Looking at the Spacetime Uncertainty Relation as an Approach to Unify Gravity

[Dubbelosix]

Also another question is, I know the Ricci tensor is symmetric in torsion free theories, is this so in the non-zero torsion case?

[Mordred]

anti symmetric for the ricci tensor see Einstein Cartan

"The Ricci tensor is no longer symmetric because the connection contains a nonzero torsion tensor;"

https://en.m.wikipedia.org/wiki/Einstein–Cartan_theory

 

A little hint look at conservation of angular momentum and how torsion is involved in a specifically closed system. Also look specifically at the mathematical definition of a conserved quantity in terms of symmetry ie invariance.

Edited November 13, 2017 by Mordred

[Dubbelosix]

I said earlier I wanted the non-zero torsion tensor to represent the non-commutivity of the phase space - it seems natural to consider a non-commutativity of the torsion tensor playing the role of a Von Neumann phase space. With much reading done today, I do have some new stuff to write up on, such as treating the torsion like an operator, which was done by Heisenberg. It seems, this attempt to view torsion as the ''quantum part'' of Einstein's equations is not actually new, as I discovered tonight:

 

''Here torsion measures the noncommutativity of displacement of points in the flat spacetime in the teleparallel theory and the noncommutativity scale is given by the Planck length. So this approach is very much akin to our present approach if we consider that torsion is connected tointernal space which makes the spacetime noncommutative such that the gravitational constant fixes the scale of noncommutativity.''

 

https://www.academia.edu/7020453/Gravitational_Constant_and_Torsion

 

 

The non-commutativity plays a role in the Planck domain for the spacetime uncertainty relationship as well, keep in mind. 

[Dubbelosix]

I do learn a lot from my exchanges with Matti, he even gets to correct me on anything I have wrong 

what is this ''quantum symmetrization'' on the torsion connections? It's in chapter II. He says he uses a symmetric R_{mu \nu} and I am guessing this has to do with his dot notation denoting the symmetrization?
http://sci-hub.bz/10.1103/PhysRevD.3.2325

sci-hub.bz

sci-hub.bz

02:53

Christoffel symbols are not symmetric in lower indices if there is torsion. This might well give analog of Ricci tensor which is not symmetric anymore. It can be symmetrized. The antisymmetric part is problematic concerning generalization of Einstein equations since energy momentum tensor is symmetric.

I know the curvature tensor is not symmetric when torsion is non-zero, I am not sure what quantum symmetrization means in this case. Non-commutivity plays a big if not central role in describing the quantum phase space, so what does symmetrization mean, are you restoring a symmetry in the indices?

07:55

Ricci tensor is not symmeric for non-vanishing torsion. Yes. Symmetrization is purely algebraic operation: R_ab+R_ba. Nothing specially quantal is involved. Tensors can be decomposed to symmetric, antisymmetric and part proportional to metric. This is standard thing. Energy momentum tensor must couple to the symmetric part: otherwise one obtains nonsense.

And the product of a symmetric and an antisymmetric matrix has zero trace.., to be expected since trace is linear. So looking at Einstein's equations in full, tells us that it ha symmetric and antisymmetric parts. An antisymmetric matrix can square to identity though, I am thinking of Pauli matrices as such an example. They are traceless as well?

so many complicated questions with little information on the net!

The product of antisymetric and symmetric matrix indeed vanishes but Tr(AB)= Tr(BA) rather than linearity.

Pauli spin matrices sigma_i, i=1,2,3 are traceless but not antisymmetric. sigma_0 is unit matrix and has non-vanishing trace.

Einstein' equations have only symmetric part if torsion is not present. I do not know what the situation is in presence of torsion.

 

Don't know, he ended up corrected himself this time 

 

''Trace is also linear Tr(aA+bB)= aTr(A)+bTr(B). Tr(AB)= Tr(BA) is easy to prove and means that trace is symmetric. I was sloppy: Pauli spin matrices sigma_x,sigma_y,sigma_z are of course antisymmetric. I was referring to sigma_+ = sigma_x+isigma_y and its conjugate sigma_- which are not antisymmetric and have only non-diagonal component 12 or 21.''

I know I had things to write on a torsion operator but been busy reading on the Berry curvature for geometric phases in the Hilbert space. Here was a nice paper I came by 

https://sci-hub.bz/https://doi.org/10.1016/0378-4371(96)00092-1

 

and 

 

http://sci-hub.bz/10.1209/epl/i1999-00508-7

 

The more reading I do, the more I can explore the possibilities of describing geometry in the Hilbert space and get a better understanding of it.

