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Is Gravity Complex?


Dubbelosix

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I'll be investigating for this model, linearized gravity as there are many similarities hidden in these equations which can be re-written in the form of gravielectromagnetism. But for now I will be investigating the mass currents as related to the wave operator on the gravitational scalar.

[math]J^{\mu} = \partial^{\mu} \phi[/math]

The covariant derivative [math]\nabla_{i}[/math] and the scalar [math]\phi[/math] do not commute [math](\nabla \phi - \phi \nabla) \ne 0[/math]

and a mass current is conserved

[math] \partial_{\mu} J^{\mu} = \partial^{\mu}\partial_{\mu} \phi = \Box \phi [/math]

And using a substitution of identity we can combine this definition with the stress energy:

[math]-4 \pi \mathbf{T}_{matter} = \Box \phi[/math]

 

we get

 

[math]- 4 \pi \mathbf{T}_{energy} = \mathbf{U}_{\mu}\mathbf{U}^{\mu} (\partial_{\mu} J^{\mu}) [/math]

Edited by Dubbelosix
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There's loads of different approaches I could take to describe how the particle becomes entangled with the empty wave, but sometimes a logical argument alone can suffice. Below, I use some notation some people may not be so acquainted with so I give it some explanation. It does involve the concepts of entanglement.

 

[math]H(\Psi) = H(\psi_{empty}|\psi_{occup}) + H(\psi_{empty} : \psi_{occup})[/math]

 

Here, [math]H(\psi_{empty}|\psi_{occup})[/math] is the entropy in [math]\psi_{empty}[/math] (after) having measured the system that became correlated in the occupied wave function [math]\psi_{occup}[/math].

 

Further, [math]H (\psi_{empty} : \psi_{occup})[/math] is the information gained about [math]\psi_{empty}[/math] by measuring the occuped state [math]\psi_{occup}[/math]. This reveals a conservation in the second law not so dismilar to the Shannon entropy in terms of information theory. Likewise from information theory, we can talk about the empty wave and occupied wave in terms of the upper bound of correlation which is given as:

 

[math]H(\Psi) = H(\rho_{\psi_{empty}}) + H(\rho_{\psi_{occup}}) - H(\rho_{\psi_{empty,\ occup}})[/math]

 

Which is a quantum discord attempt to unify an empty wave with an occupied wave.

 

Going back to this theoretical set up for the gravitational pilot wave within a framework of entanglement, the next question is, why does this mean the empty wave should be in phase with the occupied wave?

 

A very simple explanation for entanglement, is that two apparently ''individual'' systems, are not individual at all and in fact act as if they are ''one system.'' This is what would happen between the empty wave and the occupied one, but there is dominating effect from the occupied wave - what I mean by this is that the presence of the particle would forbid the empty wave to move at its preferred speed in classical physics (light speed). Therefore, entanglement would give rise to an in-phase effect. I suspect, the absence of a particle in either wave would allow it to move at the speed of light, the same kind of empty wave I predicted from the absence of gravitons in a quantum theory of gravity.

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Of course, it explains an additional part of the double slit experiment, since if it travels definitely in one trajectory through one slit, how does it reach the outermost edge of the other slit so that it manages to reach all places? The waves can become out of phase, meaning they will appear in different places with different values.

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To explain properly what I mean, I have took the liberty of finding a quote on quora:

 

''Two coherent waves with the same wavelength or a constant phase difference will interfere with one another. If their phase difference is a whole number multiples of the waves' wavelength, the waves will add. If their phase difference is some whole number plus one half wavelengths, they will cancel each other out. If the waves are not the same wavelength or their phase difference something else, then they will add differently in different places. This is called wave interference, it is responsible for the fringe pattern you see when you perform a double slit experiment.''

 

 

In must the same sense, while we are capable of explaining the empty wave entanglement with the occupied wave, as soon as they hit those slits, it seems like they would have to break their entanglement only temporary since the empty wave will interfere with the occupied wave, which as has been explained a few times, are in this model being treated as gravitational pilot waves.

