Kartazion Metric

[Kartazion]

Hello.

Here is a topic which is not one of the simplest. Here is the proposal for a new metric which has the principle of being able to take into account cosmic inflation in relation to the expansion of the universe. Indeed its mechanism is that of the interpretation and for a gravitational singularity of a single-particle entropic convergence. IOW what is defined as a singularity with infinite value would then only be an interpretation of a paricle at rest. The expansion of the particle from its singular point which expresses its rest will therefore propagate asymmetrically in space-time. At first, the entropy of a supermassive black hole is only the particle's alternation between singularity (singular point) and the galactic halo. When the correlated particle explained by chromodynamics, is expressed in mass, while its flow between singularity and correlated matter is expressed in energy by the prince of mass-energy equvalence (Figure 1). The next post will be fed with a more explicit explanation.


Figure 1
 

Reference:
[1] Kerr–Newman metric 
[2] Friedmann-Robertson-Walker

[Bufofrog]

6 minutes ago, Kartazion said:

Here is the proposal for a new metric which has the principle of being able to take into account cosmic inflation in relation to the expansion of the universe

So this metric is involves the very early universe?

[Kartazion]

4 minutes ago, Bufofrog said:

So this metric is involves the very early universe?

Yes of course. It corresponds to the particle at rest of an immeasurable value of its mass, and as well as the interpretation of the true vacuum in quantum field theory.

Thereafter the extension of the particle is interpreted by the principle of mass-energy equivalence.

The purpose of this experiment is to be able to precisely add the explanation of inflation up to the Friedmann-Robertson-Walker metric which is already very good one.

The problem with Kerr–Newman metric is the consideration of symmetric entailment.

But Newman found a way out with Axial symmetry.

In this reference I base myself on an energy density of the particle as a convergent point thanks to the principle of mass-energy equivalence, until its rest:

Quote

The Kerr–Newman solution is a special case of more general exact solutions of the Einstein–Maxwell equations with non-zero cosmological constant. Einstein's cosmological constant, is the constant coefficient of a term Albert Einstein temporarily added to his field equations of general relativity. He later removed it. Much later it was revived and reinterpreted as the energy density of space, or vacuum energy, that arises in quantum mechanics. It is closely associated with the concept of dark energy. Wikipedia

IOW Kerr-Newman metric describes a black hole solutions to the Einstein-Maxwell equations of general relativity.

Very simply and for a black hole considering a potential energy and a kinetic energy of the particle by Hawking radiation determines the following clock Figure 2.
 

Figure 2.

[Markus Hanke]

2 hours ago, Kartazion said:

Here is the proposal for a new metric

Ok, so what is the metric? You haven’t written it down yet.

[Kartazion]

26 minutes ago, Markus Hanke said:

Ok, so what is the metric? You haven’t written it down yet.

I don't know yet demonstrated a metric tensor. But the metric (maybe the term is badly chosen) in the proper sense of the term does not change the orthonormal basis. On the other hand, the interpretation of the curvature of space-time of the black hole is affected by the innovation that I support. To do this, and at first, a simple matrix can be used. But I haven't finished writing about it yet. I already have all the technical elements but the pofinage remains to be determined. This remains a complex subject in its interpretation of what is a space-time curvature from the inflation. In a second time the tensor will be able to generate the shape of our universe in its continuum, while the matrix will be the mechanism which constitutes it through the particle.

[joigus]

No metric to be seen.

[joigus]

What is a "pofinage"?

And how can a basis remain orthogonal irrespective of the metric? The metric is what tells you whether a basis any given set of vectors is orthogonal or not.

7 hours ago, Kartazion said:

IOW what is defined as a singularity with infinite value would then only be an interpretation of a paricle at rest.

How can rest (which is an observer-dependent concept) be associated with a singularity? Does a frame change remove the singularity?

Infinite value of what? Singularities in the metric are meaningless. It's only singularities in the Riemann tensor that are physically significant.

Also, a 3pi angle points in the negative x direction, not in the negative y direction. I think you meant 3pi/2. Whatever x and y mean. There we go again.

 

Edited August 11, 2022 by joigus
minor correction

[Markus Hanke]

3 hours ago, Kartazion said:

I don't know yet demonstrated a metric tensor.

Well, without a metric we can’t really have a discussion about this.

The thing here is that you don’t start with a metric - you begin with an energy-momentum tensor plus boundary conditions, then you use these to solve the Einstein equations. That gives you the metric.

All solutions to the Einstein equations are metrics, but not all metrics are valid solutions to the Einstein equations.