 

[Dubbelosix]

So, I think I have come to some conclusion about a valid approach now. But it will require even more studying, because I like to be meticulous in a study.

 

In my approach I wanted to consider the non-symmetric part of Einstein's equations to describe a phase space solution for quantum gravity for particles (involving a non-zero torsion). This was conjectured, because you may have noticed, the Einstein equations are actually the product of a symmetric and antisymmetric tensor, which is well known to produce no trace, which is connected to a linear algebra in the theory. 

Interestingly, the Berry curvaure is an antisymmetric (second rank) tensor derived from the Berry connection - it was my aim to describe gravity in the phase space equally by using an antisymmetric part of the curvature tensor. More interestingly relationships have found linking torsion to the Berry geometric phase

Berry and geometrical phase induced by torsion field

S. Capozziello1,3(∗), G. Lambiase1,3(∗∗) and C. Stornaiolo2,3(∗∗∗)

So my wonder right now, is whether the Eisntein equations can be summed up in full, as predicting both a classical and non-classical counterpart in the form of the non-zero torsion tensor. If this is the case, then it makes only sense dealing with the symmetric part for the classical dynamics in the low energy regime. 

Going to the paper

Berry's phase and Aharonov-Anandan's phase
Zhaoyan Wu a,*, Jingxia Wang b 

It is proposed in chapter 2, that the constraint exists such that

[math]<\psi(t)|\psi(t)>[/math]

is a constant, no matter whether the Hamiltonian in the theory [math]H(t)[/math] is hermitian or not - I argue this [may not always be true].

Defining the wave function with a spatial derivative, allows us to make a curve equation from it. 

[math]\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>}\ = \int\  |W(q,p)|^2\ \sqrt{<\psi|R_{ij}|\psi>}\ dqdp\ \geq \frac{1}{\hbar} \sqrt{<\psi|H|\psi>}[/math]

So the ''phase'' is not too far from some (arguably non-trivial manipulations) to form a curvilinear metric - moreover, if the above spoken about before, it may not need to even consider a Hermitian form the curvature. As noted, the wave function above has a spatial wave function, as a result, this involves the tangent vector [math]\dot{\psi}[/math] which with the time derivative, has lengths that is the velocity which travels in the Hilbert space. 

 

Edited November 15, 2017 by Dubbelosix

[Dubbelosix]

The Berry connection is 

[math]A_i = <n(\lambda)|\frac{\partial}{\partial \lambda_i}|n(\lambda)>[/math]

It needs to be understood in geometric terms as a connection encoding how to compare the phase of the state [math]|n>[/math] at nearby points of the parameter space [math]\mathcal{M}[/math].

And Berry curvature is constructed as a set of antisymmetric commutation relationships,

[math]F_{ij} = [\partial_i, A_j] - [\partial_j, A_i] = \frac{\partial A_j}{\partial R^j} - \frac{\partial A_i}{\partial R^j}[/math]

The Berry curvature only becomes a gauge invariant theory if it has an additional commutator

[math]F_{ij} = [\partial_i, A_j] - [\partial_j,A_i] + [A_i,A_j][/math]

This looks like, a very similar object we have already encountered, we calculated those gravitational connections earlier and that took the form

[math]-[\partial_j, \Gamma_i] + [\partial_i, \Gamma_j] + [\Gamma_i, \Gamma_j][/math]

Is this almost identical structure save notation, an indication of something more deep in the theory?

It seems, just from this analysis I have made, the Berry curvature in the gauge representation is completely analogous to the full Einstein equations and it seems the last commutator would be antisymmetric iff it had connections to the torsional part of the field equations. 

Another big bonus for my model, since I just wrote in a PM, 

 

I was reading from finding this, that being related to gauge-invariance could make it possibly related to physical observables.

I wanted geometry to be observable in the model I was investigating.

 

Again, I wanted time to become observable in the final theory, as it is observably manifest as the curvature of three dimensional space from first principles of relativity. 

''

REF

 

https://arxiv.org/pdf/1701.05587.pdf

Edited November 15, 2017 by Dubbelosix

[Dubbelosix]

https://physics.stackexchange.com/questions/258053/calculating-the-berry-curvature-in-case-of-degenerate-levels-non-abelian-berry

[Dubbelosix]

ahem! I got the subscripts mixed up, I did them again, just a simple error. Fixing it gives you the same indice order as the Berry curvature, they really should be the same. So, here is a really good question, if the Einstein field equations (without torsion) are symmetric and the Berry curvature is antisymmetric but (appears to be an object) much the same as the ordinary Einstein representation, then what makes the Berry curvature antisymmetric? 

