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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.