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

[Mordred]

No wrong again the axiom of infinity is

[math]\mathbb{N}\not=\mathbb{O}\mathbb{N}[/math]

Now you prove to me you know What that means.I literally mean mathematical proof.

Edited December 10, 2017 by Mordred

[Vmedvil]

7 minutes ago, Mordred said:

No wrong again the axiom of infinity is

N≠ON

Now you prove to me you know What that means.I literally mean mathematical proof.

What you want me to prove that 1/0 is not 0, but undefined which aleph null or not zero but nothing or undefined.

Edited December 10, 2017 by Vmedvil

[Mordred]

Because I want you to prove you understand number set theory in the first place. It has nothing to do with what you wrote in your last post. 

Edited December 10, 2017 by Mordred

[Vmedvil]

53 minutes ago, Mordred said:

Because I want you to prove you understand number set theory in the first place

Von neumann hierarchy. 

Which can be deined by ∑_{ω = 0} ^N=ω ∑_{ω = 0}^N=ω

 

and you want me to prove that  -∞ = -ω₀  which is the smallest cardinal number....... ya Mordred is there anyone that doubts that, which are both -∅

What is the difference between negative infinity and total non existence, they are no different. 

Edited December 10, 2017 by Vmedvil

[Mordred]

Thank you now apply that, to the above as the majority of the above uses operators, 

Then under the same lemmas and axioms under the majority of the above.

a) define the boundary conditions of each field ( including their range of force and curvature terms with stress tensor.

b) once you have defined the operators above then define the geometry (under time dependant metrics) 

c) Define locality of each field your using under time dependence.

treat each of the above as scalar spin zero. First then do spin 1, followed by spin two.

Those are your 3 principle symmetry groups for the metric terms of the bosonic family.

(write down and apply the applicable Hamiltons) they are in natural units.)

d) Now apply all applicable axioms that depend on each other to the above Number set theory (that apply to the axioms of infinity) as defined to the above as this is the new territory your trying to add.

(it not extra universes but geometry manifolds to form our Atlas from each stage under Manifold boundary confinement as per locality time dependency. 

For programming use Natural units then convert...

Quite frankly you broke too many math rules in your equation ie dividing by tensors and matrixes in the above.

 

Edited December 11, 2017 by Mordred

[Vmedvil]

46 minutes ago, Mordred said:

Thank you now apply that, to the above as the majority of the above uses operators, 

Then under the same lemmas and axioms under the majority of the above.

a) define the boundary conditions of each field ( including their range of force and curvature terms with stress tensor.

b) once you have defined the operators above then define the geometry (under time dependant metrics) 

c) Define locality of each field your using under time dependence.

treat each of the above as scalar spin zero. First then do spin 1, followed by spin two.

Those are your 3 principle symmetry groups for the metric terms of the bosonic family.

(write down and apply the applicable Hamiltons) they are in natural units.)

d) Now apply all applicable axioms that depend on each other to the above Number set theory (that apply to the axioms of infinity) as defined to the above as this is the new territory your trying to add.

(it not extra universes but geometry manifolds to form our Atlas from each stage under Manifold boundary confinement as per locality time dependeat ncy. 

 

a) so where the field = -infinity, so it needs to be solved for g_uv then, which is going to take forever as 

b) So I would have to do a cross product of every part which it is solved for dot product which would take forever.

c) well, that is based on the M_i R_i  or M_iR_i² terms, which would mean solving for I_s which would take forever.

cII), is easy as it is just plugging in S terms. 

well, there is just that one which is when a dimension does not exist, so that is done.

 

Edited December 11, 2017 by Vmedvil

[Mordred]

Are you done I don't think so. This isn't cut and paste.

Describe using that equation each class of solution I mentioned.

After all it does you no good , not knowing yourself how they are properly used instead of relying on software....

Yes it takes time, of course it does to properly model a new GUT.

Especially if you want success. Ie those math rules

I keep reminding  you matrix and tensors are not algebraic variables. As long as you keep making that fundamental mistake. I will keep hounding you to fix that math violation...

You need vectors to describe fields that have directional components. You Cannot model a force without vectors.....

If you can"t correcty model under classical then quantum how do you expect to model less than planckian units?

Edited December 11, 2017 by Mordred

[Vmedvil]

32 minutes ago, Mordred said:

Are you done I don't think so. This isn't cut and paste.

Describe using that equation each class of solution I mentioned.

After all it does you no good , not knowing yourself how they are properly used instead of relying on software....

Yes it takes time, of course it does to properly model a new GUT.

Especially if you want success. Ie those math rules

I keep you matrix and tensors are not algebraic variables. As long as you keep making that fundamental mistake. I will keep hounding you to fix that math violation...