Back the order of the indices, what was really intended was

 

[math][\partial_i, \Gamma_j] - [\partial_j, \Gamma_i] + [\Gamma_i, \Gamma_j][/math]

 

So the Einstein curvature with torsion really looks like the gauge invariant Berry curvature, why the latter is antisymmetric when the former isn't, I'll need to look into.

 

Edited November 16, 2017 by Dubbelosix

[Dubbelosix]

I've ended up posting the question on physicsstack, who knows, maybe someone will answer, actually someone already has, but doesn't seem to have understood what object I was talking about and seems a bit confused. 

 

 

https://physics.stackexchange.com/questions/369233/berry-curvature-and-curvature-tensor

 

[Dubbelosix]

Read the very last informative post, this reveals that there is something deep between the two theories, that in effect, arise from the same ''beautiful' concept in physics. 

 

https://physics.stackexchange.com/questions/369233/berry-curvature-and-curvature-tensor

[Dubbelosix]

See the question I have is that the Riemann tensor is antisymmetric in the last two indices. The Berry curvature uses a two indice representation of the field strength. Christoffel symbols are actually related to their own definition of force albeit in a loose way, but could write a bit on that later. I suppose the Berry curvature indices would be analagous to the two antisymmetric indices of the Riemann tensor. 

I said I would show why the Christoffel symbols are loosely represented as force field strength terms.

In Newtons theory, the force is

 

[math]F = -\frac{\partial \phi}{\partial x}[/math]

 

and in Einstein's it is

[math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}[/math]

 

The relationships between the metric [math]g_{00}[/math] and the gravitational potential [math]\phi[/math] are well-known in literature to be related through some constants, and with it, you can see how Einstein's theory is analogous to Newton's. 

[Mordred]

yeah never worked with Berry curvature. Also didn't have time to examine it glad you found the difference

[Dubbelosix]

Probably best I write it up anyway, I have been tempted to look at the curve equation in terms of the covariant derivative. 

The curve related to the geometry of the curvature tensor, further related to the Hamiltonian was suggested as

[math]\sqrt{<\dot{\psi}|\dot{\psi}>} = \int \int\ |W(q,p)^2| \sqrt{<\psi|\Gamma^2|\psi>}\ dqdp \geq \frac{1}{\hbar}\sqrt{<\psi|H^2|\psi>}[/math]

(where we have used the Christoffel symbol, you'll see why).

A curve in general relativity involves the proper time

[math]\frac{dx^{\mu}}{d\tau^n}[/math]

and in fact, the covariant derivative acting on the curve is [math]\nabla \frac{dx^{\mu}}{d\tau}[/math] and would give rise to an acceleration term. The term also needs to be squared in our approach to match terms - this is so that you can decompose the equation into Schodinger solutions. 

[math]\frac{d}{d\tau} \frac{dx^{\mu}}{d\tau} + (correcting\ term)[/math]

The correcting term is a connection [math]\Gamma[/math]. 

The covariant derivative acting on our connection must obey a rank 2 tensor,

[math]\nabla_n\Gamma^{ij} = \frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i}[/math]

It also follows then the stress energy tensor responds in much the same way

[math]\nabla_nT^{ij} = \frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i}[/math]

Removing the square root we have for the curve equation

[math]<\dot{\psi}|\dot{\psi}>\ = \int \int\ |W(q,p)^2| <\psi|\Gamma^2|\psi>\ dqdp \geq \frac{1}{\hbar}\ <\psi|H^2|\psi>[/math]

(there are also two such terms now in the Wigner function, no need to write  it out)

 in which, we may come to think of the Hamiltonin being replaced with a stress energy tensor definition - just keep in mind there will be a density to get rid of. It allows us to decompose it into bra-ket solutions, with a covariant derivative acting on it all - the ket solution is

[math]\nabla_n|\dot{\psi}>\ = \int \int\ |W(q,p)^2|\ (\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i}|\psi>)\ dqdp \geq \frac{1}{\hbar}\ (\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})|\psi>[/math]

It can also be noticed that the Schrodinger equation is 

 

[math]H|\psi> = i \hbar| \dot{\psi}>[/math]

 

and so it satisfies for solutions for

 

[math]\frac{1}{\hbar}H|\psi> = i | \dot{\psi}>[/math]

 

Except the equation has to be non-linear now with our additional terms describing the corrections to the acceleration. 

 

And when you have a solution like this above, I can't help but think of the time dependent Schrodinger equation for the Berry Phase

https://courses.cit.cornell.edu/ece5390/7_berry_phase.pdf

And uses an imaginary part. 

