Vmedvil Posted November 28, 2017 Author Share Posted November 28, 2017 (edited) 1 hour ago, SuperPolymath said: Yeah, just beware that there's an understanding of the equations @Mordred has been exposed to so I try and understand everything he shows you first. In math, you have to not only cover all the angles, but know all the angles you have to cover. @swansont, this is my last post in this thread, I promise. Good luck, Vmedvil. Thanks Superpolymath for your assistance. me = (Ar(e) / mu) , Ar(e) = (Ar(X) + (Ebx /muC2) - Ar(XZ+)/Z) , In any case, that adds the residual SNF. Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji.) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - (Isωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(2ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) Where aji is the last one left undefined. Which will get summed along with ∑iIs =(∑iMiR2i) across space as that didn't effect it. Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji.) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - ((∑iMiR2i)ωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(∑iMiR2i)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(2ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) Where aji = (∑j∑iaji) Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - ((∑iMiR2i)ωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(∑iMiR2i)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(2ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) i = Number of objects and j = I dunno. Edited November 28, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 28, 2017 Author Share Posted November 28, 2017 (edited) 1 hour ago, Vmedvil said: Thanks Superpolymath for your assistance. me = (Ar(e) / mu) , Ar(e) = (Ar(X) + (Ebx /muC2) - Ar(XZ+)/Z) , In any case, that adds the residual SNF. Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji.) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - (Isωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(2ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) Where aji is the last one left undefined. Which will get summed along with ∑iIs =(∑iMiR2i) across space as that didn't effect it. Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji.) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - ((∑iMiR2i)ωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(∑iMiR2i)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(2ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) Where aji = (∑j∑iaji) Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - ((∑iMiR2i)ωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(∑iMiR2i)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(2ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) i = Number of objects and j = I dunno. j = Number of Probability currents, All of it is Metric system for units. Edited November 28, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Mordred Posted November 28, 2017 Share Posted November 28, 2017 (edited) Ok here is a simple fact, no offense but I cannot work with that garbage equation above. There is no basis to even start with. So here is what I would like from you. 1) Clearly define what unit system you are going to work in. The above has such a mix of units, both Natural and otherwise that it is obvious you never did a single unit conversion to anything above (prime example kiloparsecs in the above mixed with units expressed in Natural units.) 2) Clearly define your goal. 3) define the system metric you wish to use, You still have curvature mixed with non curvature expressions above. Lets start with that. Toss away that mess above. Lets do this properly. Lets start with a coordinate system that is generic enough to apply above, (after all your going to find the symmetry relations of a fundamental use.) do this before we worry about such things as curvature and frame dragging. Those dynamics won't apply at all the measurement scales you have above. (ie curvature is useless at Planck scales) You would never be able to determine a curvature below the Planck length so absolutely pointless at that scale). My recommendation is that you stick to Natural units as it will greatly simplify things. Here is Natural units. [math]\hbar[/math] is one unit of action [math]ML^2/T[/math] c is one unit of velocity [math]L/T[/math] [math]\hbar=\frac{h}{2\pi}=1.055^{-34} J/s[/math] c you already know. ( least I assume so) so the unit of energy becomes [math]ML^2/T^2[/math] so you can be a bit lazy and speak of mass (m), momentum (mc) and energy (mc^2) all in units of Gev=[math] 10^9[/math] electron volts. Length becomes [math] \hbar/mc[/math] time as [math]\hbar/mc^2[/math] these two are units of [math]GeV^{-1}[/math] lets start with that. This will correspond to [math]c=\hbar=1[/math] Had some RL to deal with so lets add some conversion factors to the above. 1 kg=[math]5.61^{26}[/math] GeV 1 metre= [math]5.07^{15}[/math] [math]GeV^{-1}[/math] 1 sec= [math]1.52^{24}[/math] [math]GeV^{-1}[/math] Edited November 29, 2017 by Mordred 2 Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) 10 hours ago, Mordred said: Ok here is a simple fact, no offense but I cannot work with that garbage equation above. There is no basis to even start with. So here is what I would like from you. 1) Clearly define what unit system you are going to work in. The above has such a mix of units, both Natural and otherwise that it is obvious you never did a single unit conversion to anything above (prime example kiloparsecs in the above mixed with units expressed in Natural units.) 