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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PART ONE

The concept was relatively simple to understand - I had an idea that consisted of an argument along the lines of ''if the curvature tensor must describe spacetime at all scales, maybe a non-trivial spacetime commutation could describe how the curvature can be larger than a specific wave. In some way, I argued, the curvature has to be proportional to a set of terms:

$\mathbf{G} = \frac{2 - n}{2}\mathbf{R}g = \mathbf{R}^{ij}g_{ij} - \frac{1}{2}n\mathbf{R}g_{ij}g^{ij} = r^2 \frac{1}{r^4} + r^2\ sin^2\ \theta \frac{1}{r^4\ sin^2\ \theta} = \frac{2}{r^2} \approx \frac{1}{\Delta L\Delta t}\ \geq \frac{c^4}{G \hbar}$

This early attempt was hoping to find some understanding in terms of the curvature of a sphere - I soon dropped this idea and concentrated more on how those Christoffel symbols and the commutation should be interpreted. After some talks with my friend Matti on the subject, it struck me that the same commutation relationship that already exists for Riemann geometry could already explain the commutation.

This did not come so quick though, I knew some preliminaries but it took to understand how the commutation can be described by two connections would be the easiest way to approach this. Certain things were clear: For instance, I knew crucially in its structure, we have a derivative of $\Gamma$ known as the covariant derivative or the connection of a gravitational field - it seemed there must have been commutation happening between the derivatives of the field and between the field itself.

$\frac{\partial \Gamma(x)}{\partial x} + \Gamma \Gamma$

The only way to assume a spacetime commutation property, is to insist they follow the usual rules implied:

$\frac{2}{\Delta r^2} = \frac{1}{\Delta X_a \Delta X_b} \geq \frac{1}{\ell^2}$

Its like there are two Gamma matrices with commutative properties! It was by understanding this, we could understand maybe the relationship being described by the following commutation relationship:

$[\nabla_x, \nabla_0]$

It appears, this would be the most natural way to translate the dyamics of the curvature tensor $G_{ij} = Rg_{ij}$ to look for that wanted spacetime uncertainty relationship. The only natural continuation then, would be to see how to construct the commutation and also expand it for the results of its Christoffel symbols. The commutation relationship is (in a usual convention)

$R_{x,0} = [\nabla_x, \nabla_0] = \nabla_x \nabla_0 - \nabla_0 \nabla_x \geq \frac{1}{\ell^2}$

Here we have explicitly wrote out the connections as having commutative properties satisfying our desired inequality. Writing the whole commutation out to find the christoffel symbols, (using differential notation) reveals the following and using general indices:

$[\nabla_i,\nabla_j] = (\partial_i + \Gamma_i)(\partial_j + \Gamma_j) - (\partial_j + \Gamma_j)(\partial_i + \Gamma_i)$

$= (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i \Gamma_j) - (\partial_j \partial_i + \partial_j \Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)$

$= -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]$

Pulling it out of its differential notation form, what we really have is

$[\nabla_i,\nabla_j] = \frac{\partial \Gamma_i}{\partial x^j} + \frac{\partial \Gamma_j}{\partial x^i} + \Gamma_i \Gamma_j$

and is just the Riemann tensor, but we hope we have specified, it one that follows the spacetime relatioship
$\Delta x\ c \Delta t = \Delta X_i \Delta X_j \geq \ell^2$.

What this unveils is that my initial idea of there being some commutator between the connection and the Christoffel symbols was true - but we also have another term, another commutator arising relativistically. The final mathematical model proposed to pave the way towards a unification of two concepts, a non-trivial spacetime uncertainty principle with the concept of geometry:

$G_{ij} = \mathbf{R}_{ij} - \frac{1}{2}Rg_{ij} \approx [\nabla_i,\nabla_j] = (\nabla_i\nabla_j - \nabla_j \nabla_i) =\frac{\partial \Gamma_i}{\partial x^j} + \frac{\partial \Gamma_j}{\partial x^i} + \Gamma_i \Gamma_j = \frac{1}{\Delta L\ c \Delta t} \geq \frac{c^3}{G \hbar}$

3. GeonRegistered Member

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190
PART TWO

So... what is the whole point?

Yes, maybe it can describe some way to capture a particle in its own gravitational field, what is the signifiance? This is a question that will require a longer explanation.

Essentially a Geon is a type of black hole - but a very special kind of one. Its not like your usual black hole which consists of a many-system body which radiates thermal energy - we don't know whether a Geon would radiate, but what we do know, is that it may describe all states of matter, according to A. Wheeler, who created the Geon hypothesis, but could not prove it using quantum mechanics.

Is matter really just trapped states of waves?

