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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In post two, there is a part two, which explains it more in detail and summarises most of it - it details some aspect of what we are attempting to achieve. There is a better model to follow now which is one that will directly look into a quantum approach to gravity by studying how uncertainty/commutation effects the structure of a spacetime. To understand that, we need to understand why the commutation smears classical space out and into a quantum phase space.

3. James RJust this guy, you know?Staff Member

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sciforums is not your blog, Geon.

Do you have something you'd like to discuss with our membership?

I assume all your work is hosted on some other site of your own? How about you just provide us with a link to that and we can take a look if interested?

Have you published any of your work in peer-reviewed journals?

5. James RJust this guy, you know?Staff Member

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Moderator note: I have merged several of Geon's threads regarding his "quantum theory of gravity". Still trying to decide whether any of this is appropriate for our Physics forum....

7. GeonRegistered Member

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190

Of course, but you have bundled all my posts together, how will that cause any clarity with what is being discussed? It will just confuse everything, you've combined posts involving general relativity, the friedmann equation...

I understand it is not a blog, if anyone had asked any questions I am quite sure I would have answered them.

8. exchemistValued Senior Member

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Then report it: please do not start another argument on the science threads.

9. geordiefValued Senior Member

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@ Geon .based on post#2
If ,as seems to be the common assumption our understanding of physical observations relies on their geometric behaviour ,what might this tell us?

(not how the geometry works but what the implications might be that it does-or can we say that the geometric models are only models that may themselves be thoroughly reworked /abandoned at some stage)

Interesting to think that gravity could become the dominant actor at very small(but not Planckian) scales. Is there any discussion of this idea to be found elsewhere?

Usual disclaimer: I am all questions and not any help to anyone looking for answers

Oh and the maths is a foreign country to me.....

10. GeonRegistered Member

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It tells us that geometry can be understood to have the classical limit and then a quantum limit - in this case, the quantum case is just a special case of classical mechanics, because the quantum limit requires a ''correction.''

Many types of corrections to the quantum scale have been suggested, but one that pretty much seems cornerstone to geometry is an application of commutation principles - or non-commutating as the case often is - it is the non-commuting connections of a gravitational field that would require these ''corrections'' I have been speaking about.

It is actually predicted by string theory and quantum loop gravity that a non-trivial spacetime relationship could offer such a path - I simply understand it as the antisymmetric part of a Reimann tensor $R_{ij}$. It is this object specifically, when you understand the antisymmetric properties as related to the spacetime non-commutation relationship, that possibly non-trivial relationship which screams to be investigated further.

You can find further information on dynamics at the Planck scale, by reading into Wheelers more generic concept of quantum foam, which is basically the suspected particle activity associated to the bubbling of fluctuations on the Planck scale.

11. GeonRegistered Member

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190
On my most best idea's on describing the phase transition

The model I use depends on a phase of Bose/or/Fermi statistics and the pre-big bang phase was super cool, similar in that respect to Ekyprotic theory. An all-matter state would consist of some state of matter more primordial than the matter we deal with today. The result of the phase transition from a supercool, all matter liquid state is what heated the universe giving rise to a bath of radiation, mostly of which we detect as the background temperatures.

Why Bose or Fermi statistics?

The thermal de Broglie wavelength is denoted as $\lambda_{T}$. When the thermal de Broglie wavelength is smaller than the inter-particle distance, the gas can be considered to be a classical. On the other hand, when the thermal de Broglie wavelength is on the order of or larger than the inter-particle distance, quantum effects will dominate and the gas must be treated with Fermi or Bose statistics.

$\ddot{\lambda}_T \approx \frac{3}{2} \frac{\Delta T \Delta t}{p} \geq \frac{\hbar}{p}$

The phase transition obeys the following considerations:

$\frac{V}{N\lambda^3_T} \leq 1$

and

$\frac{V}{N\lambda^3_T} > \lambda_T$

The reason why it is important the pre-big bang phase was supercool (but not zero Kelvin), is because a process of lowering the temperature of a liquid or a gas (which are both fluids) beyond its freezing point - normally when a fluid reaches freezing point it becomes a solid: in this case it will freeze below this point without becoming a solid! Strangely enough, solid state physics as a phase state is considered a third phase state below liquid!