 

Edited November 18, 2017 by Dubbelosix

[Dubbelosix]

For those not familiar with dimensions yet we have set a few constants in the equation to natural units - we will undo this so you can see the full structure of the equation

 

[math]\nabla_n|\dot{\psi}>\ = \int \int\ |W(q,p)^2|\ \frac{c^4}{8 \pi G}(\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i})|\psi>\ dqdp =\int\  \frac{1}{\hbar}(\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})dV|\psi>[/math]

 

But we also have a correction volume term on the stress energy tensor (this is because the stress energy tensor is an energy density by definition). 

Since this was a curve equation (which has a covariant derivative) attached to it, it is in some sense [like] a geodesic wave equation. You can follow up on how to construct  the geodesic equation with the christoffel connections which act as a correction (which was mentioned a few posts back) by following the link

 

 http://www.blau.itp.unibe.ch/newlecturesGR.pdf

 

A very detailed, and nicely written piece of work. 

Edited November 19, 2017 by Dubbelosix

[Mordred]

One of my favourite references hence its been a link on my website ever since I first read that lecture note. One can learn a ton on GR as well as how it applies to cosmology in its later chapters.

Edited November 20, 2017 by Mordred

[Dubbelosix]

I didn't know it was on your website, but great minds and all that

[Dubbelosix]

The Christoffel symbol can be 'loosely' though of as being analogous to a force in Newtons equations (where mass has been set to 1 to denote that it is a constant in this formulation):

[math]\Gamma = \frac{1}{2} \frac{\partial g_{00}}{\partial x}[/math]

Newtonian formulation of this acceleration is

[math]F = -\frac{\partial \phi}{\partial x}[/math]

However as mentioned, the gravitational force is not actually a true definition of a force as we come to expect say, in the proposed fundamental fields of nature, which are inherently complex (when quantum gravity is not) and that require quantization of field particles acting as mediators of the force (something which gravity is expected to use to form the unification theory in the opinions of many scientists). It's actually a crucial component of many theories, most notably string theory. 

Gravity is a pseudo force and can be understood in the following (neat) and (concise and short) way:

[math]\frac{d^2x^{\mu}}{d\tau^2} + \Gamma^{\mu}_{\nu \lambda} \frac{dx^{\nu}}{d \tau} \frac{dx^{\lambda}}{d\tau} = 0[/math]

where

[math]\Gamma^{\mu}_{\nu \lambda} = \frac{\partial x^{\mu}}{\partial \eta^{a}}\frac{\partial^2 \eta^a}{\partial x^{\nu}\partial^{\lambda}}[/math]

or more compactly 

[math]\Gamma^{\mu}_{\nu \lambda} = J^{\mu}_{a} \partial_{\nu} J^{a}_{\lambda} = J^{\mu}_{a} \partial_{\lambda} J^{a}_{\nu} \equiv J^{\nu}_{a} J^{a}_{\nu \lambda}[/math]

which represents a pseudo force for gravity which makes it in the same league as the Coriolis and the Centrifugal forces.

Also keep in mind, when you take the bra solution [math]<\dot{\psi}| \nabla[/math] and it;s product with its conjugate above, you could understand the product as producing an object like [math]<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>[/math] which shows the non-commutation between covariant derivatives

[Dubbelosix]

Minimum Distance in Relativity and the Non-Commutation of the Phase Space

First... a blow by blow account of what led up to the proposal of the curve equation: Anandan proposes an equation

[math]E = \frac{k}{G} (\Delta \Gamma)^2[/math]

and I offer also true

[math]E = \frac{c^4}{G} \int \Delta \Gamma^2\ dV = \frac{c^4}{G} \int \frac{1}{R^2} \frac{d\phi}{dR}(R^2 \frac{d\phi}{dR})\ dV[/math]

That was after correcting the constant of proportionality, and after I derived a following inequality by making use of the Wigner function - The Mandelstam-Tamm inequality for instance I have shown can be written in the following way, with [math]c = 8 \pi G = 1[/math] (as usual), we can construct the relationship: (changing notation only slightly)

[math]|<\psi(0)|\psi(t)>|^2 \geq \cos^2(\frac{[<\Gamma^2> - <\psi|\Gamma^2|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{[<H> - <\psi|H|\psi> ]\Delta t}{\hbar}) = \cos^2(\frac{\Delta H \Delta t}{\hbar})[/math]

This linking of geometry to the energy of the system can be understood through a curve equation I derived using the same principles