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I've looked at this possibility of entanglement between empty waves and particles. Ultimately to describe them as states where one is occupied and one is empty will require similar arguments that exist for the theory of entanglement. Because matter can entangle with space itself, not just matter to matter, it even re-opened a question concerning microscopic wormholes within ER=EPR. Before, as Susskind said, as elegant as the idea is, he would have rolled his eyes, before the talk of fire walls and the new considerations that try to avoid it (entanglement of matter to space). The relevant equation to describe such an entanglement, will come to the last equation, adopted from Markopoulou who provided the equation as a key feature to her own toy model of gravity.

 

Now, we can demonstrate the relevant equations for this post and I will take you through the definitions. The equation which can describe the maximal entangled state is:

 

[math]|\Psi >_{\psi_{emp},\psi_{occ}}\ = \sum^{N}_{n = 0}\ <\psi_{emp}|\mathbf{P}_n|\psi_{occ}>\ |\psi_n>_{\psi_{emp},\psi_{occ}}[/math]

 

Here, [math]\mathbf{P}_n[/math] is the projection operator onto the state with n-particles. The entropy is related to one such term in the equation as:

 

[math]|\psi_n>_{\psi_{emp},\psi_{occ}>} = -\mathbf{Tr}(\hat{\rho}\ log_2\ \hat{\rho})[/math]

 

In which we have used what is known as the density operator [math]\hat{\rho}[/math]. A direct interpretation is that this implies it is a local theory. If we call each density representing a state of particles then we can transform it slightly:

 

[math]|\psi_n>_{\psi_{emp},\psi_{occ}} = -\mathbf{Tr}[(\frac{1}{N}\ log_2\ \frac{1}{N}) = S(\hat{\rho})[/math]

 

How do you make sense of entropy like this, especially the middle term?

 

In this entropy equation, you will find it is summing over all the possible states of [math]N[/math]. As you will see, this will reduced to [math] -\ln(N)[/math]. Lets show why this is: In order to solve it, you must multiply fractions using the equation

 

[math]a \cdot \frac{b}{c} = \frac{a \cdot b}{c}[/math]

 

in the key expression [math] \frac{1}{N}\log(\frac{1}{N})[/math] and this gives

 

[math]= \frac{\log(\frac{1}{N})}{N}[/math]

 

The summation will also yield a value of N which is why when you cancel all the terms out you get

 

[math]N\frac{\log(\frac{1}{N})}{N} = -\log(N)[/math]

 

A note to remember, that the log of the inverse of the variable has the relationship:

 

[math]\log(\frac{1}{N}) = -\log(N)[/math]

 

Which is known as the log rule. So at least in theory, gravitational wave functions may be able to follow the ordinary rules of entropy, which we would come to expect for consistency.

Edited by Dubbelosix
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Primer on Possible Lorentz Violations and Particle Creation

 

 

A Lorentz violating vector field is actually part of a formalism to obtain absolute accelerations.



In particle physics, a scalar field getting a vacuum expectation is the most common among scientists since it also can givbe rise to spin zero particles (ie. any class of Higgs boson).

If a vector field, like the notation we have been using for velocity [math]\mathbf{U}^{\mu}[/math] has a non-zero expectation value, it will violate Lorentz invariance. Sean Carrol mentions

''this is ok, but the subject to interesting experimental constraints''

Alternatively, an expectation of a gravitational scalar, produces particles -

[math]<\phi>\ =\ <\frac{Gm}{r}>\ = \sum_n [a_nf_n + a^{\dagger}_{n}f^{*}_{n}][/math]

I investigated this type of quantization and suggested in my Friedmann investigations that the particle production in highly curved spaces (like the early universe) could happen in an irreversible way (short duration of conservation law breakdown).

