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

[Vmedvil]

2 minutes ago, SuperPolymath said:

This is still finite toric geometry. There will always be a variant of you don't apply infinite fractal geometry.

There aren't enough dimensions for that eventually it will cap out when all variables are defined it won't be SU(n) it will be less than that.

 

 

Edited November 25, 2017 by Vmedvil

[Mordred]

5 minutes ago, Vmedvil said:

There aren't enough dimensions for that eventually it will cap out when all variables are defined it won't be SU(n) it will be less than that.

 

 

that's one way to describe it, it may generate pretty pictures but you won't be able to calculate the path integral of a photon.

[SuperPolymath]

4 minutes ago, Vmedvil said:

There aren't enough dimensions for that eventually it will cap out when all variables are defined it won't be SU(n) it will be less than that.

 

 

No, remember what I was saying about what makes such things undefined?

You even said it would take me 500,000,000 years because of how undefinable the asymptote of the infinite contour is as you approach the 0 space-time center of the heterotic superstring

But those last three links provide some background on how to define those variables invariantly

[Vmedvil]

4 minutes ago, SuperPolymath said:

No, remember what I was saying about what makes such things undefined?

You even said it would take me 500,000,000 years because of how undefinable the asymptote of the infinite contour is as you approach the 0 space-time center of the heterotic superstring

Yes, but humans are not that smart in math, this is something a Hypercore would have to do and I am not saying Binary Hyper-threading either like I said what can stack a quantum computer this equation.

Edited November 25, 2017 by Vmedvil

[SuperPolymath]

11 minutes ago, Vmedvil said:

Yes, but humans are not that smart in math, this is something a Hypercore would have to do and I am not saying Binary Hyper-threading either like I said what can stack a quantum computer this equation.

Not if you only do it for like the space a meter. You have you lasers in a Cern lab, you predict the way the particles will interact, & you design a small qe gate just enough for a vast number of simple transmissions for one complex calculation. That qe gate could be one of many doing faster calculations.

[Vmedvil]

2 minutes ago, SuperPolymath said:

Not if you only do it for like the space a meter. You have you lasers in a Cern lab, you predict the way the particles will interact, & you design a small qe gate just enough for a vast number of simple transmissions for one complex calculation. That qe gate could be one of many doing faster calculations.

Something like that which in Quantum Machinery is called a String Computer.

Edited November 25, 2017 by Vmedvil

[Mordred]

 

 

here is an example the image on the left is a global vector field, the right is a local vector field. You want a set of equations to get from image left to image right. This is just U(1)

image from Quantum Field theory Demystified, page number on image

Edited November 25, 2017 by Mordred

[Vmedvil]

Just now, Mordred said:

 

 

here is an example the image on the left is a global vector field, the right is a local vector field. You want a set of equations to get from image left to image right. This is just U(1)

thats why I keep adding them as I defined or someone else does.

[Mordred]

You want to be able to plot every vector at every coordinate when you add

[math]g_{\mu\nu}=\eta_{\mu\nu}+h_{\mu\nu}[/math] that is where group symmetry comes into play every coordinate must be plotted under transformation. The number of coordinates increases the complexity.

Edited November 25, 2017 by Mordred

[Vmedvil]

2 minutes ago, Mordred said:

You want to be able to plot every vector at every coordinate when you add

Gμν+ημν+hμν that is where group symmetry comes into play every coordinate must be plotted under transformation. The number of coordinates increases the complexity.

Yes, which is why ∇E_b(x,y,z,t,ω,M,I,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_s² + 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) + (Λ/3))^1/2(Δx/3.08567758128*10¹⁹)))²/C²))^1/2))M_bC²

says ∑∑∑∑∑∑∑∑∑∇

Edited November 25, 2017 by Vmedvil

[Mordred]

great I see several matrixes in the above can you identify which terms are actually matrixes in the above? or better yet can be simplified by using matrixes to organize?

[Vmedvil]

1 minute ago, Mordred said:

great I see several matrixes in the above can you identify which terms are actually matrixes in the above? or better yet can be simplified by using matrixes to organize?

