# Towards Ideas on a Quantum Theory of Gravity

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

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1. ### GeonRegistered Member

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And then when you are finished with that, you can read up on Carrols notes, who even discusses the commutator relationship in general relativity. Note what he writes

''The antisymmetry of $R^{\rho \sigma}_{\mu \nu}$ in its last two indices is immediate from this formula and its derivation.''

https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

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3. ### NotEinsteinValued Senior Member

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1,131
You know, if you had just done that yourself and used his notation, we wouldn't have had to go through all of this.

You started working in 2 dimensions, because "the curvature of a sphere is 2-dimensional". This is of course a huge thing: even if the curvature of a sphere-surface is that, the embedding space is not in general, so you setting the dimensionality of your metric to that totally changes what you're doing. All further derivations are not directly relevant for our universe, because we don't live a 2-dimensional one.

You then calculate $\mathbf{g}^{\theta\theta}$, $\mathbf{R}^{\theta\theta}$, and $\mathbf{R}_{\theta\theta}$.
Next, you calculate the Ricci scalar, by contracting the indices on the Riemann curvature tensor:
$\mathbf{R}g =$ ... $= \frac{2}{r^2}$

(See, this is something I can follow. Why can't you just point me to that derivation you posted earlier, in a different place?)

You seem to have missed a couple, dealing with the time-coordinate, etc. So I assume those were deduced to be zero?

And when you finally reach the first equation of this thread, you are not stating the tensor G, but you're giving the expression of the tensor G contracted over the indices! So the bold G in that equation s not a tensor after all: it's a number. Now at least that makes sense.

I'll go over everything again with this new insight on my end, and see if this helps me understand the rest.

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5. ### GeonRegistered Member

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190

You are cherry picking things and making them statements about my work which simply isn't true... such as this

''You started working in 2 dimensions, because "the curvature of a sphere is 2-dimensional". This is of course a huge thing: even if the curvature of a sphere-surface is that, the embedding space is not in general, so you setting the dimensionality of your metric to that totally changes what you're doing. All further derivations are not directly relevant for our universe, because we don't live a 2-dimensional one.''

How do you make this stuff up, nowhere in the work have I stated a two dimensional case extends to anything.... !!!!

The sphere was applied to a geon model, has nothing to do with the universe. You're actually wasting my time.

edit* and yes, you have been told three times now I dropped a sphere model for an investigation into a more general case involving christoffel symbols.

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7. ### NotEinsteinValued Senior Member

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Yes, of course I'm cherry picking, I'm only asking about things I don't understand in your derivations. That you then explain them with even more confusing or seemingly contradictory things doesn't help.

For example, you said that the bold G was a tensor, while it really is a scalar. Check it out:
Post 61: I said that bold G cannot be a tensor, and I asked you about it.
Post 62: You start adding indices, clearly signaling it's a tensor (because tensors have indices, scalars don't).
Post 63: You add indices to the middle term, making it a tensor as well.
Post 64: I (among things) ask if indices need to be added to other terms as well; especially r.
Post 65: I re-inforce that bare indices cannot signify scalars.
Post 66: This now makes sense:
I will admit, I should have read that better, and understood what was going on.
Post 67: You agree with me that you were talking about tensors. (Remember that I asked about bold G, as used in the first equation of the opening post.)

Can you understand that I came to the conclusion that you were calling that bold G a tensor? (The first two sentences of post 74 certainly didn't help.)

Anyway, water under the bridge. Let's move on!

$\frac{2}{r^2} \approx \frac{1}{\Delta L\Delta t}$
r is the radius of your spherical surface, and thus some (fixed?) number. What are $\Delta L$ and $\Delta t$? Differences of length and time of what, with respect to what? And why do they relate to r in that manner?

(Also note that when "finding your inspiration" in material from other sources, you don't copy the mistakes in them too. The ones I found are (hopefully) inconsequential, but you should still have spotted and corrected them!)

I'm sorry, it's just when you said:
I thought you were talking about, you know, "spacetime". That thing in our universe. Next time please clearly indicate you're not working with our spacetime, but some hypothetical one.

Additionally, since your goal is to describe gravitational wave-bubbles, you are planning to extend it to our universe. Which is impossible, as I've already pointed out, so if you were never planning to do that, why even start with this approach in the first place?

I know, I know, it's not relevant anymore, but why even bring it up then? I mean, the very line of text under the first formula in the first post basically says: "disregard all that; it's irrelevant". You brought it up, so now you oughtn't complain I'm talking about it!

