# My Favorite Equations I Have Derived Over The Years.

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### #1 Dubbelosix

Dubbelosix

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Posted 27 May 2019 - 10:44 AM

The gravimagnetic field defined for magnetic coupling to orbit is, as the master equation we will work from:

$\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{me}\frac{a}{G} \mathbf{J} = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}$

where we have used the relationships I derived for clarity:

$\frac{1}{e}\frac{\partial U}{\partial r} = \frac{c^2}{e} \frac{\partial m}{\partial r}$

$\frac{1}{r}\frac{\partial m}{\partial r} = \frac{\omega^2r}{G} = \frac{a}{G}$

We derive this from the ordinary spin coupling equation which has dimensions of:

$\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r}\frac{\partial U}{\partial r} \mathbf{J}$

With an electric field identified as:

$\mathbf{E} = \frac{1}{e} \frac{\partial U}{\partial r}$

The gravielectric formula is just

$\nabla \cdot \mathbf{E} = -\frac{\rho}{\epsilon_G} = -4 \pi G \rho$

Comparing this with Maxwell’s equation for electric field,

$\nabla \cdot \mathbf{E} = \frac{\rho}{\epsilon}$

There is no real formal difference, with

$\frac{1}{\epsilon_G} = 4 \pi G$

$\frac{1}{\mu_G} = \frac{c^2}{4 \pi G}$

The units though for the potential are the same as the gravitational potential, which we find the relationship:

$\mathbf{E} = -\nabla \phi = \frac{1}{r} \frac{Gm}{r}$

and so units become obvious when:

$\nabla \cdot \mathbf{E} = -G \frac{m}{r^3} = -G \rho$

In our equations, the factor of G will drop out of the definition for the gravielectric field . The gravielectric field from the master equation requires only a distribution of a factor of the speed of light

$\mathbf{E} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J}c = \frac{1}{me}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J}c = -\frac{1}{e}\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}c = -\frac{1}{me}\frac{a}{G} \mathbf{J}c = -\frac{1}{me}\frac{\omega^2 r}{G} \mathbf{J}c = -\frac{1}{m e} \frac{m}{r^2} \mathbf{J}c$

The gravimagnetic field can written as:

$\mathbf{B} = -\frac{1}{2m e} \frac{m}{r^2} \mathbf{J} = -\frac{1}{m }\frac{\mathbf{J}}{2e} \frac{m}{r^2}$

So that it has the same solution as the main gravimagnetic field we opened with, but in this case, we retrieve the definition of the Josephson constant with the wanted coefficient of 2 attached to the charge.

$\Phi = \frac{h}{2e} = 2.067833 \times 10^{15}Wb$

...this quantity arising exactly like it has, in context of electromagnetism, is no coincidence in my eyes.

The gravimagnetic field was given as:

$\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J}$

With an electric field

$\frac{1}{e}\frac{\partial U}{\partial r} \rightarrow \frac{c^2}{e} \frac{\partial m}{\partial r}$

Which just demonstrates how the gravielectric field is part of the dynamics of the gravimagnetic field

$\mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{mc^2} \frac{\mathbf{E}}{r}\mathbf{J} = \frac{r \times p}{mc^2} \frac{\mathbf{E}}{r} = \frac{r \times p}{me} \frac{\partial^2 m}{\partial r^2}$

and the curl of the field is

$\nabla \times \mathbf{B} = \frac{1}{mc^2 e} \frac{1}{r} \frac{\partial^2 U}{\partial r^2} \mathbf{J}$

Where $\frac{\mathbf{J}}{e^2}$ is the Von Klitzing constant. Distribution of the charge is just:

$e \mathbf{B} = \frac{1}{mc^2 } \frac{1}{r} \frac{\partial U}{\partial r} \mathbf{J} = \frac{1}{m}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{m}\frac{a}{G} \mathbf{J} = -\frac{1}{m}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m} \frac{m}{r^2} \mathbf{J}$

and with the curl gives a spin density:

$e (\nabla \times \mathbf{B}) = -\frac{\mathbf{J}}{r^3}$

and a relationship to the gyromagnetic ratio:

$\gamma(\nabla \times \mathbf{B}) = \frac{e}{2m}(\nabla \times \mathbf{B}) = \frac{\mathbf{J}}{2e^2} \frac{1}{m} \frac{\partial^2 U}{\partial r^2} = \frac{\mathbf{J}}{2e^2} \frac{1}{mc^2} \frac{\partial^2 U}{\partial t^2} = -\frac{1}{2m^2} \frac{m}{r^3} \mathbf{J}$

identifying some terms for clarity

$mc^2\ r = Gm^2 = \hbar c = \mathbf{J}c = e^2$

$e \mathbf{B} = \frac{ \mathbf{J}}{e^2 } \frac{\partial U}{\partial r}= \frac{1}{m}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{Gm}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{m}\frac{a}{G} \mathbf{J} = -\frac{1}{m}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{m} \frac{m}{r^2} \mathbf{J}$

And the torsion-like field:

