravity  - ∇Dark Energy)

So a Field with 9.9023511969154288921026543960449 * 1059 cubes that are a Planck length  with states of (+1/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) if the proton is at rest.

The Strong Nuclear Force or color Map will look something like this which is the only thing over the 3-D field that varies in a proton. If the Proton is in motion let's say moving in a particle accelerator at 8 Tev then the State is 7.6171932283964837631558879969576 * 1057 (+1/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (8000000/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (8000000/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+1/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (8000000/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

All of the Information being within the equation with a smaller color field of the same picture being less Planck Lengths within the particle due to length contraction. The Graphing Equation displays all possible properties of the particle or substance to an qauntized amount of a Planck Length being exact without error, I could write the entire Tensor for each substance but it would take the big number amount of states. These were done assuming Dark Energy was not existent and a non expanding universe which are the zero terms. There is only one unknown in these equations which is the Spin number of Dark Energy particles being the final zero in the spin term, the graph is over d/dx + d/dy  + d/dz the big number shows the number of planck lengths that the fields manifest for a proton at rest versus in motion for these examples.

This shows this equation to be in working order and accurate to reality.

This equation is actually more complex than the long equation as it gives a single state for everything rather than a large number of multiple Planck States like this one.

Edited by VictorMedvil, 14 April 2019 - 05:01 AM.

### #3 VictorMedvil

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Posted 15 April 2019 - 02:33 AM

If you wanted more detail of the Quarks within the Proton you could graph the equation with the same set of coordinates including the quarks with the same result.

For the Rest proton with quarks in finer detail.

9.9023511969154288921026543960449 * 1059  (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

Now the charges varies given the details of the quarks within the proton which as now the charges vary you will have two varing graphs one for the Strong Nuclear Force or Color and one for the Electromagnetic Force or Charge being the (+2/3/(dx + dy +dz),R/(dx + dy +dz))  + (+2/3 /(dx + dy +dz), B/(dx + dy +dz)) + (-1/3/(dx + dy +dz), G/(dx + dy +dz)) = (+1/(dx + dy +dz),RGB/(dx + dy +dz))

Charge Map Color Map The equation can be used to whatever detail you would like it to be this being a more exact map of the proton next would be to add gluons if you wanted or even more protons and neutrons to construct an atom, but it is always exact to the planck length, no matter what detail is used.

Overlapped Charge and Color Map, (+2/3/(dx + dy +dz),R/(dx + dy +dz))  + (+2/3/(dx + dy +dz), B/(dx + dy +dz)) + (-1/3/(dx + dy +dz), G/(dx + dy +dz)) = (+1/(dx + dy +dz),RGB/(dx + dy +dz)) Which solves perfectly making the graphing equation even physically correct next we will try something more challenging like a Feynman diagram using this equation, it should be able to graph anything in the universe to the planck length is the test.

Edited by VictorMedvil, 15 April 2019 - 03:07 AM.

### #4 VictorMedvil

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Posted 20 April 2019 - 06:24 AM

The Feynmann Diagram we are going to test this on is Beta Decay of Carbon 14 into Nitrogen 14 to start off with the calculations need to be done for the planck state of an Electron and Neutron as beta decay is P+ > N + e- + Ve , so we willl start with mapping the quarks within the proton which a proton's state is  9.9023511969154288921026543960449 * 1059  (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

Then the neutron can be described as a Planck State too which is

9.9023511949154288921026543960449 * 1059  (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

The electron has a smaller state 1.1998578848809383445875560276978 * 1051 (-1/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .511/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

The Neutrino has State of 28722.600151171579743008314436886(0/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .2/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))  being much smaller than all of them

This Completes the Feynmann Diagram for Beta minus decay and satisfies  P+ > N + e- + Ve

9.9023511969154288921026543960449 * 1059  (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

9.9023511949154288921026543960449 * 1059 (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz),938.28/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

1.1998578848809383445875560276978 * 1051 (-1/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .511/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

28722.600151171579743008314436886(0/(dx + dy +dz),0/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 0/(dx + dy +dz), .2/(dx + dy +dz)- 0/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))\

All properties have been conserved. This shows the volume of the neutron to be slightly smaller in size to the proton by .0000002%.

Edited by VictorMedvil, 20 April 2019 - 06:50 AM.