The wonderful thing about maths is that it has to be correct, even if you've dropped the model it describes. All I'm doing is trying to understand the maths you're using; I haven't even started interpreting your model yet. And when I do, I'll be sure to use the more general case involving christoffel symbols.

8. ### GeonRegistered Member

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190
The geon model was not a model of spacetime, you are confusing the early attempt as a model for gravity in a general and fundamental sense. It wasn't... I started from the idea of an uncertainty principle which would allow a deviation of curvature from some [given] wavelength/radius. The idea I had for quantum gravity was different, I looked at the spacetime uncertainty relationship as an antisymmetric tensor, or part of a tensor with antisymmetric parts.

9. ### NotEinsteinValued Senior Member

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1,131
Yes, and now that I've found all the material you didn't link in your first post, it makes a lot more sense. I figured out that the length and time you're using for the uncertainty are the size of the bubble.

Anyway, now that that's all cleared up, I'll start reading about your new approach! (Or more precisely, I'll do that some other day; it's getting late here.)

By the way, is it ok if I call you Reiku? I mean, seeing as you are him; why deny it? Be proud of who you are, Gareth!
(Oh, and don't worry: I'm not going to report you. @Mods: I do have evidence for this assertion; just ask me if you want it.)

10. ### GeonRegistered Member

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190

No you won't you created this account this other day, only to have some kind of ability to come in here and start arguing with me. You have no intentions studying the rest of the work, you struggled listening to me in the first place which would have made your day a lot easier.

11. ### NotEinsteinValued Senior Member

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1,131
Even if that's true, wouldn't that suggest I'd keep reading your work? Also, doesn't the fact that your first post is so inadequate that it leaves this much room to argue clueing you in as to your lack in clarity in your communications?

I was listening, just not understand what you were saying. If you hadn't completely mislead me (see the start of post 84), everyone's day would've indeed been a lot easier.

Reiku, I can't help it if you are unable to explain basic terminology to a person that's obviously confused by you leaving out the entire start of your derivation. Heck, even James R pointed this out in his post 20! And your inability to straightforwardly state what variable means what isn't helping either.

And just you wait, I will continue studying your work. (Unless this thread gets locked, because then there'd be no place for me to reply.)

12. ### GeonRegistered Member

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190
You didn't even qualify reading the work correctly first time around, what makes you think it will be any different second?

''And your inability to straightforwardly state what variable means what isn't helping either.''

Errr... I think if you check back when you asked me what the ''variables'' meant, I answered you. It was you who got ''confused'' as you keep saying. You didn't get confused at all, you thought you spotted a mistake and you were wrong.

13. ### NotEinsteinValued Senior Member

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1,131
I could've read it a thousand times: you not defining your terms and then using a non-familiar (as least to me) notation made sure I stood no chance.

Post 84 already described the process we had to go through for G, so I've already checked back, and demonstrated this statement to be wrong.

But let's indulge ourselves. Let's check back when I asked about r (post 50).
In post 54 you say it stands for radius.
In post 56 I asked what radius r is referring to. (no answer to that one!)
In post 64 we're in the middle of the "everything is a tensor"-phase. I point out r cannot be a radius if it's a tensor.
In post 66 you claim that you wrote $(\Delta r^2)^{-1}$, which isn't the case for the first formula of the first post, where r gets introduced. (You don't answer the posed problem of r being a tensor.)
In post 69 I reiterate that if r is a tensor, that would violate everything we know about radii.
In post 71 I summarize the current situation.
In post 77 I already posted a simplified version of this list.
In post 82 I finally figure it all out, after having found the start of your derivation elsewhere myself.

So no, you didn't answer me what I asked you what r stood for. In fact, your "answers" lead me even further astray.

I did get confused, because your answers were confusing. Why was it too much to just explain you were talking about a spherical surface, define r as its radius, and then not change everything into a tensor?

And even if I didn't spot a mistake in the equation, your inability to correctly answer the questions does strongly suggest you at the very least don't understand how that particular equation works.

14. ### GeonRegistered Member

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190
You're a waste of time. Also, a troll, who created a thread to get a free pass at trying attack me, or make statement about physics without his other account being tarnished... something along those lines.

Last edited: Sep 4, 2017
15. ### NotEinsteinValued Senior Member

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1,131
And you've wasted mine when you mislead me with your wrong answers.