Above, we have used the creation and annihilation operators, they follow the standard rules:

[math]a|n>\ \sqrt{n_k + 1}|n_k + 1>[/math]

[math]a^{\dagger}|n> = \sqrt{n_k + 1}|n_k - 1>[/math]

The expectation for the particle number is equivalent to:

[math]<N>\ = \sum_k aa^{\dagger}[/math]

The operators also follow a commutation rule

[math][a,a^{\dagger}] - aa^{\dagger} - a^{\dagger}a = \mathbf{I}[/math]

and so this can be arranged like so:

[math]aa^{\dagger} = a^{\dagger}a + 1[/math]

The operations on the left hand side is basically the same as the summation of the particle number:

[math]<N>\ = \sum_{k} aa^{\dagger}[/math]

 

It's well known that particles and their antiparticles are entangled. We have a gravitational field potential coefficient in the equation we derived in the Planck phase space:

 

[math]\mathbf{G} = \kappa\ \mathbf{T} = \kappa\ g_{\mu \nu}\ T^{\mu \nu} \equiv \frac{2 - D}{2}\ \mathbf{R} \geq \frac{8 \pi G}{c^4}\ (\frac{1}{\ell^2_{planck}}) \mathbf{U}_{\mu}\mathbf{U}^{\mu}\ \phi[/math]

 

It's interesting to note that it is possible also to take the operator form of this potential to validate particle production in the Planck space.

 

Primer on the Projection Operator

 

The projector in 2 dimensions can be written in the form

[math]\mathbf{P}_n = \frac{(\mathbf{I} + n \cdot \sigma )}{2} = |\psi_n><\psi_n|[/math]

This does introduce the traceless Pauli spin matrices where [math]n \in R^3[/math] which is known as the Bloch sphere.

So n has to be the real unit three vector [math](n_x,n_y,n_z)[/math]. The Pauli matrices are

[math]\sigma_3 = \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix}[/math]

and has eigenvalues of +1 and -1.

[math]\sigma_1 = \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}[/math]

A property of these two matrices is that if you square them, you just get the identity back. The last one is

[math]\sigma_2 = \begin{pmatrix} 0 & -i \\i & 0 \end{pmatrix}[/math]

and

[math]n \cdot \sigma = \begin{pmatrix} n_3 & n_{-} \\ n_{+} & -n_3 \end{pmatrix}[/math]

Which explains one component of this projective space. This has some importance to the projection operator in which we found makes an appearance in an equation we derived as:
 

[math]|\Psi >_{\psi_{emp},\psi_{occ}}\ = \sum^{N}_{n = 0}\ <\psi_{emp}|\mathbf{P}_n|\psi_{occ}>\ |\psi_n>_{\psi_{emp},\psi_{occ}}[/math]

Edited by Dubbelosix
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Primer on Possible Lorentz Violations and Particle Creation

 

 

A Lorentz violating vector field is actually part of a formalism to obtain absolute accelerations.

 

 

In particle physics, a scalar field getting a vacuum expectation is the most common among scientists since it also can givbe rise to spin zero particles (ie. any class of Higgs boson).

 

If a vector field, like the notation we have been using for velocity [math]\mathbf{U}^{\mu}[/math] has a non-zero expectation value, it will violate Lorentz invariance. Sean Carrol mentions

 

''this is ok, but the subject to interesting experimental constraints''

 

Alternatively, an expectation of a gravitational scalar, produces particles -

 

[math]<\phi>\ =\ <\frac{Gm}{r}>\ = \sum_n [a_nf_n + a^{\dagger}_{n}f^{*}_{n}][/math]

 

I investigated this type of quantization and suggested in my Friedmann investigations that the particle production in highly curved spaces (like the early universe) could happen in an irreversible way (short duration of conservation law breakdown).

 

Above, we have used the creation and annihilation operators, they follow the standard rules:

 

[math]a|n>\ \sqrt{n_k + 1}|n_k + 1>[/math]

 

[math]a^{\dagger}|n> = \sqrt{n_k + 1}|n_k - 1>[/math]

 

The expectation for the particle number is equivalent to:

 

[math]<N>\ = \sum_k aa^{\dagger}[/math]

 

The operators also follow a commutation rule

 

[math][a,a^{\dagger}] - aa^{\dagger} - a^{\dagger}a = \mathbf{I}[/math]

 

and so this can be arranged like so:

 

[math]aa^{\dagger} = a^{\dagger}a + 1[/math]

 

The operations on the left hand side is basically the same as the summation of the particle number:

 

[math]<N>\ = \sum_{k} aa^{\dagger}[/math]

 

It's well known that particles and their antiparticles are entangled. We have a gravitational field potential coefficient in the equation we derived in the Planck phase space: I

 

[math]\mathbf{G} = \kappa\ \mathbf{T} = \kappa\ g_{\mu \nu}\ T^{\mu \nu} \equiv \frac{2 - D}{2}\ \mathbf{R} \geq \frac{8 \pi G}{c^4}\ (\frac{1}{\ell^2_{planck}}) \mathbf{U}_{\mu}\mathbf{U}^{\mu}\ \phi[/math]

 

It's interesting to note that it is possible also to take the operator form of this potential to validate particle production in the Planck space.