In the case the pre-big bang phase consisted of spin 1/2 particles, then the all matter liquid state may be considered equivalent to a degenerate supercool gas.

12. GeonRegistered Member

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190
Now, this above isn't all the story - this presentation above just give the interesting physics for an early universe where temperature and phase transitions play a role. The interesting thing, this physics talks about phase transitions only in the post era. The pre-era is considered a little different - for instance, consider the pre-big bang phase as a system which has almost no thermal degree's of freedom?

If we speak loosely of deBroglies wavelength as the general statement of the wave function of matter, then we have a general statement:

1) if the thermal wavelength of matter is larger than the distance between particles in a system, it will follow new statistics and quantum corrections.

2) if the thermal wavelength is smaller than the interparticle distance, then the system follows classical rules

What can we say about the pre-big bang state, does it follow similar physics? Can the post physics describe the pre state? Yes, I suspect it should be able to since there has to be physics making sense of the phase transition in itself. What we can say in the mathematical model we chose, that the universe has been suspended from collapsing to a point - normal theories of the big bang generally tend to deal with large densities and small spaces - this may even be true of the pre-big bang models... we also have to be open to the idea matter-creation was a continuous event during the early stages which actually gives an alternative to inflation. Small spaces also tends to mean particles are closely packed together, much closer than any metal, or solid here on earth.

summary

What we might find is that a pre-big bang state may have been controlled like a small system obeying classical laws - only if, the thermal properties of the pre-big bang particles have wavelengths smaller than their separation. This is a difficult question to answer, because how small can you probe spacetime without seeing those quantum effects (?)... well... generally speaking most scientists think those quantum effects are visible at and around the Planck scale, which is much smaller than the thermal interpretation of a wavelength and we have already established, we do not want a universe to collapse to a point... maybe the size of a football would do (roughly).

As you can see this look on things may lead to an interesting pre-big bang model. It seems quantum effects would be directly related to the temperature of the system (universe). When a universe heats up, the wave functions related to particles can have physics that describes a quantum vacuum.

I'll be giving the pre-big bang phase a break, and returning back to general relativity and ideas towards its quantization. I hit on some ideas last night that need to be typed.

Last edited: Aug 29, 2017
13. GeonRegistered Member

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190
Just one last thing to add, some crucial things to be noted about the pre-big bang phase:

1) In this model the pre-big bang was supercool but not frozen (solid/quantum crystal)

2) The pre big bang phase does not need to contain the observable count of particles in the post big bang horizon

3) A matter of non-conservation interpretation as particle-creation in curved spaces giving rise to a diabatic phase transition

4) The thermal properties of the wavelength of the pre-big bang gas of particles can be described using quantum corrections so long as their wavelengths exceed the pre-big bang inter-particle distance.

5) Any thermal properties of the system [must] be related to thermal or kinetic movement. This can be interpreted to mean there was very little movement of any pre-big bang particles - we have a system sitting on the precipice of absolute zero.

6) Simply put now, a new understanding of how we approach this must be taken to figure out the correct way to work this model. Can the thermodynamic deBroglie wavelength have any significance in the physics concerning any quantum corrections taken in this phase, when we have already considered that the thermodynamic properties have almost been smeared out (corresponding to little kinetic motion in the system) it must also be considered what kind of gas we are dealing with in terms of statistics - Fermi or Bose, or are we looking at a system that has to be described, with a new statistic, or one that has to be modified? Interesting stuff.

Is this a case where temperature is related directly to quantum corrections? When a cold universe with little degree of freedom heats up, the wave functions related to particles can have physics that describes a quantum vacuum! The question is, does the pre-state ''deviate to create'' or is it also described by quantum mechanics?

Last edited: Aug 29, 2017
14. GeonRegistered Member

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190

There is a seventh stipulation - that the thermal interpretation of the wavelength breaks down to the classical vacuum - but then the classical vacuum is about geometry and any geometry at the scales we are talking about, requires a quantum interpretation.

