# The Equivalence Principle And The Relativity Of Charges

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### #18 Dubbelosix

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Posted 01 February 2019 - 06:37 AM

We can even go as far to describe it through a Schrodinger-like equation. Doing so you would replace the energy and momentum by their respective operators:

$\hat{E} = i \hbar \frac{\partial}{\partial t}$

And the momentum operator

$\hat{p} = -i \hbar \frac{\partial}{\partial x}$

This is how you obtain a Klein Gorden equation from the relativistic:

$E^2 = m^2c^4 + p^2c^2$

identify the operators

$\hat{E}^2\psi = (c^2 \hat{p}^2 + m^2c^4)\psi$

Which makes

$\hat{E}^2\psi = (c^2 \hat{p}^2 + m^2c^4)\psi$

$(i \hbar \frac{\partial}{\partial t})^2 \psi = (-i \hbar \frac{\partial}{\partial x})^2c^2\psi + m^2c^4\psi$

remember the rules of complex numbers, (ie. $i^2 = -1$ and $-i^2 = 1$) and we end up with

$- \hbar^2 \frac{\partial^2}{\partial t^2} \psi = \hbar^2 \frac{\partial^2}{\partial x^2}c^2\psi - m^2c^4\psi$

and divide off the Planck constant squared creates the famous equation:

$\frac{\partial^2}{\partial t^2} \psi = (\frac{\partial^2}{\partial x^2}c^2 + \frac{m^2c^4}{\hbar^2})\psi = (\nabla^2 c^2 + \frac{m^2c^4}{\hbar^2})\psi$

Or more commonly encountered:

$\frac{1}{c^2}\frac{\partial^2}{\partial t^2} \psi = (\frac{\partial^2}{\partial x^2} + \frac{m^2c^2}{\hbar^2})\psi = (\nabla^2 + \frac{m^2c^2}{\hbar^2})\psi$

In the case of the corrections, we also had Plancks correction, which we hypothesized a form of:

$E = \sqrt{(mc^2 + pc + pV)^2} = \sqrt{\frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2 + P^2V^2}$

It is possible that the true form to write is:

$E = \sqrt{\frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2 } + PV$

and this needs to be investigated. Either way, to describe the rest mass from these equations in terms of the charges is not easy. Usually when we do second quantization process, we attempt to replace terms by their respective operators - how to do it in our equation I am uncertain about. I do know charge operators do exist out there:

https://physics.stac...r-action-in-qft

We replace for operators in an attempt to quantize an equation, but as you may notice, one charge term already comprises quantum aspects, notably $\frac{\hbar c}{R}$. While I personally do not know of any pressure operators, I am also aware that there does exist a volume operator:

https://en.wikipedia...Volume_operator

It is possible to write the terms in an operator form, including a mass operator. But more interesting is that there may be a way to represent this equation however in terms of the second quantization method - we just need to keep track of what we define. We already recognized that the rest reduced mass-energy was

$E_0 = \frac{Gm^2}{R}$

This term is not exactly quantum - it does contain aspects of gravitational theory and the important square of the mass term which does in fact arise from quantum theory and string theory. Since the term however is nothing more than $mc^2$, then this term alone can be left as it is. The correction term, though related to the total mass, is related to the quantum theory. This takes the form, removing the pressure from the equation:

$E^2 = (mc^2 + pc)^2 = \frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2$

then

$(i \hbar \frac{\partial}{\partial t})^2 = (-i \hbar \frac{\partial}{\partial x}c + mc^2)^2 = (-i \hbar \frac{\partial}{\partial x})^2c^2 + \frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2}$

And this produces:

$\hbar^2 \frac{\partial^2}{\partial t^2} = \hbar^2 \frac{\partial^2}{\partial x^2}c^2 + m^2c^4 = \hbar^2 \frac{\partial^2}{\partial x^2}c^2 + \frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2}$

Insert the wave function and divide through by Planck constant squared and we have:

$\frac{\partial^2 \psi}{\partial t^2} = ( \frac{\partial^2}{\partial x^2}c^2 + \frac{m^2c^4 }{\hbar})\psi= (\frac{\partial^2}{\partial x^2}c^2 + \frac{G^2m^4}{R^2\hbar^2} + \frac{ c^2}{R^2})\psi$

and simplify further

$\frac{1}{c^2}\frac{\partial^2 \psi}{\partial t^2} = (\nabla^2 + \frac{m^2c^2}{\hbar^2})\psi= (\frac{\partial^2}{\partial x^2} + \frac{1}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{ 1}{R^2})\psi$

Edited by Dubbelosix, 01 February 2019 - 09:19 AM.