If so, what happens when particles exhibit wave properties? How does the wave escape, does it escape? As great as these questions are, one at a time... and even then, I cannot answer them all. I am new to this Wheeler model of particles and I am literally, still learning about what he had in mind. We may think it has appealing properties, since it would mean a wave can have a definite radius, avoiding singularity problems. It also stops natural divergence problems of a wave without invoking quantum fluctuations to dampen it.
Was Wheeler a bit of a prophet, or have I subliminally took in where he may have been heading? Here's an interesting quote from Susskind who was remarking on a conversation and things Wheeler had told him:
''The first time I met Wheeler was in 1961, I was an undergraduate... with a somewhat unorthodox academic record. ...The hope was that ...I would be admitted as a graduate student... At the time I was working as a plumber... I was completely enthralled. John was enthusiastically describing his vision of how space and time would become a wild, jittery, foamy world of quantum fluctuations when viewed through a tremendously powerful microscope. He told me that the most profound and exciting problem of physics was to unify Einstein's two great theories—General Relativity and Quantum Mechanics. He explained that only at the Planck distance would elementary particles reveal their true nature, and it would be all about geometry—quantum geometry. ''

The spacetime uncertainty, predicted from both string theory unification attempts and quantum loop gravity attempts, also links it to Planckian dynamics. It's true within our own formalization above, that in the region of the Planck length, commutation is revealing it's true properties related to quantum mechanics, making a quantum geometry. Wheelers vision was much more rich though, he believed quantum fluctuations where part and parcel of the vacuum, arguing for a hypothetical, but reasonable, quantum foam (analogous to how we think about fluctuations arising in the vacuum, making the vacuum not truly empty) and so related to the zero point energy concept, which is nothing more than the continuous residual ground state of the field.
Since gravity is about geometry, this feels like a very natural path to investigate and I would never had considered it, if it was not by linking these two concepts together.

The problem with considering this though, that this is something that can be applied to all particles, should we not be able to detect large curvatures associated to the required curvature at such distances? Interestingly, $P^{-4}$ propagators can be described entirely in terms of their spacetime curvature which gives rise to speculation whether gravity is a direct alternative to the strong binding force and indeed, that confinement itself arises from spacetime geometry - which means also that the phenomenon of quark duality, that is, that no quark is ever found alone, also has to be described by such a principle.

Such a model where curvature plays the role of the strong force was explored in depth by a number of authors, with the idea originating with Abdus Salam, though I am skeptical of the model he chose. But who am I, he's a nobel prize winner right?

For such a model to work, the scale in which the curvature needs to be measured would be the femtometer range - at this scale, gravity can be of magnitude 137 times stronger then electromagnetism matching the strength of the strong force. Even though the femtometer range is below the nanometer range, it is still nothing compared to the Planck scale - though it may seem conjecture to think gravity increases between systems as you reach particle scale, but it not so much rooted in hypothesis as one may suspect!
General relativity already predicts, in a way, for spherical systems, curvature increases as the radius of a system decreases! What does this really tell us about the nature of curvature? It means that the gravitational force produced from a small spherical system could actually be large! So in a way, there maybe hints, from this argument, curvature has two scales, a macroscopic scale and a microscopic one.

If it did increase as a systems radius decreased, it would make the investigation into how a Planck system could capture itself in its own gravitational field, much more easier - after all, if all you need is a femtometer range theory of gravity to describe imagine the kind of curvatures you will be invoking for a Planck like system, which is also spherical?

What about this application to all particles, do I believe in Wheelers vision? I think his vision is nice, but it may suffer that such systems can easily have thermodyamic properties leading to inexorably Hawking radiation. I showed such a possibe relationship. My earliest idea's about cosmology was to study it under a microscope, which led to my idea about how commutation could imply a geometry larger than it's specified wave length. That early study looked at a deBroglie wavelength and how the commutation properties may be seen in light of a time-temperature relationship, which may be seen with a Boltzmann constant or be seen as related to the entropy of system, entropy seems fitting if we find it follows thermodynamic principles related to black hole dynamics - in fact, it wasn't ad hoc at all, this following interpretation of a wavelength inequality, as it not only related it to the deBroglie wavelength, but rather, a special interpretation of the wavelength in thermal properties allows it to be written in the following way - in this version, we chose to write it as related to entropy ~

$\lambda_T \approx \frac{3}{2}S \frac{\Delta T \Delta t}{p} \geq \frac{\hbar}{p}$

The thermal energy is obtained from the kinetic interpretation given by the equipartition theorem as

$E = \frac{3}{2}k_BT$

So yes waves trapped in their own geometry could follow temperature relatioships as well that could imply that generally-speaking that these kinds of objects are not stable: At least, this is the stance I take. You can make your own mind up. Anyway, this small essay, summing up concisely my recent investigations is coming to a close. I am not happy about gauge theories of gravity, I think they are doing the wrong thing, in regards to how relativity treats gravity from first principles; its not a real force (pseudo force) which may be a non-trivial indication that it cannot actually be quantized, like a real force ie (electromagnetism).

If gravity really does describe confinement, we may come to understand something strange about further attempts at unification, would this mean gravity would no longer be the seat of importance because it is not a real, fundamental field? No of course not, the very investigation into how to treat gravity, on the quantum scale, is of great importance. No one questions curvature and accelerations could have quantum interpretations, the question is how to do it.

5. DaveC426913Valued Senior Member

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Thread reported to get it moved to Alt Theories Forum.

7. GeonRegistered Member

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Well, non-commutative relativistic geometry is not a new theory - you might call this an alternative theory... but it does prove something fundamental to the spacetime relationship to the Einstein tensor. If it's not recognized as such... then so be it. I am sorry.