Here is what I will do, enjoy now plug it all back in.

(((1+ (∇'(x',y',z')/∇)²C²)^1/2 - (2M_bG / R_s) - V)-2M_b)/(ω_s1ω_s2 ) = I_s

The first one says they are infinite in range all of them.

And the Cross product is never going to happen.

I can read my own equation, I don't care if you cannot.

Use the dunce hat version then.

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)

(Universe Volumetric Planck State @ size of universe in radius) =(3/4)π ((1/(t_pC²)) L_universe R_Universe)³

Edited December 11, 2017 by Vmedvil

[Vmedvil]

It is just like the standard model in dunce form.

 

Edited December 11, 2017 by Vmedvil

[Vmedvil]

Ya, another post made me think about this which applies to this not directly but is a concept that helps my cause that they are infinite in range the WNF and SNF just very small in magnitude, did you know the impedance of a Superconductor is not actually zero but near zero and can be measured by heat released when AC current is run through YBCO. Maybe, the experiments just cannot detect it below that level, Too bad there isn't AC Strong Nuclear Force or Weak Nuclear Force. The Theoretical on that is supposed to be zero but in reality it is not.

Edited December 11, 2017 by Vmedvil

[Mordred]

Why follow an equation that I already pointed out is wrong You don't treat tensors and matrix's as  per algebra rules no division of either.

[Vmedvil]

8 minutes ago, Mordred said:

Why follow an equation that I already pointed out is wrong You don't treat tensors and matrix's as  per algebra rules no division of either.

I told you how that works.

Edited December 11, 2017 by Vmedvil

[Mordred]

Oh yay you can wiki so now do you understand what I mean by the right hand rule. What does that mean for your indices? Ie contravariant and covariant rules

here Google the answer to this

let X be a vector in Minkoskii space, expressed in the time space basis of the tensor space.

[math]\mathbb{x}\bullet\mathbb{x}=? [/math] and what does ?

[math]\mathbb{x}\bullet\mathbb{x}[/math] provide with regards to time like, space like and lightlike

Edited December 11, 2017 by Mordred

[Vmedvil]

11 minutes ago, Mordred said:

Oh yay you can wiki so now do you understand what I mean by the right hand rule. What does that mean for your indices? Ie contravariant and covariant rules

here Google the answer to this

let X be a vector in Minkoskii space, expressed in the time space basis of the tensor space.

x∙x=? and what does ?

\mathbb{x}\bullet\mathbb{x}[/math] provide with regards to time like, space like and lightlike

and actually I was wrong about one thing, see I thought it would change g_uv but it doesn't, it has the same state inverse and normal.

[Mordred]

very good at least your studying the complex nature behind a matrix and tensors. Please note those are simply unity units not the actual vector expressions of each element.

[Vmedvil]

6 minutes ago, Mordred said:

very good at least your studying the complex nature behind a matrix and tensors. Please note those are simply unity units not the actual vector expressions of each element.

No, this is all stuff I already knew. You just assumed I was always uneducated on math like Polymath. In Either case, those dimensions are still zero which still = does not exist.

Edited December 11, 2017 by Vmedvil

[Mordred]

10 hours ago, Vmedvil said:

(G_uv - R_uv/-g_uv)^1/2

why do you have this expression then when it is clearly wrong with regards to the above?

[Vmedvil]

3 minutes ago, Mordred said:

why do you have this expression then when it is clearly wrong with regards to the above?

that was on the K side and not the matrix side, it was algebra rules until it hit 1/2 R

Edited December 11, 2017 by Vmedvil

[Mordred]

excuse me come again?

[Vmedvil]

4 minutes ago, Mordred said:

excuse me come again?

exactly what I said (k²  + m² )^1/2        k²= K = 1/2 R , which then I set equal to that which was already there, It would break dimensional analysis to change it.

Edited December 11, 2017 by Vmedvil

[Mordred]

Oh my talk about not knowing how the tensors work with 4 momentum 3 velocity Precisely what the question mark is asking for. More precisely the applicable Sylvester formula to the [math] G_{ij}[/math] tensor (Minkowskii) .

 

1 hour ago, Mordred said:

let X be a vector in Minkoskii space, expressed in the time space basis of the tensor space.

x∙x=? and what does ?

x∙x provide with regards to time like, space like and lightlike

Here is a hint [math]\mathbb{x}[/math] uses coordinates [math] {x_o, x_1,x_2, x_3}[/math]  ie under [math]G_{ij}[/math] and correlates to the Kronecker

1 hour ago, Vmedvil said:

exactly what I said (k²  + m² )^1/2        k²= K = 1/2 R , which then I set equal to that which was already there, It would break dimensional analysis to change it.

this expression is only under 1 type of coordinate system. Not applicable to Cartesian. See quoted section below.