### #19 Dubbelosix

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Posted 01 February 2019 - 07:16 AM

Ignoring pressure again, we saw how the rest mass could be represented by a gravitational charge with an electric charge correction:

$E^2 = (mc^2 + pc)^2 = \frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2$

It has been noted by Feynman that there are additional energies which complicate the whole picture, such as contributions from magnetic moment. The way to talk about a magnetic moment in terms of energy is relatively easy since the Hamiltonian is in respect to a Bohr magneton interacting with a magnetic field:

$H = \mu_B \cdot \mathbf{B}$

And so a charge can be expressed through a cross product $\mu_B \cdot (\mathbf{B} \times R)$, however the cross product will become insignificant because the entire term is weighted by an inverse squared length such that:

$E^2 = (pc + mc^2 + \mu_B \cdot \mathbf{B})^2 = p^2c^2 + \frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + \mu_B^2 \cdot \mathbf{B}^2$

Which in theory adds the contribution from magnetic moments.

Edited by Dubbelosix, 01 February 2019 - 07:19 AM.

### #20 VictorMedvil

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Posted 01 February 2019 - 08:16 AM

Insightful as always dubbel keep it up!

### #21 Dubbelosix

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Posted 01 February 2019 - 12:04 PM

If we introduce the gravitational scalar potential $\phi$ we would get some terms that look like a diffusion equation.

$\frac{\phi}{c^2}\frac{\partial^2 \psi}{\partial t^2} = (\nabla^2\phi + \frac{m^2c^2}{\hbar^2}\frac{Gm}{R})\psi= (\frac{\partial^2 \phi}{\partial x^2} + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

The reason why I say it looks like a diffusing equation for $\phi$ simply because I obtained solutions from similar problems I sought, called a Ricci flow. Here I take an excerpt from the work:

$\frac{\partial \phi}{\partial t} = \alpha \Box^2 \phi = \alpha \partial^{\mu}\partial_{\mu}\phi = \alpha g^{\mu \nu} \partial_{\nu} \partial_{\mu}\phi$

The idea would invoke a relativistic Newton-Poisson equation of the form

$\frac{\partial \phi}{\partial t} = \alpha \Box^2 \phi = - \alpha (\frac{\partial^2 \phi}{\partial \tau^2} - \frac{\partial^2 \phi}{\partial x^2} + \frac{\partial^2 \phi}{\partial y^2} + \frac{\partial^2 \phi}{\partial z^2}) = \alpha ( \frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2\phi)$

And this equation in curved spacetime looks like:

$\partial_0 \phi \equiv \alpha g^{\mu \nu} \nabla_{\mu} \nabla_{\nu} \phi = \alpha g^{\mu \nu} \nabla_{\mu} (\partial_{\nu} \phi) = \alpha(g^{\mu \nu} \partial_{\mu}\partial_{\nu} \phi + g^{\mu \nu} \Gamma^{\sigma}_{\mu \nu}\partial_{\sigma}\phi) = \alpha \frac{-1}{\sqrt{-g}}\partial_{\mu}(g^{\mu \nu} \sqrt{-g} \partial_{\nu} \phi)$

Which is a diffusion equation with respect to $\phi$  in the language of general relativity. It turned out after some more investigation that such an idea was already suggested, the saving grace was I haven’t found anyone write it in this form. The solutions show [strongly] the ability to express a wave equation, like the Schrodinger equation, into a heat equation for a Riemann manifold.