[Kartazion]

1 hour ago, joigus said:

What is a "pofinage"?

Tweaking. Sorry. 'pofinage' was a French word.
 

1 hour ago, joigus said:

And how can a basis remain orthogonal irrespective of the metric? The metric is what tells you whether a basis any given set of vectors is orthogonal or not.

I wrote orthonormal and not orthogonal.
 

1 hour ago, joigus said:

Infinite value of what? Singularities in the metric are meaningless. It's only singularities in the Riemann tensor that are physically significant.

A gravitational singularity is a region of spacetime in the vicinity of which certain quantities describing the gravitational field become infinite regardless of the coordinate system used.
 

1 hour ago, joigus said:

Singularities in the metric are meaningless. It's only singularities in the Riemann tensor that are physically significant.

The Kerr metric is also a vacuum solution and describes a black hole and the shape of a ring can be called a ring singularity or ringularity.
 

1 hour ago, joigus said:

Also, a 3pi angle points in the negative x direction, not in the negative y direction. I think you meant 3pi/2. Whatever x and y mean. There we go again.

I corrected. Thanks.

1 hour ago, Markus Hanke said:

 - you begin with an energy-momentum tensor plus boundary conditions, then you use these to solve the Einstein equations. That gives you the metric.

I don't want to be clumsy in my answer. But already the mass of the particle at rest located at the level of the singularity represents the total mass of the universe. I therefore let you imagine the amount of energy to be applied to the particle in order to be able to simulate inflation through an oscillating movement.

[joigus]

25 minutes ago, Kartazion said:

I wrote orthonormal and not orthogonal.

The argument still stands. Orthonormal is a particular case of orthogonal. Orthonormal=orthogonal and normalised.

25 minutes ago, Kartazion said:

A gravitational singularity is a region of spacetime in the vicinity of which certain quantities describing the gravitational field become infinite regardless of the coordinate system used.

More specifically, a gravitational singularity is a region of spacetime in which any components of the Riemann curvature tensor become infinite.

Read carefully @Markus Hanke's previous post. You do not invent the properties of the metric. You postulate the other (non-gravitational) fields. Then you obtain the energy-momentum tensor. Then you symmetrise it (with techniques like, eg, Belinfante's symmetrisation technique), because the canonical energy-momentum tensor is generally non-symmetric, and the source of the gravitational field must be symmetric in the space-time indices. Then you postulate boundary conditions, as Markus told you. Then you solve for your metric.

Having done all that, you're still not home-free, because the particular coordinates that you use to solve for the metric can have false singularities, ie, singularities of your coordinate map that are not physical. So you must obtain the Riemann tensor and try to identify the singularities there.  You have a lot of ground to cover still before you can meaningfully talk about your singularity.

I hope the comments here you find helpful. The metric is not gauge-invariant. It's the Riemann tensor that's gauge invariant. This is in close analogy to electromagnetism. The vector potential in EM doesn't really give you the physics (except for the Aharonov-Bohm effect or the "holonomy" of the field). Infinitely many vector potentials give you the same physics. It's Faraday's tensor plus the holonomy which gives you the complete physics of electrodynamics. There is only one Faraday tensor (the E's and the B's) that define the physics.

Gravity displays remarkable mathematical similarities to EM. It has a huge gauge arbitrariness. In modern GR we say space-time is not defined by a metric, but by an equivalence class of infinitely many metrics, all gauge-equivalent to each other.

The matter is even more subtle. Sometimes you find a coordinate map that solves the equations. But the map has singularities of itself --fictitious.

Then you introduce a change of coordinate maps that fixes the coordinate singularities. Example: Kruskal-Skezeres coordinates being a well-known example.

Edited August 11, 2022 by joigus
minor correction

[Kartazion]

Here in Figure 3 is an archaic representation, really very basic and non-operational yet, but which already gives an idea of how one could interpret the flow/flux of mass-energy equivalence related to the impulssion/momentum energy ; And integrated into the gravitational attraction. A clear distinction must be made between the mass-energy equivalence and the momentum energy given to the particle.

The depletion of momentum energy occurs as the particle picks up speed. IOW the exhaustion of the energy of the impulssion is related (proportinnal?) to the speed of the particle. I have to complete the missing parameters. I'm working on it. Your help has been invaluable to me.

I'm not sure I have correctly identified the problem in relation to the tool called energy–momentum tensor or else it represents a different vision.


Figure 3.

