Wormhole Metric...... How is this screwed up.

[Vmedvil]

Furthermore.

Which I using that form. k_s = (K(8π/C²)) , K = ((-D_uv(g) -kt_uv)T_uv^-1)

∇'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s((((-D_uv(g) -kt_uv)T_uv^-1)(8π/C²))²)^1/2 + (S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s((((-D_uv(g) -kt_uv)T_uv^-1)(8π/C²))²)^1/2 + (S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) - A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((((-D_uv(g) -kt_uv)T_uv^-1)(8π/C²))²)^1/2(ΔKiloparsec)))²/(C²)))^1/2)

Edited December 11, 2017 by Vmedvil

[Mordred]

7 hours ago, Vmedvil said:

Okay, I don't see how that changes anything, I used Gaussian Curvature and set it equal to (1/2)R in EFE

I didn't go to T_uv but I could have, stopped at G_uv

Oh, I see what you are whining about that is for g_ab, which the first time I used for this part.

Mordred you caused this to be that way, I changed it because of you from the original form which was g_ab because I took it on oh Mordred must know what he is talking about but this version is worse.

 

(1/2)R = ((-8πGT_ab' - Λg_ab + R_ab) * -g_ab^-1)

 

 

sigh I give up, keep making  mistakes. You have looked over dozens of papers that mention Covarient and contravariant terms and indices  as per Einstein summation.

Yet you replace this with Gaussian curve fitting. Replacing the entire metric tensor with  a central potential (covariant term).

Please study the ***^&UUGFTYYGHGD  Eienstein summation rules. Then apply the *&^&^&%^&$ right hand rules to the covariant and contravariant indices. In that equation above. How can you be so bleeding blind (right hand rule=covariant)

[math]\frac{8\pi G}{c^4}[/math] is only central potential force there is examples where this is not the case.

 

 

Edited December 12, 2017 by Mordred

[Vmedvil]

13 minutes ago, Mordred said:

sigh I give up, keep making  mistakes. You have looked over dozens of papers that mention Covarient and contravariant terms and indices  as per Einstein summation.

Yet you replace this with Gaussian curve fitting. Replacing the entire metric tensor with  a central potential (covariant term).

Please study the ***^&UUGFTYYGHGD  Eienstein summation rules. Then apply the *&^&^&%^&$ right hand rules to the covariant and contravariant indices. In that equation above. How can you be so bleeding blind (right hand rule=covariant)

 

 

Well, it has to be _uv and not _ab  has to be up to my standards of a non screwed up equation too, I won't use the _ab form.

[Mordred]

No indices

Einstein summation

[math]G_{\mu\nu}, G^{\mu\nu}, G^{\mu_\nu}[/math] study what each of those means per the tensor your using.

Edited December 12, 2017 by Mordred

[Vmedvil]

24 minutes ago, Mordred said:

No indices

Einstein summation

Gμν,Gμν,Gμν study what each of those means per the tensor your using.

Ya, Contravariant and covariant are no different, from everything I have looked at.

[Mordred]

Then identify the central potential Gaussian under Newtons laws then do this under the Newton limit under GR

 

[Vmedvil]

4 minutes ago, Mordred said:

Then identify the central potential Gaussian under Newtons laws then do this under the Newton limit under GR

 

Yes, it is co-variant my curvature is co-variant why does this matter Co-variant  k_s²  = K = (1/2)R Co-variant

Edited December 12, 2017 by Vmedvil

[Mordred]

So is your equation valid when

[latex]g_{\mu\nu}=g_{\alpha\beta}=\frac{dx^{\alpha}}{dy^{\mu}}\frac{dx^{\beta}}{dy^{\nu}}[/latex]

Metric tensor [latex]dx^2=(dx^0)^2+(dx^1)^2+(dx^3)^2[/latex]

 

[latex]G_{\mu\nu}=\begin{pmatrix}g_{0,0}&g_{0,1}&g_{0,2}&g_{0,3}\\g_{1,0}&g_{1,1}&g_{1,2}&g_{1,3}\\g_{2,0}&g_{2,1}&g_{2,2}&g_{2,3}\\g_{3,0}&g_{3,1}&g_{3,2}&g_{3,3}\end{pmatrix}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