I also showed how a mass term enters and it begins to look a lot more like the equations we have looked at so far:

would look like with a mass parameter. That takes the form of:

$\alpha (g^{\mu \nu}\partial_{\mu} \partial_{\nu} + |\mathbf{k}|^2 + \frac{\omega^2}{c^2})\phi = \alpha \Box^2 \phi + \alpha \frac{m^2c^2}{\hbar^2}\phi = \alpha (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi + \frac{m^2c^2}{\hbar^2}\phi)$

In which

$- |\mathbf{k}|^2 + \frac{\omega^2}{c^2} = \frac{m^2c^2}{\hbar^2}$

Is the dispersion relation, in which $\mathbf{k}$ is the wavenumber and $\omega$ the angular frequency. This is a relativistic relationship. We should now compare the two equations for a closer look:

$\frac{\phi}{c^2}\frac{\partial^2 \psi}{\partial t^2} = (\nabla^2\phi + \frac{m^2c^2}{\hbar^2}\frac{Gm}{R})\psi= (\frac{\partial^2 \phi}{\partial x^2} + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

$\alpha (g^{\mu \nu}\partial_{\mu} \partial_{\nu} \phi + |\mathbf{k}|^2 + \frac{\omega^2}{c^2}) = \alpha \Box^2 \phi + \alpha \frac{m^2c^2}{\hbar^2}\phi = \alpha (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi + \frac{m^2c^2}{\hbar^2}\phi)$

To make the first equation a true heat equation, it requires a diffusion constant $\alpha$, but this is a trivial matter right now as it can be placed in at any time. One thing to notice, when I could, I wrote the potential as a ratio of $\frac{\phi}{c^2}$ intentionally and we will return to this soon.

Edited by Dubbelosix, 01 February 2019 - 05:27 PM.

### #22 Dubbelosix

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Posted 01 February 2019 - 01:12 PM

The two equations of interest was:

$\frac{\phi}{c^2}\frac{\partial^2 \psi}{\partial t^2} = (\nabla^2\phi + \frac{m^2c^2}{\hbar^2}\frac{Gm}{R})\psi= (\frac{\partial^2 \phi}{\partial x^2} + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

$\alpha (g^{\mu \nu}\partial_{\mu} \partial_{\nu} + |\mathbf{k}|^2 + \frac{\omega^2}{c^2})\phi = \alpha \Box^2 \phi + \alpha \frac{m^2c^2}{\hbar^2}\phi = \alpha (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi + \frac{m^2c^2}{\hbar^2}\phi)$

When you look at the heart of  the equations, you can see that they are pretty much identical. This means we can write a four dimensional case now:

$(\frac{Gm}{R})\ \Box \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi + \frac{m^2c^2}{\hbar^2}\frac{Gm}{R})\psi = (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

Here we recognize $\tau$, not specifically as a proper time, but better worded yet, it is the spacelike time coordinate. We learn from the equivalences of the equations that it is also mathematically true to state the relationship:

$\alpha (g^{\mu \nu}\partial_{\mu} \partial_{\nu} \phi + |\mathbf{k}|^2 + \frac{\omega^2}{c^2}) = \alpha (\Box^2 \phi + \frac{m^2c^2}{\hbar^2})\phi = \alpha(\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

Remember, the very last term $\frac{\phi}{R^2}$ is the second part of a correction to the mass of the system, related to $\frac{m^2c^2}{\hbar^2}$ as the relic left over from the mass rest term, so it is this term that contains two more terms ~

$\frac{m^2c^2}{\hbar^2}\frac{m}{R} = \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2}$

Edited by Dubbelosix, 01 February 2019 - 05:30 PM.

### #23 Dubbelosix

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Posted 01 February 2019 - 01:48 PM

From now on, we will need to know some mics. knowledge. We already identified

$\frac{m^2c^2}{\hbar^2}\frac{m}{R} = \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2}$

But let us recognize $\frac{\phi}{c^2} = \frac{1}{G}$ - this means we would be working in cgs units of Newtons constant and will generally fall out of the definition, as if it is independent of it (Sciama's approach). It is also interesting to note, that this future physics we will look at involves a definition of the gravielectric field:

$\mathbf{E} \approx \frac{m}{r^2}$

The fact that this is essentially the last term on the right hand side, I see as no coincidence, since this too was the part associated to the charge contribution to rest mass.

