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

[Vmedvil]

2 hours ago, Vmedvil said:

Where I still forgot G in ab to uv

 

Split and detransform back to ω_s  to make it the same as Cosmo

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Which Cut E= MC² and (1/f((x,y,z,t,M,I,k,φ,ρ,m)))

∇'(x,y,z,t,M,I,k,φ,ρ,m) = ∇((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/C²))^1/2)

Reverse Solve

∇(x,y,z) = ∇'(x,y,z,t,M,I,k,φ,ρ,m)((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/C²))^1/2)

Absorb Schrodinger Equation.  iħ(d/dt)Ψ(r,t) - V(r,t)Ψ(r,t) / (ħ²/2M)Ψ(r,t) = ∇²

∇'(x,y,z,t,ω,M,I,k,φ,ρ,m,Ψ) = (iħ(d/dt)Ψ(r,t) - V(r,t)Ψ(r,t) / (ħ²/2M_b)Ψ(r,t))^1/2((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/C²))^1/2)

Reaquire  E= MC² and (1/f((x,y,z,t,M,I,k,φ,ρ,m)))

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,Ψ) = (iħ(d/dt)Ψ(r,t) - V(r,t)Ψ(r,t) / (ħ²/2M_b)Ψ(r,t))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where 

 

Then C² = (1/ε₀ μ₀)

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,Ψ) = (iħ(d/dt)Ψ(r,t) - V(r,t)Ψ(r,t) / (ħ²/2M_b)Ψ(r,t))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Energy Eighenvalue transform.

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,Ψ) = (ÊΨ(r,t) - V(r,t)Ψ(r,t) / (ħ²/2M_b)Ψ(r,t))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Ψ Transform.

 

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)e^iS(r,t)/ħ)) - V(r,t)((|φ²|)e^iS(r,t)/ħ))) / (ħ²/2M_b)((|φ²|)e^iS(r,t)/ħ)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

S Quantum number transform.  

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,s) = (Ê((|φ²|)e^i(s(s+1))1/2)) - V(r,t)((|φ²|)e^is(s+1)1/2))) / (ħ²/2M_b)((|φ²|)e^is(s+1))))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Where I cannot do that already a exponent. 

Split ħ into h/2π

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)e^i2πS(r,t)/h)) - V(r,t)((|φ²|)e^i2πS(r,t)/h))) / (ħ²/2M_b)((|φ²|)e^i2πS(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

 

Transform 

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)-e^2S(r,t)/h)) - V(r,t)((|φ²|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|φ²|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

 

 

Edited November 26, 2017 by Vmedvil

[Vmedvil]

1 hour ago, Vmedvil said:

Energy Eighenvalue transform.

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,Ψ) = (ÊΨ(r,t) - V(r,t)Ψ(r,t) / (ħ²/2M_b)Ψ(r,t))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Ψ Transform.

 

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)e^iS(r,t)/ħ)) - V(r,t)((|φ²|)e^iS(r,t)/ħ))) / (ħ²/2M_b)((|φ²|)e^iS(r,t)/ħ)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

S Quantum number transform.  

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,s) = (Ê((|φ²|)e^i(s(s+1))1/2)) - V(r,t)((|φ²|)e^is(s+1)1/2))) / (ħ²/2M_b)((|φ²|)e^is(s+1))))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Where I cannot do that already a exponent. 

Split ħ into h/2π

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)e^i2πS(r,t)/h)) - V(r,t)((|φ²|)e^i2πS(r,t)/h))) / (ħ²/2M_b)((|φ²|)e^i2πS(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

 

Transform 

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)-e^2S(r,t)/h)) - V(r,t)((|φ²|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|φ²|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

 

 

Transform again

So, (M_bω_s((k_sG_uv./g_uv8πG)^1/2 + m_k²))R_s²/2) = V(r,t) = Mω²r²/2

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = (Ê((|φ²|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|φ²|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|φ²|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

 

Which badly needs to be simplified, In any case, where 

Transform

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|φ²|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|φ²|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|φ²|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

which looks right, which d³p = 2E = 2ħω_s  being a negative probability in the Klein Gordan equations.