$\omega = -\frac{\Omega}{2} = \frac{e \mathbf{B}}{2m} = \frac{\mathbf{J}}{2e^2} \frac{1}{m}\frac{\partial U}{\partial r} = \frac{1}{2m^2}(\frac{\phi}{c^2})\frac{\partial v}{\partial t} \mathbf{J} = -\frac{1}{Gm^2}\frac{\partial v}{\partial t} \mathbf{J}= -\frac{1}{2m^2}\frac{a}{G} \mathbf{J} = -\frac{1}{2m^2}\frac{\omega^2 r}{G} \mathbf{J} = -\frac{1}{2m^2} \frac{m}{r^2} \mathbf{J}$

With

$\frac{m\Omega}{2} = \frac{e \mathbf{B}}{2} = \frac{\mathbf{J}}{2e^2}\frac{\partial U}{\partial r}$

With the curl of the field:

$\nabla \times \Omega = \frac{e}{2m} \frac{\partial \mathbf{B}}{\partial r} = \frac{\mathbf{J}}{2e^2} \frac{1}{mc^2}\frac{\partial^2 U}{\partial t^2}$

$\gamma = \frac{e}{2m} =\ gyromagnetic\ ratio$

The gravimagnetic field in standard physics is defined as:

$\mathbf{B} = \Omega$

And is known as the torsion field. The definition of the gravimagnetic field then under standard approach is to treat it with the same dimensions as frequency. The frequency field is related to a central potential:

$\Omega^2 = \frac{1}{mc^2}(\frac{d^2U}{dt^2})$

In which case, since the gravimagnetic field is defined by the torsion field, the torsion field itself, has dimensions of inverse time which is a deviation from the units we have been using because we have tried to identify the gravimagnetic field under conventional units for the magnetic field.

Spin coupling interaction is described by a Hamiltonian as:

$H = \Omega \cdot \mathbf{J}$

(see standard spin-orbit coupling and torsion related works by Arun and Sivaram to find reference to the coupling equation)

Though related to gravimagnetism, $\Omega$  a notation kept for torsion field to distinguish it from frequency which has same units. Another relationship

The second work I am most proud of, but has a lot of details I summarize that will make the model work, is a very... simple set of arguments towards the quantization of a black hole, ultimately coming to the conclusion there are no black hole particles in nature. If they do exist, they cannot exist for long and probably on Planck scales which makes them unobservable, (for now). Also it led to  a theoretical argument in which accelerated black holes may be subject to the weak equivalence of accelerated charges.

Quantization of the Black Hole Particle

To derive a quantization of the black hole, in terms of discrete quantum mechanical processes, I defined the Rydberg constant in terms of the gravitational coupling constant we get:

$\mathbf{R} = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$

Even though the Rydberg constant was first applied to hydrogen atoms, it could be derived from fundamental concepts (according to Bohr). In which case we may hypothesize energy levels:

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

Plugging in the last expressions we get an energy equation:

$\Delta E_G = \frac{n\hbar c}{\Delta \lambda} = \frac{1}{4 \pi }\frac{m_0v^2}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{1}{4 \pi \lambda_0}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2}) = \frac{p}{4 \pi \hbar}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2})$

The relativistic gamma appears from the definition of the deBroglie wavelength. The ultimate goal of creating this equation, was to explore an idea I can trace back to Lloyd Motz suggesting a stable black hole. Black holes, even the microscopic kind, cannot be mathematically ruled out from physics (but later I will show in a logical sense why the would be unstable) - and if the work of Hawking is to be taken seriously, including analysis provided by Motz, then a black hole system really is deduced from a discrete set of quantum processes - those discrete processes always lead to an increase in the entropy of a system like a black hole.

The earliest model of a black hole stated that the entropy of a black hole was exactly zero - Hawking challenged this and stated if it has an entropy, then a black hole must possess a temperature and later we came to understand it as a thermal property of the black body radiation of the black hole. It may still be true, at least for the ground state black hole, that entropy may be essentially constant.

So how can we have an increasing entropy but a system with an infinite lifetime?

Something needs to give. Entropy is a measure of disorder, and if a system is infinitely stable, then the disorder remains a constant. It could be that the black hole particle at near zero temperatures exhibits a behaviour like a zeno effect. The zeno effect is when an atom is incapable of giving off energy suspended in an otherwise, infinite animation and it will remain in a ground state because its wave function is incapable of evolving from the ground state.

It is therefore similar to how an atom, ready to give up energy can be suspended infinitely in time - and the reason why it cannot give up radiation is essentially the same for the micro black hole, since the zeno effect is all about keeping an atom in the ground state. To put in a summary for clarity I have came to some conclusions: The main one being that the entropy of a stable black hole would immediately render the black hole particle as a system at absolute zero! The reason why is because if a black hole particle did not have an entropy, by first principles, it consequently cannot give up any radiation (consequence of the third law of thermodynamics) and as far as we know, systems can never actually reach absolute zero temperatures, there is reasonable argument here that the black hole particle must evaporate. In fact, I wondered immediately if there are applications of this with the micro black hole and maybe in regards to its energy levels. We start with the equation

$S = k_B \frac{Gm^2}{\hbar c}$

Where the entropy has taken on the dimensions of the Boltzmann Constant. If you plug in the mass for a black hole:

$m = \frac{c^2R}{2G}$

You will retrieve the Bekenstein entropy. It is often said that entropy has dimensions of the Boltzmann constant but it's entirely plausible, even with better arguments for, a dimensionless case of entropy in the form