The divergence gives:

$\nabla \cdot \mathbf{E} \approx \frac{m}{r^3} = 4 \pi \rho$

To incorporate all this, we need to involve the new definitions into the original derivation. This will be next.

Edited by Dubbelosix, 01 February 2019 - 05:28 PM.

### #24 Dubbelosix

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Posted 01 February 2019 - 03:26 PM

Working in the units we require, we can see the gravitational charge holds onto the gravitational constant since mass is an invariant in the theory and is essentially a charge definition, not related to the potential specifically. The four dimensional wave equation would need to be re-defined ~

$(\frac{m}{R})\ \Box \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi + \frac{m^2c^2}{\hbar^2}\frac{m}{R})\psi = (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

Due to a set of equivalences I developed when investigating gravielectromagnetism, I ended up finding a relationship to the Coriolis field (a natural development from a gravimagnetic theory)

$\mathbf{E} = \frac{m}{r^2} \equiv \frac{(\omega \times R) \times R}{G} = \frac{a}{G} = \frac{\phi}{c^2}\frac{\partial v}{\partial t}$

where again we identify the gravielectric field as $\mathbf{E}$ and again we obtain a density:

$\nabla \cdot \mathbf{E} = 4 \pi \rho \equiv \frac{\partial}{\partial R}\frac{(\omega \times R) \times R}{G} = \frac{1}{G} \frac{\partial a}{\partial R} = \frac{\phi}{c^2}(\frac{1}{c}\frac{\partial^2 v}{\partial t^2})$

Let's take a look at the main result we had from our derivations:

$= (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2})\psi$

We can notice immediately that $\frac{\phi}{R^2}$ is equivalent to the density term as noted above, so we can write that in ~

$= (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{m}{R^3})\psi$

$becomes\ \rightarrow (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + 4 \pi \rho)\psi$

$\phi = \frac{c^2}{G}$ related to the gravitational permeability the analogue of its respective electric interpretation. However, this would make it only an approximated equation, there would be a correction on order of:

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

I derived this when I investigated gravielectromagnetism - it truly was nice to see this simplify to a direct consequence of the potential related to the inverse of permeability. Inverse relationships exist in such a way, that when $\phi$ becomes larger, the permeability would have to be smaller. And vice versa.

$(\frac{m}{R})\ \Box \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi + \frac{m^2c^2}{\hbar^2}\frac{m}{R})\psi = (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + 4 \pi \rho)\psi$

Now... we will take a look at this expression:

$\frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2}$

and notice that if we plug in it's definition $\frac{\phi}{c^2} = \frac{1}{G}$ we obtain:

$(\nabla \cdot \mathbf{E})\ \psi = (\frac{m}{R})\ \Box \psi = (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{1}{G}\frac{G^2m^4}{R^2\hbar^2} + 4 \pi \rho)\psi$

and simplifies it to

$(\nabla \cdot \mathbf{E})\ \psi = (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{Gm^4}{R^2\hbar^2} + 4 \pi \rho) \psi = (\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi + \frac{Gm^2}{c^2R^4} + 4 \pi \rho) \psi$

and this reveals something interesting, this expression specifically $\frac{Gm^2}{R^4}$ because it is the energy density of the system. The final bit for now, involves how we can be creative in how we interpret the following sum of terms:

$\frac{\partial^2 \phi}{\partial x^2} + \nabla^2\phi$

and predict it could explain the connection of the gravitational field with its correction as given through the covariant derivative ~

$\nabla = \partial + \Gamma$

$(\nabla \cdot \mathbf{E})\ \psi = (\frac{\partial^2 \phi}{\partial x^2} + \Gamma^2\phi + \frac{Gm^2}{c^2R^4} + 4 \pi \rho) \psi$

Such approaches have been used before in literature and gives it a distinct gravitational feature.

Edited by Dubbelosix, 01 February 2019 - 04:19 PM.