 

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|φ²|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|φ²|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|φ²|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Edited November 26, 2017 by Vmedvil

[Vmedvil]

On 11/15/2017 at 3:25 PM, Vmedvil said:

Where is places the graviton as something very odd Time-Space Radiation with a spin of ω_s² = 4π²f_s²  where E_p = hf_p then E_s = h(ω_s/2π) which does something screwed up change the causality of space where  C is negative because it travels sideways in time being space which is gravity if 

Where

E_s = h(ω_s/2π), Never took the simple solution huh isn't much easier like that, not being like this, which are both the graviton, S = 2

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|φ²|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|φ²|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|φ²|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

back to  ≈

E_b(t,ω,R,M,I) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s²/2M_b) + V)²/C²))^1/2))M_bC²

Edited November 26, 2017 by Vmedvil

[Mordred]

Oh my Here lets save you a ton of effort. You can scream at me later.

[latex] \stackrel{Action}{\overbrace{\mathcal{L}}} \sim \stackrel{relativity}{\overbrace{\mathbb{R}}}- \stackrel{Maxwell}{\overbrace{1/4F_{\mu\nu}F^{\mu\nu}}}+\stackrel{Dirac}{\overbrace{i \overline{\psi}\gamma_\mu\psi}}+\stackrel{Higgs}{\overbrace{\mid D_\mu h\mid-V\mid h\mid}} +\stackrel{Yukawa-coupling}{\overbrace{h\overline{\psi}\psi}} [/latex]

There ya go Dirac, Higgs, Yukawa coupling, Higgs and Maxwell action all under a single equation

[Vmedvil]

34 minutes ago, Mordred said:

Oh my Here lets save you a ton of effort. You can scream at me later.

LAction∼Rrelativity−1/4FμνFμνMaxwell+iψ¯¯¯γμψDirac+∣Dμh∣−V∣h∣Higgs+hψ¯¯¯ψYukawa−coupling

solve thourgh Yukawa Coupling where φ²  or Dirac not sure of which or both?

Edited November 26, 2017 by Vmedvil

[Mordred]

 

Or study this equation from the Sean Carrol lecture

 

Edited November 26, 2017 by Mordred

[Vmedvil]

9 minutes ago, Mordred said:

 

Or study this equation from the Sean Carrol lecture

 

Yes that has to be pulled through the higgs based on that image, D_u²φ²  where φ² = φ² , So there will be a  logarithm of some sort 

Edited November 26, 2017 by Vmedvil

[Mordred]

yes the lecture is primarily on what the Higg's revealed to us

However both equations clearly demonstrate a key aspect. The methodology to correctly understand any and all field theories is to understand how they affect kinematic motion under ACTION. The correct methodology to understand any lie group is to study the action path integrals for each Lie group. This includes relativity....

Once you understand how Action is involved in any and all lie groups the rest falls into place. ALL interactions can be described under action.

This includes the Feyman diagrams themselves under S_matrix... The external lines of the Feyman diagrams requires quanta of action while the propagators (virtual particles) are the internal lines on the diagrams.

Edited November 26, 2017 by Mordred

[Vmedvil]

Log_{(DgDaDψDφ-W)}((id⁴xg^1/2  (M_p²/2)R - id⁴xg^1/2(1/4)F^a_μvF^aμv + id⁴xg^1/2i(ψ-bar)γ^μD_μψⁱ + id⁴xg^1/2((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - V(φ)/-id⁴xg^1/2D_μ²) = φ²

 

Transform.