Edited August 12, 2022 by Kartazion

[joigus]

5 hours ago, Kartazion said:

Here in Figure 3 is an archaic representation, really very basic and non-operational yet, but which already gives an idea of how one could interpret the flow/flux of mass-energy equivalence related to the impulssion/momentum energy ; And integrated into the gravitational attraction. A clear distinction must be made between the mass-energy equivalence and the momentum energy given to the particle.

The depletion of momentum energy occurs as the particle picks up speed. IOW the exhaustion of the energy of the impulssion is related (proportinnal?) to the speed of the particle. I have to complete the missing parameters. I'm working on it. Your help has been invaluable to me.

I'm not sure I have correctly identified the problem in relation to the tool called energy–momentum tensor or else it represents a different vision.


Figure 3.

What does this have to do with a metric in curved space-time?

And how could energy-momentum be "depleted" when the particle picks up speed?

[Kartazion]

8 minutes ago, joigus said:

What does this have to do with a metric in curved space-time?

The ratio between the "impulssion energy" and the velocity of the particle gives the curvature by "?".
 

9 minutes ago, joigus said:

And how could energy-momentum be "depleted" when the particle picks up speed?

Dark energy can maintain constant or accelerated kinetics. But in the fugure case presented the particle has reached the end of the expansion of the universe and falls back into a big-crunch.

[joigus]

1 hour ago, Kartazion said:

The ratio between the "impulssion energy" and the velocity of the particle gives the curvature by "?".

That's not how curvature is defined. Do you know what curvature is?

[Kartazion]

On 8/12/2022 at 11:43 AM, joigus said:

That's not how curvature is defined.

Curvature occurs only by under the influence of mass or energy. Mass being that represented at rest in Figure 3 is then substituted by energy. The curvature is therefore produced by this energy distribution.

For example for inflation similarities are to be noted between the shape of the charge/discharge curve of an RC circuit through Vc.

[Bufofrog]

1 hour ago, Kartazion said:

For example for inflation similarities are to be noted between the shape of the charge/discharge curve of an RC circuit through Vc.

You should probably give up on this, you really don't seem to have any idea what you are talking about.

[joigus]

On 8/12/2022 at 11:43 AM, joigus said:

That's not how curvature is defined. Do you know what curvature is?

3 hours ago, Kartazion said:

Curvature occurs only by under the influence of mass or energy.

No. Curvature is defined by going from one point to another by applying equal changes in 2 coordinates in different order, and thereby learning if there's a difference in going through different paths across your space. If there is, your space is curved.

You can also have curvature as solutions to Einstein's vacuum field equations.

It so happens that there are 20 numbers that give you the curvatures in dimension 4 or more.

Such is the nature of Riemannian curvature. That's the curvature we talk about when we do GR.

Mind you, you  can also define a special curvature for a curve, but that's not Riemannian curvature. It's based on a moving reference frame embedded in a flat space of 3 dimensions. The Riemannian curvature of a curve is 0. That's because Riemannian geometry is intrinsic: not embedded in a higher-dimensional space.

Language and intuitive notions by themselves are misleading: A curve is not curved, a plane is not curved (makes sense, but it's true nonetheless), and a cilinder is not curved. A cone is not curved, but at a point, where it has infinite curvature.

I can't tell which one is your curvature. And I don't think that's a good thing. It all sounds like you lack a basic understanding of curvature.

A historical note:

Einstein found his famous equations only after a long correspondence with David Hilbert, and consultation with mathematicians, and from there he postulated a certain combination of the 20 curvatures of space-time (reducing them to just 10) to be proportional to the different components of energy-momentum densities. This checks with 4(4+1)/2=10 independent components of the energy-momentum tensor in dimension 4.

He achieved that only after several faltering attempts. If my memory serves, it took him the best part of a decade to come up with that after his initial intuitions. He later proved that in the limit of low velocities and weak gravitational fields you get Newton's law of gravitation. He predicted light to be bent by gravitational fields.

Lastly, he did not coin the name "Einstein field equations". It was other people who named them after him.

To my mind, it belittles the genius of Einstein when people try to emulate or better this mind-blowing feat based on some loose notions and a couple of graphs.

Don't take this personally but, at what point are you going to decide it's time to go back to the drawing board?

 

5 minutes ago, joigus said:

Such is the nature of Riemannian curvature. That's the curvature we talk about when we do GR.

To be more precise: pseudo-Riemannian, because of the difference in sign for time and space in the metric.

Edited August 13, 2022 by joigus
minor correction

[Phi for All]

!

Moderator Note

The OP has not engaged in discussion with those patient members who bothered to supply some science, and instead has chosen to stand on a soapbox and wave a metric into existence with their hands. This fails all the parameters for Speculations. Thread closed.