The God likes this.
8. DaveC426913Valued Senior Member

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I owe you an apology.

I saw your username was Geon, and I saw you make reference to a type of black hole called a Geon - a term I've not encountered before.
I assumed that your username came first, and you were naming your new object after yourself - when, in fact, it was other way around.

Anyway, all I did was suggest it get moved to the Alt Forum. The mods will make that decision and move it or not.

Geon likes this.
9. GeonRegistered Member

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PART THREE

This part is reserved for idea's towards this kind of picture/model. One such approach, was to view the curvature in terms of Gaussian curvature - this is not illegal, such curvature is already relativistic.

The mean curvature is

$H = \frac{1}{2}(R_1 + R_2)K$

And a mean curvature of regular surface in $\mathcal{R}^3$ at some point $\mathbf{p}$ is given in standard form as

$H(\mathbf{p}) = \frac{1}{2} Tr(\mathcal{S}(\mathbf(p))$

Where $\mathcal{S}$ is the shape operator.It is said that when the shape operator is seen in terms of tensor analysis, it will describe the curvature of a system - this is without saying, since that type of mathematics involves the Christoffel symbols and in our investigation, we are only concerned with sphere dynamics - we did show those dynamics in the previous post. It is also standard that the Gaussian curvature which is just $K = \frac{1}{\ell^2}$ and the mean curvature satisfies an inequality

$H^2 \geq K$

That's interesting because it has formal similarities to (this time the exact dimensional form) of an inequality I derived concerning a Geon particle model

$G_{ij} \geq \frac{c^3}{G\hbar}$

And suggested this principle goes for all kinds of waves and there is even possibilities that it could have thermal consequences as well.
When the mean curvature is equal to the Gaussian curvature, it means the curvature is the same in any direction. That is true for a 2-dimensional surface of course.
The Gaussian curvature of a three dimensional hypersphere equal to the Planck wave is

$K = \frac{1}{k_1k_2} = \frac{c^3}{G\hbar}$

Where $\frac{1}{k_1k_2}$ is the formal definition of a Gaussian curvature involving the principles bundles. Motz notes that the three dimensional hypesphere of a radius, the Guassian curvature is $\frac{6}{R^2}$ so we also have

$\frac{1}{R^2} = \frac{c^3}{6G\hbar}$

This already implies dimensional possibilities of invoking the spacetime uncertainty on the LHS! The Gaussian curvature further can be represented in terms of general relativity as

$K = 4 \pi \rho_0 (\frac{G}{c^2})$

We use this in the following form, where $\rho_0$ is the proper density. Relating this to a function of smooth Riemannian manifold with a boundary $\partial M$ and $K$ is the Gaussian curvature of $M$ and $k_g$ is a geodesic curvature of the boundary.

$4 \pi \int_M\ \rho_0(\frac{G}{c^2})\ dA + \frac{1}{2} \int_{\partial M}\ k_g\ ds = \chi(M)$

where $dA$ is some differential change of area on the surface and $ds$ is the metric of the boundary.

Only if a boundary is ''piece-wise smooth'' then we can interpret the integral $\int_{\partial M}\ k_g\ ds$ as corresponding to the sum of the angles by which they turn the corners of the boundary.

The value of the Euler characteristic is simply $2$ for a sphere. This last equation is known as the Gauss-Bonnet Theorem, or at least, another form of it. In the case of the exact dimensions of the sphere solution $\frac{2}{r^2}$ found from previous Christoffel symbols investigation for a certain topology, this reveals that in Gaussian curvature there is a correction

$\frac{2}{r^2} = \frac{c^3}{3G\hbar}$

10. GeonRegistered Member

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190
CONCLUSIONS

We have looked at a very possible non-trivial spacetime commutation relationship and looked at it as it may be under the light of a geometric interpretation.
Since quantum loop gravity and string theory predict the same relationship, it seems like a strong argument/case to begin from such unifying principles. Susskind has promoted the vague but non-trivial suggestion of $GR = QM$, but as a physicist friend (Andrei) has equally agreed upon, this may not mean $GR = QFT$. Certainly, we should investigate possible non-trivial commutation relationships, especially if they relate to reinterpretations involving the geometry of spacetime.
Susskind himself, recognizes why such investigations are important, but he probably stresses too much on gauge theory as the way towards that.

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Reiku?

12. The GodValued Senior Member

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Look how these two DaveC and Origin are attempting to derail this thread! Some credit to former for retracting but then Origin is back to one word or one liner trolling.

13. river

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Hmm...is QFT the same as ZPF ?

14. GeonRegistered Member

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They are certainly linked. Zero point energy is the ground state of a quantum field.

15. GeonRegistered Member

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It was noted from my friend Matti, that the Einstein tensor when written as $G_{ij}$ is symmetric - I tried to argue a proportionality of terms from the tensor - but we must keep in mind what Matti has said because it is important to remember.