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|(Log_{(DgDaDψDφ-W)}((id⁴xg^1/2  (M_p²/2)R - id⁴xg^1/2(1/4)F^a_μvF^aμv + id⁴xg^1/2i(ψ-bar)γ^μD_μψⁱ + id⁴xg^1/2((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - id⁴xg^1/2V(φ)/id⁴xg^1/2D_μ²)|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|(Log_{(DgDaDψDφ-W)}((id⁴xg^1/2  (M_p²/2)R - id⁴xg^1/2(1/4)F^a_μvF^aμv + id⁴xg^1/2i(ψ-bar)γ^μD_μψⁱ + id⁴xg^1/2((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - id⁴xg^1/2V(φ)/-id⁴xg^1/2D_μ²)|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|(Log_{(DgDaDψDφ-W)}((id⁴xg^1/2  (M_p²/2)R - id⁴xg^1/2(1/4)F^a_μvF^aμv + id⁴xg^1/2i(ψ-bar)γ^μD_μψⁱ + id⁴xg^1/2((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - id⁴xg^1/2V(φ)/-id⁴xg^1/2D_μ²)|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Like I said what can Absolutely never be calculated that. where it wouldn't even let put the symbol for ψ bar

Where id⁴xg^1/2  cancels. 

Log_{(DgDaDψDφ-W)}((id⁴xg^1/2  (M_p²/2)R - id⁴xg^1/2(1/4)F^a_μvF^aμv + id⁴xg^1/2i(ψ-bar)γ^μD_μψⁱ + id⁴xg^1/2((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - id⁴xg^1/2V(φ)/-id⁴xg^1/2D_μ²) = φ²

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R -(1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - V(φ)/-D_μ²)|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +((ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.)) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

 

Edited November 26, 2017 by Vmedvil

[Mordred]

As long as you keep jumping to the end points and try to modify and apply the end point equations without understanding the proofs behind those equations you will never be able to calculate a single thing. You must take the time to properly understand how those equations got developed in the first place.

For example try this question which I will later post how to solve.

Find the energy of the vacuum for a real scalar field ?

 

[Vmedvil]

46 minutes ago, Mordred said:

As long as you keep jumping to the end points and try to modify and apply the end point equations without understanding the proofs behind those equations you will never be able to calculate a single thing. You must take the time to properly understand how those equations got developed in the first place.

For example try this question which I will later post how to solve.

Find the energy of the vacuum for a real scalar field ?

 

I dunno but it is better now after it canceled.

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R -(1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

What does h.c. mean?

38 minutes ago, Vmedvil said:

I dunno but it is better now after it canceled.

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R -(1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (h.c.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

What does h.c. mean?

Figured it out. a_ji

E_b(x,y,z,t,ω,M,I,k,φ,ρ,m,S) = ((ħω_s)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji.) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)) - (M_bω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)R_s²/2)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R_s -(1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - V(φ)/-D_μ²)|)-e^2S(r,t)/h))) / (ħ²/2M_b)((|(Log_{(DgDaDψDφ-W)}((  (M_p²/2)R_s - (1/4)F^a_μvF^aμv + i(ψ-bar)γ^μD_μψⁱ +(ψ-bar)ⁱ_LV_ijφψ^j_r + (a_ji) - V(φ)/-D_μ²)|)-e^2S(r,t)/h)))^1/2(1/((1-(((2M_bG / R_s) - (I_sω_s(k_sG_uv./g_uv8πG)^1/2 + m_k²)/2M_b)+ (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)(2ħω_s)) - (k_s(1/ε₀ μ₀)/R_s²) + (k_sG_uv/g_uv8πG ))^1/2(ΔKiloparsec)))²/(1/ε₀ μ₀)))^1/2))M_b(1/ε₀ μ₀)

Mordred you have turned my equation into literal gravity hell now which is probably right but still.

Edited November 26, 2017 by Vmedvil

[Mordred]

d Precisely because the details you need lie in how the creation and annihilation field operators apply to those equations. From those operators you can literally calculate the particle number density etc...