 

Which corresponds to

[latex]\frac{dx^\alpha}{dy^{\mu}}=\frac{dx^\beta}{dy^{\nu}}=\begin{pmatrix}\frac{dx^0}{dy^0}&\frac{dx^1}{dy^0}&\frac{dx^2}{dy^0}&\frac{dx^3}{dy^0}\\\frac{dx^0}{dy^1}&\frac{dx^1}{dy^1}&\frac{dx^2}{dy^1}&\frac{dx^3}{dy^1}\\\frac{dx^0}{dy^2}&\frac{dx^1}{dy^2}&\frac{dx^2}{dy^2}&\frac{dx^3}{dy^2}\\\frac{dx^0}{dy^3}&\frac{dx^1}{dy^3}&\frac{dx^2}{dy^3}&\frac{dx^3}{dy^3}\end{pmatrix}[/latex]

The simplest transform is the Minkowskii metric, Euclidean space or flat space. This is denoted by [latex]\eta[[/latex]

Flat space [latex]\mathbb{R}^4 [/latex] with Coordinates (t,x,y,z) or alternatively (ct,x,y,z) flat space is done in Cartesian coordinates.

In this metric space time is defined as

[latex] ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}[/latex]

 

[latex]\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}[/latex]

do you see where the (c^2 1 1 1) diagonal terms comes from ? in the diagonal components yet?

Edited December 12, 2017 by Mordred

[Vmedvil]

24 minutes ago, Mordred said:

So is your equation valid when

gμν=gαβ=dxαdyμdxβdyν

Metric tensor dx2=(dx0)2+(dx1)2+(dx3)2

 

Gμν=⎛⎝⎜⎜⎜⎜g0,0g1,0g2,0g3,0g0,1g1,1g2,1g3,1g0,2g1,2g2,2g3,2g0,3g1,3g2,3g3,3⎞⎠⎟⎟⎟⎟=⎛⎝⎜⎜⎜−1000010000100001⎞⎠⎟⎟⎟

 

Which corresponds to

dxαdyμ=dxβdyν=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜dx0dy0dx0dy1dx0dy2dx0dy3dx1dy0dx1dy1dx1dy2dx1dy3dx2dy0dx2dy1dx2dy2dx2dy3dx3dy0dx3dy1dx3dy2dx3dy3⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟

The simplest transform is the Minkowskii metric, Euclidean space or flat space. This is denoted by η[

Flat space R4 with Coordinates (t,x,y,z) or alternatively (ct,x,y,z) flat space is done in Cartesian coordinates.

In this metric space time is defined as

ds2=−c2dt2+dx2+dy2+dz2=ημνdxμdxν

 

η=⎛⎝⎜⎜⎜−c2000010000100001⎞⎠⎟⎟⎟

Yes, but it is in that ds solved form being Laplace prime, XYZ being Laplace that you see.

Edited December 12, 2017 by Vmedvil

[Mordred]

NO you follow the chain rule at each infinitisimal when curve fitting via the tangent vector at each coordinate of particle travel through spacetime. That is the curve fitting under GR

[latex]\frac{dx^\alpha}{dy^{\mu}}=\frac{dx^\beta}{dy^{\nu}}=\begin{pmatrix}\frac{dx^0}{dy^0}&\frac{dx^1}{dy^0}&\frac{dx^2}{dy^0}&\frac{dx^3}{dy^0}\\\frac{dx^0}{dy^1}&\frac{dx^1}{dy^1}&\frac{dx^2}{dy^1}&\frac{dx^3}{dy^1}\\\frac{dx^0}{dy^2}&\frac{dx^1}{dy^2}&\frac{dx^2}{dy^2}&\frac{dx^3}{dy^2}\\\frac{dx^0}{dy^3}&\frac{dx^1}{dy^3}&\frac{dx^2}{dy^3}&\frac{dx^3}{dy^3}\end{pmatrix}[/latex]

