# A Quantum Gravity Approach

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

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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

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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 Senior Member

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792
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

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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

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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
11. ### NotEinsteinRegistered Senior Member

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792
Well, seems the gods have not permitted me to take a closer look at part two of the other thread ( http://sciforums.com/threads/towards-ideas-on-a-quantum-theory-of-gravity.159827/page-7#post-3474063 ).

I might as well tackle this thread instead. There's an interesting link to an arxiv-paper at the start of post #2, which Geon indicates does something similar to his work. This could be interesting; we may be able to gain additional insight into what Geon meant!

Of course, if the moderation thinks this is superfluous or downright unfair (kicking a perma-banned horse, so to speak); feel free to lock this thread without notification.

12. ### NotEinsteinRegistered Senior Member

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792
Remember the odd notation: The nabla stands for coordinated ($x$ in the referenced paper; equation 1), the indices i and j are called $\mu$ and $\nu$ in the paper, and R is called $\theta$ in the paper. This $\theta$ is explicitly anti-symmetric in the paper, while the Ricci tensor is traditionally symmetric. This suggests that Geon's R is not the Ricci tensor, even though it is.

This is obviously wrong. Equation 2 in the paper is talking about $\Delta$'s of the coordinates, not the commutation of the coordinates. In other words, Geon's equation does not match the one given in the paper.

The equations are at least trivially wrong: in flat space, they do not hold. Obviously the operators in them should (at least) be $\geq$.

Looking further down below, there's a reference. That's where this is taken from, so let's compare Geon's version with the original reference.
We note immediately that the reference is talking about "ordinary vector space", which is not what we have here. So let's be careful...

This is obviously wrong: the operator is an equal sign, while we started with an inequality! Let's assume this is a typo for now...

(Again with the wrong operator.)

Which is thus also wrong.

And that is complete nonsense. It has nothing to do with uncertainty, and to even suggest that is proof of a fundamental misunderstanding of it.

Notice that this equation (right above section 3.4.1 in the PDF) has the operator the right way around. It's magically fixed!

(We jump ahead a few pages to page 68 in the PDF.)
You can see the PDF introducing the new vectors, but Geon has (accidentally) left that part out, leading to a misleading notation: $X_A$ and $X_B$ have nothing to do with coordinates anymore.

I see no derivative in that equation. Nor a mean, or a mean of the derivative. So this statement is plain wrong.
We can read in the PDF what the sentence originally was; that sentence actually referred to terms that are present in the equation...

(Summary, no comment.)
Notice that equation 2 has suddenly picked up a factor of $i$ that wasn't there before. Typo?

So we take equation 2, insert equation 1 (and the mysterious factor $i$ goes missing!). Then we equate coordinates with arbitrary vectors in Hilbert space (equating nabla with X), and insert into equation 3 our previous result.

Great, and now?

Nothing, because that's where this part of the story ends, and this (seemingly important) result is not brought up again.

13. ### NotEinsteinRegistered Senior Member

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792
Identical? So this object you speak of is: $[i, j]$

Sure, considering $R^{\rho}_{\sigma\ [i, j]}$ doesn’t affect the fundamental structure of the of the Riemann tensor.

This is an incomplete sentence.

This is compatible with the initial assumption, yes.

The grammar of this sentence is broken.
It is not clear what is meant here. The covariant derivative of what? Probably what is meant is the covariant derivative of the metric. In that case, the statement is true: if the Riemann tensor vanishes (i.e. all its elements are zero), you have flat spacetime, and the covariant derivatives of the metric are indeed zero.

I’m not familiar with the term "quantum Bianchi identities", and "non-zero interpretation" is a nonsensical term. I suppose this is just a statement that the covariant derivatives of the metric are zero, but the equation doesn’t match that, because neither is there a metric in it, nor covariant derivatives. But let’s see what happens next anyway...

Oh, so it’s not covariant derivatives, but the normal derivatives after all?

Yes, because that would have major implications for all the derivations in GR as well.

This statement has nothing to do with anything.

Obviously not, because the physicality of "points" has no bearing on anything. And the given equation has no derivatives in it.

Aha, we’re still using Carroll! This is one of his exact statements, from: https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

This has nothing to do with Carroll’s statement. However, it does indeed touch upon a hard fact: what’s physically true in one coordinate system, must be true in all. So if there is a "spacetime uncertainty" in one coordinate system, there must be in all.

It is not clear why this is a model of discretized spacetime, nor how this equation can hold under coordinate transformations. Note that I already asked Geon about this earlier:
which, of course, he never gave an answer to.

This is, as far as I can understand what Geon it trying to say, nonsense. A non-zero value of "the derivatives" does not imply a quantum scale gravitational description.

Translation: A flat spacetime is related to the properties of the Riemann tensor. Well obviously!

Wait, the entire metric vanishes? But then you literally have no spacetime! You have no dimensions left to work with!
And what "action"? What does the metric have to do with a physical action?

This is rubbish. If an entity is invariant under coordinate transformations and vanishes due to that fact, it never represented a physical entity to begin with. So this definition of "action" is not only non-standard, it is truly incompatible with the standard definition of the word! It has no physical meaning whatsoever.

The last bit is trivially true. The laws of nature cannot depend of the used coordinate system.

Another broken sentence. And the last part is wrong. In fact, Geon his/her-self knows this, as (s)he in this very post was talking about how tensor equations must hold in all coordinate systems.

This seems to refer to Heisenberg uncertainties, which has nothing to do with coordinate systems.

Another broken sentence. And this statement seems to refer to something that wasn’t mentioned. It is unclear what classical space time and classical objects inside of it are so different from.