$\mathbf{S} = \frac{S}{k} = \frac{Gm^2}{\hbar c}$

In which case, it is possible to identify the entropy in a dimensionless form

$\mathbf{S} = \frac{\hbar c}{k_BT}\frac{p}{4 \pi \hbar} = \frac{\hbar c}{k_BT}\frac{\sqrt{2mE}}{4 \pi \hbar} = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

In which the wave number is identified with the deBroglie wavelength

$k \equiv \frac{p}{\hbar} = \frac{\sqrt{2mE}}{\hbar}$

Because of the equivalence of the wave number the definition of the deBroglie wavelength. The SI value for the Boltzmann constant is given in units of the Hartree energy divided by the Kelvin

$\frac{E}{K} = k_B = 3.1668114(29)\times 10^{−6}$

In which the Hartree energy is simply

$E = 2\mathbf{R}\hbar c$

and so we can write it with temperature T which is also measured in Kelvin,

$S = k_B = \frac{E}{T} = \mathbf{R}\frac{\hbar c}{T}$

This can be written in such a way to introduce those quantized atomic levels ~ by making use of the formula

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

giving

$k_B = \frac{1}{T} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1T} - \frac{Gm^2}{n^2_2T})$

In which $\mathbf{R}$ is the usual Rydberg constant. The Boltzmann constant can also be expressed entirely in terms of Planck units ~

$S = \frac{m_PL^2_P}{k_BT_P t^2_P}$

$k_B = \frac{\sqrt{\frac{\hbar c}{G}} \frac{\hbar G}{c^3}}{\sqrt{\frac{\hbar c^5}{Gk^2_B}} \frac{\hbar G}{c^5}}$

The Transition of Conducting Black Holes

Another theory in literature involves a theoretical suggestion which asserts the surface of a black hole [may] be considered as a conducting surface; the argument goes like this:

‘’ Introducing an effective thickness \Delta r of the membrane, and assuming the same resistivity $R_H$ within $\Delta r$, one can relate total resistance of the black hole with the conductivity of the membrane using a model of two concentric spheres with radii $a_1 = r_s$, $a_2 = r_s + \Delta r$ where the spherical shell is filled by homogeneous conducting matter is given by

$R_H = \frac{1}{4 \pi \sigma}(\frac{1}{r_1} - \frac{1}{r_2})$

Where $\sigma$ is the conductivity. ‘’

Early on it struck me how similar this equation is, compared to our transition equation using the Rydberg constant:

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

Fortunately for both equations, the inverse length features and for conductivity, (which is the inverse of electrical resistivity), the resistance of a system increases with the length of the system. This must mean that in each transition from one energy to another, the conductance must weaken, as shown:

$R_H = \frac{1}{\Delta( \sigma \lambda)} = \mathbf{R}(\frac{1}{\sigma n^2_1} - \frac{1}{\sigma n^2_2})$

From the last equation, you can calculate conductance G as

$G = \rho (\frac{A_{BH}}{R_1} - \frac{A_{BH}}{R_2}) = \frac{\mathbf{E}}{J}(\frac{A_{BH}}{R_1} - \frac{A_{BH}}{R_2})$

Where \rho is the resistance and is related to conductance by $\frac{1}{\sigma}$ and $\mathbf{E}$ is the electric field where $J$ is the current. Also it is good to note, the shape operator for a sphere is mathematically described under the Weingarten equations. However, literature hints that the equation does not depend on the geometry of the system, since the resistivity and conductivity are proportionality constants; in such a case it is said it depends only on the material the system is made of but not the geometry, which robs any importance with viewing it under a shape operator. The black hole area is defined by

$A_{BH} = \frac{G^2m^2}{c^4}$

And the angular momentum per unit area is

$\mathbf{J} = \frac{J}{A_{BH}} = \frac{c^3}{G}$

Which enters the resistivity equation like;

$R_H = \frac{c^4}{G^2m^2}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2}) = \frac{\mathbf{J} c}{Gm^2}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2})= \alpha_G\mathbf{R}(\frac{1}{\sigma n^2_1} - \frac{1}{\sigma n^2_2})$

We have retrieved the gravitational charge (squared) in the denominator. The ratio

$\alpha_G = \frac{Jc}{Gm^2} \approx 1$

can be interpreted as a fine structure constant, as following the same idea but reached a different derivation of the same argument from Arun and Sivaram from their paper ‘’some unique constants associated to extremal black holes.’’

The equation can further be distributed with a factor of \hbar c, similar to the Rydberg formula to find the energy. we get

$\hbar c\ R_H = \frac{c^4}{G}(\frac{R_1}{\sigma n^2_1} - \frac{R_2}{\sigma n^2_2}) = \alpha_G(\frac{E_1}{\sigma n^2_1} - \frac{E_2}{\sigma n^2_2})$

With this time, a classical gravitational upper limit of the force $\frac{c^4}{G}$. According to Arun and Sivaram for a black hole, the angular momentum and the square of the charge is related to the speed of light via ~

$\frac{J}{Q^2} = \frac{1}{c}$

It has crossed my mind whether it would be related more fundamentally to the gravitational aether which is defined in this work by the variable parameters $\sqrt{\epsilon_G \mu_G} = \frac{1}{c}$. Either way it was noted by the astrophysicists that it is remarkably similar to the Von Klitzig constant $\frac{\hbar}{e^2}$ or the Josephson constant in superconductors $\frac{e^2}{\hbar}$ - resistivity decreases in the case of superconductors, while it increases for metals as they increase temperatures.

