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

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Furthermore.

Which I using that form. k= (K(8π/C2)) , K = ((-Duv(g) -ktuv)Tuv-1)

'(x,y,z,t,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ /(2Erest/C2)) 3a =1 (d2/d((C2/Erest)Ni = 1 MiRi)2) + (1/2)3a,β = 1  μaβ(PΠa)(Pβ Πβ) + U - (ħ2/2)3N-6s=1(d2/dq2) + V)((|(Log(DgDaDψDφ-W)(((2ħGC2))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμ(((Lghost QE  - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)ψi +(ψ-bar)iLVijφψjr + (aji) - (μ2((φ-Dagger)φ) + λ((φ-Dagger)φ)2)/-(((Lghost QE   - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)2)|)-e2S(r,t)/h)) - ((Erest/C2)ωs((((-Duv(g) -ktuv)Tuv-1)(8π/C2))2)1/2 + (S/ (((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp ))))Rs2/2))) / (ħ2/2(Erest/C2))))1/2(((1-(((2(Erest/C2)G / Rs) - (Isωs((((-Duv(g) -ktuv)Tuv-1)(8π/C2))2)1/2 + (S/(((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp )))))/2(Erest/C2))+ (((8πG/3)((g/(2π)3)∫(((Erelativistic- Erest2 / C2) + ((Ar(X) + (ENucleon binding SNF εμ/mu) - Ar(XZ±)/Z) / mu)2)(1/2)(1/e((ERelativistic  - μchemical)/TMatter)±1)(ħω + ħωs) - ((ksC2)/ Rs2) + (((((-Duv(g) -ktuv)Tuv-1)(8π/C2))2)1/2(ΔKiloparsec)))2/(C2)))1/2)

Edited by Vmedvil

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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 Tuv but I could have, stopped at Guv

Oh, I see what you are whining about that is for gab, 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 gab because I took it on oh Mordred must know what he is talking about but this version is worse.

(1/2)R = ((-8πGTab'Λgab + Rab) * -gab-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) $\frac{8\pi G}{c^4}$ is only central potential force there is examples where this is not the case. Edited by Mordred #### Share this post ##### Link to post ##### Share on other sites 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.

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No indices

Einstein summation

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

Edited by Mordred

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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.

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

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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  ks = K = (1/2)R Co-variant

Edited by Vmedvil

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So is your equation valid when

$g_{\mu\nu}=g_{\alpha\beta}=\frac{dx^{\alpha}}{dy^{\mu}}\frac{dx^{\beta}}{dy^{\nu}}$

Metric tensor $dx^2=(dx^0)^2+(dx^1)^2+(dx^3)^2$

$G_{\mu\nu}=\begin{pmatrix}g_{0,0}&g_{0,1}&g_{0,2}&g_{0,3}\\g_{1,0}&g_{1,1}&g_{1,2}&g_{1,3}\\g_{2,0}&g_{2,1}&g_{2,2}&g_{2,3}\\g_{3,0}&g_{3,1}&g_{3,2}&g_{3,3}\end{pmatrix}=\begin{pmatrix}-1&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$

Which corresponds to

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

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

Flat space $\mathbb{R}^4$ 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

$ds^2=-c^2dt^2+dx^2+dy^2+dz^2=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$

$\eta=\begin{pmatrix}-c^2&0&0&0\\0&1&0&0\\0&0&1&0\\0&0&0&1\end{pmatrix}$

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

Edited by Mordred

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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 by Vmedvil

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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

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

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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 by Vmedvil

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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

the second link is the ODE as per QM QFT.

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

Edited by Mordred

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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

the second link is the ODE as per QM QFT.

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-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2

is like saying,

d/dx(1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2 + d/dy(1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2 + d/dz((1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2)

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

Edited by Vmedvil

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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

$\partial_i,j$ 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 by Mordred

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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 by Vmedvil