[Vmedvil]

8 minutes ago, Mordred said:

NO you follow the chain rule at each infinitisimal when curve fitting via the tangent vector at each coordinate of particle travel through spacetime. That is the curve fitting under GR

dxαdyμ=dxβdyν=⎛⎝⎜⎜⎜⎜⎜⎜⎜⎜dx0dy0dx0dy1dx0dy2dx0dy3dx1dy0dx1dy1dx1dy2dx1dy3dx2dy0dx2dy1dx2dy2dx2dy3dx3dy0dx3dy1dx3dy2dx3dy3⎞⎠⎟⎟⎟⎟⎟⎟⎟⎟

See, this is probably why I don't understand this Welcome to the parts of Calculus I, that I hated. you said the magic word "Chain Rule" No I don't understand. If you follow that by product rule along with quotient rule, I will literally just leave the part for GR blank as not worth it.

Edited December 12, 2017 by Vmedvil

[Mordred]

yeah variational Calculus can be a pain in the arse but no matter what direction your particle changes path at each infinitisimal this provides the required derivitaves. Your indice values give the entry required at each coordinate. This is where the term locality becomes important. As not the entire field affects the above at once due to speed of information exchange. Only the local portion of the field does at each coordinate. Local range defined by speed of information exchange from field to particle at each time dependent coordinate.

 

I have really bad news for you Every Symmetry group uses the chain rule, so does every tensor. ODE and PDE is part of the chain rule under Taylor expansions. So does all Langrene's and Hamilton's once you understand them properly prime example Feyman path integrals

http://wwwf.imperial.ac.uk/~jdg/AECHAIN.PDF

the second link is the ODE as per QM QFT.

http://www1.phys.vt.edu/~ersharpe/6455/ch1.pdf

all these examples are detailed under MULTI-Variational calculus ie two or more variables at each coordinate.

Edited December 12, 2017 by Mordred

[Vmedvil]

45 minutes ago, Mordred said:

yeah variational Calculus can be a pain in the arse but no matter what direction your particle changes path at each infinitisimal this provides the required derivitaves. Your indice values give the entry required at each coordinate. This is where the term locality becomes important. As not the entire field affects the above at once due to speed of information exchange. Only the local portion of the field does at each coordinate. Local range defined by speed of information exchange from field to particle at each time dependent coordinate.

 

I have really bad news for you Every Symmetry group uses the chain rule, so does every tensor. ODE and PDE is part of the chain rule under Taylor expansions. So does all Langrene's and Hamilton's once you understand them properly prime example Feyman path integrals

http://wwwf.imperial.ac.uk/~jdg/AECHAIN.PDF

the second link is the ODE as per QM QFT.

http://www1.phys.vt.edu/~ersharpe/6455/ch1.pdf

all these examples are detailed under MULTI-Variational calculus ie two or more variables at each coordinate.

See, but I get Multi-variable calculus, ya Simple chain rule makes no sense but the one of multi-variables perfect, oh wait, that wasn't the chain rule, I was thinking of there is another one. Product and Quotient rule, whatever that is caused from.

Wait ya, I am pretty sure that is there as (i,j,k)

I hated calculus I in general.

17 minutes ago, Vmedvil said:

See, but I get Multi-variable calculus, ya Simple chain rule makes no sense but the one of multi-variables perfect, oh wait, that wasn't the chain rule, I was thinking of there is another one. Product and Quotient rule, whatever that is caused from.

Wait ya, I am pretty sure that is there as (i,j,k)

I hated calculus I in general.

Where 

∇'(x',y',z') = ∇(1-(((2M_bG / R_s) - (I_sω_s²/2M_b))²/C²))^1/2

is like saying,

d/dx(1-(((2M_bG / R_s) - (I_sω_s²/2M_b))²/C²))^1/2 + d/dy(1-(((2M_bG / R_s) - (I_sω_s²/2M_b))²/C²))^1/2 + d/dz((1-(((2M_bG / R_s) - (I_sω_s²/2M_b))²/C²))^1/2)

Where dx,dy,dz or whatever is that equation.