We deducted not long ago that for a black hole, the conductance decreases (increase in resistivity) for a black hole as it gets smaller, this will concur nicely with the later investigation, because the equations so far tell us that in the ground state, conductance must be zero (to ensure shielding from environment) and will complement the idea of a ‘perfect insulator’ as the smallest possible radiator. While it remains curious science may predict a perfect insulator, the physics systematically argues against it due to the third law.

Differences Between The Holeum Model and Mine

Holeums were proposed by L. K. Chavda and Abhijit Chavda in 2002 and is suggested to be a form of cold ‘’dark’’ matter. The system itself is not a black hole and they consequently radiate gravitational waves as opposed to the usual electromagnetic Hawking radiation - later we will see why I do not believe a Planck particle (black hole particle) is stable at all, concerning contradictions that arise from the laws of thermodynamics, never mind two bound stable black holes.

The suggested binding energy of a bound black hole particle-pair is:

$E_n = - \frac{mc^2 \alpha_G^2}{4n^2}$

I came across the Holeum model after I constructed my approach to the quantization transition formula. I began from arguments concerning the Rydberg constant and asserted this would be a true analogous model to a ground state hydrogen atom in which the radiation inside the black hole attains its lowest quantum bound state.

Interestingly, it is claimed their model too is analogous to the ground state hydrogen atom, but notice, the models are totally different fundamentally. My approach is direct towards the treatment of a single black hole particle, not a bound pair. Notice also they used the gravitational fine structure, which was of similar contention to my own derivation that came from a modification of the Rydberg constant

$\mathbf{R} = \frac{\alpha_G^2}{4 \pi \lambda_0} = \frac{1}{\hbar c}\frac{Gm^2}{4 \pi \lambda_0} = \frac{1}{\hbar c} \frac{Gm^3c}{4 \pi \hbar} = \frac{Gm^2}{\hbar c}\frac{p}{4 \pi \hbar}$

or as the gravitational Rydberg-Hartree energy ~

$\hbar c\ \mathbf{R} = \frac{\hbar c\alpha_G^2}{4 \pi \lambda_0} = \frac{Gm^2}{4 \pi \lambda_0} = \frac{Gm^3c}{4 \pi \hbar} = Gm^2\frac{p}{4 \pi \hbar}$

• this is logical for a true analogue model since the constant is fundamental to the Bohr model derivation of the hydrogen atom it is a statement and Bohr maintained the derivation itself, is fundamental to many systems.

If for instance, we can prove later it doesn’t make sense to talk about one stable black hole particle, it would certainly not make any sense to think any stable bound black holes exist in nature.

An Argument For Temperature Variations With Black Hole Acceleration

A postulate in this work is that black holes when accelerated will obey their own radiation power law similar to an accelerated charge following the Larmor radiation law. It’s not an ad hoc hypothesis, since a micro black hole with a charge would have to obey the weak equivalence principle and give up radiation in a gravitational field in the form of the usual Hawking process.

In fact, early theories attempted to explain the electron as a black hole, and they must have crossed the question of what Larmor radiation would be: And if an electron had been a black hole, the radiation from the accelerated charged black hole would have had to expressed itself through the usual Hawking emission process.

A charge in a gravitational field should give off an electromagnetic radiation. This can be formulated in the following way:

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{\sqrt{g_{tt}}(r)}{\sqrt{g_{tt}}(s)})}$

The metric $g_{tt}$ is made up of terms of $1 + \frac{\phi}{c^2}$. Higher terms then of the field are calculated like so:

$1 + \phi_1 + \phi_2 - \frac{1}{2}\phi^2_1 - \phi_1\phi_2... higher\ terms$

To first order it is just the difference formula

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

However, employing a connection made by Motz, there is a correction to the denominator of the first equation,

which he notes should be squared because there is one such term to account for in the metric ds. In our case it is simply the removal of the square root signs. A second difference between the approach I took and the one they have taken is that their denominator consists of only one such term of the metric $(1 - \frac{R}{r})$ whereas mine considered the ratio of the two signals in a relativistic way that satisfies the red shift.

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2$

His interpretation of the squared denominator ensures that the radiation increases as something falls towards a singularity. Of course, modern approaches are to involve situations which don't involve singularities at all, as we mentioned before.