### #5 VictorMedvil

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Posted 26 April 2019 - 06:48 AM

This calculator can also be used to find the effects of Dark Energy on the particle in question for a proton you could solve the amount of Dark Energy on the particle on Nucleon, we can find that Dark Energy has a velocity currently of 54 meters per second using a simple equation E = (1/2)MV2 , V = 54 m/s . The Mass of the Dark Energy Particles are unknown so I will use a mass of electron or mass of proton. Giving each section of space a energy of 1.458 Kev outward with the push of Dark Energy if mass of electron or mass of proton it would be 1.313 Mev , now we can write the proton effected by Dark Energy.

9.9023511969154288921026543960449 * 1059  (+2/3/(dx + dy +dz),R/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 1.45/(dx + dy +dz),938.28/(dx + dy +dz)938.28/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (+2/3/(dx + dy +dz),B/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 1.45/(dx + dy +dz),938.28/(dx + dy +dz)938.28/(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz))) , (-1/3/(dx + dy +dz),G/(dx + dy +dz),I/(dx + dy +dz), (0/(dx + dy +dz) - 1.45/(dx + dy +dz),938.28/(dx + dy +dz)938.28 /(dx + dy +dz),1/2/(dx + dy +dz)- 0/(dx + dy +dz)))

Now the Proton is displaying the expansion of Dark Energy upon the Proton.

Edited by VictorMedvil, 28 April 2019 - 10:04 AM.

### #6 VictorMedvil

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Posted 01 June 2019 - 03:15 AM

It has been shown that this graphing tool can be used to graph anything that is contained with the universe using the information about its dimensions, so this test has been concluded about the graphing equation as successful, but I wanted to note that (dx2 + dy2 +dz2) = (Planck State) being R2 in Planck lengths which is why the dimensions are divided by (dx + dy +dz) and that the planck state( C ) data is used being the dimensions that the field is over being the Complex Manifold. The manifold of space (Euclidean Space) is being used as  (dx + dy +dz) which can also be (dx' + dy' +dz') if you wanted to directly start to use special relativity (Minkowski space) on it where as the Field dimensions are from Quantum field theory to be put over the manifold which is a type of quantum gravity when used that way. Next will be a proof of the big equation which will take longer to test which will give a ds2 value based on a complex system which can be used with the graphing equation to graph the actual state of the entire universe exactly without error based on a complex set of 18 variables or kept in its natural state for a ds2 value which is a Grand Unified Field equation that takes in account the Strong Nuclear Force, Weak Nuclear Force, Gravity and Electromagnetism all in one equation yielding E8 Killing Vectors. This Metric takes in account General Relativity, Special Relativity, Quantum Mechanics, and Quantum Field Theory to arrive at the solution in Killing Vectors which are then placed in Minkowski space.

∇'(x,y,z,t,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (d2/((ħ /(2Erest/C2)) 3a =1 (d2/d((C2/Erest)Ni = 1 MiRi)2) + (1/2)3a,β = 1  μ(PΠa)(Pβ - Πβ) + U - (ħ2/2)3N-6s=1(d2/dq2) + V)((|(Log(DgDaDψDφ-W)(((2ħGC2))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμ(((Lghost QE  - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)ψi +(ψ-bar)iLVijφψjr + (aji) - (μ2((φ-Dagger)φ) + λ((φ-Dagger)φ)2)/-(((Lghost QE   - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)2)|)-e2S(r,t)/h)) - ((Erest/C2)ωs((((8πGTab/C4) + Λgab  - Rab) * gab-1))1/2 + (S/ (((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp ))))Rs2/2))) / (ħ2/2(Erest/C2))))1/2(((1-(((2(Erest/C2)G / Rs) - (Isωs((((8πGTab/C4) + Λgab  - Rab) * gab-1))1/2 + (S/(((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp )))))/2(Erest/C2))+ (((8πG/3)((g/(2π)3)∫(((Erelativistic2 - Erest2 / C2((Ar(X) + (ENucleon binding SNF εμ/mu) - Ar(XZ±)/Z) / mu)2)(1/2)(1/e((ERelativistic  - μchemical)/TMatter)±1)(ħω + ħωs) - ((ksC2)/ Rs2) + (((8πGTab/C4) + Λgab  - Rab) * gab-1))1/2(ΔxKiloparsec)))2/(C2)))1/2)

(d2/∇') - (Ctp) = ds2

One solved solution for this equation already is for ∇' being d2/dx'2 + d2/dy'2 + d2/dz', The original solution for the equation was LGhost QE  Which states that Quantum Entanglement is the same as creating a wormhole between two spaces or universes, and that theoretically if you did quantum entanglement on matter between universes you can transmit matter just like is often done across space during standard Quantum Entanglement experiments.  Edited by VictorMedvil, 01 June 2019 - 04:07 AM. 