2) Clearly define your goal. 3) define the system metric you wish to use, You still have curvature mixed with non curvature expressions above. Lets start with that. Toss away that mess above. Lets do this properly. Lets start with a coordinate system that is generic enough to apply above, (after all your going to find the symmetry relations of a fundamental use.) do this before we worry about such things as curvature and frame dragging. Those dynamics won't apply at all the measurement scales you have above. (ie curvature is useless at Planck scales) You would never be able to determine a curvature below the Planck length so absolutely pointless at that scale). My recommendation is that you stick to Natural units as it will greatly simplify things. Here is Natural units. ℏ is one unit of action ML2/T c is one unit of velocity L/T ℏ=h2π=1.055−34J/s c you already know. ( least I assume so) so the unit of energy becomes ML2/T2 so you can be a bit lazy and speak of mass (m), momentum (mc) and energy (mc^2) all in units of Gev=109 electron volts. Length becomes ℏ/mc time as ℏ/mc2 these two are units of GeV−1 lets start with that. This will correspond to c=ℏ=1 Had some RL to deal with so lets add some conversion factors to the above. 1 kg=5.6126 GeV 1 metre= 5.0715 GeV−1 1 sec= 1.5224 GeV−1 There is nothing sub planck there and all those natural units have SI values like I said this is no for you to use but a computer, ultra simple version since it never changed Photonlike Es = ℏωs Particlelike annihilation. Ea = 2ħωs = ℏωs + ℏωs Inverse Moment of Inertia Curvature and Inverse Mass Curvature ωs , which has a Irreducible EFE in it. ωs = -1/(Is(Guv - Ruv/-guv)1/2 + Mb(Guv - Ruv/-guv)1/2) Those are the All the ωs Terms in the equation, but it is inverse so it is gonna look weird. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) Furthermore and do you see this term repeated, which says it is the same as C2 (2ħG/ε0 μ0) and I noticed that typo in the last picture. Intertia should be Inertia. There we go. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) Defined 1 as Reduced Planck's constant and. Defines as 2ħ Now Defined that as this. Split 2ħωs = ℏωs + ℏωs Eb(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((∑j∑iaji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - ((∑iMiR2i)ωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(∑iMiR2i)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(ħωs + ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2))(Mb/ε0 μ0) Lol, I just looked at this picture in a very different way, did you know that mother nature has almost perfect breasts? I had to say that, cause it looks like it with that curve. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) Okay, expanded to a wave which we will now always call Graviton wave "Mother nature's Massive boobs" *zooms in* I feel like a peeping tom somehow, well now we know how those are made Inverse sine or Inverse cosine cellular reproduction curve ladies and mother nature has many more than human females. In any case, if that is too sexualized then we can just say "Mother Nature's Massive Hershey kisses" or Graviton wave. Hold on I gotta post something somewhere where kids wouldn't see this. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Mordred Posted November 29, 2017 Share Posted November 29, 2017 (edited) Ok fair enough, if your going to stick to SI thats your choice. The math exression above is still an extemely messy jumble where you have Newtonian approximations mixed with relativity equations. Tensors mixed with complex conjugates, vectors mixed with scalar values. So good luck getting the above fully reducible. Here is the thing many of those equations you have above simplify to common kinematics. You would have been far better off taking the final equations and examining the proofs of each to find the common variables that apply to each expression then developing your formula from those variables common to each equation instead of having the same variables repeated numerous times on the RHS of the equal sign. However your doing the work so have fun with that lol. Edited November 29, 2017 by Mordred Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) 1 hour ago, Mordred said: Ok fair enough, if your going to stick to SI thats your choice. The math exression above is still an extemely messy jumble where you have Newtonian approximations mixed with relativity equations. Tensors mixed with complex conjugates, vectors mixed with scalar values. So good luck getting the above fully reducible. Here is the thing many of those equations you have above simplify to common kinematics. You would have been far better off taking the final equations and examining the proofs of each to find the common variables that apply to each expression then developing your formula from those variables common to each equation instead of having the same variables repeated numerous times on the RHS of the equal sign. However your doing the work so have fun with that lol. well, you know what at-least this way I don't have to proof it and wolfram will do it, what did you think Fusing it all into a single field would be pretty, wolfram said that my equation worked without all this and the Einstein part worked.((x,y,z,t,ωs) Function as intended. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) Even in Parts in Big O notation this is Rank 6 as nested to a high degree, I dunno what rank the entire EQ is. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Mordred Posted November 29, 2017 Share Posted November 29, 2017 (edited) Let me ask you one simple question. Why do you think you need to mix claasical, with relativity and QFT when everything can be done under a single theory? Both relativity and QFT can describe any Newton equation. QFT can describe anything under relativity with extremely good approximation as QFT includes relativity. Edited November 29, 2017 by Mordred Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) 29 minutes ago, Mordred said: Let me ask you one simple question. Why do you think you need to mix claasical, with relativity and QFT when everything can be done under a single theory? Both relativity and QFT can describe any Newton equation. QFT can describe anything under relativity. GR is incompatible with QM in the present state. SR can be merged with QFT not GR, along with GR with SR which makes SR a good bridge for all of it once put in Laplace form being the same Laplace used in QM, this is all about how screwed up QM is while still being correct. The real test will be not all the other variable but these two (φ,S) , Those others are being checked just for shits and giggles, I kinda knew that GR would work in SR. (ωs) that why I took EFE to irreducible they should not ever change, the others can but that one cannot to join QM at Laplace, which is why you were right when I should not even touch the EFE's state besides solving it for Gaussian curvature which is ((1/2) R = K), but making it irreducible I did the equivalent of observer effect it. So, as I screw with QM when it doesn't work in that state I can change it without worrying about GR., I did the same to QM. as ∇ = (d2/dx2 + d2/dy2 + d2/dz2)1/2 Irreducible. Edited November 29, 2017 by Vmedvil 1 Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) Part of S works. and S totally works but Anderson's Equation has roots but whatever, it worked in a Taylor Series and Laurent Series, I was replaced by F, which I am convinced is because S has no value without the other EQ. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Mordred Posted November 29, 2017 Share Posted November 29, 2017 (edited) No GR and QM are not incompatible, the only issue is renormalization which your not even addressing on this thread. As far as Heisenburg is concerned the effect is essentially average or washed out as some literature will describe it as. Its a common misconception the two are incompatible but they simply have different aporoaches. QM being canonical while GR is conformal. This involves different transposed vectors. Rather I should say QFT not QM as they use different operators entirely though they both use p,q. for those operators. The Poincare group itself is involved under both regimes. Both use the same group and symmetry relations. Ie both use SO(1.3) which is the Poincare group. Anyways if that is your goal you should really be looking at QFT vs GR and not QM and SR. That is where the renormalization issues crop up. Edited November 29, 2017 by Mordred Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) ks Works, but has a odd parity. Mordred's equation got an exact result. E was replaced with G Which even with my Residual SNF modification got an exact result, many variables were changed to other symbols. And I am not going to test Fermilab's equation there is no way it is wrong. State Change Unified Gravity Field in a time slice. ∇t'(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((aji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(1/((1-(((2MbG / Rs) - (Isωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(Is)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(ħωs + ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2)) Forever ∑∇t' = ∇t'1 + ∇t'2 + ∇t'3 + ... + ∇t'n ∑∇t' = Δt' Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) And I pulled a error-ed version of Time Dilation does anyone not know the proper form. Which I screwed up that state change, forgot to removed 1/(), it has to be length contract or it mess with QM, how did I know you were going to give me problems Schrodinger. ∇'(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((aji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(((1-(((2MbG / Rs) - (Isωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(Is)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(ħωs + ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2)) Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) Even Irreducibly Phase Locked with the observer effect it gives problems when I switch that. You know what, I will do it this way then ∇'(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((aji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(((1-(((2MbG / Rs) - (Isωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(Is)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(ħωs + ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2) ∑∇'/Δt = ∇1/Δt1 + ∇2/Δt2 + ∇3/Δt3 + ... + ∇n/Δtn ∑∇'/Δt = (Δ'/Δt) It's a Frequency so QM doesn't implode, have it your way. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) 1 hour ago, Vmedvil said: Even Irreducibly Phase Locked with the observer effect it gives problems when I switch that. You know what, I will do it this way then ∇'(x,y,z,t,ωs,ωp,M,I,k,φ,S,X,Z,u) = ((ħωs)((|(Log(DgDaDψDφ-W)(((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h)) - (Mbωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+GIs/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2)Rs2/2)((|(Log(DgDaDψDφ-W)((((2ħG/ε0 μ0))Rs -(1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + (aji) - V(φ)/-Dμ2)|)-e2S(r,t)/h))) / (ħ2/2Mb)((|(Log(DgDaDψDφ-W)(( ((2ħG/ε0 μ0))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμDμψi +(ψ-bar)iLVijφψjr + ((aji)) - V(φ)/-Dμ2)|)-e2S(r,t)/h)))1/2(((1-(((2MbG / Rs) - (Isωs(Guv - Ruv/-guv)1/2 + (S2/ (3GMb/2(1/ε0 μ0)2Rs3(Rp x vp)+G(Is)/(1/ε0 μ0)Rs(3Rp/Rs2(ωpRp) - ωp))2))/2Mb)+ (((8πG/3)((g/(2π)3)∫(p2 + ((Ar(X) + (Ebx ε0 μ0 /mu) - Ar(XZ+)/Z) / mu)2)(1/2)(1/e((E - μ)/T)±1)(ħωs + ħωs)) - ((ksRs2/ε0 μ0)) + (Guv - Ruv/-guv))1/2(ΔKiloparsec)))2/(1/ε0 μ0)))1/2) ∑∇'/Δt = ∇1/Δt1 + ∇2/Δt2 + ∇3/Δt3 + ... + ∇n/Δtn ∑∇'/Δt = (Δ'/Δt) It's a Frequency so QM doesn't implode, have it your way. So, I suppose it is actually uncertain and not a error in the equation and Einstein was actually wrong. Do you know who was right John Ford. Where that makes Matter Editation impossible in our universe, which was a concept from a video game that I though would be interesting to do. Matter Editation Sci Fi Game Something like Neutronium is impossible to stabilize which will decay in about 30 min as you cannot change how WNF interacts with it. Neutronium Edited November 29, 2017 by Vmedvil -1 Link to comment Share on other sites More sharing options...