Which thread are you referring to?

First of all, "his"?

But you're right that one of us is a sockpuppet of a banned member. Go on then, I invite you to have the admins figure out which one.

Keep on thinking; you'll get to the truth eventually.

Now, as for my promise to look at the rest of your work. Let's skip the remaining questions I have about the first equation of the first post, and focus on what's stated next.

I'm not sure that a Christoffel symbol has a derivative, but I'm pretty sure it's not the covariant derivative. But there certainly is a connection between the two, so I'll just mark this up to sloppy language. Let's jump ahead to the formula.

The Christoffel symbols do not depend on some coordinate x, so right of the bat, this expression is non-sense. Note that x isn't defined at all; I'm just using the mainstream interpretation of its use. The second term contains two Christoffel symbols "applied to each other", so that probably means two Christoffel symbols and three metrics to up-/down-stair the indices. It is never explained what the significance of this expression is, so we can skip interpreting it.

As explained earlier, this formula is garbage. Its first term is a number, its second term appears to be a tensor, and the third term is again a number. Objects a and b are not explained, so it is possible these are normal subscripts, and not indices. If that's the case, the second term could be a number, but it's unclear what a and b signify. In addition to that, l is undefined, and the capital X is also undefined. In other words, this formula is meaningless, because the second and third terms have undefined meaning.

Note as well that the grammar of the second part of the sentence is unclear. "rules" is plural, but only one formula is given. Are a and b indices after all? Additionally, the word "implied" doesn't grammatically fit, but this could be a simple typo/mistake.

The "they" refers back to "the derivatives of the field and between the field itself", neither of which shows up in the formula. But perhaps that "they" is another typo?

So to conclude: once again we're missing crucial information (both definitions and derivation steps), and we have to conclude it's impossible to understand what is being meant.

It is unclear (to me) how this follows from the formula. I see nothing that deals with commutation in it, let alone some property of the gamma-matrices.

And that's a commutation of the covariant derivatives. It appears small x could be used as an index, but that would make its meaning in the expression we saw earlier even more confusing. It's thus safe to assume it's a coordinate instead, and that we probably are working with Cartesian spatial coordinates.

Note that while the text talks about a "relationship", the stated expression is nothing of the sorts, as it doesn't relate anything to anything else.

The definition given for the curvature tensor is incorrect; see https://en.wikipedia.org/wiki/Einstein_tensor#Definition . Let's assume this is typo.

In the formula, a previous unseen usage of R appears. It carries two subscripts, one a coordinate and the other index zero. They are separated by a comma, strongly suggesting they are not used as indices. In other words, a new and previously undefined R that's not the Ricci scalar, Ricci tensor, or the Riemann tensor appears out of nowhere. Since it appears to go unused for the rest of the post, it's safe to ignore this confusingly named unknown variable.

The second term is the commutation relationship we're looking at, the third is that same expression but then expanded.

However, no derivation of the fourth term is given. If we look back up above, we find two similar expressions, but neither involves the covariant derivatives. In fact, in the first instance (first formula) we get a single number as output, not a variable, so it's most similar to the second formula (which would make the most sense too, seeing as the first formula is related to the dropped bubble model). However, the second instance uses Delta X, where X is still undefined (as is l, by the way). Even if those can be replaced with the covariant derivative, we're still an inverse off.

Conclusion: crucial derivation steps are missing, and the last inequality of this formula cannot be verified and appears to be plainly wrong.

This raises the question what the "desired inequality" actually is. What are we trying to accomplish here? Remember, we're looking for commutation relationships; the only inequality that has been introduced was in that formula earlier that cannot be understood as it is currently written.

But hey, let's say it's just me, assume this step to be true, and continue on.

This doesn't look unreasonable, so I didn't verify it. Note though that the Christoffel symbols are written without two of their indices.

16. ### NotEinsteinValued Senior Member

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1,131
Aha, this is where the derivative of the Christoffel symbol enters the game! However, what is missing due to the notation here is that the derivatives are not taken on just the Christoffel symbols, but on some object that is unstated in this expression. Remember those missing indices I mentioned earlier? That are supposed to be contracted with a vector field that the covariant derivative is operating on. In other words, this unexplained rewritten formula is not taking derivatives of the object it's operating on into account, making it very suspect as to its correctness.

