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Everything posted by Dubbelosix

You mean the observation of gravitational waves? This is not a surprise, they were predicted by the classical theory of general relativity any way, predating any foundation of graviton physics.

Because if more scientists recognized gravity as a pseudo force, it would become...evidently clear that quantization of the psuedo field into spin 2 gravitons is complete nonsense. If it was recognized more, it would never have been an issue. I would say more scientists today accept the idea of gravitons, than not.

Very detailed, very good paper. It even goes into some detail why gravity is classed as a pseudoforce, something a bit rare these days but pertinent to unification attempts, or failure, using a better word. http://www.blau.itp.unibe.ch/newlecturesGR.pdf

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.

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{<\psiH^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}\ <\psiH^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 braket 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 nonlinear 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.

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 wellknown in literature to be related through some constants, and with it, you can see how Einstein's theory is analogous to Newton's.

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/berrycurvatureandcurvaturetensor

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.

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 gaugeinvariance 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

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 nonsymmetric part of Einstein's equations to describe a phase space solution for quantum gravity for particles (involving a nonzero 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 nonclassical counterpart in the form of the nonzero 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 AharonovAnandan'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{<\psiR_{ij}\psi>}\ dqdp\ \geq \frac{1}{\hbar} \sqrt{<\psiH\psi>}[/math] So the ''phase'' is not too far from some (arguably nontrivial 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.

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://scihub.bz/10.1103/PhysRevD.3.2325 scihub.bz scihub.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 nonzero, I am not sure what quantum symmetrization means in this case. Noncommutivity 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 nonvanishing 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 nonvanishing 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 nondiagonal 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://scihub.bz/https://doi.org/10.1016/03784371(96)000921 and http://scihub.bz/10.1209/epl/i1999005087 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.

I said earlier I wanted the nonzero torsion tensor to represent the noncommutivity of the phase space  it seems natural to consider a noncommutativity 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 noncommutativity plays a role in the Planck domain for the spacetime uncertainty relationship as well, keep in mind.

Space (s) the third form of matter
Dubbelosix replied to Dr. Charles Michael Turner's topic in Speculations
Of what exactly? 
What I am trying to figure out, is whether the classical domain has (this antisymmetric part zero) whereas it may be nonzero in the quantum domain, just as noncommutation gives nonzero results. It's funny, because literature is not very clear sometimes. I have been confused from several different papers. This paper will be beneficial for me when I get back to the pure spin space, it has loads of useful identities in it http://peeterjoot.com/archives/math/pauli_matrix.pdf

Symmetry under interchange of the first pair of indices with second pair [math]R_{abcd} = R_{cdab}[/math] Antisymmetry [math]R_{abcd} = R_{bacd} = R_{abdc} = R_{bacd}[/math] Cylicity is the sym of the permtutations of the last three indices which vanish, [math]R_{abcd} + R_{acdb} + R_{adbc} = 0[/math] A contraction can give the Ricci tensor [math]R_{cd}[/math] It should be antisymmetric in its last two indices, but as a consequence of the Bianchi identities the Ricci tensor is symmetric for a Riemannian manifold. Thus it is the cyclicity which implies the curvature part vanishes for the totally antisymmetric part of the Riemann tensor. The antisymmetric part is actually just [math]T^{e}_{cd} = \Gamma^e_{cd}  \Gamma^e_{dc} = 2\Gamma^e_{[cd]}[/math] There does exist a gauge symmetry between curvature and torsion ''Gauge fixing torsion to 0, we end up with general relativity, whereas fixing curvature to 0, we end up with its teleparallel equivalent.'' https://physics.stackexchange.com/questions/103576/whycanweassumetorsioniszeroingr

Space (s) the third form of matter
Dubbelosix replied to Dr. Charles Michael Turner's topic in Speculations
Are you for real? Why are you entertaining a bogus idea? The whole terminology ''space as a state of matter'' is ridiculous. A few years ago, the community got excited over new buzzwords that didn't make any sense, like considering consciousness as a state of matter, and yet thinking of consciousness as a state of matter is one thing, but space is an arena where fields and their particles exist, it is not a state of matter per se. So thinking of the two on equal footing, makes no sense to me. And I assure you, will make little sense with anyone else here. 
So when taking the trace relationships for the curvature it is important to realize how limited those relationships really are since those traces will not be satisfied with the Ricci tensor since it is not completely symmetric when the torsion + curvature is nonzero. I went to see if anyone has constructed a theory for quantum Bianchi identities based exsctly on the issue of a nonvanishing antisymmetric part (as I have linked this in the past) to the nonzero phase space of noncommutivity (quantum domain) and it seems at least one author I have found seems to liken the issues correctly as I have and speaks about the role of the commutator in regards to the symmetries. And behold, he uses the key word I searched for, ''quantum Bianchi identities.'' ''I ?;r the classical domain the existence of any gauge group in a field theory gives rise to a set of identities which the field variables satisfy. A convenient method of obtaining these socalled Bianchi identities' for any gauge theory is to invoke the invariance of the action integral under the given gauge transformations of the theory. This procedure was used by Schrodinger3 to obtain the contracted Bianchi identities for the classical theory of general relativity. Even if the action integral of a gauge is invariant under a group of cnumber gauge transformations, the above procedure does not in general hold in the quantum domain because of the need to maintain operator orderings. Essentially, the basic problem resides in the dual role the commutator plays in the quantum theory.'' http://scihub.bz/10.1103/PhysRevD.3.2325