There for, the wavelengths are corrected gravitationally as well into the commuting lengths of the spacetime triangle, so not entirely through its thermodynamics which is supposed to be vanishingly small but non-zero. This means the pre-big bang phase really would require a quantum interpretation of the geometry which in my picture, would not require a Boson field (gravitons). Any particles existing in this pre-big bang state in this kind of model, looks largely gravitational in nature. They may even be considered as gravitational waves themselves!

15. arfa branecall me arfValued Senior Member

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A quantum theory of gravity will need to include a theory of quantum information (of gravity and its entropy).

To paraphrase Leonard Susskind: "the emergence of space behind an event horizon needs to be understood in terms of quantum information."

What I mean to say, I guess, is a theory will have to explain away the appearance of this "quantum information (of entanglement)", at dual places on a timelike slice through the interior. I conjecture that a part of this problem is our understanding of the information itself, we say there is quantum information, but that by itself is a bit of a contradiction, really it should just be called entanglement, much less confusing that way . . .

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16. GeonRegistered Member

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An Approach to Spacetime Triangulation as the Benchmark towards Gravitational Unification

It is well known from Pythagoras' theorem that there exists the spacetime inequality ~

$AB + BC > AC$

$AB + AC > BC$

$AC + BC > AB$

Is it possible to apply a spacetime commutator inside of this inequality? Yes I think so! Or at least, this occurred to me.

For a scalar product defind on a vector space the length of vector is determined by

$|X_a|^2 = X_a \cdot X_a$

With some invesigation (see references) a spacetime inequality can indeed satisfy the following relationship

$|X_a| + |X_b| \geq |X_a + X_b|$

Squaring both sides also yields

$|X_a + X_b|^2 = |X_a|^2 + |X_b|^2 + 2|X_a| |X_b|$

$(X_a + X_b) \geq ...$

$|X_a|^2 + |X_b|^2 + 2X_a \cdot X_b \geq |X_a|^2 + |X_b|^2 + 2|X_a| |X_b|$

from which it follows

$|X_a \cdot X_b| \geq |X_a||X_b|$

which is known as the Cauchy Schwartz inequality which can be thought of as a direct interpretation of a spacetime uncertainty. Another important identity whicch further can be identified from the spacetime relationships is

$|X_a||X_b| \geq \frac{1}{2} |<X_a|X_b> + <X_b|X_a>|$

If you have trained your eye on all my previous work into gravity, this looks like the structure of commutators!

In a Hilbert space, you can define new vectors

$\sqrt{|<\Delta X_A^2>< \Delta X_B^2>|} \geq \frac{1}{2} i< \psi|X_AX_B|\psi > + i<\psi|X_BX_A|\psi> = \frac{1}{2} <\psi|[X_A,X_B]|\psi>$

The left hand side calculates the deviation of the derivative from the mean of the derivative, at least, that is how it would be interpreted in the approach we took to the quantization of gravity. We will use these solutions as a benchmark into how to treat this commutivity in spacetime from the connections we solved.

ref http://rocs.hu-berlin.de/qm1415/resources/Lecture_Notes_10_11_12.pdf

Last edited: Aug 30, 2017
17. GeonRegistered Member

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Yes, a quantum theory of gravity will indeed involve a concept of information. The problem is, scientists are not too sure what information really is. In terms of the physics we deal with, we tend to think of information as bits of information -- but some scientists have argued the role of information may be much richer. The question of entropy is such an important question, especially considering questions of whether the universe had a high entropy or a low one, when it came into existence, becomes pivotal questions.

18. GeonRegistered Member

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190
And yes... you are right. Quantum entanglement requires explanation in a model like mine. I expect there to be unification somehow - that requires more investigations on my part. Though well-read on the subject, I am clearly not well-read enough to find the idea's I need.

19. GeonRegistered Member

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190

To give a hint in how to do this unification attempt, we have three key equations,

1.

$R_{i j} = [\nabla_i,\nabla_j] = -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]$

These are the exact Christoffel symbols of the antisymmetric tensor indices $R_{ij}$.