Let compare this with Reiku's own provided source, https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html About halfway down (before eq 3.65), Carroll says "The actual computation is very straightforward. Considering a vector field $V\scriptstyle \rho$, we take", followed by a derivation. But note that Carroll gets a different answer than we have here.

If we assume the derivative-of-the-Christoffel-symbol clash is just a notational issue, we're still left with:
- A missing minus sign
- A missing Christoffel symbol pair (and its minus sign)
- A missing second term

In other words, this formula is plain wrong, as demonstrated by Reiku's own source. But, let's assume this is just some tex-mixup, and continue!

This is not just the Riemann tensor. Corroll is quite clear that only part of this expression is the Riemann tensor. Reiku mistakenly dropped various Riemann and all non-Riemann tensor contributions. In other words, this sentence is unambiguously incorrect. The formula looks like an inverse of an earlier formula, but in that case, the inequality is the wrong way around. Let's assume this is (another) typo, and soldier on!

Let's focus on the formula. We start with the Einstein tensor, with its indices explicitly stated.
The second term is the definition of the Einstein tensor.
The third term is an approximation that we haven't seen before. We recognize the commutator, but nowhere was that argued to be approximately equal to the Einstein tensor. I can only guess that this follows from the "and is just the Riemann tensor" statement (which is wrong). Worse, what was a tensor now suddenly has two unevaluated covariant derivatives! This is of course mathematically impossible.
The fourth term is just an expansion of the third term.
The fifth term follows from an earlier formula.
The sixth term is straight up non-sense. We've switch from a tensor to a scalar; no indices are present, and they cannot be, as all the terms in the expression are scalar-valued.
The seventh term follows from that totally-no-really-irrelevant formula at the start of the post. Let's just assume it's true, even though it looks like the Heisenberg uncertainty principle and we haven't been doing any quantum mechanics up to this point.

So in conclusion, what do we have? We've approximately related through an inequality the Einstein tensor (with indices noted) to a scalar value. Since a tensor is not a number, let's interpret this the mainstream way (I don't see any alternative?): we aren't talking about the tensor itself, but the value of all the elements in it. What consequences does this have?

Immediately obvious should be the lack of any non-positive values (term seven is a positive constant scalar, and each element is larger than that). If we assume a zero cosmological constant, we can translate the Einstein tensor directly into the stress-energy tensor. (See https://en.wikipedia.org/wiki/Einstein_field_equations#Mathematical_form ) All its elements will be positive too. So we end up with a spacetime where even empty space has a minimum energy density. We end up with a constant momentum density and momentum flux, even in empty space. We end up with a space that is always exerting pressure. But how large are these contributions? Well, I cannot tell, as the units don't match up. Term seven evaluates to $1.896\times 10^{-19}\dfrac{m^8}{s^6}$, which is no energy density units I've ever heard of. We can only conclude that various constants are missing in the last term of the formula, even though the presence of $c$ suggests they were put back in, so we can't trust the value that's given for term seven as it is written.

Note that playing with the cosmological constant won't help you much, as by compensating for one term, you're making the others worse due to the minus sign(s) in the metric.

So what was the goal of all this? Let's look ahead to the first bit of the second post...

Maybe, maybe not, but the question bares no relation to the derivation we just went through. No particle has been postulated; we're still working with empty spacetime. Additionally, there's nothing that suggests any kind of self-gravitation: all we've done is given the elements of the Einstein-tensor a required minimum value. So this statement doesn't help us here, and we must conclude that part two is about something else, and part one stands on its own.

And as it stands, we've got a huge pile of probable typo's, undefined terms, copy-paste mistakes, and a bunch of incorrect statements. We're missing huge steps in the derivation, and thus it looks like assumptions are made left and right. In the end, we end up with a formula whose derivation is so obscured by all these issues, we can only take it as a postulate or assumption. Especially note how the critical "uncertainty relation" was just assumed to be true in the first place. Since all this post was trying to accomplish was relating that uncertainty to something metric-y, the entire post might just as well have succinctly stated:

"Let's assume: $G_{ij} = [\nabla_i,\nabla_j] \approx [\nabla_i,\nabla_j] \geq \frac{c^3}{G \hbar}$."

As the post currently stands, this is fully equivalent to it.

17. ### GeonRegistered Member

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190

I really don't have time for a troll-account which you created fresh the other day to come here, trolling. I am way moved on from all that if you are still struggling with anything sorry, I can be no help from now on. I actually want to move on and discuss my work and where it is heading, not have endless debates with you over the simple matters that I cannot be bothered going over a several time.