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

Edited by Mordred

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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,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ /(2Erest/C2)) 3a =1 (d2/d((C2/Erest)Ni = 1 MiRi)2) + (1/2)3a,β = 1  μaβ(PΠa)(Pβ Πβ) + U - (ħ2/2)3N-6s=1(d2/dq2) + V)((|(Log(DgDaDψDφ-W)(((2ħGC2))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμ(((Lghost QE  - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)ψi +(ψ-bar)iLVijφψjr + (aji) - (μ2((φ-Dagger)φ) + λ((φ-Dagger)φ)2)/-(((Lghost QE   - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)2)|)-e2S(r,t)/h)) - ((Erest/C2)ωs(((8πGTab' + Λgab  - Rab) * gab-1))1/2 + (S/ (((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp ))))Rs2/2))) / (ħ2/2(Erest/C2))))1/2(((1-(((2(Erest/C2)G / Rs) - (Isωs(((8πGTab' + Λgab  - Rab) * gab-1))1/2 + (S/(((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp )))))/2(Erest/C2))+ (((8πG/3)((g/(2π)3)∫(((Erelativistic- Erest2 / C2) + ((Ar(X) + (ENucleon binding SNF εμ/mu) - Ar(XZ±)/Z) / mu)2)(1/2)(1/e((ERelativistic  - μchemical)/TMatter)±1)(ħω + ħωs) - ((ksC2)/ Rs2) + (((8πGTab' + Λgab  - Rab) * gab-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 by Vmedvil

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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.

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

"Lie Groups and Diﬀerential Equations"

Edited by Mordred

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

L2ghost QE  = L1ghost QE

All values equal 2 or 1, besides constants.

2'(x,y,z,t,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ /(2Erest/C2)) 3a =1 (d2/d((C2/Erest)Ni = 1 MiRi)2) + (1/2)3a,β = 1  μaβ(PΠa)(Pβ Πβ) + U - (ħ2/2)3N-6s=1(d2/dq2) + V)((|(Log(DgDaDψDφ-W)(((2ħGC2))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμ(((Lghost QE  - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)ψi +(ψ-bar)iLVijφψjr + (aji) - (μ2((φ-Dagger)φ) + λ((φ-Dagger)φ)2)/-(((Lghost QE   - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)2)|)-e2S(r,t)/h)) - ((Erest/C2)ωs(((8πGTab' + Λgab  - Rab) * gab-1))1/2 + (S/ (((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp ))))Rs2/2))) / (ħ2/2(Erest/C2))))1/2(((1-(((2(Erest/C2)G / Rs) - (Isωs(((8πGTab' + Λgab  - Rab) * gab-1))1/2 + (S/(((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp )))))/2(Erest/C2))+ (((8πG/3)((g/(2π)3)∫(((Erelativistic- Erest2 / C2) + ((Ar(X) + (ENucleon binding SNF εμ/mu) - Ar(XZ±)/Z) / mu)2)(1/2)(1/e((ERelativistic  - μchemical)/TMatter)±1)(ħω + ħωs) - ((ksC2)/ Rs2) + (((8πGTab' + Λgab  - Rab) * gab-1))1/2(ΔKiloparsec)))2/(C2)))1/2)

1'(x,y,z,t,ωsp,M,I,k,φ,S,X,Z,μ,Y,q,a,β) = (((ħ /(2Erest/C2)) 3a =1 (d2/d((C2/Erest)Ni = 1 MiRi)2) + (1/2)3a,β = 1  μaβ(PΠa)(Pβ Πβ) + U - (ħ2/2)3N-6s=1(d2/dq2) + V)((|(Log(DgDaDψDφ-W)(((2ħGC2))Rs - (1/4)FaμvFaμv + i(ψ-bar)γμ(((Lghost QE  - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)ψi +(ψ-bar)iLVijφψjr + (aji) - (μ2((φ-Dagger)φ) + λ((φ-Dagger)φ)2)/-(((Lghost QE   - gfabc(δμ (c-bar)a)Aμbcc) / (c-bar)aδμca) + ig(1/2)τWμ + ig'(1/2)YBμ)2)|)-e2S(r,t)/h)) - ((Erest/C2)ωs(((8πGTab' + Λgab  - Rab) * gab-1))1/2 + (S/ (((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp ))))Rs2/2))) / (ħ2/2(Erest/C2))))1/2(((1-(((2(Erest/C2)G / Rs) - (Isωs(((8πGTab' + Λgab  - Rab) * gab-1))1/2 + (S/(((3G(Erest/C2))/2C2Rs3)(RpVp) + (GIs/C2Rs3)((3Rp/Rs2)(ωRp) -ωp )))))/2(Erest/C2))+ (((8πG/3)((g/(2π)3)∫(((Erelativistic- Erest2 / C2) + ((Ar(X) + (ENucleon binding SNF εμ/mu) - Ar(XZ±)/Z) / mu)2)(1/2)(1/e((ERelativistic  - μchemical)/TMatter)±1)(ħω + ħωs) - ((ksC2)/ Rs2) + (((8πGTab' + Λgab  - Rab) * gab-1))1/2(ΔKiloparsec)))2/(C2)))1/2)

which it is.

Edited by Vmedvil