The commutation relationship is (in a usual convention)

$R_{\mu,\nu} = [\nabla_x, \nabla_0] = \nabla_x \nabla_0 - \nabla_0 \nabla_x \geq \frac{1}{\ell^2}$

Here we have explicitly wrote out the connections as having commutative properties satisfying our desired inequality. Writing the whole commutation out to find the christoffel symbols, (using differential notation) reveals the following and using general indices:

$[\nabla_i,\nabla_j] = (\partial_i + \Gamma_i)(\partial_j + \Gamma_j) - (\partial_j + \Gamma_j)(\partial_i + \Gamma_i)$

$= (\partial_i \partial_j + \Gamma_i \partial_j + \partial_i \Gamma_j + \Gamma_i \Gamma_j) - (\partial_j \partial_i + \partial_j \Gamma_i + \Gamma_j \partial_i + \Gamma_j \Gamma_i)$

$= -[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j]$

Pulling it out of its differential notation form, what we really have is

$[\nabla_i,\nabla_j] = \frac{\partial \Gamma_i}{\partial x^j} + \frac{\partial \Gamma_j}{\partial x^i} + \Gamma_i \Gamma_j$

It is one that follows the spacetime relationship

$\Delta x\ c \Delta t = \Delta X_i \Delta X_j \geq \ell^2$

An application of the commutator can be found from parallel transport, in which one can obtain the identity

$dV = \nabla_{\nu} \nabla_{\mu} dx^{\nu}dx^{\mu} V - \nabla_{\mu} \nabla_{\nu} dx^{\mu} dx^{\nu}V$

$= dx^{\mu} dx^{\nu} V (\nabla_{\mu} \nabla_{\nu} - \nabla_{\nu} \nabla_{\mu}) = dx^{\mu} dx^{\nu} V [\nabla_{\nu},\nabla_{\mu}]$

Consider now, the Riemann tensor with torsion. It is basically a commutator acting on some vector field $V^{\rho}$

$[\nabla_{\nu},\nabla_{\mu}]V^{\rho} = (\partial_{\mu} \Gamma^{\rho}_{\nu \sigma} - \partial_{\nu} \Gamma^{\rho}_{\mu \sigma} + \Gamma^{\rho}_{\mu \lambda}\Gamma^{\lambda}_{\nu \sigma} - \Gamma^{\rho}_{\nu \lambda}\Gamma^{\lambda}_{\mu \sigma})V^{\sigma} - 2\Gamma^{\lambda}_{[\mu \nu]} \nabla_{\lambda}V^{\rho}$

The full Reimann tensor is

$R^{\rho}_{\sigma \mu \nu} = \partial_{\mu} \Gamma^{\rho}_{\mu \sigma} + \Gamma^{\rho}_{\nu \lambda} \Gamma^{\lambda}_{\nu \sigma} - \Gamma^{\rho}_{\nu \lambda} \Gamma^{\lambda}_{\mu \sigma}$

The last two terms in $R^{\rho}_{\sigma \mu \nu}$ display antisymmetry in the commutators. Again the commutation arises from the derivatives of the connections,

$[\nabla_{\mu},\nabla_{\nu}] = -[\partial_{\nu}, \Gamma_{\mu}] + [\partial_{\mu}, \Gamma_{\nu}] + [\Gamma_{\mu}, \Gamma_{\nu}]$

This was the exact identity of the two commutators for derivatives concerned in space and derivatives in time - it's much more simple in the construct than the full Riemann tensor would suggest. The Riemann tensor vanishes, if we are in any coordinate system

$\partial_{\sigma}g_{\mu \nu} = 0$

Then the Christoffel symbols are zero

$\Gamma^{\rho}_{\mu \nu} = 0$

and

$\partial_{\sigma} \Gamma^{\rho}_{\mu \nu} = 0$

and so

$R^{\rho}_{\sigma \mu \nu} = 0$

Clearly, we are looking into cases though in which

$R^{\rho}_{\sigma \mu \nu} \ne 0$

In which curvature does not vanish, but has a specific meaning and relationship with spacetime.

So far, what we have learned, is the last two terms in Riemann tensor is antisymmetric: $R^{\rho}_{\sigma \mu \nu}$. We want to deal in spaces, maybe even two dimensional cases that are not locally Euclidean - ie. a vanishing Riemann tensor concerned with a commutation between connections describing the space and time derivatives. Hopefully we have clarified any misunderstanding of the commutators role in this and how they are interpreted as an anti-symmetric part of the Riemann tensor. I will continue the work with new ideas later.

Last edited: Aug 27, 2017
16. GeonRegistered Member

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The possible non-trivial spacetime uncertainty relationship, predicted by both quantum loop gravity and string theory, can be thought of as an analog of a quantum phase space.