You have to reverse how I constructed to see the different parts, the first seven are SR/GR, First four are four current SR

Edited November 25, 2017 by Vmedvil

[Mordred]

I can already deconstruct it, I am asking you to

start with the [math]\frac{8\pi G}{3}[/math]

start with a scalar field then do a vector field

Edited November 25, 2017 by Mordred

[Vmedvil]

10 minutes ago, Mordred said:

I can already deconstruct it, I am asking you to

start with the [math]\frac{8\pi G}{3}[/math]

start with a scalar field then do a vector field

i(That) + j(That) + k(That) - Ct(That) past that you gotta know the signs but ..... (-/+).....n

Edited November 25, 2017 by Vmedvil

[Mordred]

I know how those unit vectors work, no they don't work the way you think they do ie you need the Jacobian matrix to use them correctly

ps I posted this a few days ago ie the Kronecher delta and Levi-Cevita connections section I posted earlier this thread.

Edited November 25, 2017 by Mordred

[Vmedvil]

13 minutes ago, Mordred said:

I know how those unit vectors work, no they don't work the way you think they do ie you need the Jacobian matrix to use them correctly

ps I posted this a few days ago ie the Kronecher delta and Levi-Cevita connections section I posted earlier this thread.

Okay, I got that but I really don't want to construct one but once.

[Mordred]

http://www.hri.res.in/~debsadhukhan/HRI web/pdf/Units & Vectors .

the above is part of it, here go through this then think back to Dubbelsix mentioning of dimensionality. which he was quite correct to do so

Those three unit vectors only ever have 1 value (hint unity)

 

https://www.khanacademy.org/math/precalculus/vectors-precalc/unit-vectors/v/intro-unit-vector-notation

for those that don't like reading math lol

Edited November 25, 2017 by Mordred

[Vmedvil]

42 minutes ago, Mordred said:

http://www.hri.res.in/~debsadhukhan/HRI web/pdf/Units & Vectors .

the above is part of it, here go through this then think back to Dubbelsix mentioning of dimensionality. which he was quite correct to do so

Those three unit vectors only ever have 1 value (hint unity)

 

and why doesn't = R

[Mordred]

[math]\mathbb{R}[/math] is the set of real numbers, a unit vector only ever has a single value of 1. its not a set.

To answer the next possible question the reason we square the unit vectors in probability density functions is to take a negative number and make it positive as probabilities are always positive integers.

unfortunately there is no agreed upon symbol for the set of imaginary numbers so many use [math]i\mathbb{R}[/math] or just [math]\mathbb{I}[/math]

Edited November 26, 2017 by Mordred

[Mordred]

Here i,j,k Vector Calculus by Michael Corral

 

 

note distribution of this is under the GNU free documentation license which gives permission to copy paste portions

 

one has to be careful on copyright notices lol

 

Anyways here you go Schwartzchild with the QFT treatments... including the Einstein-Rosen bridge.

https://arxiv.org/pdf/1708.00748.pdf

should be enough that if you study it properly should take a month to understand just how each equation is derived. (to properly understand it) when your not that familiar with QFT or vector calculus...

Edited November 26, 2017 by Mordred

[Vmedvil]

10 hours ago, Mordred said:

Here i,j,k Vector Calculus by Michael Corral

 

 

note distribution of this is under the GNU free documentation license which gives permission to copy paste portions

 

one has to be careful on copyright notices lol

 

Anyways here you go Schwartzchild with the QFT treatments... including the Einstein-Rosen bridge.

https://arxiv.org/pdf/1708.00748.pdf

should be enough that if you study it properly should take a month to understand just how each equation is derived. (to properly understand it) when your not that familiar with QFT or vector calculus...

Yes, where the Langrangians agree with the coordinates used upon density in four current like I had thought you were gonna do which is why I took the equation then stuck matter density for that rather than moment of inertia being for that in L form when the Hubble's Constant was inserted.