Edited December 12, 2017 by Vmedvil

[Mordred]

Well its required to understand how tensors work. For example Kronecker Delta (parallel transport of two vectors)(equivalence Principle. Levi_Cevita loss of parallel transport of two vectors (tidal forces due to geometry curvature). Kronecker linear coordinates Calculus 1. Levi_Cevita is more applied under calculus 2. curvilinear coordinates

[math]\partial_i,j[/math] Kronecker

[math\partial_{i,j,k}[/latex] Levi

inner product of two vectors returns a scalar

cross product of two vectors a vector

outer product is the product of two coordinate tensors ie kronecker and will return a tensor product as such

 

Edited December 12, 2017 by Mordred

[Vmedvil]

8 minutes ago, Mordred said:

Well its required to understand how tensors work. For example Kronecker Delta (parallel transport of two vectors)(equivalence Principle. Levi_Cevita loss of parallel transport of two vectors (tidal forces due to geometry curvature)

Oh, I realize what you did now okay got it. So, your saying I need to solve for u,v from a,b doing a reverse chain rule or just a chain rule in the opposite direction as that.

Edited December 12, 2017 by Vmedvil

[Mordred]

should give you a new understanding of groups and tensors. As well as how to use them.

Edited December 12, 2017 by Mordred

[Vmedvil]

well, I didn't know you could transfer _ab to _uv, So, it will stay like this until I can solve that mess to transfer between them.

∇'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s(((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s(((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) - A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2(ΔKiloparsec)))²/(C²)))^1/2)

Ya, ab doesn't properly define it in free space is why it could not be used, This defines it all in non free space currently.

Edited December 12, 2017 by Vmedvil

[Mordred]

Every expression you have above can be solved with Calculus of variations. How is that for a hint, Tensors allow us to organize them under group symmetries.

Here Calculus of variations for shortest path.

http://www-users.math.umn.edu/~olver/ln_/cv.pdf

Here example article that is jumping ahead a great deal but it will prove my last statement

"Lie Groups and Differential Equations"
 

http://www.physics.drexel.edu/~bob/LieGroups/LG_16.pdf

 

Edited December 12, 2017 by Mordred

[Vmedvil]

Let's see if this is still a polymorphic code.

Polymorphic

L_{2ghost QE}  = L_{1ghost QE}

All values equal 2 or 1, besides constants.

∇₂'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s(((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s(((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) - A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2(ΔKiloparsec)))²/(C²)))^1/2)

∇₁'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s(((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s(((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) - A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((8πGT_ab' + Λg_ab  - R_ab) * g_ab^-1))^1/2(ΔKiloparsec)))²/(C²)))^1/2)

which it is.

Edited December 12, 2017 by Vmedvil

[Vmedvil]

Oh, I wanted to post this before someone reads this far into the post as my final change.

∇'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s((((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s((((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) - A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2(ΔKiloparsec)))²/(C²)))^1/2)

∇' - (Cdt)²  = ds²

which dt in this case is t_p

Lastly, this was a typo (Universe Volumetric Planck State @ size of universe in radius) =(3/4)π ((1/(t_pC²)) L_universe R_Universe)³

It was supposed to be 

(Universe Volumetric Planck State @ size of universe in radius) =(3/4)π ((R_Universe/(t_pC)) L_universe )³

L_universe = (∇_Charge,∇_Color,∇_flavour,∇_gravity  - ∇_{Dark Energy})

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,G,B,R)

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

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

as 

t_pC = l_p

 

Every detail.