SuperPolymath Posted November 29, 2017 Share Posted November 29, 2017 31 minutes ago, Vmedvil said: So, I suppose it is actually uncertain and not a error in the equation and Einstein was actually wrong. Do you know who was right John Ford. Where that makes Matter Editation impossible in our universe, which was a concept from a video game that I though would be interesting to do. Matter Editation Sci Fi Game You're not going to be able to determine anything if you don't account for efe interactions beneath the planck length. Stop not doing that! Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) 4 minutes ago, SuperPolymath said: You're not going to be able to determine anything if you don't account for efe interactions beneath the planck length. Stop not doing that! Well, it does account for EFE, but QM causes divergence of it. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
SuperPolymath Posted November 29, 2017 Share Posted November 29, 2017 1 minute ago, Vmedvil said: Well, it does account for them, but QM causes divergence of it. QM is based on action at distance. EFE governs the behavior of commutative quantum automation once you've applied the respective lorentz transformations in scale relativity. I liked your comments on teleparallelism & QM's incompatibility with it, the mathematical incompatibility itself should tell you something. You have to define sub-planckian curves by assuming smaller components such as the pseudo-particles or fractional-photon, generating the waves of fractional-graviton (abrupt accelerations cause frame dragging of the EM field linking polarities) when a solid particle goes solid to wave. It's called particle scattering. Link to comment Share on other sites More sharing options...
Vmedvil Posted November 29, 2017 Author Share Posted November 29, 2017 (edited) In any case, Replicators are impossible below a certain point which is the quantum limit. like the ones the Star Trek completely impossible, which would have been cool to turn like Energy into food or something. Quantum Limit So, Ribosomes are about as small as you can go then on a assembler being molecular. You cannot let's say make a Quantum Assembler, Nano is about as small as you can get. Edited November 29, 2017 by Vmedvil Link to comment Share on other sites More sharing options...
SuperPolymath Posted November 29, 2017 Share Posted November 29, 2017 Lorentz transformations arise from c propagating over a distance of 1/n planck length at 1/n planck time. These small changes between adjacent qubit cells occur at that rate & eventually effect large scale polarity over any distance Link to comment Share on other sites More sharing options...
Mordred Posted November 29, 2017 Share Posted November 29, 2017 26 minutes ago, SuperPolymath said: You're not going to be able to determine anything if you don't account for efe interactions beneath the planck length. Stop not doing that! There is no possible way to determine anything below Planck scales. A planck is a unit of action. No action no possible influence on any possible or even hypothetical perfect detector. 21 minutes ago, SuperPolymath said: QM is based on action at distance. EFE governs the behavior of commutative quantum automation once you've applied the respective lorentz transformations in scale relativity. I liked your comments on teleparallelism & QM's incompatibility with it, the mathematical incompatibility itself should tell you something. You have to define sub-planckian curves by assuming smaller components such as the pseudo-particles or fractional-photon, generating the waves of fractional-graviton (abrupt accelerations cause frame dragging of the EM field linking polarities) when a solid particle goes solid to wave. It's called particle scattering. garbage Link to comment Share on other sites More sharing options...
SuperPolymath Posted November 29, 2017 Share Posted November 29, 2017 (edited) 14 minutes ago, Vmedvil said: In any case, Replicators are impossible below a certain point which is the quantum limit. like the ones the Star Trek completely impossible, which would have been cool to turn like Energy into food or something. Quantum Limit So, Ribosomes are about as small as you can go then on a assembler being molecular. No, it's a matter of mathematical complexity having to account for number of interactions everything is governed by relativity 2 minutes ago, Mordred said: There is no possible way to determine anything below Planck scales. A planck is a unit of action. No action no possible influence on any possible or even hypothetical perfect detector. garbage Prove that there's no action Edited November 29, 2017 by SuperPolymath Link to comment Share on other sites More sharing options...
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