The Motz-Kraft model is for the calculation of the luminosity from charges accelerating in gravitational fields. Quasars are essentially black holes with a dense electrically charged material around it and they produce a large luminosity for these reasons. In our case, we are saying however, the weak equivalence applies to all kinds of systems with electric charge: Including the black hole itself. In such a case, the only way a black hole could radiate is through Hawking radiation suggesting an analogue case of the Larmor radiation ~ and did in fact satisfy such a case with an uncanny analogue to the classical electron radiation par a factor of $\frac{2}{3}$ which is only an adjustable parameter - the radiation (I derived) given up by a black hole is:

$\frac{Q^2a^2}{ c^3} + \frac{a^2}{Gc}(\frac{J}{m})^2 \leq \frac{a^2mr_g}{c}$

I obtained this nice simple solution from the known inequality satisfying a black hole with mass, charge and angular momentum (in natural units):

$Q^2 + (\frac{J}{m})^2 \leq m^2$

Notice this last term in the inequality is formally identical to the classical equation describing the radiation for an electron, with the exceptions that neither the mass or radius are associated to the electron but instead the black hole:

$P = \frac{2}{3}\frac{a^2m_er_e}{c}$

Since this too describes the power emitted it would not be wrong to set it in the same physics, in a gravitational field - doing so is childsplay, I get while adopting the coefficient for a true analogue theory:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3}\frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2}$

It is also true that the luminosity is a simple calculation which measures the power emitted by an astrophysical object, but we were looking for a more general case above for any charged object. The relationships for luminosity are briefly covered ~

$L = \frac{f}{A(1 + z)^2} = \frac{\sigma T^4}{4 \pi R^2(1 + z)^2}$

You can calculate the two different luminosities as

$\frac{L_1}{L_2} = \frac{(\frac{\sigma T^4_1}{4 \pi R^2_1})}{(\frac{\sigma T^4_2}{4 \pi R^2_2})}= \frac{T^4_1 R^2_1}{T^4_2R^2_2}$

So an alternative formula to calculate power is suggested as:

$P = \frac{2}{3}\frac{a^2}{(1 - \frac{R}{r})^2}(\frac{Q^2}{c^3} + \frac{J^2}{Gm^2c}) \leq \frac{2}{3} \frac{mr_g}{c}\frac{a^2}{(1 - \frac{R}{r})^2} = \frac{2}{3} \frac{r_g}{mc}\frac{1}{(1 - \frac{R}{r})^2}(\frac{dp}{dt})^2$

The last term arises from:

$P = \frac{2}{3}\frac{Q^2}{m^2c^3} (\frac{dp}{dt})^2$

And can be found by replacing the acceleration in our equation for $(\frac{1}{m}\frac{dp}{dt})^2$ - this is Lorentz invariant but it isn't generally relativistic.

The true generalized formula is

$P = -\frac{2}{3} \frac{Q^2}{m^2c^3}\frac{dp_{\mu}}{d\tau}\frac{dp^{\mu}}{d \tau}$

Since there are less dynamics in the luminosity relationship, the deduction from our equation (which involves the black hole mass, electric charge and angular momentum) is that a charged black hole will radiate more energy than one at rest: This partially makes sense of my stipulation that a moving black hole will appear hotter than one at rest, though the presence of red shift would mean it is also direction dependent. In units of $G = c = 1$ our equation to first order becomes:

$P = \frac{2}{3}\frac{a^2}{(1 - \Delta \phi)^2}(Q + \frac{J}{m})^2 \leq \frac{2}{3} mr_g\frac{a^2}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2} = \frac{2}{3} \frac{r_g}{m}\frac{1}{(1 - \frac{m}{r_1r_2}(r_2 - r_1))^2}(\frac{dp}{dt})^2$

with

$\Delta \phi = \phi_1 - \phi_2$

and

$\phi_1 - \phi_2 = \frac{m}{r_1r_2}(r_2 - r_1)$

Let’s just say a bit about the standard equations you might encounter which also attempt to calculate the power emitted by black holes. We know now that accelerated matter aroud the black hole contributes significantly to the light that comes from superluminous objects and so, this has offered a novel retake of the theory that Motz suggested first and later collaborated with Kraft.

ref: https://quantizedblackhole.quora.com/

Edited by Dubbelosix, 23 June 2019 - 04:52 PM.

### #2 SaxonViolence

SaxonViolence

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Posted 28 May 2019 - 02:58 PM

Friend,

Those are beautiful!

I only wished that I had the Mathematical ability to understand them.

When I was in college I used to collect books of higher math simply because all the arcane formulas were beautiful to behold.

Sadly, all of my math books were stolen during a move. It really puzzles me. They filled a large plastic cooler. If the thief wanted the cooler, why not dump the 50 or 60-pounds of books? I doubt that a random sneak-thief also had a fondness for higher math.

Remember the old ditty:

"Tom Tom the Piper's Son;

"Stole a Pig and Away He Run!"

If my burglar gabbed the cooler and ran like Tom in the ditty...

He must have been quite disappointed when he got home and opened his prize!

Anyway

Saxon Violence

### #3 Dubbelosix

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Posted 28 May 2019 - 04:50 PM

I only wished that I had the Mathematical ability to understand them.

When I was in college I used to collect books of higher math simply because all the arcane formulas were beautiful to behold.

Sadly, all of my math books were stolen during a move. It really puzzles me. They filled a large plastic cooler. If the thief wanted the cooler, why not dump the 50 or 60-pounds of books?

You are assuming they had enough time to check what they are stealing. The immediate response of a psychopath or sociopath in the same circumstances, would reveal identical reactions. If a cooler contains a large amount something including doing so in haste, would simply remove it for a multiple set of reasons, the first being  a container of frozen foods, possibly hoping it would be quality meat.

This indicates to me, your robbers where local and only after the second-class goods they expected, instead, they ended up with valuable books, being contained in the freezer I suppose to remarkably halt the process of decay. My one question would be, how dry are these books? I mean, even in a frozen environment with the slightest moister can cause damage to a papyrus of any kind, which is why the dead sea scrolls lasted for so long, not because the environment was cold, but because conditions in the far east are far more hotter securing the notion how written language on paper had lasted for so long... which is anything between 3000-2000 years old.