Spacetime non-commutativity is defined by replacing the canonical variables with commutation relationships - it seems also non-trivial that the connections of the gravitation field in question have dimension of $1/length^2$ and so, it seems very natural to assume maybe gravity will follow the same dynamics on the Planck scale. This application of uncertainty into the equations requires a full interpretation. The preliminary investigation which leads to this idea of some unification between gravity and the quantum structure of spacetime came from an investigation into a quantum interpretation of the Geon particle. This required an interpetation where the geometry could be larger than a specified wavelength imposed by the spacetime uncertainty. These uncertainties in spacetime, which is just a reinterpretation of the usual uncertainty principle between energy and time, can be thought of as corrections on a manifold that are deviating it from the classical world. Using this understanding, nothing seems more natural than to look for such non-trivial spacetime relationships and see how they would (directly) relate to gravity - if they can. This may be a key point of how we may be thinking about it wrongly; as Susskind suggested, $GR = QM$ though it has been suggested not to be taken as a literal equality, there may already be cases which hint that gravity already has commutation possibilities to describe why these corrections are imposed at a Planck length.

We know what that relationship is, we defined it as taking the form of an antisymmetric tensor

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

How do we interpret this, without delving into the mathematics too deep this time around?

One interpretation may come from Von Neumann actually suggested that a point in quantum phase space is meaningless because of the uncertainty principle - there may be undertones or preferrably, direct relationships of this with regards to relativity. In a way, general relativity already had concluded from base concepts that points in themselves, are not actually physical. General relativity found a solution by treating not only the interaction of a system but also their worldlines. Notice though, in the Von Neumann universe, this becomes a non-problem, because points in space themselves are simply not physical systems when you appropriately correct them using the commutation.

We can intrepret maybe further, as you converge into the quantum Planck scales, you diverge from the classical theory and must be described by the corrected theory, which does involve a concept of the non-commutation. There are other things we must remain vigilant about such theories: Such as any possible unitary violations that can be understood simply as

$R^{ij}R_{ij} > 0$

After some reading, I found an author https://arxiv.org/pdf/hep-th/0007181v2.pdf who has applied the uncertainty identical to me, but hasn't looked at the theory from the same perspective. They find for the equality term ~

$[\nabla_i,\nabla_j] = iR_{ij}$

And they identify

$[\Delta_i,\Delta_j] \geq \frac{1}{2}|R_{ij}|$

Which is a very useful construction to remember. Eventually I might find an author who has investigated it directly as I have. My idea was based on dimensional grounds only, so we investigated likely only one small corner of this field.

Last edited: Aug 27, 2017

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18. GeonRegistered Member

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By the way, the last equation obviously should contain the nabla operators, for some reason I have typed in Delta (the change operator) not sure why but just correct this in your mind since I cannot edit posts any more.

In fact I know why, I mistaken the physics for $\Delta x \Delta t$ which is also the spacetime algebra.

19. GeonRegistered Member

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It was suggested by Arun (et al.) That rotation enters the Friedmann equation like

$(\frac{\ddot{R}}{R})^2 = \frac{8 \pi G}{3}\rho + \omega^2$

[1]. see references

It's proposed the correct derivation is not only longer, but in this form, should have a sign change for the triple cross product. We're going to prove how to get the proper form - really what is in implied by the centrifugal term is a triple cross product.

We will write the Friedmann equation in the following way:

$\ddot{R} = \frac{8 \pi G R}{3}\rho + \omega \times (\omega \times R)$

Even though the centrifugal force is written with cross products $\omega \times (\omega \times R)$ it is not impossible to show it in a similar form by using the triple cross product rule

$a \times (b \times c) = b(a \cdot c) - c(a \cdot b )$

Using $\omega \cdot R = 0$ because of orthogonality we get

$\ddot{R} = \frac{8 \pi G R}{3}\rho - \omega^2R$

which justifies this form of writing it as well. If the last term is the centrifugal acceleration then the acceleration in the two frame is just, while retaining the cross product definition with positive sign,

$a_i \equiv \frac{d^2R}{dt^2}_i = (\frac{d^2R}{dt^2})_r + \omega^2 \times R$

Expanding we get

$\ddot{R} = (\frac{d}{dt} + \omega \times)(\frac{dR}{dt} + \omega \times R)$

$= \frac{d^2R}{dt^2} + \omega \times \frac{dR}{dt} + \frac{d\omega}{dt} \times R + \omega \times \frac{dR}{dt}$

$= \frac{d^2R}{dt^2} + \omega \times \frac{dR}{dt} + \frac{d\omega}{dt} \times R + \omega \times ([\frac{dR}{dt}] + \omega \times R)$

$= \frac{d^2R}{dt^2}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)$

Which is the four-component equation of motion which describes the pseudoforces. This gives us an equation of motion as

$\ddot{R}_i = \frac{8 \pi G R}{3}\rho + \frac{d^2R}{dt^2}_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)$

Or simply as

$\ddot{R}_i = \frac{8 \pi G R}{3}\rho + a_r + \frac{d\omega}{dt} \times R + 2\omega \times \frac{dR}{dt} + \omega \times (\omega \times R)$

In the rotating frame we have

$\ddot{R}_r = \frac{8 \pi G R}{3}\rho + a_i - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)$

Since inertial systems are only a local approximation, the inertial frame of reference here may been seen to go to zero leaving us the general equation of motion for a universe

$\ddot{R} = \frac{8 \pi G R}{3}\rho - \frac{d\omega}{dt} \times R - 2\omega \times \frac{dR}{dt} - \omega \times (\omega \times R)$

Using

$a_{eul} = -\frac{d\omega}{dt} \times R$

$a_{cor} = -2 \omega \times \frac{dR}{dt}$

$a_{cent}= -\omega \times (\omega \times R)$

we get

$\ddot{R} = \frac{8 \pi GR}{3}\rho + a_{eul} + a_{cor} + a_{cen}$

This a purely classical Friedmann universe with all the classical pseudoforces encoded in it.

references

[1].https://www.researchgate.net/public..._Momentum_of_a_wide_rangeof_Celestial_Objects

20. GeonRegistered Member

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Part one

In this work, we adopt a solution from Motz and Kraft for a reversible isothermal Gibbs-Helmholtz phase change of an all matter liquid degenerate gas into a radiation vapor. It is assumed in our model, that a realistic phase change involves neither reversible or isothermal phase changes.