### #25 Dubbelosix

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Posted 01 February 2019 - 06:04 PM

The dispersion relation was shown to be:

$- (|\mathbf{k}|^2 + \frac{\omega^2}{c^2})\frac{m}{R} = \frac{m^2c^2}{\hbar^2}\phi = \frac{m^2c^2}{\hbar^2}\frac{c^2}{4 \pi G}$

and this last expression would have encoded an energy term, which through a series of derivations we ended up stipulating

$\frac{m^2c^2}{\hbar^2}\frac{m}{R} = \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2}$

We also recognize that the main equation also has such a term inside of it:

$(\nabla \cdot \mathbf{E})\ \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi - |\mathbf{k}|^2\phi + \frac{\omega^2}{c^2}\frac{m}{R})\psi$

Or rearranged as:

$(\nabla \cdot \mathbf{E} + 4 \pi \rho + \frac{\omega^2}{c^2}\frac{m}{R})\ \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi)\ \psi$

A similar density term condenses into the equation like before, but has arisen from different physics talking about the same things. For instance, because the wave number has dimensions of inverse length, the product of $\frac{m}{R}|\mathbf{k}|^2$ has units of density, which for historical reasons, is denoted as $4 \pi \rho$. From identifying the permeability as:

$\frac{m}{R} \equiv \frac{c^2}{4 \pi G} = \frac{1}{\mu_g}$

We also obtain:

$(\nabla \cdot \mathbf{E} + 4 \pi \rho + \frac{\omega^2}{c^2}\frac{c^2}{4 \pi G})\ \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi)\ \psi$

which when simplified reveals:

$(\nabla \cdot \mathbf{E} + 4 \pi \rho + \frac{\omega^2}{4 \pi G})\ \psi = (\frac{\partial^2 \phi}{\partial \tau^2} + \nabla^2 \phi)\ \psi$

From the stipulations, it also seems natural to look at the combined equations of the first part of this post:

$- (|\mathbf{k}|^2 + \frac{\omega^2}{c^2})\frac{m}{R} = \frac{\phi}{c^2}\frac{G^2m^4}{R^2\hbar^2} + \frac{\phi}{R^2}$

And we will look at all this later.

### #26 Dubbelosix

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Posted 03 February 2019 - 05:17 AM

The diffusion of \phi in any coordinate system is:

$\nabla \cdot \mathbf{E} = \frac{4 \pi G}{c^2}(\rho + \frac{\omega^2} {c^2} + \frac{\partial^2 }{\partial \tau^2} + \alpha \nabla^2)\phi$

Where $\alpha$ is the diffusion constant. This is a gravielectromagnetic equation and though has very little to do with the mass of the system with respect to charges, it was a nice equation to fall upon and reminded me of its importance when visioned as a diffusion equation. We will come back to these solutions at some point.

As stated in the opening post, Planck introduced a variant of Einstein’s mass-energy equivalence of the form

$E_0 = mc^2 + PV$

The pressure is related fundamentally to a number of terms related to the temperature and the gas constant:

$PV = N k_BT = \frac{N}{N_A}\mathbf{R}T$

$\mathbf{R} = N_Ak_B$

In which N_A is the number of moles in the gas. As it can be noticed, when speaking about the pressure we can also talk about the internal thermodynamics of the system - intended by Planck. The relationships to the pressure would imply the rest energy as:

$E_0 = mc^2 + PV = mc^2 + Nk_BT = mc^2 + \frac{N}{N_A}\mathbf{R}T$

Let’s talk a bit about electromagnetic mass and how different it is to the suggestion I have made about corrections to relative charges and measurable inertia.

When electromagnetic mass was talked about, it was tended to be done so in terms of an electrostatic energy and the mass of an electron at rest:

$E_{em} = \frac{1}{2}\frac{e^2}{R}$

$m_{em} = \frac{2}{3}\frac{e^2}{c^2R}$

Where in such cases, the charge is uniformly distributed, either over the sphere itself, or perhaps through the sphere itself. The radius of the electron has to be non-zero to avoid non-trivial singularities that arise within the self-energy of the system.