∇E_b(x,y,z,t,ω,M,I,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_s² + m_k²)^(1/2)/2M_b) + (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (Λ/3))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where ρ(x,y,z,t) = r(x,y,z,t), I assume the function itself is φ being d³p

Edited November 26, 2017 by Vmedvil

[Vmedvil]

1 hour ago, Vmedvil said:

Yes, where the Langrangians agree with the coordinates used upon density in four current like I had thought you were gonna do which is why I took the equation then stuck matter density for that rather than moment of inertia being for that in L form when the Hubble's Constant was inserted.

∇E_b(x,y,z,t,ω,M,I,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_s² + m_k²)^(1/2)/2M_b) + (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (Λ/3))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where ρ(x,y,z,t) = r(x,y,z,t), I assume the function itself is φ being d³p

So, I am missing a coordinate 

∇E_b(x,y,z,t,ω,M,I, φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_s² + m_k²)^(1/2)/2M_b) + (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (Λ/3))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Then Λ =

Edited November 26, 2017 by Vmedvil

[Mordred]

Your missing more than that.

Are you familiar with performing dinensional analysis via replacing each term with the SI unit to test that the lefthand side does equal the RHS of the equal sign?

 

https://www.google.ca/url?sa=t&source=web&rct=j&url=http://web.mit.edu/2.25/www/pdf/DA_unified.pdf&ved=0ahUKEwitheutlNzXAhUB8mMKHZnJBjgQFggdMAA&usg=AOvVaw3hTYBubBOwEQX7596gEKFR

Test the viability of your equation via the procedure in that link. Make sure your LHS equals the RHS

 

[Vmedvil]

1 hour ago, Vmedvil said:

So, I am missing a coordinate 

∇E_b(x,y,z,t,ω,M,I, φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_s² + m_k²)^(1/2)/2M_b) + (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (Λ/3))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Then Λ =

Ya, there is a problem with using that for the Cosmo k and p do not equal zero that would be devoid of matter, but it does = k

∇E_b(x,y,z,t,ω,M,I, φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_s² + m_k²)^(1/2)/2M_b) + (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_s²))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where now I am going to fully transform ω_s to k_s² + m_k² because of that.

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_s² + 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_s²))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where it does literally say that it is caused by Spin/Curvature DE.

and Gaussian Curvature = 1/r²  in a sphere, which is k₁k₂ = K = k²

Where K = R_ab/g_ab

Where R_ab = kT_ab

Where this is about to go through a transform. 

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_sT_ab./g_ab) + 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_sT_ab/g_ab ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where 

Now Another Transform.

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_sG_ab./g_ab8π) + 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_ab/g_ab8π ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

(a,b) >(u,v) transform and 

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_sG_uv./g_uv8π) + 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_uv8π ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Edited November 26, 2017 by Vmedvil

[Vmedvil]

1 hour ago, Vmedvil said:

Ya, there is a problem with using that for the Cosmo k and p do not equal zero that would be devoid of matter, but it does = k

∇E_b(x,y,z,t,ω,M,I, φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_sω_s(k_s² + m_k²)^(1/2)/2M_b) + (((8πG/3)((g/(2π)³)∫(p² + m_p²)^(1/2)(1/e^{((E - μ)/T)}±1)d³p) - (k_sC²/R_s²) + (k_s²))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where now I am going to fully transform ω_s to k_s² + m_k² because of that.

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_s² + 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_s²))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where it does literally say that it is caused by Spin/Curvature DE.

and Gaussian Curvature = 1/r²  in a sphere, which is k₁k₂ = K = k²

Where K = R_ab/g_ab

Where R_ab = kT_ab

Where this is about to go through a transform. 

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_sT_ab./g_ab) + 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_sT_ab/g_ab ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

Where 

Now Another Transform.

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_sG_ab./g_ab8π) + 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_ab/g_ab8π ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

(a,b) >(u,v) transform and 

∇E_b(x,y,z,t,M,I,k,φ,ρ,m) = ∇(1/((1-(((2M_bG / R_s) - (I_s(k_sG_uv./g_uv8π) + 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_uv8π ))^1/2(ΔKiloparsec)))²/C²))^1/2))M_bC²

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)

Edited November 26, 2017 by Vmedvil