To help explain a diabatic anisothermal phase change from some super cool pre-big bang phase, we introduce a Friedmann equation which has been rewritten in the style of a Gibbs equation - this specific equation can be argued in a number of different ways: The basic way to view it is that the Friedmann equation is related to the entropy of a universe insomuch that it consists of two parts, a reversible and irreversible particle creation dynamics.

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\Gamma) = \mathbf{k}nT \dot{S}$

where $S$ has dimensions of $k_B$ and $\mathbf{k}$ is the Einstein factor.

in which

$nk_B T \dot{S} = \dot{\rho} + (\frac{\rho + P}{n})n \frac{\dot{T}}{T}$

which also justifies the following form as a fully thermodynamic interpretation:

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\frac{\dot{T}}{T}) = nk_B T \dot{S}$

$P_{irr}$ is known as the irreversible pressure, and inside of it, we can talk about the Gibbs-Helmholtz free energy equation for an irreversible phase change from a liquid particle creation phase to vapor for some infinitesimal change in volume,

$(P_{irr}(\frac{1}{n}))\dot{n} = -(\frac{1}{4 \pi R^2}\frac{dU_2}{dR})\frac{\dot{n}}{n} = -(\frac{dU_2}{dV}(\frac{1}{n}))n\Gamma$

Where, $n$ is the particle number density. The pressure is irreversible because particle production through the phases has happened in this case, in an irreversible way in which we take $n\Gamma \ne 0$. The irreversibility of particle creation is expected to happen when the term $\frac{kc^2}{a}$ is large, implying a large cosmological curvature which would have been present during these phase changes; it is a result of particle creation in curved spacetimes that yields possibilities for non-conservation in the universe. Any reversible dynamics $\dot{\mathbf{q}}_{rev}$ in the universe I speculate as probably a post big bang phenomenon, whereas I expect this conversion from the two phase states as related to chaotic and irreversible dynamics.

In this model, the universe is expected to have a cold dominated era (the pre big bang phase) as a degenerate gas of particles in a very high condensed gravity-dominated region with little thermodynamic freedom. Some collapse underwent in the cold dominated region leading to the radiation vapor phase - which can be thought of as the heating of the universe leading to a big bang scenario. We live inexorably, in this vapor phase of the universe.

Let's try and understand how you derive this. The energy of a universe is related to the reversible and irreversible terms:

$\frac{d}{dt}(\rho V) + P \frac{dV}{dt} = (\frac{dQ}{dt})_{rev} + (\frac{\rho + P}{n} \frac{d}{dt}(nV))_{irr}$

$\frac{d}{dt}(nV) = (\dot{n} + 3\frac{\dot{R}}{R}n) V$

This last equation is formally similar to the equation of state for a universe.

$\frac{d}{dt}(\rho V) + P \frac{dV}{dt} = (\frac{dQ}{dt})_{rev} + (\frac{\rho + P}{n} (\dot{n} + 3\frac{\dot{R}}{R}n) V)_{irr}$

since

$\dot{n} + 3\frac{\dot{R}}{R}n = n\Gamma$

$\frac{d}{dt}(\rho V) + P \frac{dV}{dt} = (\frac{dQ}{dt})_{rev} + ((\frac{\rho + P}{n})n V \Gamma)_{irr}$

Dividing the volume off, we get

$((\frac{d\mathbf{q}}{dt})_{rev} + ((\frac{\rho + P}{n})n\Gamma)_{irr}$

which fits the construction (once you treat it as heat per unit particle)

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\Gamma)$

So the bolded $\mathbf{q}$ means we have taken $\mathbf{q} = \frac{Q}{V}$.and can even be written as,

$(\frac{\ddot{R}}{R} + \frac{kc^2}{a})\Theta = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\Gamma)$

This is a definition of a Friedmann equation I think is soley associated to me - the fluid expansion implies a third derivative, imples that energy varies as a sphere with some homogeneous distrubution of matter and energy expands. We notice this from using a simple definition of particle number related to spacetime expansion:

$\Theta = (\frac{\dot{n}}{n}) = (\frac{1}{n})n\Gamma$

(this really means particle production happens and energy varies as a spacetime expands - its only irreversible in strongly curved spacteimes)**

The equation is also equivalent to the local distribution of matter for small systems when you make the following replacement:

$\frac{\ddot{R}}{R} = \dot{H} + H^2$

What you end up with is

$(\dot{H} + H^2 + \frac{kc^2}{a})\Theta = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\Gamma)$

For the diabatic Raychauduri equation.