Edited December 21, 2017 by Vmedvil

[Vmedvil]

So, people know what it says

 

                                                                              (==== (Nuclei Vibration and Molecular Coupling Hamiltonian Energy Stress  =================================)(=============

∇'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}

====Maxwell Electromagnetism=Dirac========= QE Virtual Ghost particles======= Hypercharged Higgs===========Particle Coupling======Standard Higgs==

(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-

============= Virtual Particles Gravity =================== and Hypercharged Higgs=Spin Q Number=)(== Energy Stress Tensor Rest Mass and Rest Energy

Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s((((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2 +

=Angular precession of moving Rest Mass and Rest Energy================Quantum Action of it)(=================Energy Stress angular momentum==

(S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s((((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2 +

Angular Precession of Angular momentum================================)  (==========Energy Stress of Linear momentum== Energy Stress of Strong

(S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) -

Interaction and electron orbitals= Energy Stress of Temperature= Energy Stress of Particle Annihilation===Inverse Stress of Dark Energy=======)

A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2(ΔKiloparsec)))²/(C²)))^1/2)

Edited December 23, 2017 by Vmedvil

[Strange]

On 21/12/2017 at 9:42 AM, Vmedvil said:

Oh, I wanted to post this before someone reads this far into the post as my final change.

∇'(x,y,z,t,ω_s,ω_p,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ²  /(2E_rest/C²)) ∑³_{a =1} (d²/d((C²/E_rest)∑^N_{i = 1} M_iR_i)²) + (1/2)∑³_{a,β = 1}  μ_aβ(P_a - Π_a)(P_β - Π_β) + U - (ħ²/2)∑^3N-6_s=1(d²/dq²) + V)((|(Log_{(DgDaDψDφ-W)}(((2ħGC²))R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μ(((L_{ghost QE}  - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)ψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - (μ²((φ-Dagger)φ) + λ((φ-Dagger)φ)²)/-(((L_ghost QE   - gf^abc(δ^μ (c-bar)^a)A_μ^bc^c) / (c-bar)^aδ^μc^a) _{+ ig(1/2)}τW_μ + ig'(1/2)YB_μ)²)|)-e^2S(r,t)/h)) - ((E_rest/C²)ω_s((((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/ (((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p ))))R_s²/2))) / (ħ²/2(E_rest/C²))))^1/2(((1-(((2(E_rest/C²)G / R_s) - (I_sω_s((((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2 + (S/(((3G(E_rest/C²))/2C²R_s³)(R_pV_p) + (GI_s/C²R_s³)((3R_p/R_s²)(ω_p R_p) -ω_p )))))/2(E_rest/C²))+ (((8πG/3)((g/(2π)³)∫(((E_relativistic² - E_rest² / C²) + ((A_r(X) + (E_{Nucleon binding SNF} ε₀ μ₀ /m_u) - A_r(X^Z±)/Z) / m_u)²)^(1/2)(1/e^{((ERelativistic  - μchemical)/TMatter)}±1)(ħω_s  + ħω_s) - ((k_sC²)/ R_s²) + (((8πGT_ab/C⁴) + Λg_ab  - R_ab) * g_ab^-1))^1/2(ΔKiloparsec)))²/(C²)))^1/2)

∇' - (Cdt)²  = ds²

which dt in this case is t_p

Lastly, this was a typo (Universe Volumetric Planck State @ size of universe in radius) =(3/4)π ((1/(t_pC²)) L_universe R_Universe)³

It was supposed to be 

(Universe Volumetric Planck State @ size of universe in radius) =(3/4)π ((R_Universe/(t_pC)) L_universe )³

L_universe = (∇_Charge,∇_Color,∇_flavour,∇_gravity  - ∇_{Dark Energy})

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,G,B,R)

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

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

as 

t_pC = l_p

 

Every detail.

Why are you posting screen shots of Wikipedia pages that no one can read, instead of just a link? What is wrong with you.

[Vmedvil]

Just now, Strange said:

Why are you posting screen shots of Wikipedia pages that no one can read, instead of just a link? What is wrong with you.

Click them to enlarge.

[Strange]

Just now, Vmedvil said:

Click them to enlarge.

They are still unreadable. And WHY post an image in the first place. It is an idiotic thing to do. Please stop. PROVIDE LINKS.

[Vmedvil]

1 minute ago, Strange said:

They are still unreadable. And WHY post an image in the first place. It is an idiotic thing to do. Please stop. PROVIDE LINKS.

Can you not read english strange, I thought the Chinese were taught English as a second language, assuming a "Coffee Shop" in china 珈琲店