Here is the process taken from Quantum Field theory Demystified. Though its not the greatest book as it has errors its written simple enough for the average person to get a handle on QFT. page 132. To find the energy of the vacuum we need to compute
[math]\langle|\hat{H}|\rangle[/math]

[math]\langle|\hat{H}|\rangle=\langle 0|\int d^3k\omega_k[\hat{N}\vec{k}+\frac{1}{2}]|0\rangle[/math]

[math]\langle|\hat{H}|\rangle=\langle 0|\int d^3k\omega_k[\hat{a}^\dagger(k)\hat{a}(\vec{k}+\frac{1}{2}]|0\rangle[/math]

[math]=\langle 0|\int d^3k\omega_k(\hat{a}^\dagger(\vec{k}\hat{a}(\vec{k}))|+\langle 0|\int d^3k\omega_k(\frac{1}{2}|0\rangle[/math]

[math]=\frac{\omega_k}{2}\int\langle 0|0\rangle[/math]

[math]=\frac{\omega_k}{2}\int d^3k[/math]

which is reminiscent of the quantum harmonic oscilator

[math]\frac{1}{2}\hbar\omega[/math]

however this will lead to

[math]\int d^3 k\longrightarrow\infty[/math] over all of momentum space so one must renormalize the Hamilton via subtraction of the term that leads to the infinity term and set the ground state at zero to arrive at the renormalized Hamilton.

[math]\hat{H}_R=\hat{H}-\int d^3 k[/math]

[math]=\int d^3k\omega\hat{N}(\vec{k})=\int d^3k\omega_k\hat{a}^\dagger(\vec{k})\hat{a}(\vec{k})[/math]

 

 

 

Edited November 26, 2017 by Mordred

[Vmedvil]

6 minutes ago, Mordred said:

Precisely because the details you need lie in how the creation and annihilation field operators apply to those equations. From those operators you can literally calculate the particle number density etc...

Here is the process taken from Quantum Field theory Demystified. Though its not the greatest book as it has errors its written simple enough for the average person to get a handle on QFT. page 132. To find the energy of the vacuum we need to compute
[math]\langle|\hat{H}|\rangle[/math]

[math]\langle|\hat{H}|\rangle=\langle 0|\intD^3k\omega_k[\hat{N}\vec{k}+\frac{1}{2}]|0\rangle[/math]

[math]\langle|\hat{H}|\rangle=\langle 0|\intD^3k\omega_k[\hat{a}^\dagger(k)\hat{a}(\vec{k}+\frac{1}{2}]|0\rangle[/math]

[math]=\langle 0|\int D^3k\omega_k(\hat{a}^\dagger(\vec{k}\hat{a}(\vec{k}))|+\langle 0|\int d^3k\omega_k(\frac{1}{2}|0\rangle[/math]

[math]=\frac{\omega_k}{2}\int\langle 0|0\rangle[/math]

[math]=\frac{\omega_k}{2}\int d^3k[/math]

which is reminiscent of the quantum harmonic oscilator

[math]\frac{1}{2}\hbar\omega[/math]

however this will lead to

[math]\int d^3 k\longrightarrow\infty[/math] over all of momentum space so one must renormalize the Hamilton via subtraction of the term that leads to the infinity term and set the ground state at zero to arrive at the renormalized Hamilton.

[math]\hat{H}_R=\hat{H}-\intd^3 k[/math]

[math]=\intd^3k\omega\hat{N}(\vec{k})=\int d^3k\omega_k\hat{a}^\dagger(\vec{k})\hat{a}(\vec{k}[/math]

 

 

 

If this is true no wonder Gravitons are so screwed up having meditate gravity across all this shit which I have an idea what it is but ya.

Edited November 26, 2017 by Vmedvil

[Mordred]

Welcome to the complexity of physics, corrections applied to the above...

There is no easy way to simply skip this stuff, the real physics behind how particles arise for the number of different particles and field treatments is a complex topic that takes years of study to understand.

That is the reality of nature.

 

[Vmedvil]

13 minutes ago, Mordred said:

Welcome to the complexity of physics, corrections applied to the above...

There is no easy way to simply skip this stuff, the real physics behind how particles arise for the number of different particles and field treatments is a complex topic that takes years of study to understand.

That is the reality of nature.