Edited by Dubbelosix, 28 May 2019 - 04:55 PM.

### #4 Dubbelosix

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Posted 30 May 2019 - 06:33 AM

Since there has been much discussion on relative velocities and especially the velocity addition formula, I want to highlight that it is a very important feature which also decodes the mathematical idea's behind the red shift. We can rewrite the modified equation I adjusted to suit a more appropriate formula suggested from Motz and Kraft. That modified equation took the form in the opening post as

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2$

But we can rewrite this in the following way:

$P = \frac{2}{3} \frac{Q^2}{c^3} \frac{a^2}{(\frac{g_tt (r)}{g_tt(s)})} = \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(1 + z)})^2= \frac{2}{3} \frac{Q^2}{c^3} (\frac{a}{(\frac{c + v}{c - v})})^2$

All this does is re-express the red shift according to the velocity additions. But during my thoughts the other day, I did come to realize there is a possible extension to the quantization of the black hole. I will write this up later. It involves a postulated correction to the aberration of light around the black hole system.

Edited by Dubbelosix, 30 May 2019 - 06:37 AM.

### #5 SaxonViolence

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Posted 31 May 2019 - 12:46 AM

Friend,

'

The books didn't live in the cooler. The house was flooded with sewage when the sewer backed up.

All books on the first shelves of my personal library were ruined. I grabbed the books and shoved them into the cooler as a temporary expedient simply because it was water-tight and convenient.

Those were all textbooks that the bookstores used to sell for a buck or two after they weren't used anymore—so although they're precious to me, they weren't "Valuable."

I failed out of Purdue University 3 times because I simply could not grasp calculus—and calculus turned into a dinosaur that ate all of my study time and caused me to fail all my other classes.

Over half my books were various calculus texts. Even though my college career was already over, I hoped one day to find a book about calculus that I could understand.

……Saxon Violence

### #6 Dubbelosix

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Posted 31 May 2019 - 07:19 AM

Don't be dismayed by the title, but a ''Dummies guide to calculus'' would be my suggestion.

### #7 VictorMedvil

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Posted 31 May 2019 - 07:56 AM

Don't be dismayed by the title, but a ''Dummies guide to calculus'' would be my suggestion.

I hated Calculus

Friend,

'

The books didn't live in the cooler. The house was flooded with sewage when the sewer backed up.

All books on the first shelves of my personal library were ruined. I grabbed the books and shoved them into the cooler as a temporary expedient simply because it was water-tight and convenient.

Those were all textbooks that the bookstores used to sell for a buck or two after they weren't used anymore—so although they're precious to me, they weren't "Valuable."

I failed out of Purdue University 3 times because I simply could not grasp calculus—and calculus turned into a dinosaur that ate all of my study time and caused me to fail all my other classes.

Over half my books were various calculus texts. Even though my college career was already over, I hoped one day to find a book about calculus that I could understand.

……Saxon Violence

I hated Calculus 2 , I passed with a really low grade in calculus 2 it was the class i did the worst on during college and failed it once, but my favorite class was biochemistry 401 which uses calculus 2 for the math in it, it is hard doing pure math, once it is applied to a object or something it is much easier I found as you know what the expected answer should be and can draw a picture of the math at hand, but pure math sucks which is why you don't see me use anything but differential equations and algebra, I hated Calculus 1 and Calculus 2 so much, but Calculus 3(3-D Calculus) and differential equations wasn't bad, but my favorite math course was Differential geometry.

Edited by VictorMedvil, 31 May 2019 - 08:08 AM.

### #8 Dubbelosix

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Posted 02 June 2019 - 10:39 AM

Now I have to do some schizophrenic moves between posts, to understand how the development of the transition equation involves the dragging coefficient, we will recognize

$S = \frac{1}{k_BT} \frac{\hbar c}{\lambda_0} = \mathbf{R}(\frac{Gm^2}{n^2_1k_BT} - \frac{Gm^2}{n^2_2k_BT})$

$\frac{1}{n^2_1} - \frac{1}{n^2_2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

And note the derivation for the second formula was obtained from my derivations found here:

http://www.sciencefo...-of-the-aether/

So let's combine these equations, we'll first distribute the Rydberg constant - to understand the following formula we will obtain, we first define the transition levels

$\frac{1}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2})$

So by distribution we get:

$\mathbf{R}(\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}) = \frac{1 + f}{\Delta \lambda}$

Let's flip the equation for clarity

$\frac{1 + f}{\Delta \lambda} = \mathbf{R}(\frac{1}{n^2_1} - \frac{1}{n^2_2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

Distribution of the gravitational charge $e^2 = n\hbar c = Gm^2$ will give:

$n\hbar c\ (\frac{1 + f}{\Delta \lambda}) = \mathbf{R}(\frac{Gm^2}{n^2_1} - \frac{Gm^2}{n^2_2} + e^2\frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

Now we divide through by the thermodynamic energy which gives back an equation similar to the original dimensionless entropy

$\frac{n\hbar c}{k_BT}\ (\frac{1 + f}{\Delta \lambda}) = \mathbf{R}(\frac{Gm^2}{n^2_1 k_BT} - \frac{Gm^2}{n^2_2k_BT} + \frac{e^2}{k_BT} \cdot \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda})$

A drag coefficient is often described under the terms:

$f = \frac{2F}{\rho v^2 A}$

Where the large $F$ is the dragging force, $\rho$ is our case would be the density of the gravitational field and $v^2$ is the speed of the object relative to the fluid.