14. ### NotEinsteinRegistered Senior Member

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792
The left hand side is zero, per Carroll's equation 3.64 from here: https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll3.html

That is nonsense; there are not extra indices in this equation. Compare to equation 3.67: the anti-symmetric brackets are plainly wrong.
In other words, the equation as given only hold in flat spacetime, because only there is the right hand side zero.

Compare to Carroll's equation 3.16. Again, the brackets are wrong (or there is a factor of 2 missing).

This is sloppy language. The Riemann tensor is always describing the torsion (and other things). For example, look at: http://www.slimy.com/~steuard/teaching/tutorials/GRtorsion.pdf
Equation 3.2.3 and the text around it. That's the definition of the Riemann tensor when spacetime isn't torsion-free. Notice how the torsion tensor is part of it.

I've never heard of a "non-vanishing theory". I'm going to assume it should read: "theory of non-vanishing Riemann curvature".
False. Any curved spacetime will have a non-vanishing Riemann curvature.

At no point in this derivation did any quantum domains play any role. This whole statement is irrelevant.

"Transcend into phase space"? And what "same value"? Also, "hypothesis"; don't you mean speculation? I ask, because the extrapolation is simply unwarrented: that the laws of physics have to be true in every coordinate frame has nothing to do with whether they are true at every point in phase space. Coordinate frames are an abstraction we use in our descriptions, while phase spaces are based on real properties. The comparison makes no sense.

This person called Einstein did the same thing with his theory of relativity, but he didn't need non-commuting coordinates to do it!

Wrong. Relativity only deals with spacetime curvature, not the curvature of a particle's surface.

Yes, the surface's curvature will approach infinity. A strong hint that a particle with radius zero doesn't exist in nature.

Yes, that's why every scientist knows that point particles are only approximations.

This is simply wrong. Point particles do not have infinite energy.

Again, since a point particle has no infinite energy, there is no need to avoid it.

No, you were using R to denote the particle radius. Or is this Orwellian notation again?

Watch out for the sloppy language. The curvature of the particle's surface has nothing to do with the Riemann geometry of spacetime.

At no point (pun intended) did we talk about phase space, nor was anything related to spacetime said in the entire particle-related discussion. So no, this is a big non-sequitur, and thus can be disregarded as unimportant.

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It appears our friend has you on ignore....Akin to sticking his fingers in his ears, so he'll at least see this.

16. ### NotEinsteinRegistered Senior Member

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792
Actually, I think (s)he was banned. But I promised I'd respond to the rest of the posts, so that's what I'm doing.

17. ### LongitudeRegistered Member

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Yes I was banned, and accused of writing nonsense, while the moderation team supported this as motive to ban me and close my account down. If you want to see how actual discussions go here,

h tt p://
ww
w.scienceforums.net/topic/110072-looking-at-the-spacetime-uncertainty-relation-as-an-approach-to-unify-gravity/?page=1

Fix obviously to get access, and if you want to follow my work as it updated

h
ttp:
//ww
w.physicsgre.com/viewtopic.php?f=10&t=127412&p=198855#p198855

Anyone who has dealt with Einstein on a technical level (not likely to happen here anyway) but if they ever do, they'll understand why I ignored him. He was a total troll, and he even fooled mods like Kitamaru who clearly shouldn't be moderating as he liked to abuse his power. Don't worry, I haven't returned for good, just not letting my fans down.

Or at least, I thought best to come here and expose what an actual discussion in science goes like.

18. ### NotEinsteinRegistered Senior Member

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792
According to the ban message it was because you were (and again are) a sockpuppet of a previously banned member: http://sciforums.com/threads/towards-ideas-on-a-quantum-theory-of-gravity.159827/page-7#post-3475455

Unfortunately, we never really got to actual discussions here, because you refused to explain your terms, or admit and fix obvious mistakes. It was you who prevented "actual discussions" from taking place.

Perhaps you want to address all the mistakes I've pointed out in that post?

(That is not my name.)

Indeed unlikely to happen as you seem unable to even address the smallest of mistakes you've made.

Please give your definition of "trolling", because I do not consider "pointing out mistakes" to be trolling.

But aren't you letting your fans down by not even fixing the glaring mistakes that were pointed out to you?

Does it involve admitting and fixing your mistakes? Because you demonstrated a distinct lack of that on this forum so far.

19. ### LongitudeRegistered Member

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And yet, you haven't demonstrated in any intelligible way any mistakes. You didn't point out mistakes, because you are incapable of doing so. wish not to speak to you any more.

I was rudely accused by Kitamaru of not answering your questions because ''I couldn't'' is utter bullshit. The people here, have their noses so far up their ass they are smelling their own shit, I won't accept people accusing my work for not making sense, when for some reason, people who have done work in the field agree I haven't made mistakes and was clear I was putting effort in.

These facts were ignored, because of your rhetoric. This is the last time I speak to you, these are exceptional circumstances.

20. ### NotEinsteinRegistered Senior Member

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Except that one where you indeed admitted that you missed a minus-sign, after I pointed it out repeatedly. The rest: I'm sorry you apparently cannot understand them.

Then why don't you address all the issues I've brought up? If you have made no mistakes, that shouldn't be so difficult.

If merely my pointing out your mistakes is enough to make you throw insults left and right, perhaps the field of science isn't for you.

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22. ### LongitudeRegistered Member

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Ok ok... I give up. How about you hold a petition here to see whether I SHOULD COME BACK? If I loose, I NEVER COME BACK. And until decision is made, I will not speak here at all?

23. ### LongitudeRegistered Member

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Actually he is wrong on that one, but I will discuss his with james, no one else.