 

I get it man, which is why it is probably like that. I always noticed they are/were/will be a very odd particle which is why E_s ≈ h(ω_s/2π) E_s being energy of graviton don't dig deeper than that or you will regret it.

Edited November 26, 2017 by Vmedvil

[Mordred]

Understanding a spin 2 boson such as the graviton is also rather tricky as you are dealing with two individual polarity states which will correspond to creation and annihilation operators but would take me forever to latex how [math]h_+,h_x[/math] would be described in terms of path integrals. the above is one of the simpler examples being scalar spin 0.

LOL you should see how bad the quaternion Higg's field looks under the above

Edited November 26, 2017 by Mordred

[Vmedvil]

10 minutes ago, Mordred said:

Understanding a spin 2 boson such as the graviton is also rather tricky as you are dealing with two individual polarity states which will correspond to creation and annihilation operators but would take me forever to latex how h+,hx would b e described in terms of path integrals. the above is one of the simpler examples being scalar spin 0.

There are some serious infinities I had to dodge to get it like that which was why D_u²φ² was chosen to join it at via φ² = φ² try to join it at Dirac or Yukawa and see what happens.

Edited November 26, 2017 by Vmedvil

[Mordred]

 

Yes infinities arise quite often when dealing with any field treatments this is where one must apply renormalization and IR and UV cutoffs to avoid infinity issues. LQC does an interesting technique by applying Wicks rotation to avoid some of these issues. Under LQC renormalization is avoided via Wicks rotation. However that is just one of many numerous techniques.

Another technique being killing vectors which will correspond to you IR and UV cutoffs

Edited November 26, 2017 by Mordred

[Vmedvil]

24 minutes ago, Mordred said:

Yes infinities arise quite often when dealing with any field treatments this is where one must apply renormalization and IR and UV cutoffs to avoid infinity issues. LQC does an interesting technique by applying Wicks rotation to avoid some of these issues. Under LQC renormalization is avoided via Wicks rotation. However that is just one of many numerous techniques.

You know how I said punch a wall and stuff..... I retract that .... DON'T FUCKING MOVE FROM THAT SPOT Not even a Planck Action, it will make it that much more hell for these particles, you think you have to work hard these little guys have to work zillions of times harder even if you had to build this moving Big stones all day.

Edited November 26, 2017 by Vmedvil

[Mordred]

Umm this is a forum where we have readers of all age groups, so lets avoid the language. It is one of our forum rules, you might note Resident experts is part of the Moderator staff

Edited November 26, 2017 by Mordred

[Vmedvil]

2 minutes ago, Mordred said:

Umm this is a forum where we have readers of all age groups, so lets avoid the language. It is one of our forum rules

Sorry, but it is a quote from something I wanted to use. This definitely debunks a "Fine Tuned Universe"

Edited November 26, 2017 by Vmedvil

[Mordred]

lol no prob I understand the sentiment

In all honesty It took me decades of active study to reach the understanding I now have. I at one time tried to shortcut much like your tried.

However after so many failed attempts I finally sat down and studied how every common formula was developed and derived via their individual proofs. After a while you will start to see some very common patterns. These patterns will apply to any application or theory.

One of the most common patterns being kinematic motion under the action principle. The other is group theory itself and how to organize all this complexity.

I would start with the U(1) guage then goto SU(2) then SU(3) followed by SO(1.3)

take the time to fully understand the notation Dubbelosix used is an extremely handy set of mathematics.

Unfortunately I haven't studied much on the extended groups as for the most part I never need them.

Edited November 26, 2017 by Mordred

[Vmedvil]

7 minutes ago, Mordred said:

lol no prob I understand the sentiment

Meh, it is not like anyone is ever going to read this anyways.

[Mordred]

You would be amazed how many readers we get that don't even register an account or post.

[Vmedvil]

6 minutes ago, Mordred said:

You would be amazed how many readers we get that don't even register an account or post.

Yes, but like 265 posts deep.

Edited November 26, 2017 by Vmedvil