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An Attempt Of Quantum Gravity


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#52 exchemist

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Posted 14 September 2017 - 03:38 AM

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, 14 September 2017 - 05:22 AM.


#53 Dubbelosix

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Posted 14 September 2017 - 03:24 PM

Are you having a joke....

The laplacian is not dimensionless. This is so basic. It has dim 1/over length squared

Edited by Dubbelosix, 14 September 2017 - 03:24 PM.


#54 Dubbelosix

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Posted 14 September 2017 - 03:29 PM

The square of the divergence is the laplacian. The divergence has dim 1/length

Edited by Dubbelosix, 14 September 2017 - 03:29 PM.


#55 exchemist

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Posted 15 September 2017 - 02:34 AM

The square of the divergence is the laplacian. The divergence has dim 1/length

Yes on reflection you are quite right, since x in the case of the Laplacian is specifically a spatial dimension, not just a mathematical variable. Sorry.  



#56 Dubbelosix

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Posted 15 September 2017 - 04:27 AM

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. 

 

 

You're just an angry little man. 

 

What purpose did it serve to get me banned? I was one of the only competent people at the place capable of talking about science. You see that post you linked everyone to...? Thanks for doing that, it serves the same purpose as this thread I am making now. Take a look at that thread and take a look at the constant arguing from posters about certain statements I am 100% certain about...

 

 

...Such as, an atom can be theoretically infinitely stable, so long as you ware willing to measure a particle in a particular way. Just before Kittamaru rounds it up with his nonsense, there was actually some good physics being discussed in that thread, par the one the person who couldn't get certain things into his head. That's not for the lack of me trying, as you will see.

 

Then I get immature comments from Kittamaru like ''well if you are that good, why did you do a runner with NotEinstein.'' He once again, misrepresented the actual facts. The facts was that I had spent considerable time arguing with the troll and decided eventually to put him on ignore. Had NOTHING to do with intellectual prowess, NotEinstein was about as dumb as they come at that site.



#57 Dubbelosix

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Posted 15 September 2017 - 04:28 AM

And yes, you would be right about the actual intellectual honesty of the moderation, especially Kittamaru who said, ''the work is clearly nonsense.''

 

When I challenged him to provide evidence, he couldn't. That's how honest the place really is.


Edited by Dubbelosix, 15 September 2017 - 06:33 AM.


#58 Dubbelosix

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Posted 15 September 2017 - 04:30 AM

And by the way, if no one reads my work, why does my post at nakedscientists have hundreds and hundreds of views? Even the moderator there comes in on the odd time and comments...

 

You want to pierce me so bad, are you jealous or something?


Edited by Dubbelosix, 15 September 2017 - 04:36 AM.


#59 Dubbelosix

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Posted 15 September 2017 - 04:31 AM

Yes on reflection you are quite right, since x in the case of the Laplacian is specifically a spatial dimension, not just a mathematical variable. Sorry.  

 

Well, at least you can recognize when something is stated to you as a mistake. Which is more progress than I ever made with NotEinstein. So you have one bonus point.


Edited by Dubbelosix, 15 September 2017 - 04:32 AM.


#60 exchemist

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Posted 15 September 2017 - 07:03 AM

Well, at least you can recognize when something is stated to you as a mistake. Which is more progress than I ever made with NotEinstein. So you have one bonus point.

I'm always looking to learn things. Even from Reiku! :)



#61 Dubbelosix

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Posted 15 September 2017 - 07:26 AM

Whatever.

 

Anyway, looking back, an equation was missing a term, so just stating it now - there was a missing volume term there.

 

 
[math]E = \frac{c^4}{G} \int (\nabla \Gamma)^2\ dV = \frac{c^4}{G} \int \frac{1}{R^2} \frac{d\phi}{dR} (R^2 \frac{d\phi}{dR})\ dV[/math]
 
 
This part 
 
[math]\frac{1}{R^2} \frac{d\phi}{dR} (R^2 \frac{d\phi}{dR})[/math]
 
Is just another way to write a squared product [math]\frac{d\phi}{dR} \cdot \frac{d\phi}{dR}[/math]. And of course, this is just [math]\nabla^2 \phi^2[/math]. We've stated this identity before in an equation
 
[math]\Delta E =  \frac{c^4}{G} \int (\nabla \phi)^2\ dV[/math]
 
and we also stated it has noncommutative properties in the squared terms. 


#62 Vmedvil

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Posted 28 September 2017 - 11:50 PM

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, 20 December 2017 - 03:15 AM.


#63 Dubbelosix

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Posted 04 October 2017 - 01:51 PM

No gravitons in my universe I am sorry to say.

 

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



#64 Vmedvil

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Posted 06 October 2017 - 01:11 AM

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

 

img74.png

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


Edited by Vmedvil, 06 October 2017 - 01:21 AM.


#65 Dubbelosix

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Posted 11 October 2017 - 07:13 AM

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.



#66 Vmedvil

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Posted 11 October 2017 - 12:29 PM

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

 

hyp.gif

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.

 

mixb.png

 

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, 05 January 2018 - 08:29 AM.