Vmedvil5 Posted April 13, 2022 Report Share Posted April 13, 2022 (edited) I have decided to continue my Quaternionic Equation from the original Wormhole Metric Thread = https://www.scienceforums.net/topic/111607-wormhole-metric-how-is-this-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,ωs,ωp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) ∇'(x,y,z,t,ω_{s},ω_{p,}M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (d^{2}/((ħ^{2 } /(2E_{rest}/C^{2})) ∑^{3}_{a =1} (d^{2}/d((C^{2}/E_{rest})∑^{N}_{i = 1 }M_{i}R_{i})^{2}) + (1/2)∑^{3}_{a,}_{β = 1 }μ_{a}_{β}(P_{a }- Π_{a})(P_{β}_{ }- Π_{β}) + U - (ħ^{2}/2)∑^{3N-6}_{s=1}(d^{2}/dq^{2}) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC^{2}))R_{s} - (1/4)F^{a}_{μ}_{v}F^{a}^{μv} + i(ψ-bar)γ^{μ}(((L_{ghost QE } - gf^{abc}(δ^{μ }(c-bar)^{a})A_{μ}^{b}c^{c}) / (c-bar)^{a}δ^{μ}c^{a})_{ + ig(1/2)}τW_{μ} + ig'(1/2)YB_{μ})ψ^{i} +(ψ-bar)^{i}_{L}V_{ij}φψ^{j}_{r} + (a_{ji}) - (μ^{2}((φ-Dagger)φ) + λ((φ-Dagger)φ)^{2})/-(((L_{ghost QE } - gf^{abc}(δ^{μ }(c-bar)^{a})A_{μ}^{b}c^{c}) / (c-bar)^{a}δ^{μ}c^{a})_{ + ig(1/2)}τW_{μ} + ig'(1/2)YB_{μ})^{2})|)-e^{2S(r,t)/h})) - ((E_{rest}/C^{2})ω_{s}((((8πGT_{ab}/C^{4}) + Λg_{ab } - R_{ab}) * g_{ab}^{-1}))^{1/2} + (S/ (((3G(E_{rest}/C^{2}))/2C^{2}R_{s}^{3})(R_{p}V_{p}) + (GI_{s}/C^{2}R_{s}^{3})((3R_{p}/R_{s}^{2})(ω_{p }R_{p}) -ω_{p} ))))R_{s}^{2}/2))) / (ħ^{2}/2(E_{rest}/C^{2}))))^{1/2}(((1-(((2(E_{rest}/C^{2})G / R_{s}) - (I_{s}ω_{s}((((8πGT_{ab}/C^{4}) +_{ }Λg_{ab } - R_{ab}) * g_{ab}^{-1}))^{1/2} + (S/(((3G(E_{rest}/C^{2}))/2C^{2}R_{s}^{3})(R_{p}V_{p}) + (GI_{s}/C^{2}R_{s}^{3})((3R_{p}/R_{s}^{2})(ω_{p }R_{p}) -ω_{p} )))))/2(E_{rest}/C^{2}))+ (((8πG/3)((g/(2π)^{3})∫(((E_{relativistic}^{2 }- E_{rest}^{2} / C^{2})^{ }+ ((A_{r}(X) + (E_{Nucleon binding SNF}_{ }ε_{0 }μ_{0 }/m_{u}) - A_{r}(X^{Z}^{±})/Z) / m_{u})^{2})^{(}^{1/2)}(1/e^{((ERelativistic - }^{μchemical)}^{/TMatter}^{)}±1)(ħω_{s } + ħω_{s}) - ((k_{s}C^{2})/ R_{s}^{2}) + (((8πGT_{ab}/C^{4}) + Λg_{ab } - R_{ab}) * g_{ab}^{-1}))^{1/2}(ΔKiloparsec)))^{2}/(C^{2})))^{1/2}) (d^2/∇') - (Ctp)^2 = ds^2 https://www.wolframalpha.com/input/?i=(d^2+%2F+∇')+-+(C+t)^2 (Universe Volumetric Planck State @ size of universe in radius) =(4/3)π((RUniverse/(tpC))^3 Luniverse https://www.wolframalpha.com/input/?i=(4%2F3)+π+L+((R%2F(t+C))+)^3 https://www.wolframalpha.com/input/?i=∇+d+(4%2F3)+π+((R%2F(t+C))^3 Luniverse = (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy) https://www.wolframalpha.com/input/?i=(∇'Q%2C+∇'u%2C+∇'F+%2C∇'g++-+∇'D) https://www.wolframalpha.com/input/?i=∇+d https://www.wolframalpha.com/input/?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. 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))^3 (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy) RUniverse = RProton = 10^-15 meters The Equation Yields a Planck State of 9.9023511969154288921026543960449 * 10^59 (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy) So a Field with 9.9023511969154288921026543960449 * 10^59 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 * 10^57 (+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 quantized 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. 