$H$ is the Hubble parameter and in fact, $\dot{H} + H^2$ is an approximation of the local distribution of matter. We can understand why the local distribution of matter is significant for a very small universe without global implication and further we may understand why these solutions where chosen by Hawking and Penrose in the development of the singularity theorems in which the pressure has an irreversible sign change collapsing to a point, without quantum correction.

The fact is, this is described here specifically as a Raychauduri equation, is often considered as the cornerstone description of matter having attractive influence on each other, as expected from the classical theory. The symbol $\Theta$ as we have explained, is related to the particle production of a spacetime, meaning that as a spacetime expands, new particles enter the metric ~ and implies the formation of the fluid expansion in the following form

$\Theta^{n;}_{n;} = \frac{\dot{n}}{n}$

Again, this is a measure of how particle number is counted as a spacetime expands and the following equivalences also hold in Friedmann cosmology:

$\Theta^{\mu;}_{\mu;} = \frac{\dot{n}}{n} = \frac{\dot{a}}{a} = \frac{\dot{T}}{T} = \frac{\dot{R}}{R}$

$\mu; = n; = R; = a; = T;$

Often, equations like the equation of state, can introduce and reassemble the equations to fit different descriptions of an expanding universe, in terms of either the particle number density $n$, the Hubble radius $R$ or the scale factor $a$ where the final term $T$ is associated to temperature. The $\frac{kc^2}{a}$ associates to the strength of the curvature $\frac{1}{R^2}$ of an early universe --- and an important key thing to note here, is that it is the only realistic term in the model which is/has, isodensity between both phase states, that is, there is little deviation between the curvature states of the two phases of the universe.

To sum up, the left hand side of this equation then explains the local distribution of gravity and curvature when a universe was young - its non-conservation in the form of particle production and fluid expansion coefficient leads nicely to a definition of how non-conserved particle production in spacetime is implied when expansion and large curvature are producing these effects. Let's take a look at the right hand side of the equation now.

This side of the equation was argued from at least, three different articles. It is related to entropy, since it contains reversible and irreversible dynamics. It has elements of Gibbs thermodynamics since the right hand side satisfied a diabatic first law thermodynamic equation for the heat per unit particle (density). The equation can satisfy easily a definition of the Motz-Kraft solution for some irreversible pressure term. The right hand side once again, can accommodate a nice thermodynamic interpretation as well.

Conclusions

It is taken from this work that expanding spacetime as an exact solution to non-conservation is linked to the concept of field theory and zero point fields in terms equally of expanding spacetime - in other words, you cannot separate the concept of new energy from the notion of new spacetime since there is no classical Newtonian vacuum; we have given this up for a new concept of spacetime, which is not empty but filled with quantum activity. Sean Carrol has noted a similar argument in his own blog. I also take from this work, that the pre-big bang phase to the phase we live in today, may be an irreversible process.

References

http://sci-hub.bz/10.1111/j.1749-6632.1992.tb17071.x

https://www.researchgate.net/public...horizon_of_the_universe_in_entropic_cosmology

Last edited: Aug 28, 2017
21. GeonRegistered Member

Messages:
190
Part Two

In case you are wondering why the final reversible and irreversible components are important, the entire RHS has been constructed in such a way that our Friedmann equation is nothing more than an effective density which treats the heat per unit particle resulting in a Gibbs equation, or alternatively, the same arguments can be obtained from a modification of the first law of thermodynamics.

Heat per unit particle is

$dq = \frac{dQ}{dN}$

which changes the thermodynamic law into

$d(\frac{\rho}{n}) = dq - qd(\frac{1}{n})$

This leads to a Friedmann equation of the form

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}[(\frac{\rho}{n}) + 3P(\frac{1}{n})]\dot{n}$

Where the third derivative on the LHS denotes non-conservation (of some form) and on the RHS, the non-conservation is expressed through non-conserved particle number $\frac{\dot{n}}{n}$. And so, you will get the same reversible and irreversible dynamics when you consider the full equations featured in the OP. It certainly makes sense in the context of having it when entropy is considered.

How is entropy related to a Friedmann equation?

Many different papers have looked into this question, in this work, we will look at it from a set of arguments in which we define first the thermodynamic equation of state in the following way:

$k_B T \dot{S} = \frac{\dot{\rho}}{n} + (\frac{\rho + P}{n}) \frac{\dot{T}}{T}$

Rearranging, and using previous equations as a justification in the extended Friedmann equation for reversible and irreversible components, it can be identified in the following way

$k_B T \dot{S} = ((\frac{\rho}{n})_{Rev} + (\frac{\rho + P}{n})_{Irr})\frac{\dot{T}}{T}$

The irreversible and reversible components where argued from the entropy of a universe in terms of reversible and irreversible dynamics. If this is introduced into a Friedmann equation, you can find solutions like the following which describe the dimensionless entropy of the universe in a thermodynamic way:

$\mathbf{k} \dot{S} = \frac{8 \pi G}{3}((\frac{\rho}{nk_BT})_{Rev} + (\frac{\rho}{nk_BT} - 3P(\frac{1}{nk_BT}))_{Irr})\frac{\dot{T}}{T}$

Where $k$ is the Einstein conversion factor and $k_B$ is the Boltzmann constant. As an entropy equation alone, this is nice, considering we can see the entropy consists clearly of two parts $S = S_{rev} + S_{irr}$ as intended.