Sylvesters theorem as applied to metric signature which the above expression relates to...

https://en.wikipedia.org/wiki/Metric_signature

mainly the following.

"The signature (p, q, r) of a metric tensor g (or equivalently, a real quadratic form thought of as a real symmetric bilinear form on a finite-dimensional vector space) is the number (counted with multiplicity) of positive, negative and zero eigenvalues of the real symmetric matrix g_ab of the metric tensor with respect to a basis. Alternatively, it can be defined as the dimensions of a maximal positive, negative and null subspace. By Sylvester's law of inertia these numbers do not depend on the choice of basis. The signature thus classifies the metric up to a choice of basis. The signature is often denoted by a pair of integers (p, q) implying r = 0 or as an explicit list of signs of eigenvalues such as (+, −, −, −) or (−, +, +, +) for the signature (1, 3) resp. (3, 1).^[1]

The signature is said to be indefinite or mixed if both p and q are nonzero, and degenerate if r is nonzero. A Riemannian metric is a metric with a positive definite signature (p, 0). A Lorentzian metric is a metric with signature (p, 1), or (1, q)."

 

Edited December 11, 2017 by Mordred

[Vmedvil]

11 hours ago, Mordred said:

Oh my talk about not knowing how the tensors work with 4 momentum 3 velocity Precisely what the question mark is asking for. More precisely the applicable Sylvester formula to the Gij tensor (Minkowskii) .

 

Here is a hint x uses coordinates xo,x1,x2,x3   ie under Gij and correlates to the Kronecker

this expression is only under 1 type of coordinate system. Not applicable to Cartesian. See quoted section below.

Sylvesters theorem as applied to metric signature which the above expression relates to...

https://en.wikipedia.org/wiki/Metric_signature

mainly the following.

"The signature (p, q, r) of a metric tensor g (or equivalently, a real quadratic form thought of as a real symmetric bilinear form on a finite-dimensional vector space) is the number (counted with multiplicity) of positive, negative and zero eigenvalues of the real symmetric matrix g_ab of the metric tensor with respect to a basis. Alternatively, it can be defined as the dimensions of a maximal positive, negative and null subspace. By Sylvester's law of inertia these numbers do not depend on the choice of basis. The signature thus classifies the metric up to a choice of basis. The signature is often denoted by a pair of integers (p, q) implying r = 0 or as an explicit list of signs of eigenvalues such as (+, −, −, −) or (−, +, +, +) for the signature (1, 3) resp. (3, 1).^[1]

The signature is said to be indefinite or mixed if both p and q are nonzero, and degenerate if r is nonzero. A Riemannian metric is a metric with a positive definite signature (p, 0). A Lorentzian metric is a metric with signature (p, 1), or (1, q)."

 

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.

Detransform

∇'(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((1/2)R)^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((1/2)R)^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) + ((1/2)R)^1/2(ΔKiloparsec)))²/(C²)))^1/2)

This is g_ab for Calabi Yau Manifold which this was a String Theory Equation so it is.

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

And I even changed it to inverse matrix instead of 1/g_ab  so you would stop bitch about how I handled EFE even though they are exactly the same matrix.

 

∇'(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)

 

Edited December 11, 2017 by Vmedvil

[Mordred]

When you study it long enough you will come to agree 

tensors under GR takes considerable time to fully appreciate and understand. 

[Vmedvil]

7 minutes ago, Mordred said:

When you study it long enough you will come to agree 

tensors under GR takes considerable time to fully appreciate and understand. 

Whatever that adds more to my case, that being the right form saying Energy stress and Dark Energy.

Which I am going to take that part through negative -1, so it flips the signs.

∇'(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)

Edited December 11, 2017 by Vmedvil

[Vmedvil]

2 hours ago, Vmedvil said:

Whatever that adds more to my case, that being the right form saying Energy stress and Dark Energy.

Which I am going to take that part through negative -1, so it flips the signs.

∇'(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)

And where the Electromagnetic Stress tensor is useless, so That will get reverted again back to k_s²

∇'(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((k_s²)^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(k_s²)^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) + ((k_s²)^1/2(ΔKiloparsec)))²/(C²)))^1/2)

has to be _uv, _ab is useless.

Which K goes to Chistoffel symbols which is the wrong direction.

So, now I am going to Einstein's Original papers as human hands have corrupted it to see how he did this.

1900-1909 Einstein papers

Starting from k 

Luckily I can read german.

1912-1914 Writing of Einstein

Here

Edited December 11, 2017 by Vmedvil