[math]\frac{ds}{dt} \equiv \sqrt{<\dot{\psi}|\dot{\psi}>}\ = |W(q,p)| \sqrt{<\psi|\Gamma^4|\psi>} \geq \frac{1}{\hbar} \sqrt{<\psi|H^2|\psi>}[/math]

By understanding that the curve equation when squared provides solutions to the time-dependent Schrodinger equation (in the following way) ~

[math]\frac{1}{ i \hbar}H|\psi>\ = |\dot{\psi}>[/math]

Then you can construct a more serious equation that may be seen as a gravitational analogue to the Schrodinger equation when the covariant derivative acts on the tensor components and I calculate it as:

[math]\nabla_n|\dot{\psi}>\ =  \frac{c^4}{8 \pi G} \int \int\ |W(q,p)^2|\ (\frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i})|\psi>\ dqdp \geq \int\  \frac{1}{\hbar}(\frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i})dV|\psi>[/math]

Also keep in mind, when you take the bra solution [math]<\dot{\psi}| \nabla[/math] and it;s product with its conjugate above, you could understand the product as producing an object like [math]<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>[/math] which shows the non-commutation between covariant derivatives - the wave functions, in its most simplest form can be understood as

[math]|\psi>\ =  e^{iHt}|q>[/math]

[math]<\psi| =\ <q| e^{-iHt}[/math]

When we take the product of the bra and ket solutions, we get an analogous identity found in General relativity - we simply take the tangent vector [math]\frac{dx^{\mu}}{d\tau}[/math] and allow the covariant derivative to act on this (and further set it to zero) and defines the minimum curve, or better yet, as a geodesic for a minimum distance,

[math]\nabla_n\frac{dx^{\mu}}{d\tau} \equiv\ min\ \sqrt{<\dot{\psi}|[\nabla,\nabla]|\dot{\psi}>}[/math]

We know how the covariant derivative acts on the curve from the following equation

[math]T_{nm}(y) = \nabla_n V_m = \frac{\partial V_m}{\partial y^{n}} + \Gamma_{nm} V_{r}(x)[/math]

The interesting thing about setting this to zero and using this as the definition of the minimum distance is that non-commutation between the covariant derivatives in a phase space is generally not zero [math][\nabla_i \nabla_j] \ne 0[/math]! The Von Neumann algebra insists that deviation from the classical vacuum relies on the non-commutative  properties of the quantum phase space. 

The covariant derivative also acts on rank 2 tensors in the following way

The covariant derivative acting on our connection must obey a rank 2 tensor,

[math]\nabla_n\Gamma^{ij} = \frac{d}{dx^n}\Gamma^{ij} + \Gamma^{i}_{n\rho} \Gamma^{\rho j} + \Gamma^{j}_{n \rho}\Gamma^{\rho}_{i}[/math]

It also follows then the stress energy tensor responds in much the same way

[math]\nabla_nT^{ij} = \frac{d}{dx^n}T^{ij} + \Gamma^{i}_{n\rho} T^{\rho j} + \Gamma^{j}_{n \rho}T^{\rho}_{i}[/math]

which were the primary tools we used to construct our form of the non-linear Schrodinger equation.

Also, it has also been established that the square of the curve can be seen as related to the kinetic energy:

[math]K_BT = (\frac{dx^{\mu}}{d\tau} \cdot \frac{dx^{\mu}}{d\tau}) \equiv\ <\dot{\psi}|\dot{\psi}>[/math]

Where again, this is a massless form (m = 1) for a constant mass. This linking of temperature to geometry could be lucrative. 

https://en.wikipedia.org/wiki/Maupertuis'_principle

Edited November 20, 2017 by Dubbelosix

[Dubbelosix]

Actually, I knew there had to be an inconsistency with setting the geodesic equation to zero but finding that it could satisfy non-commutation. The answer I have figured out relied on the fact both connections have derivatives in time. I have been so fixated on the non-commutative properties of spacetime, I forgot in this one simple case, it doesn't hold. 

 

The question was, was the geodesic equation a case where non-commutivity means it is non-zero? The answer it seems, there is actually commutation between them since they are both covariant derivatives in (time). In the case of a covariant derivative with both space and time indices, there would be spacetime non-commutivity. So in this case, the order of the covariant derivatives do not matter and they actually commute which preserves the definition of the geodesic being zero [math][\nabla_j,\nabla_j] = 0[/math]

Edited November 22, 2017 by Dubbelosix

[Dubbelosix]

http://www.physicsgre.com/viewtopic.php?f=10&t=127878

[Mordred]

I always have to spend time on your posts lol. They are precise enough to warrent the proper time to reply

[Dubbelosix]

I do try lol