 

Primer on the Projection Operator

 

The projector in 2 dimensions can be written in the form

 

[math]\mathbf{P}_n = \frac{(\mathbf{I} + n \cdot \sigma )}{2} = |\psi_n><\psi_n|[/math]

 

This does introduce the traceless Pauli spin matrices where [math]n \in R^3[/math] which is known as the Bloch sphere.

 

So n has to be the real unit three vector [math](n_x,n_y,n_z)[/math]. The Pauli matrices are

 

[math]\sigma_3 = \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix}[/math]

 

and has eigenvalues of +1 and -1.

 

[math]\sigma_1 = \begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}[/math]

 

A property of these two matrices is that if you square them, you just get the identity back. The last one is

 

[math]\sigma_2 = \begin{pmatrix} 0 & -i \\i & 0 \end{pmatrix}[/math]

 

and

 

[math]n \cdot \sigma = \begin{pmatrix} n_3 & n_{-} \\ n_{+} & -n_3 \end{pmatrix}[/math]

 

Which explains one component of this projective space. This has some importance to the projection operator in which we found makes an appearance in an equation we derived as:

 

[math]|\Psi >_{\psi_{emp},\psi_{occ}}\ = \sum^{N}_{n = 0}\ <\psi_{emp}|\mathbf{P}_n|\psi_{occ}>\ |\psi_n>_{\psi_{emp},\psi_{occ}}[/math]

 

 

 

 

 

Now, we can expand a little on some of the equations we have presented, many of them just based on the standard rules found in quantum mechanics. As I stated:

 

'' [math][a,a^{\dagger}] - aa^{\dagger} - a^{\dagger}a = \mathbf{I}[/math]

 

and so this can be arranged like so:

 

[math]aa^{\dagger} = a^{\dagger}a + 1[/math]

 

The operations on the left hand side is basically the same as the summation of the particle number:

 

[math]<N>\ = \sum_{k} aa^{\dagger}[/math] ''

 

The last equation here is of course a number operator. The number operator can also act on Hermitian whose action on the number state is

 

[math]aa^{\dagger}|n>\ = n|n>[/math]

 

For this reason, it is also called a number operator because it will count how many particles you are dealing with in your apparatus. Even though it is valid to say an anticommutation applies to the first equation, all other commutations are zero:

 

[math]{a, a^{\dagger}} = {a^{\dagger},a^{\dagger}} = 0[/math]

 

And I suppose these could be constraints for classical physics alone, since anticommutation is the deviation from classical to quantum definitions.

Edited by Dubbelosix
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As I said, we can take the operator value of the gravitational field potential, this leads to particle creation in the Planck phase space:

 

[math]\hat{\mathbf{G}} = \kappa\ \hat{\mathbf{T}} = \kappa\ g_{\mu \nu}\ \hat{T^{\mu \nu}}  \geq \frac{8 \pi G}{c^4}\ (\frac{1}{\ell^2_{planck}}) \mathbf{U}_{\mu}\mathbf{U}^{\mu}\ <\phi>\ = \frac{8 \pi G}{c^4}\ (\frac{1}{\ell^2_{planck}}) \mathbf{U}_{\mu}\mathbf{U}^{\mu}\ \sum_{k} [af,a^{\dagger}f^{\dagger}][/math]

 

In which we have used:

 

[math]<\phi>\ = \sum_{k} [af,a^{\dagger}f^{\dagger}] = afa^{\dagger}f^{\dagger} - a^{\dagger}f^{\dagger}af [/math]

 

Perhaps also the last expression can be further simplified for the Schwarzschild constant:

 

[math]\frac{8 \pi G}{c^2}\ (\frac{1}{\ell^2_{planck}})  \sum_{k} [af,a^{\dagger}f^{\dagger}][/math]

