# A Quantum Gravity Approach

Discussion in 'Alternative Theories' started by Geon, Sep 5, 2017.

1. ### GeonRegistered Member

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190
I don't see any reason why I should just stop my investigations and publishing and talking about them with whoever desired, and if people do abuse the position of coming in here without honest questions, you can go on block like NotEinstein will be getting from now on.

I'll continue updating my work when I literally have it fresh.

Spacetime Commutation from Antisymmetric Part of Riemann Tensor.

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

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

All these things, which comes from standard general relativity, use the antisymmetric indices [ij].

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, it 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.

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

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

3. ### GeonRegistered Member

Messages:
190
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

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

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.

An Approach to Spacetime Triangulation as the Benchmark towards Gravitational Unification

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

$AB + BC > AC$

$AB + AC > BC$

$AC + BC > AB$

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

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

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

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

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

Squaring both sides also yields

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

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

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

from which it follows

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

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

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

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

In a Hilbert space, you can define new vectors

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

(system here can't deal with the latex ^^^ translate it for yourself if you can in code

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

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

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

1.

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

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

2.

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

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

3.

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

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

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

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

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

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

without imaginary number on $R_{ij}$.

On Possible Quantum Bianchi Identities

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

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

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

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

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

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

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

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

The spacetime relationship

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

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

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

5. ### GeonRegistered Member

Messages:
190
The full tensor

The full Riemann equation (in usual standard form) with LHS showing commutation in indices, we have

$R^{\sigma}_{\rho\ [i,j]} = \partial_i \Gamma^{\sigma}_{j\rho} - \partial_i \Gamma^{\sigma}_{j\rho} + \Gamma^{\sigma}_{ie} \Gamma^{e}_{j\rho} - \Gamma^{\sigma}_{je} \Gamma^{e}_{i\rho}$

The extra two indices make the full equation symmetric.

The curvature and torsion is given by

$T^{\sigma}_{[i,j]} = \Gamma^{\sigma}_{ij} - \Gamma^{\sigma}_{ji}$

When torsion is non-zero, the Riemann tensor becomes a matrix/object describing the torsion. Though torsion remains a possible way to create a non-vanishing theory of the Riemann curvature, I still prefer a new ''look'' on how we view things - that ''new look'' on things was an argument that quantum domains and classical domains yields different understanding of the physics.... though the statement, ''the laws of physics are true in every coordinate frame'', is correct, I don't dispute this, but the idea that this should transcend into phase space with the same value is hypothesis. In fact, we argue there is a non-vanishing structure to spacetime itself!

For particles that do not have radii, (point particles) we can argue something happens in two ways:

1) Relativity implies as R decreases, the curvature increases and as it approaches zero, approaches infinity: Whether or not a point actually implies infinite curvature is for philosophers, for me, it just implies a non-physical situation. A singularity. You can also argue that self-energies become divergent as well.

2) If on the other hand, we are led to believe that electron particles are not actually pointlike, then this avoids infinite curvature and infinite energy and we save a definition of the electron which must have a radius-structure and therefore a curvature R and a stress energy T and a non-vanishing curvature in Riemann geometry.

Why is this important, the thing with particles? In a way it may be related to Von Neumann and his idea that points in phase space where meaningless.

7. ### NotEinsteinRegistered Member

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71
Where did the mu and nu indices go? Are you now saying all the elements of the Riemann tensor have to be at least as large as $\frac{1}{\ell^2}$, where l is not defined?

And could you at least fix the typo you admitted you have in this formula? It's quite intellectually dishonest not to.

8. ### GeonRegistered Member

Messages:
190

''The effect of torsion on Riemannian geometry is profound and far reaching. This was not fully realized until2003, when the Einstein Cartan Evans (ECE) unified field theory was inferred. It gradually became clear as the ECE series of papers progressed that the entire edifice of the Riemennian geometry collapses if torsion is forced to vanish through use of a symmetric Christoffel symbol. This means that Einsteinian general relativity becomes meaningless, because the Einstein field equation is based on torsionless Riemannian geometry. Most of the obsolete textbooks of the twentieth century do not even mention torsion, and if they do it is regarded as a removable nuisance. These textbooks are based on the arbitrary and unprovable assertion that torsion does not exist because the Christoffel connection is by definition symmetric in its lower two indices.''

This is an important statement, because it is actually true, what good are the equations if the curvature and torsion are made to vanish?

The full Riemann tensor featured in post 3. in a way, already implies torsion, even though we may not have it written there... it is encoded in the second equation connected to the antisymmetric part of the theory. The Bianchi identity in algebraic form is

$R_(X,Y)Z + R_(Z,X)Y + R_(Y,Z)X = 0$

You can think of the identity, being made of three forms of the Riemann tensor. The Bianchi identity in terms of usual Riemann notation we have

$R^{\sigma}_{\rho ij} + R^{\sigma}_{ij \rho} + R^{\sigma}_{i \rho j} = 0$

And is cyclic up to these three definitions. From here it should not be impossible to find (only those terms) associated to the dynamics of the curvature/torsion - and see those identities also as commutator relationships. Certainly, I argued that a phase space changes the physics so that the Riemann tensor had to be physically meaningless for points in space - instead of believing what physics textbooks tell us about the vanishing torsion tensor, it might be true even for quantum systems.

http://www.math.ucla.edu/~petersen/Bianchi_Ricci_Identities.pdf

So right now, I have to work out, are quantum Bianchi identities enough, or do we need the torsion tensor by default to describe curvature

EDIT* The Riemann tensor can have and does have a vanishing torsion but non-zero curvature... as is the case with the usual Christoffel connection. Which is an interesting difference. The Weitzenbock connection is one with non-zero torsion and vanishing curvature.

Last edited: Sep 5, 2017
9. ### GeonRegistered Member

Messages:
190
What is the physical interpretation of the vanishing of his tensor in all system of coordinates?

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

It just means the space is flat, that's all. It may be a misunderstanding from this to think a geometric object cannot exist at some point for this very fact - and it seems to be the only glaring statement by relativity which seems can be challenged.

Clearly there are cases, where geometry is not zero for some location in space, such fundamental objects surround us every day in the form of particles. Even electrons, arguably require radii. If they do not, they suffer real problems - in phase space, they become meaningless(because points are not physical) and in relativity they become meaningless (because of infinite curvatures) and in classical mechanics they become meaningless through infinite self energies.

How many clues do we need?

10. ### GeonRegistered Member

Messages:
190
Actually the Bianchi identity I keep reading is

$R_{\sigma \rho ij} + R_{\sigma ij \rho} + R_{\sigma j \rho i} = 0$

I will say up front though, not sure how much having sigma either covariant or contravariant would have changed the sign of the identities. It doesn't seem like changes the order of anything.

To create it is simple, contract the Riemann tensor in the following way and it becomes completely covariant.

$R_{\sigma\rho i j} = R^{k}_{\sigma i j}g_{\rho k}$

The symmetries of a Riemann tensor is what allows someone to write the Bianchi identities like they are.

Last edited: Sep 6, 2017