# Towards Ideas on a Quantum Theory of Gravity

Discussion in 'Pseudoscience' started by Geon, Aug 25, 2017.

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1. ### NotEinsteinValued Senior Member

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The very first equation of the first post in this thread. It is impossible for $\mathbf{G}$ to be a tensor. Because you said:
$\mathbf{G} \approx \frac{1}{\Delta L\Delta t}\ \geq \frac{c^4}{G \hbar}$
That middle term is a number. So the left-hand side must be too. Your $\mathbf{G}$ is a number, not a tensor.

Erm, you started the topic? On a forum? Thus inviting comments/replies?

3. ### GeonRegistered Member

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Oh I am sorry, can you not deal with this, without the dummy variables?

$G_{ij} \leq (\ell^2_{ij})^{-1}$

Of course that is without saying, those indices are making a statement about the curvature being proportional to a given length squared (Planckian in fact). $G_{ij}$ is an antisymmetric tensor and is antisymmetric in the indices.

5. ### GeonRegistered Member

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That ''middle term being a number'' follows the same principle

$(\Delta X_{i} \Delta X_j)^{-1}$

edit - the inverse, because tensor has dimensions of 1/length^2

7. ### NotEinsteinValued Senior Member

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As I said, my GR is rusty, sorry about that. I didn't realize you were silently dropping the indices.

Ah, I see. One more question: are the ij-indices just the spatial ones, or do they include time?

Also, earlier you said:
I assume I need to add the a and b indices back to the first term as well? But you said:
How can r be a radius when it has indices? A radius just a number, not a vector!

8. ### NotEinsteinValued Senior Member

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You're not summing over the indices, so it's most definitely not a number.

9. ### GeonRegistered Member

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No you don't understand, its not like that. If you look carefully, what I have written in fact is

$(\Delta r^2)^{-1}$

This uses the change operator, its a deviation from some value. And yes, when we had the contracted form of $G$ we constructed the curvature of sphere - I later drop this approach for a more direct tackle on the quantum gravity aspect. What you don't realize, the equations you are citing from, was never my quantum gravity approach, they were preliminary investigations into the curvature of a sphere in the use of some deviation of curvature for a geon.

As I said, this is all dropped, you'd be having a less hard time if you read through and take my approach in the most recent form.

10. ### GeonRegistered Member

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Right... That's why I quoted you and corrected you. This is a tensor, an antisymmetric tensor. If you read the newest work and actually follow the newest construction of the work, all that will be clear.

11. ### GeonRegistered Member

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They are spatial and temporal, hence my understanding of using two connections as deviation from classical vacuum

$[\nabla_i,\nabla_j]$

This is like seeing gravity as a Von Neumann operator.

12. ### NotEinsteinValued Senior Member

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Remember when you said:
So if it's an antisymmetric tensor, so is $\frac{2}{\Delta r^2}$, and thus r is not a radius in the normal sense of the word.

Last edited: Sep 3, 2017
13. ### GeonRegistered Member

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You poor thing, you are so lost. You are still concentrating on the radius when I told you, that isn't even the right model for my quantum gravity.

You can't talk about what I am talking about now, with being stuck on page 1. You need to read on and see how the model is given now. You just don't get this bit do you?

14. ### NotEinsteinValued Senior Member

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Erm, dear SimonsCat, I'm not the one introducing an undefined variable r, then calling it a radius, and then calling it a tensor. If your notation and terminology is so far from mainstream, how can I have any hope understanding it?

15. ### originHeading towards oblivionValued Senior Member

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Since it is obvious to me that Geon is a sock and still adding nothing to the forum but noise, I guess it is time... Click

16. ### GeonRegistered Member

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Now you are lying.

I did nothing of the sort, you did. I told you we used a contracted solution of the equations, which was solved for the Ricci curvature of a sphere - I made that early solution proportional and today, have dropped that model entirely.

It's you who isn't

1) listening to me

2) making shit up as you go along

17. ### GeonRegistered Member

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You asked me early on, what G was. I told you in the context we came to understand, was an antisymmetric tensor. Since I told you, you cannot derive that understanding from post 1, you've done everything to concentrate on post 1, as if it proves some point and it never did. I explained to you what those early solutions where and why I looked into them. I abandoned that investigation for a direct look into the christoffel symbols to find the relationships.

18. ### NotEinsteinValued Senior Member

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I am literally trying to understand the first few equations of your first post. You seem to be unable to define its terms in a clear and meaningful manner. Every time you explain a bit, it seemingly contradicts the mainstream interpretation of the terminology. Without a translation manual of some kind, how can anybody have any hope of understanding this? Without a translation manual, it's quite simply word salad, seeing as the meaning of all the terms and words are unclear.

I mean, let's look at the first equation of the first post again now that we've covered some ground.
Adding the indices for the bold G back, we get non-sense. The third term is summed, and thus a number, but the first is your G antisymmetric tensor. A tensor is not a number. (Also sloppy to use the same indices for the two terms, but okay.)

I am listening, I'm just not understanding what you are trying to say because it's so non-mainstream in its usage of terms.

19. ### NotEinsteinValued Senior Member

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Earlier you said:
The Einstein tensor is symmetric. So you're so far left-field, you're not even doing GR anymore. I strongly suggest you stop using its terminology then, because it leads to the confusion I'm expressing right now.

20. ### NotEinsteinValued Senior Member

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And just to clear my name...
Let's break that down.

"I'm not the one introducing an undefined variable r,": See the first equation in post 1.
"then calling it a radius," See the last sentence in post 54.
"and then calling it a tensor." See my conclusion in post 64. True, you never explicitly called it a tensor, but this is an inescapable conclusion from your statements earlier and in post 63; see the rest of post 64.

If applying mainstream understanding of GR is "making shit up as you go along", then I'm totally guilty, yes.

21. ### GeonRegistered Member

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No no... let me break it down for you. This is the post you want if you want the correct form of the equations. I ask you, why are you going to an early investigation which didn't even take into account the christoffel symbols? The uncertainty arises as part of the antisymmetric tensor... If you want the correct approach to understand it, why are you avoiding the final, approved form?

22. ### GeonRegistered Member

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Not always symmetric. There are non-symmetric tensor theories of $G$ out there. But that's neither here nor there, since G is related to the Riemann tensor, we can certainly say the tensor $R_{ij}$ is antisymmetric in those two indices.

To understand symmetric and antisymmetric in terms of indices, I suggest you do some reading http://www.physicspages.com/2013/03/22/tensors-symmetric-and-anti-symmetric/. Especially from this part

''A higher rank tensor can be symmetric or anti-symmetric in any pair of its indices, provided both indices are either upper or lower.''

23. ### GeonRegistered Member

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As you can see we have only been concerned with two connections $[\nabla_i,\nabla_j]$. Which means we have only been concerned with an antisymmetric object.