Now that we have a definition of the entropy, we can argue the reversible and irreversibility in another type of form, the Clausius entropy production equation will provide such a possible form:

$N = S - S_0 - \int \frac{dQ}{k_BT}$

When $N=0$, it means a thermodynamic process was reversible and $n > 0$ for some irreversible process where $S$ is the final state and $S_0$ is the initial state of entropy. So yes, it would be possible to write a Freidmann equation to satisfy this equation as well. In fact, this is doing the same thing as our initial approach in a way as it is measuring the irreversible dynamics, if there is any. In the previous form, we looked at the entropy per unit particle and the heat per unit particle.

It can be written as

$\dot{S} = \sum_k \frac{\dot{Q}_k}{T_k} + \sum_k \dot{S}_k + \sum \dot{S}_{ik}$

with $\dot{S}_{ik} \leq 0$

we can see how this is formally similar to how we treat the reversible and irreversible dynamics separately within the same equation. If any terms in the last equation have a subscript of ''i'' indicates it is an irreversible process, so it can be written in the following way to express which parts are reversible and which are not

$\dot{S} = (\sum_k \frac{\dot{Q}_k}{T_k} + \sum_k \dot{S}_k)_{rev} + (\sum \dot{S}_{ik})_{irr}$

Again, the structure of the equation reveals a global entropy must consist of reversible and irreversible dynamics. Irreversible dynamics occur in strongly curved spacetimes.

Related to all these equations, is a relationship derivable from similarities that have been noticed between the modified Friedmann equation and something related to the Gibbs Duhem relationship. The Gibbs-Duhem relation is

$dp = (\rho+ p)\frac{dT}{T} + nTd(\frac{\mu}{T}) = (\rho+ p)d\log_{T} + nTd(\frac{\mu}{T})$

Where $\mu$ is a chemical potential. From this one can get the chemical relationship in the Gibbs-Duhem relation is

$\frac{\mu}{k_BT} = \frac{\dot{p}}{nk_BT} - \frac{\rho + p}{nk_BT} \frac{\dot{T}}{T}$

There is no formal difference between the RHS of this equation and the RHS of the modified Friedmann equation

$\frac{\dot{R}}{R}(\frac{\ddot{R}}{R} + \frac{kc^2}{a}) = \frac{8 \pi G}{3}(\dot{\mathbf{q}}_{rev} + [(\frac{\rho}{n}) + 3P_{irr}(\frac{1}{n})]n\Gamma)$

This becomes more clearer when we introduce the thermodynamic interpretation of an equation of state was given through the following equation we have featured before already:

$k_B T \dot{S} = ((\frac{\rho}{n})_{Rev} + (\frac{\rho + P}{n})_{Irr})\frac{\dot{T}}{T}$

In terms of the entropy we get after replacing the reversible and irreversible parts for the chemical potential,

$\frac{8 \pi G}{6}[\frac{\mu}{k_BT}]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}\dot{S}$

This can be extended to describe the theory of mixed gases in this model as well:

$\mu_B = -\frac{n_A}{n_B}\mu_A$

$\frac{8 \pi G}{6}[\frac{\mu}{k_BT}]\frac{\dot{T}}{T} = \frac{8 \pi G}{6}\dot{S}$

$= -\frac{8 \pi G}{6}[\frac{n_A}{n_B}\frac{\dot{\mu}_A}{k_BT}]$

Here, the RHS is negative, meaning entropy rightfully, remains positive (entropy can never be negative).

https://en.wikipedia.org/wiki/Entropy_production

https://www.researchgate.net/public...n_and_Masslike_Function_in_General_Braneworld

22. GeonRegistered Member

Messages:
190
No more work... just thoughts now.

Singularity theorems are not something we should be encapsulated by, or find them romantic in some way. Singularities or infinities in nature do not happen - some scientists attributed this to a renormalization of type and yes it ispossible. It may just be, we don't understand the theory well enough to make statements about certain things, or maybe there are aspects of the theory which is wrong. Usually, when a singularity arises in a theory, its an indication we have done something wrong. Using this as a principle - making a universe collapse to a point, is equally wrong.

Making a universe collapse to a point suffers the same divergence problems as treating an electron like a point particle - in other words, the self-energy inflates to infinity and the physics makes no sense any longer. The Motz-Kraft model that we have modified heavily suggests there is a sign change between phases - that sign change is required if you want to have a negative pressure to explain why a collapsing universe perfers $R \approx 0$ to one that is $R=o$ (pointlike, zero radius).

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

Messages:
32,024
Geon:

Your opening post jumps into the middle of a complicated discussion of general relativity, apparently on the assumption that people here will be familiar with whatever it is you're on about.

Why don't you summarise what you're actually trying to do here, and what you think you've achieved?