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As I said, we can take the operator value of the gravitational field potential, this leads to particle creation in the Planck phase space:

 

[math]\hat{\mathbf{G}} = \kappa\ \hat{\mathbf{T}} = \kappa\ g_{\mu \nu}\ \hat{T^{\mu \nu}}  \geq \frac{8 \pi G}{c^4}\ (\frac{1}{\ell^2_{planck}}) \mathbf{U}_{\mu}\mathbf{U}^{\mu}\ <\phi>\ = \frac{8 \pi G}{c^4}\ (\frac{1}{\ell^2_{planck}}) \mathbf{U}_{\mu}\mathbf{U}^{\mu}\ \sum_{k} [af,a^{\dagger}f^{\dagger}][/math]

 

In which we have used:

 

[math]<\phi>\ = \sum_{k} [af,a^{\dagger}f^{\dagger}] = afa^{\dagger}f^{\dagger} - a^{\dagger}f^{\dagger}af [/math]

 

Perhaps also the last expression can be further simplified for the Schwarzschild constant:

 

[math]\frac{8 \pi G}{c^2}\ (\frac{1}{\ell^2_{planck}})  \sum_{k} [af,a^{\dagger}f^{\dagger}][/math]

 

 

However, I shall not be so quick to accept the simplification in the last expression because I could very well find an operator of the form:

 

[math]<\mathbf{U}\mathbf{U}\phi>[/math]

 

There are such velocity operators found in physics, here is a link I will study later and implicate into my theory.

 

https://quantummechanics.ucsd.edu/ph130a/130_notes/node498.html

 

It does involve notions of zitter motion which is heavily predicted in quantum mechanics - this ''circular internal motion'' will also be present at zero temperatures.

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To do what we proposed, assuming people have read the link, will dynamically involve matrix calculations, which I will do later as it takes time latex-wise. Notice also it writes it in the form of the gamma matrices often found within Dirac's own theory. The object we will calculate later will be of form:

 

[math]i^2c^2(\gamma_4 \gamma_j)(\gamma_4 \gamma_j)[/math]

 

The negative sign which will arise from the square of the imaginary, will in fact cancel with a negative term found in the potential so the right hand side should remain positive over-all.

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The velocity operator, does in fact possess complexification using the Lie brackets,

[math]\hat{v}_i = i[\hat{H}_0, x_i][/math]

In fact, according to the Newton Wigner approach, the relationship holds [math]\hat{H}_0 = \beta p_0[/math] then it is also written as

[math]\hat{v}_i \equiv i[\hat{H}_0, x_i] = i\beta[p_0,x_i][/math]

However, because of the square root found in the momentum:

[math]p^0 = \sqrt{m^2 + |p|^2}[/math]

There then requires an additional step which requires that the square of the mass may commute with the canonical coordinate, which features as:

[math][m^2,x] = [(p^0p_0 + p^jp_j), x_i] = [p^0p_0, x_i] + 2ip_i[/math]

The Heisenberg relationship is used here [math][x_i,p_j][/math] and an expression for [math][p_0,x_i][/math] which is represented as:

[math]i[p_0,x_i] = \frac{p_i}{p_0} = v_i[/math]

This is only a step away from being a velocity operator (from standard definition) but it features a middle term which I happen to know what it is. It is called the Newton-Wigner localization, a type of position operator under normal schemes.

The reason why such a local approach was taken was so that the particle would move at the expected velocity appropriate for its given momentum and energy.

[math][x_i,p_0] = \frac{p_i}{p_0} = i \hbar[/math]

Zitter motion in fact arises from:

[math][m,x_i] = [(\gamma^0\hat{H}_0 + \gamma^j p_j), x_i] = [\gamma^0\hat{H}_0, x] + i\gamma_i[/math]

However wiki states that it is generally believed that this term vanishes! This does seem to be a statement at odds with the predictions of quantum mechanics. Also make note some constants have been normalized in case it dimensionally looks dubious. A good follow up on the position operator and why Wigner proposed them can be found in the second reference.