2.

$[\nabla_i,\nabla_j] \geq \frac{1}{2} |iR_{ij}|$

This was an equation derived by another author, finding the relationship in a different form argued from quantum mechanics. As you will see in key equation 3. the form has similarities to application of a Hilbert space ~

3.

$\sqrt{|<\Delta X_A^2>< \Delta X_B^2>|} \geq \frac{1}{2} i< \psi|X_AX_B|\psi > + i<\psi|X_BX_A|\psi> = \frac{1}{2} <\psi|[X_A,X_B]|\psi>$

Again, this is a Hilbert spacetime commutation relationship of operators which has to translate into the gravitational dynamics dictated by key equation 1.

20. GeonRegistered Member

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190

So let's put it altogether, its just like a jigsaw puzzle now. Implemented the Christoffel symbols in approach 1. into approach 2. we get

$[\nabla_i,\nabla_j] \geq \frac{1}{2} |-[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]|$

use identities

$<\nabla_i | \nabla_j> = i <\psi| \nabla_i \nabla_j | \psi>$

$<\nabla_j | \nabla_i> = - i <\psi| \nabla_j \nabla_i | \psi>$

In the framework of the Hilbert space it becomes - assuming everything has been done correct, takes the appearance of ~

$\sqrt{|<\nabla_i^2>< \nabla_j^2>|} \geq = = \frac{1}{2} i< \psi|\nabla_i\nabla_j|\psi > + i<\psi|\nabla_j\nabla_i|\psi> = \frac{1}{2} <\psi|[\nabla_i,\nabla_j]|\psi> = \frac{1}{2} <\psi | iR_{ij}| \psi > = <\psi|[\nabla_i,\nabla_j]|\psi> < \psi | [\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]| \psi >$

Yes there will be come corrections somewhere on those derivatives, not sure yet, wuld need to work it out, but generally speaking, this seems like it is probably the correct final form. (maybe).

Last edited: Aug 30, 2017
21. GeonRegistered Member

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190
So let's move on with some explanations of this.

Why the importance of doing this in a Hilbert space? As I explained very early on, this kind of commutation on spacetime was first studied by Von Neumann. Wiki gives a little history on this:

''Applications in the representation theory of groups naturally lead to examples of Hilbert algebras. Every von Neumann algebra endowed with a semifinite trace has a canonical "completed"[8] or "full" Hilbert algebra associated with it; and conversely a completed Hilbert algebra of exactly this form can be canonically associated with every Hilbert algebra. The theory of Hilbert algebras can be used to deduce the commutation theorems of Murray and von Neumann; equally well the main results on Hilbert algebras can also be deduced directly from the commutation theorems for traces. The theory of Hilbert algebras was generalised by Takesaki[6] as a tool for proving commutation theorems for semifinite weights in Tomita–Takesaki theory; they can be dispensed with when dealing with states.[1][9][10]''

As you can see, if we wanted to find a quantum interpretation of spacetime, without invoking gauge theories, left us with commutation and Hilbert spaces are great mathematical devices to reduce to the commutation relationships that smears classical space into the quantum.

22. GeonRegistered Member

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190
Sorry, equation came out wrong, the correct version is

$\sqrt{|<\nabla_i^2>< \nabla_j^2>|} \geq \frac{1}{2} i< \psi|\nabla_i\nabla_j|\psi > + i<\psi|\nabla_j\nabla_i|\psi> = \frac{1}{2} <\psi|[\nabla_i,\nabla_j]|\psi> = \frac{1}{2} <\psi | R_{ij}| \psi > = \frac{1}{2} < \psi | [\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]| \psi >$

note* Can't edit or wouldn't have made this post. (Again, I still need to study the correct sign changes if there are any.)

Second edit : Until I work out the sign changes, I will remove imaginary symbol from curvature

Last edited: Aug 30, 2017
23. GeonRegistered Member

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190
Good, as I expected, there is no sign outside of absolute value, and any signs inside the absolute value gets quashed to a positive value anyway. ie. |-10| = 10 but -|-10| = -10. So above form is correct any way you do it.