[points] Just a bit on traces, the trace of a commutator is zero. The product of a symmetric and an antisymmetric matrix has zero trace, A trace is a linear algebra example. Ricci tensor can be symmetric, but is antisymmetric in its last two indices. (seems true if we have metric compatiility) If the covariant derivative of the metric is nonzero it doesn't have to be true. Ricci tensor is not necessarily symmetric. It has been known that while an antisymmetric matrix has no trace, its square can have a trace  right? I am trying to find cases of this with the Pauli matrices, but I am guessing the relationship holds [math]n^2 \cdot \sigma^2 = \mathbf{I}[/math]. See I need to be extra careful about how the trace enters the theory and how it enters this projective space. [math]P = \frac{\mathbf{I} + n \cdot \sigma}{2} = \psi><\psi[/math] Of course, we are not just dealing with the antisymmetric parts, we are dealing with the full Ricci curvature which tends to be a symmetric object. But finding enough reading material on this is proving hard. The Ricci tensor is antisymmeric in the last two indices such that it has the property [math]R^{a}_{bcd} = R^{a}_{bdc}[/math] Common notation to denote the antisymmetry is use of brackets [math]R^{a}_{b[cd]}[/math] So from this, is a very basic question, is the Ricci tensor symmetric or skew? There are two existing answers but the far most common is that it is symmetric. (a question more directed towards Mordred) The full Riemann equation (in usual standard form) with LHS showing commutation in indices, we have [math]R^{\sigma}_{\rho\ [i,j]} = \partial_i \Gamma^{\sigma}_{j\rho}  \partial_i \Gamma^{\sigma}_{j\rho} + \Gamma^{\sigma}_{ie} \Gamma^{e}_{j\rho}  \Gamma^{\sigma}_{je} \Gamma^{e}_{i\rho}[/math] The curvature and torsion is given by [math]T^{\sigma}_{[i,j]} = \Gamma^{\sigma}_{ij}  \Gamma^{\sigma}_{ji}[/math] So it is interesting to note, while the curvature and torsion part explains it is terms of the antisymmetric indices, it is often set to zero in general relativity (which is actually the most boring case). You can obtain the skew symmetric part from the Bianchi identities as well, which this link explains https://mathoverflow.net/questions/69374/geometricalmeaningofthericcitensoranditssymmetry It seems we are ok, I finally found something which kind of answers the question https://math.stackexchange.com/questions/349817/provingthesymmetryofthericcitensor The tensor is skew symmetric in the last two indices and the tensor is symmetric from the Bianchi identities (something I suspected to look into to answer this) Reading further though, he has set the antisymmetric part zero. We want to avoid this. I wonder how those antisymmetric parts influence the first Bianchi identities if it is nonzero? Equally with an antisymmetric property which is nonzero could be beneficial to create quantum Bianchi identities.

Mordred, I have a technical question here.... my work is a bit more complicated and requires a bit more depth than I am used to  in the case of a projective space, the identity P^2 = P is said to hold  what does it mean?
In my case I am looking at, I have an antisymmetric matrix that requires to be squared in the projected space to yield the identity/unity. The projective space looks like
[math]P = \frac{\mathbf{I} + n \cdot \sigma}{2} = \psi><\psi[/math]The square of the Pauli matrix should yield an identity [math](n \cdot \sigma)^2 = \mathbf{I}[/math] (unit vectors naturally square into unity). What is the square of the dyad in such a case?
 Show previous comments 2 more

[math]\rho^2=\rho[/math] is the projector condition [math]\rho[/rho is only a projector if and only if [math]\rho^2=\rho[/math]
wiki explains it a bit better
In operator language, a density operator is a positive semidefinite, Hermitian operator of trace 1 acting on the state space.^{[7]} A density operator describes a pure state if it is a rank one projection. Equivalently, a density operator ρ describes a pure state if and only if
 ρ=ρ2{\displaystyle \rho =\rho ^{2}},
i.e. the state is idempotent. This is true regardless of whether H is finitedimensional or not.
https://en.wikipedia.org/wiki/Density_matrix

I'll be honest, I haven't looked at what units he was working in. Probably best not to see it just as ''G'' the whole object is the upper limit of the classical gravitational force and he seems to be wanting to reinterpret it in terms of some Planck dynamics. So he writes [math]\frac{8 \pi \ell}{mc^2}[/math] [math]F_{upper}\ell = mc^2[/math] and [math]F_{upper} = \frac{mc^2}{\ell}[/math] inverse is [math]\frac{1}{F_{upper}} = \frac{mc^2}{\ell}[/math] So I don't know, just looks like a dimensional argument, nothing special.

Space (s) the third form of matter
Dubbelosix replied to Dr. Charles Michael Turner's topic in Speculations
I agree, the question of what kind of doctor you are, is very relevant for instance, you sticking ''dr'' in front of your name when you are not a doctor of physics, could be easily deemed, disingenuous. Anyway, I can find you on the net. You say you have a degree in biology? That is good  but here, it will be of little good. 
The end of the quantum vacuum catastrophe ?
Dubbelosix replied to stephaneww's topic in Speculations
sorry wrong thread.