18. ### GeonRegistered Member

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190
On Possible Quantum Bianchi Identities

Since we used an antisymmetric object identical to the antisymmetric indices of a Reimann tensor, the Riemann tensor will still be symmetric under an interchange of its first two indices with its last two. We have argued that the Riemann tensor $R^{\rho}_{\sigma\ [i, j]}$ where we use a notation to denote the antisymmetric part $[i,j]$. In a valid approach, we have also assumed dynamics satisfying:

$R^{\rho}_{\sigma\ [i, j]} \ne 0$

Bianci idenities are to be studied now - they are related to the vanishing of a Reimann tensor in the sense it is related to the vanishing of the covariant derivative - the quantum Bianchi identity is to assume there is a quantum, non-zero interpretation of the commutation of the two connections,

$[\partial_i, \partial_j] \ne 0$

In general relativity, this relationship is usually given as zero - its not difficult to understand, why if we are talking about quantum deviation from a classical theory, why a non-zero theory may be important.

We can argue (maybe) that only points are unphysical and lead to the vanishing of the derivatives
$R^{\rho}_{\sigma\ [i,j]} = 0$, or even only for classical theory, or both. Of course, the statement

''if it is true in one coordinate system it must be true in any coordinate system,''

Is hard to argue with - but there is more to play with here than just coordinates - by assuming a spacetime relationship at quantum scales by satisfying the spacetime noncommutativity - so scale is an important factor here when talking about gravity, and the spacetime uncertainty needs to be a phenomenon regardless of the coordinate of the system! When we think about Von Neumann operators and phase space, we expect a deviation from the classical way of thinking anyway...

The spacetime relationship

$\Delta x \Delta t \geq \frac{1}{\ell^2}$

is a model of discretized spacetime - leading to a Planckian spacetime dynamic. The non-zero value of the derivatives implies we have a description of gravity at a quantum scale related to fundamentally to the structure of space. The vanishing of the Riemann tensor is actually related to the same idea connected to the contraction of the Bianchi identities, which are only zero if they permit a symmetric theory of gravity.

If we think about the vanishing of a metric in terms of the vanishing of an action, the action vanishes because it is invariant under general coordinate transformations - that is, general covariance means there is an invariance of the form of physical laws under any arbitrary differentiable coordinate transformations. But I note again, coordinates in a quantum domain to coordinates in the classical domain, may not change the coordinate but definitely the situation. - notably positions are affected by momentum in phase spaces - something classical space time and the classical objects inside of it are so different.

19. ### GeonRegistered Member

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190
By the way, you are so incompetent with the math, you think I've left terms out of my Riemann tensor, which is laughable.

You should have trusted your instincts, yes it is notational. You speak of two gamma matrices minus another (missing two)

Nothing missing at all, you just can't read commutator language.

20. ### GeonRegistered Member

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190
Oh and be careful with Carrol, if you are having easy problems like this, you might get tripped up with the torsion tensor that enters some of his equations. That will completely baffle you if you can understand the rest.

21. ### GeonRegistered Member

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190
For anyone struggling with notation, like Einstein-the-troll-account clearly is here- here is the full form

$-\frac{\partial \Gamma_j}{\partial x^{i}} + \frac{\partial \Gamma_{i}}{\partial x^{j}} + [\Gamma_i \Gamma_j]$

is just

$-\frac{\partial \Gamma_j}{\partial x^{i}} + \frac{\partial \Gamma_{i}}{\partial x^{j}} + \Gamma_i \Gamma_j - \Gamma_j \Gamma_i$

22. ### NotEinsteinValued Senior Member

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1,131
No, you've left out terms in an equation, some of which are part of the Riemann tensor, and some aren't. But I guess you know better than Carroll, who you brought up as an expert in the first place. Also interesting that you can't even show where these terms are hiding. Makes you wonder...

Oh, and "commutator language" isn't anything that exists, so yes obviously I cannot read it.

Me struggling with Carroll? You can't even copy his equations correctly!

So when you "[pulled] it out of its differential notation form", there's a bunch of typo's in that formula. Good, glad you are now admitting that.

23. ### GeonRegistered Member

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Are you for fucking serious? Do you not know what differential notation is?

There are no typo's here, just one mistake, and that's you. No wonder you created a fake account, I'd be really embarrassed about now.