You obtain the velocity operator by multiplying through the last equation with [math]\gamma^0[/math] and then rearranging (see reference below);

[math]\hat{v}_i = \frac{d\hat{x}}{dt} \equiv i[\hat{H}, x_i][/math]


REF

https://en.wikipedia.org/wiki/Foldy%E2%80%93Wouthuysen_transformation

http://www1.jinr.ru/Proceedings/Burdik-2004/pdf/mir-kasimov.pdf

[from the second reference, it is especially interesting they have formulated a different version of the Newton-Wigner localization which abandons the commutative properties: They conclyde that the geometry of the momentum space, which is, the manifold of realizable states of the relativistic particle of positive frequency is the Lobachevsky space. They develop a one-particle relativistic theory ~ the key to obtain the position operator is to accept that the position operator is the hermitian part of the ordinary position operator [math]\hat{x} = i \hbar \nabla_p[/math]:

[math]\hat{x} = \frac{1}{2}[\hat{x} + \hat{x}^{\dagger}] = i \hbar \nabla_p - \frac{i \hbar}{2}\frac{\mathbf{P}}{\mathbf{P}^2 + m^2c^2}[/math]

From their own logical conclusions, the model satisfies a condition which excludes discontinuous states as localized ones; this feature is essential they argue in fixing the phase of the localized state wave function! The relativistic position operator they propose is:

[math]\hat{x} = i\hbar\sqrt{1 + \frac{\mathbf{P}}{m^2c^2}}\ \nabla_p[/math]

The most important conclusion is that in the configuration space, also known as the phase space, the position operator is a non-local operator. Similar to the phase space idea which proposed that there cannot be such a thing as a point in space but in fact is smeared into what was once named a ''Planck cell'', the localized eigenfunction is not [math]\delta(x - y)[/math] as  in the non-relativistic case, but instead is a function smeared in a spatial region around the size of a Compton wave length; this conclusion is deducted from the fact that [math]\delta(x - y)[/math] cannot be constructed from positive frequency solutions only.]

https://en.wikipedia.org/wiki/Foldy%E2%80%93Wouthuysen_transformation

notes:

1. Any type of circular motion (this includes zitter motion) involves the concept of absolute acceleration, similar to how relativity states everything in the universe is in motion with everything else, this even reduces to the world of particle physics. Therefore particles can be said to be in constant acceleration relative to an internal degree of motion.

2. Restoring units for the velocity operator is relatively easy, it just looks like:

[math]\hbar \frac{d\hat{x}_i}{dt} = i[\hat{H},x_i] = \hbar c\ \alpha_i = Gm^2\ \alpha_i[/math]

Where

[math]\alpha_i = e^{\frac{i \hat{H}t}{\hbar}}\alpha_ie^{-\frac{i \hat{H}t}{\hbar}}[/math]

 

Edited by Dubbelosix
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The equation with the square of the mass interests me the most because anything with a square mass can be represented by a gravitational charge. The mass squared term is also recognized in string theory. It's also recognized in some parts of field theory.

 

I wrote a whole load about that before and I might revisit it for a commutation principle for the gravitational charge, but for now the equation of interest is:

 

[math][m^2,x] = [(p^0p_0 + p^jp_j), x_i] = [p^0p_0, x_i] + 2ip_i[/math]

 

The mass can be written as

 

[math]m = \gamma^0\hat{H}_0 + \gamma^{i}p_i[/math]

 

which implies we must factor out the squared mass term:

 

[math]m^2 = (\gamma^0\hat{H}_0 + \gamma^{i}p_i)(\gamma^0\hat{H}_0 + \gamma^{i}p_i)[/math]

 

So this will be physics we will look into as well, all standard of course. So that I am not too cryptic, the gravitational charge commutation principle would look like:

 

[math][m^2,\hat{H}\ x] = [m^2, \hat{v}][/math]

 

where [math]\hat{v}[/math] is the velocity operator because of the commutation between [math]\hat{H}[/math] and [math]x[/math]. Note we have also come across the charge of the form directly from an operator:

 

[math]\hbar \frac{d\hat{x}_i}{dt} = i[\hat{H},x_i] = \hbar c\ \alpha_i = Gm^2\ \alpha_i[/math]

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The Einstein field equation can be directly written in an alternative form:

[math]\mathbf{G}_{\mu \nu} = R_{\mu \nu} - \frac{1}{2}R (\gamma_\mu \cdot \gamma_{\nu}) = \mathbf{T}(\gamma_\mu \gamma_{\nu} + \gamma_{\nu} \gamma_{\mu})[/math]

We recognize we have written the metric in a new way using the gamma matrices under Hestene's approach:

[math]g_{\mu \nu} = \gamma_\mu \cdot \gamma_{\nu} = \gamma_\mu \wedge \gamma_{\nu} [/math]

 

Which is a bi-vector. Normally orthogonality ensures that

 

[math]\gamma_{\mu} \cdot \gamma_{\nu} = 0[/math]

A geometric product of two vectors u and v can be interpreted by decomposing it into symmetric and skew-symmetric parts:

[math]uv = u \cdot v + u \wedge v[/math]

[math]u \cdot v = \frac{1}{2}(uv + vu)[/math]

[math]u \wedge v = \frac{1}{2}(uv - vu) = -v \wedge u[/math]

This is a scalar, of course and brings about the geometric interpretations - Hestene deducts:

''Thus,  the geometric relation of orthogonality is expressed algebraically by ananticommutative geometric product.  Similarly, collinearity is expressed by a commutative geometric product.''

 

We realize from his relationships that it can be rewritten:

 

[math]\mathbf{G}_{\mu \nu} = R_{\mu \nu} - \frac{1}{2}R (\gamma_\mu \cdot \gamma_{\nu}) = \frac{8 \pi G}{c^4}\mathbf{T}(\gamma_\mu \gamma_{\nu} + \gamma_{\nu} \gamma_{\mu}) [/math]

 

 

So seemingly from Einstein's equations, you can still retrieve a new form of it which involves commutation laws under Hestenes model.

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This will follow on from the last blog post in which we introduced Hestene’s geometric algebra to the Einstein equation, which was adapted into an inequality relationship to the Planck phase space in which we have introduced particle production from taking the operator of the potential. In the natural units of [math]8 \pi G = c = 1[/math] the equation is:

 

[math]\mathbf{G}_{\mu \nu} = R_{\mu \nu} - \frac{1}{2}R (\gamma_\mu \cdot \gamma_{\nu}) = \mathbf{T}(\gamma_\mu \gamma_{\nu} + \gamma_{\nu} \gamma_{\mu})  \geq (i <[\hat{H}, x]> \frac{1}{\ell_{planck}})^2 \sum_{k} [af,a^{\dagger}f^{\dagger}][/math]

 

[math]= (\gamma_0\gamma_1\gamma_2\gamma_3\ \frac{<[\hat{H},x]>}{\ell_{planck}})^2 \sum_{k} [af,a^{\dagger}f^{\dagger}] \geq \frac{1}{4}(\frac{\hbar^2}{\ell_{planck}})^2 \sum_{k} [af,a^{\dagger}f^{\dagger}][/math]

 

Since

 

[math][\gamma_{\mu},\gamma_{\nu}] = \gamma_{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma_{\mu} = 2 \eta_{\mu \nu} \mathbf{I}_4[/math]

 

Then we also notice a simplification in the modified Einstein field equations:

 

[math]\mathbf{G}_{\mu \nu} = R_{\mu \nu} - \frac{1}{2}R (\gamma_\mu \cdot \gamma_{\nu}) = \mathbf{T}(\gamma_\mu \gamma_{\nu} + \gamma_{\nu} \gamma_{\mu}) = 2 \eta_{\mu \nu} \mathbf{I}_4\ \mathbf{T}[/math]

 

Notation simplification is our friend.