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 * 10^59 (+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)) 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. The Feynman 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 * 10^59 (+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 * 10^59 (+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 * 10^51 (-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 * 10^59 (+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 * 10^59 (+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 * 10^51 (-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%. 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)MV^2 , 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 * 10^59 (+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. 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 (dx^2 + dy^2 +dz^2) = (Planck State)^2 being R^2 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 (Makowski 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. Next will be a proof of the big equation which will take longer to test which will give a ds^2 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 ds^2 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,ω_{s},ω_{p,}M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (d^{2}/((ħ^{2 } /(2E_{rest}/C^{2})) ∑^{3}_{a =1} (d^{2}/d((C^{2}/E_{rest})∑^{N}_{i = 1 }M_{i}R_{i})^{2}) + (1/2)∑^{3}_{a,}_{β = 1 }μ_{a}_{β}(P_{a }- Π_{a})(P_{β}_{ }- Π_{β}) + U - (ħ^{2}/2)∑^{3N-6}_{s=1}(d^{2}/dq^{2}) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC^{2}))R_{s} - (1/4)F^{a}_{μ}_{v}F^{a}^{μv} + i(ψ-bar)γ^{μ}(((L_{ghost QE } - gf^{abc}(δ^{μ }(c-bar)^{a})A_{μ}^{b}c^{c}) / (c-bar)^{a}δ^{μ}c^{a})_{ + ig(1/2)}τW_{μ} + ig'(1/2)YB_{μ})ψ^{i} +(ψ-bar)^{i}_{L}V_{ij}φψ^{j}_{r} + (a_{ji}) - (μ^{2}((φ-Dagger)φ) + λ((φ-Dagger)φ)^{2})/-(((L_{ghost QE } - gf^{abc}(δ^{μ }(c-bar)^{a})A_{μ}^{b}c^{c}) / (c-bar)^{a}δ^{μ}c^{a})_{ + ig(1/2)}τW_{μ} + ig'(1/2)YB_{μ})^{2})|)-e^{2S(r,t)/h})) - ((E_{rest}/C^{2})ω_{s}((((8πGT_{ab}/C^{4}) + Λg_{ab } - R_{ab}) * g_{ab}^{-1}))^{1/2} + (S/ (((3G(E_{rest}/C^{2}))/2C^{2}R_{s}^{3})(R_{p}V_{p}) + (GI_{s}/C^{2}R_{s}^{3})((3R_{p}/R_{s}^{2})(ω_{p }R_{p}) -ω_{p} ))))R_{s}^{2}/2))) / (ħ^{2}/2(E_{rest}/C^{2}))))^{1/2}(((1-(((2(E_{rest}/C^{2})G / R_{s}) - (I_{s}ω_{s}((((8πGT_{ab}/C^{4}) +_{ }Λg_{ab } - R_{ab}) * g_{ab}^{-1}))^{1/2} + (S/(((3G(E_{rest}/C^{2}))/2C^{2}R_{s}^{3})(R_{p}V_{p}) + (GI_{s}/C^{2}R_{s}^{3})((3R_{p}/R_{s}^{2})(ω_{p }R_{p}) -ω_{p} )))))/2(E_{rest}/C^{2}))+ (((8πG/3)((g/(2π)^{3})∫(((E_{relativistic}^{2 }- E_{rest}^{2} / C^{2})^{ }+ ((A_{r}(X) + (E_{Nucleon binding SNF}_{ }ε_{0 }μ_{0 }/m_{u}) - A_{r}(X^{Z}^{±})/Z) / m_{u})^{2})^{(}^{1/2)}(1/e^{((ERelativistic - }^{μchemical)}^{/TMatter}^{)}±1)(ħω_{s } + ħω_{s}) - ((k_{s}C^{2})/ R_{s}^{2}) + (((8πGT_{ab}/C^{4}) + Λg_{ab } - R_{ab}) * g_{ab}^{-1}))^{1/2}(ΔKiloparsec)))^{2}/(C^{2})))^{1/2}) (d^2/∇') - (Ctp)^2 = ds^2 One solved solution for this equation already is for ∇' being d2/dx'2 + d2/dy'2 + d2/dz'2 , 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. I am changing the (dx,dy,dz) parameters to display a special relativistic 4 current, now including the evolution of the state over time and not just in a static point. Luniverse = (∇Charge,∇Color,∇flavour,∇gravity - ∇Dark Energy) , ∇' (x,y,z)= d'(x,y,z)∇ , d(x,y,z)' = d(x,y,z) (1-(V(x,y,z)^2 /C^2))^1/2 , E(x,y,z) = (1/2)MV(x,y,z)^2 https://www.wolframalpha.com/input/?i=(∇'Q%2C+∇'u%2C+∇'F+%2C∇'g++-+∇'D) This shows the parameters as a function of kinetic energy in a direction or velocity in a direction now, now space properly dilates in the presence of energy at a given time giving a value for L that is special relativistic. The original equation was relativistic however this equation was not. L'universe = (∇'Charge,∇'Color,∇'flavour,∇'gravity - ∇'Dark Energy) https://www.wolframalpha.com/input/?i=∇+x(1-(V^2+%2FC^2))^(1%2F2) https://www.wolframalpha.com/input/?i=∇+y(1-(V^2+%2FC^2))^(1%2F2) https://www.wolframalpha.com/input/?i=∇+z(1-(V^2+%2FC^2))^(1%2F2) The time coordinate can be ignored but I am still doing the A.I. analysis of it anyways, which shows that our analysis of Dark Energy and Gravity are Valid with (C+V, C-V) https://www.wolframalpha.com/input/?i=∇+t+(1-(V^2+%2FC^2))^(1%2F2) This is the proof that the space contraction equation does not interfere with the "Higher Dimensions" such as gravity or dark energy or charge. https://www.wolframalpha.com/input/?i=∇+Q+(1-(V^2+%2FC^2))^(1%2F2) https://www.wolframalpha.com/input/?i=∇+g-d+(1-(V^2+%2FC^2))^(1%2F2) https://www.wolframalpha.com/input/?i=∇+q+u+f+(g-d)+(1-(V^2+%2FC^2))^(1%2F2) However this does prove that it takes the dilation effect upon dimension x upon Q or space upon charge that the space is changing however not charge. https://www.wolframalpha.com/input/?i=∇+Q+x+(1-(V^2+%2FC^2))^(1%2F2) https://www.wolframalpha.com/input/?i=∇+Q+y+(1-(V^2+%2FC^2))^(1%2F2) https://www.wolframalpha.com/input/?i=∇+Q+z+(1-(V^2+%2FC^2))^(1%2F2) Edited April 17, 2022 by Vmedvil5 Quote Link to comment Share on other sites More sharing options...

Vmedvil Posted August 18 Report Share Posted August 18 (edited) I recently attempted to put the "big equation" into wolfram alpha and the computation engine could not compute the equation as it is too large. The Result I recieved from wolfram alpha is in the picture below. It seems the technology does not exist yet to compute the "Big Equation" that I wrote. The Computation by wolfram alpha for the time space of the "big equation" is computable though which the result is in the link below. (d^2/∇') - (Ctp)^2 = ds^2 Wolfram Alpha link = (d^2/∇') - (Ct)^2 = s^2 - Wolfram|Alpha (wolframalpha.com) Edited August 18 by Vmedvil Quote Link to comment Share on other sites More sharing options...

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