Edited by Dubbelosix, 02 June 2019 - 12:27 PM.

### #9 Dubbelosix

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Posted 03 June 2019 - 01:36 PM

Before we proceed, as I always like to do, is define the formula's for clarity for further investigation:

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda} = 1 + f$

$f = \frac{2F}{\rho v^2 A}$

Plugging it in simply gives us the formula, as we flip the equation

$1 + \frac{2F}{\rho v^2 A} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

This means we also have an alternative formula to investigate at a later date:

$1 + \frac{A_b}{A_f} \frac{B}{Re^2_L} = \frac{1}{n^2} - \frac{1}{n^2} + \frac{\lambda}{n^2} \cdot \frac{dn^2}{d\lambda}$

The drag equation itself is simply

$F = \frac{1}{2} \rho v^2 f A$

And has been interpreted as a statement that the drag force on [any] object is proportional to the density of the fluid medium and porportional to the relative flow speed.

Where $Re^2_L$ is the Reynolds number related to the fluid path length. The Reynolds number is defined specifically as:

$R_e = \frac{\rho v L}{\mu} = \frac{v L}{\nu}$

We have to define the notation here, where $v$, $L$ is a linear dimension which in the case of gravity, we would expect it to be gravielectromagnetic path which is by definition the linearized gravity.$\mu$ is the dynamic viscosity, in which interpreted within gravitational physics, defines the ''thickness of the gravitational field,'' or ''stickiness of the gravitational aether. Finally $\nu$ is the kinematic viscosity of the medium itself - in a simple way of visualizing viscosity, the gravitational field is a fluid and in previous investigations for a primordial spin of the universe, is much like how you have a dense object submerged in water - what happens when you include spin in the fluid, it drags the objects inside of the medium.

References:

https://en.wikipedia...Reynolds_number

https://en.wikipedia...rag_coefficient

Edited by Dubbelosix, 03 June 2019 - 02:18 PM.

### #10 Dubbelosix

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Posted 06 June 2019 - 08:57 AM

$\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2} = 1 + f$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + f = 1$

Plug in:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

$f = \frac{2F}{\rho v^2 A}$

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + \frac{2F}{\rho v^2 A} = 1$

Plug in the definition of the standard drag,

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} + \frac{2}{\rho v^2 A} \cdot \frac{1}{2}\rho v^2 f A = 1$

Or as

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{2}{\rho v^2 A} \cdot \frac{1}{2}\rho v^2 f A$

Simplify by removing constants:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A$

Since generic dimensions of the density is simply a mass divided by the volume, we can obtain, after distributing the velocity squared term to obtain the energy density

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{1}{A} \cdot \rho v^2 f A$

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

We may even write our constructed equation using the same front and back area terms:

The drag equation is a function of the Bejan number $B_e$, the Reynolds number $R_e$ and the ratio between the back and front area:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} \cdot \rho v^2 f$

Since the density is related to the stress energy tensor through $T^{00} = \rho v^2$ we can re-write it as:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} T^{00} f$

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

We can define the stress energy, as either gravitationally related, or by electromagnetism. In some sense, gravity and electromagnetism may in certain theories be related, maybe even to the point they are the same thing… but these are ideas to take up at a later date. Firstly, the density of electromagnetic field is given through the stress energy as:

$T^{00} = \frac{1}{c^2}(\frac{1}{2}\epsilon_0 \mathbf{E}^2 + \frac{1}{2 \mu_0}\mathbf{B}^2)$

Let’s take a few steps back, we had:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} T^{00} f$

$f = \frac{A_b}{A_f} \frac{B}{Re^2_L}$

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - \frac{A_b}{A_f} \frac{A_b}{A_f} \frac{B}{Re^2_L}\ T^{00}$

This gives us the squared quantity:

$\frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} + \frac{mv^2}{n^2V} = - (\frac{A_b}{A_f})^2 \frac{B}{Re^2_L}\ T^{00}$

And now we simplify and then solve for the stress energy:

$mv^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = - (\frac{A_b}{A_f})^2 \frac{B}{Re^2_L}\ T^{00}$

$mv^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = - T^{00}$

And let’s use a sign change to finish off for now.

$- mv^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2V} + \frac{1}{n^2V} + \frac{1}{n^2V}) = T^{00}$

And so we don’t end up with any idea’s of a negative mass we should entertain:

$\rho v^2\frac{Re^2_L }{B}(\frac{A_f}{A_b})^2(\frac{1}{n^2} - \frac{1}{n^2} + \frac{1}{n^2}) = T^{00}$

Edited by Dubbelosix, 06 June 2019 - 09:16 AM.

### #11 exchemist

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Posted 06 June 2019 - 09:39 AM

The first line implies f = 1/n² -1.