Edited by Dubbelosix
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Though it may look complicated, we did in fact simplify the approach because the true geometric meaning arises from splitting up the equation into symmetric and skew-symmetric parts, and so long as I have understood the process; decomposing the product of vectors into their scalar and vector components as found from:

 

[math]uv = u \cdot v + u \wedge v = \frac{1}{2}(uv + vu) + \frac{1}{2}(uv - vu) = \frac{1}{2}(uv + vu) - v \wedge u[/math]

 

In which we have already identified:

 

[math]u \cdot v = \frac{1}{2}(uv + vu)[/math]

 

[math]u \wedge v = \frac{1}{2}(uv - vu) = -v \wedge u[/math]

 

I suspect the same form can be provided into the Einstein equations:

 

[math]\mathbf{G}_{\mu \nu} = \mathbf{T}(\gamma_\mu \gamma_{\nu}) = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu}) = \frac{1}{2}\mathbf{T}[(\gamma_{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma_{\mu}) - \gamma_{\nu} \wedge \gamma_{\mu}][/math]

Edited by Dubbelosix
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Though it may look complicated, we did in fact simplify the approach because the true geometric meaning arises from splitting up the equation into symmetric and skew-symmetric parts, and so long as I have understood the process; decomposing the product of vectors into their scalar and vector components as found from:

 

[math]uv = u \cdot v + u \wedge v = \frac{1}{2}(uv + vu) + \frac{1}{2}(uv - vu) = \frac{1}{2}(uv + vu) - v \wedge u[/math]

 

In which we have already identified:

 

[math]u \cdot v = \frac{1}{2}(uv + vu)[/math]

 

[math]u \wedge v = \frac{1}{2}(uv - vu) = -v \wedge u[/math]

 

I suspect the same form can be provided into the Einstein equations:

 

[math]\mathbf{G}_{\mu \nu} = \mathbf{T}(\gamma_\mu \gamma_{\nu}) = \frac{1}{2}\mathbf{T}(\gamma_{\mu}\cdot \gamma_{\nu} + \gamma_{\mu} \wedge \gamma_{\nu}) = \frac{1}{2}\mathbf{T}[(\gamma_{\mu}\gamma_{\nu} + \gamma_{\nu}\gamma_{\mu}) - \gamma_{\nu} \wedge \gamma_{\mu}][/math]

 

I've had quite a lot of reading to do, basically getting primers on geometric algebra. I have read what may have appeared to be conflicting evidence that it is a general rule that you cannot add a scalar with a vector, and essentially that is what this does:

 

[math]uv = u \cdot v + u \wedge v[/math]

 

How do you determine that this is a product of a symmetric and skew symmetric matrix? A general thing to remember is that skew symmetries are expressed using cross products, It is also of interest to define the dyad:

 

''The dot product takes in two vectors and returns a scalar, while the cross product returns a pseudovector. Both of these have various significant geometric interpretations and are widely used in mathematics, physics, and engineering. The dyadic product takes in two vectors and returns a second order tensor called a dyadic in this context. A dyadic can be used to contain physical or geometric information, although in general there is no direct way of geometrically interpreting it. The dyadic product is distributive over vector addition, and associative with scalar multiplication. Therefore, the dyadic product is linear in both of its operands. In general, two dyadics can be added to get another dyadic, and multiplied by numbers to scale the dyadic. However, the product is not commutative; changing the order of the vectors results in a different dyadic.''

 

from wiki. For two vectors [math](u.v)[/math] the geometric produict is the sum of a symmetric product and antisymmetric product;

 

[math]ub = \frac{1}{2}(uv + vu) + \frac{1}{2}(uv - vu)[/math]

 

The inner product of the vectors is

 

[math]u \cdot v := g(u,v)[/math]

 

So that a symmetric product is expressed as

 

[math]\frac{1}{2}(uv - vu) = \frac{1}{2}((u + v) - u - v)^2 = u \cdot v[/math]

 

It is stated in what I have read, that conversely [math]g[/math] is completely determined by the algebra and the antisymmetric part of the exterior product of the two vectors is

 

[math]u \wedge v := \frac{1}{2}(uv - vu) = -(v \wedge u)[/math]

 

then by addition we obtain

 

[math]uv = u \cdot v + u \wedge v[/math]

 

which is the vector form of the geometric product. Taking the dot product of the Pauli vector (which is a dyad) is in fact the same thing

 

[math](\sigma \cdot u)(\sigma \cdot v) = u \cdot v + u \wedge v[/math]

 

An equivalent form of this, of course involves the cross product, which as explained, is given by its antisymmetric parts:

 

[math]u \cdot v + i \sigma \cdot (u \times v)[/math]

 



			
		
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