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An Attempt Of Quantum Gravity


Dubbelosix

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Of course I am Geon. I even acknowledged you there two days or something before I was banned. Why are you asking retarded questions like this?

You, under the pseudonym Geon, were banned (again) from sciforums for being a sock of Reiku: http://www.sciforums.com/threads/qm-randomness.159457/page-15#post-3474152

 

So report me by all means, Reiku, but I have produced supporting evidence for my assertion, in accordance with forum rules, that you are he.

Edited by exchemist
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Why are you lying?

 

About three people, origin, you and someone else kept going about calling me Reiku.

 

James is a moderator there, did he call me Reiku? He was more concerned about the difficulty of the work. Kittamaru, after reading the posts by origin and Einstein, concluded I had to be this Reiku. There was no evidence presented. 

 

It was only you and a few others and that Kittamaru (who was more than biased in closing down my posts) then only banned me because I talked back. Then when I was banned, he went around leaving posts at sciforums calling me Reiku.

 

 

It's pretty sad and pathetic - the site is **** anyway. Seen the quality of the posts? Sciforums? Pseudo-sciforums more like. 

Well, it's your word, then, against the moderation of sciforums. I have made my choice. But by all means, carry on with your knitting. Nobody reads it, any more than they do at nakedscientists. 

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And of course, it is difficult and not obvious for this reason:

 

[math]\nabla \cdot \nabla = \Delta[/math]

 

Which if it was a Laplacian, has dimensions of 1/x^2. 

The Laplacian ("del squared") is a differential operator and is dimensionless, just as d/dx or d²/dx² are. Dimensions belong to physical quantities.

Edited by exchemist
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  • 2 weeks later...

Well, tell me if you find a graviton in there, both my Quantum Schwarzchild Metric for Blackholes and Quantum Bell Paradox Solution have gravitons in them when fully solved out, which is my experience in Q relativity but I don't know much about the method you are using, I write it as a Laplace manifold set across Special Relativistic Plank Lengths making each interaction as change in the manifold structure. In any case, Good Luck!   

 

Bell Paradox Solution Quantum Relativistic Movement and Acceleration 

 

∇Ek(x,y,z) = ∇(1/(1- (u2+a2t2)/C2)1/2)MC2

 

u = Initial velocity,

a = acceleration,

m = rest mass

C= Speed of Light in vacuum

t = time taken in 0 Velocity observer clock

 

This being shifting manifold structure per interaction with energy per String. 

 

∇ΣEf(x,y,z) = ∇E1(x,y,z) (-/+) ∇E2(x,y,z) (-/+) ∇E3(x,y,z) ........ (-/+) ∇Ef⇒∞(x,y,z) 

 

 

 

Note: This is the version I constructed for the bell Paradox, so it only deals with Relativistic Kinetic energy so far on the string level, but I think you could put your ΔE in that. 
 

Then here is Spin and Gravity for a BH in my modified Schwarzchild Metric 

 

∇Eb(x,y,z) = ∇(1/((1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2))MbC2

 

Original Laplace Prime Form.

 

∇'(x',y',z') = ∇(1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2

 

 

You can play with those equations all you want for Quantum Gravity, Quantum Relativistic Spin, and Quantum Relativistic Motion/Acceleration. 

Edited by Vmedvil
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No gravitons in my universe I am sorry to say.

 

Gravity is a pseudoforce and gravitons are superlfuous in general relativity.

 

Well, I guess it should be wrote either way being that Δ = ∇ * ∇ , EΔ = E

 

So, technically ΣEf(x,y,z) = EfΔ(x,y,z)

Edited by Vmedvil
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When we speak about gravitons, we tend to generally think about a metric perturbation of the form:

 

[math]g = g + h[/math]

 

where [math]h[/math] is the metric fluctuation that can be considered as the graviton.

 

Isn't h just joule time?  then g is Newtons per kilograms. Well, I got this to Joule * time2 so are you saying that should be only time. 

 

Eb(t,ω,R,M,I) = (1/((1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2))MbC2

 

Well, removed the Laplace operator from both sides then like the cosmology post for Static Non Moving objects through space. This situation not caring about exact position or number of many objects with many parts.

 

1200px-Coord_system_CA_0.svg.png

 

Eb(t,ω,R,M,I) = (1/((1-(((2MbG / Rs) - (Isωs2/2Mb))2/C2))1/2))MbC2

 

ΣEb1Δt21⇒∞(t,ω,R,M,I) = Eb1Δt2 + Eb2Δt22+........... + EbΔt2

 

This would be Graviton Frequency for an object that is exiting or entering the universe or Teleporting not moving through time linearly that extra dimension of time has been added into this equation. Those pesky time travelers, Inter-dimensional travelers, or Teleporting objects being the "Blink". This also handles "Quantum Erasure" destruction of Energy-Mass from the universe or "Quantum Fluctuation" generation of Energy-Mass into the universe from whatever source.

 

maxresdefault.jpg

 

Σg = ΣEb(t,ω,R,M,I) /hM

 

 

If your object does move through time linearly it would be.

 

Σg= ΣEb(t,ω,R,M,I)/hM

 

which does satisfy this equation either way for fluctuation caused by gravitons from one object with one part or one object with many parts.

 

 

Σg2 = gravitons received from external object with one part or object with many parts.

 

g1 = gravitational acceleration received from self. 

 

Σg = g1Σg2, being sum of all gravitational acceleration, from self and object with one part or object with many parts.

 

Σg = g1Σg2 =  g +  ΣEb(t,ω,R,M,I) /hM = g +ΣEb(t,ω,R,M,I)/hM

 

If you had an object that was non static moving object through space then the equation would be. 

 

Eb(t,ω,R,M,I) = (1/((1-(((2MbG / Rs) - (Isωs2/2Mb)+V)2/C2))1/2))MbC2

 

Then for Non static many moving objects with many parts would be. 

 

Eb(t,ω,R,M,I,x,y,z) = (1/((1-(((2MbG / Rs) - (Isωs2/2Mb)+V)2/C2))1/2))MbC2

 

 Σg2  = gravitons received from external many objects with one part or many objects with many parts.

 

g1 = gravitational acceleration received from self. 

 

Σ∇g = g1Σg2 , being sum of all gravitational acceleration, from self and many objects with one part or many objects with many parts.

 

Σ∇g = g1Σg2 =  g + ΣEb(t,ω,R,M,I,x,y,z)/hM = g +ΣEb(t,ω,R,M,I,x,y,z) /Mh

 

Δg = Σ∇g 

 

Then

 

g1 = Δg -  Σg2  

 

 Σg2 = Δg - g1

 

So, this was already calculated into the original equation. 

 

I think these were the forms you were looking for unless g was supposed to be g(u,v) but I took those as Gravitational acceleration.

Edited by Vmedvil
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