The second line says 3/n² +f = 1, so substituting for f, we have 3/n² +1/n² -1 = 1

So rearranging, we have 4/n²= 2; so 2/n² = 1, hence n² = 2 and so n = √2.

Now we can solve for f, using the first line: f = 1/2 - 1 = -1/2.

n = √2; f = -1/2.

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### #12 Dubbelosix

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Posted 06 June 2019 - 09:48 AM

Oh... you're dabbling with my equations... that's nice, but will need to check your results later, I am making dinner.

### #13 Dubbelosix

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Posted 07 June 2019 - 09:52 AM

Oh... you're dabbling with my equations... that's nice, but will need to check your results later, I am making dinner.

So... i have a new question, what do you think of the idea of the three inverse terms coinciding with a three dimensional case?

### #14 Dubbelosix

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Posted 08 June 2019 - 08:53 PM

See, this is kind of what I was thinking about:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = - \frac{1}{\rho v^2 A} \cdot \rho v^2 f A = -f$

In  this simple case, we ignore the fancy things we can do, we just accept a simple case of removing the dimensions from the equation:

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} = -f$

Another simple thing to do is just take ($-1$) from both sides and we get

$\frac{1}{n^2} + \frac{1}{n^2} + \frac{1}{n^2} - 1 = f + 1$

If we take the first three terms of the spatial dimensions, than its last part has to be associated to the timelike dimension:

$\frac{x}{n^2} + \frac{y}{n^2} + \frac{z}{n^2} - ct = f + 1$

Notice the I have derived this in which the principal quantum numbers appear to be attached specifically to the spatial dimensions - and time has been placed in not too dissimilar to a fudge factor. Since space and time are not independent of each other, it will be under the normal sum of a three dimensional Euclidean (or Pythagorean relationship) with the negative sign indicated in the metric application on the timelike dimension. Most of us know this relationship and it is:

$s^2 = x^2 + y^2 + z^2 – (ct)^2$

If we take the derivation seriously, then the drag coefficient is dependent also on the sum of those components with the negative time component, in which the metric is simply:

$s^2 = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - c^2t^2 = \frac{2F}{\rho v^2 A} + 1$

### #15 Dubbelosix

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Posted 09 June 2019 - 07:50 AM

I was in a bit of rush yesterday, I should have stated the dragging coefficient would no longer be dimensionless, so the true form is:

$s^2 = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - c^2t^2 = \frac{2F \cdot \ell^2}{\rho v^2 A} + \ell^2$

Edited by Dubbelosix, 09 June 2019 - 07:51 AM.

### #16 Dubbelosix

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Posted 10 June 2019 - 06:34 AM

So let's see where I was heading with this:

$- \frac{2F}{\rho v^2 A} = - \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 + v^2t^2 - v^2t^2$

The negative of the drag is the thrust - though in this set up, I am entertaining that the last two terms associated to the space and time follow the anticommutation laws thus

$- \frac{2F}{\rho v^2 A} = - \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 + [x,ct]^2$

But we can retrieve the ordinary drag by the distribution of the sign again

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - [x,ct]^2$

The last term here is known as the spacetime uncertainty principle. As always, I like to teach a little bit on the physics behind this, which will use the standard deviation operators, to understand these kinds of ''strong'' or ''stronger'' uncertainties, we can start off by defining the standard deviation for two commutators.

The standard deviation is:

$\sigma^{2}_{x,p} =\ <x^2t^2> - <x^2t^2>$

$\sigma_{x} = \sqrt{<x^2> - <x^2>}$

$\sigma_{p} = \sqrt{<p^2> - <p^2>}$

As usual the standard deviations involve the expectation value, in which the operators define the commutators:

$[\hat{A},\hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

$\sigma_A \sigma_B \geq |\frac{1}{2i}[\hat{A},\hat{B}]| = \frac{1}{2}|[\hat{A},\hat{B}]|$

This will do for now, we will continue this later.

Edited by Dubbelosix, 10 June 2019 - 07:32 AM.

### #17 Dubbelosix

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Posted 13 June 2019 - 12:33 PM

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - [x,ct]^2$

under standard deviation the new representation gives:

$\frac{2F}{\rho v^2 A} = \frac{A_b}{A_f}\frac{B}{Re^2_L} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - \sigma^{2}_{x,p} = (\frac{x}{n} + \frac{y}{n} + \frac{z}{n})^2 - <x^2t^2> - <x^2t^2>$

Under quantum mechanics, time is not considered an operator, yet clearly, a spacetime commutator must require that time is manifested an observable in four dimensional space as the presence of curvature. Since we are aware that quantum mechanics treats the space dimension as an observable, it does itself have the non-trivial space operators:

$\hat{\ell} = \hat{e}_x x + \hat{e}_y y + \hat{e}_z z$

$<\frac{2F}{\rho v^2 A}>\ = \frac{1}{n^2} (\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - \sigma^{2}_{x,p} = \frac{1}{n^2}(\hat{e}_x x + \hat{e}_y y + \hat{e}_z z)^2 - <x^2t^2> - <x^2t^2>$

So in this case today, we have formulated the space derivatives and the time commutator as observables, something which is unavoidable in relativity when curvature is present, because again, it has to be manifested an observable if we detect a test particle moving in curved space.

Edited by Dubbelosix, 13 June 2019 - 12:34 PM.