The formula then proposed in literature for the electromagnetic-mass relation was to be:

$m_{em} = \frac{4}{3} E_{em}{c^2}$

Concepts that where pretty much identical before the revolution of special relativity involved transverse and longitudinal definitions of the mass. Today those important idea's that had been developed by Lorentz incorporated the famous length contraction in both space and time. It was shown by Bucherer and Langevin that an electron would be contracted in the line of motion and expands perpendicular to it so that the volume remains constant. However, it has been shown by Penrose that a perfect sphere would never be seen to be contracted, though its apparent size may seem smaller.

Erroneously by wiki, it states that eventually electromagnetic theories had to be given up, in respect to Poincare stresses. Electromagnetic theories cannot simply ''be given up'' when the contribution of electric charge seems to have measurable effects on the mass of electromagnetic bodies. The Poincare stress is not a true problem in the sense it forbids or overly complicates the issue of an electromagnetic theory of mass. Poincare indeed himself persued the electromagnetic mass theory and attempted to find the stresses that contribute to a non-electromagnetic component of energy to the electrons. He found that it contributed \frac{1}{3} of their electromagnetic energy. Poincare takes a more black and white view, believing that electromagnetic energy was the only energy to contribute to the mass of an electron.

Though, this kind of view would seem at odds with how Feynman later came to explain situation, in which it was the presence of a charge that contributed some mass to a system, not the entirity of it. This of course was the motivation for me to explore a relative concept on the charge, where the mass consisted of two parts

$\frac{Gm^2}{R} + \frac{\hbar c}{R}$

The contribution of Poincare stress became known as the \frac{4}{3}-problem simply because the contribution to whole energy does not contain the fraction. He goes on to find a solution in which the total energy in a contribution also of two terms:

$\frac{E_{tot}}{c^2} = \frac{E_{em} + \frac{1}{3}E_{em}}{c^2} = \frac{4}{3}\frac{E_{em}}{c^2} = \frac{4}{3}m = m_{em}$

The problem of ‘’how’’ much electromagnetic mass is contributed to the system from the presence of charge I think, can be more elegantly explained through the method I have chosen. If we be rash and say the entire mass of the system is provided from the electromagnetic energy, we would need to explain how a neutrino, expected to have zero charge, has a mass at all. Both terms, $(\frac{Gm^2}{R}, \frac{\hbar c}{R})$ are structurally and dimensionally similar to the electrostatic energy:

$E_{em} = \frac{1}{2}\frac{e^2}{R}$

And so we may expect correcting coefficients arising within the theory suggested involving relative charges. I find something important about the concept of the pressure term as a correction in the equation

$E_{em} = \frac{e^2}{R} + pV$

Because, while the first term on the right hand side wants to rip the system apart, the question of the role of the pressure could act as the sought-after Poincare stress. If there is a contribution, the charge only makes a particle only slightly more heavier as Feynman suggested from comparing particles on the standard model - but not so insignificant if it is noticeable.

To understand how the pressure term would balance the electrostatic repulsion will have to be something to be investigated at a later point. What would be interesting though, is if the unsuspecting term $\frac{Gm^2}{R}$ could play a vital role in the balancing of the electrostatic energy which could be encoded in $\frac{\hbar c}{R}$. Is it possible gravity is playing a role of a Poincare stress? Lloyd Motz was the first physicist I know of to entertain this idea - the basic premise relied on a scale dependent theory of gravity, or one in which discontinuities in the gravitational field change over the boundary of the particle, whether or not those approaches are required, the theory would look similar to this:

$\frac{E_{tot}}{c^2} = \frac{Gm^2 + \hbar c}{c^2R} = m_{tot}$

The non-electromagnetic Poincare stress in this case, would turn out to be the gravitational equivalent of the electrostatic repulsion/charge. For a primer on the importance of electromagnetic mass, here is a link to a Feynman lecture:

Electromagnetic Mass

Notice again, I refuse to make reference to any coefficient on the charges, but this is because it really depends on what kind of charge distribution we are speaking about. For instance, for a charge uniformly distributed throughout the volume of a sphere, the $\frac{2}{3}$ gets replaced by $\frac{4}{5}[math] so though technically there will be correcting coefficients, we won’t rush into that because the physics depends on the situation. In our case, we did explore the notion of a uniformly distributed charge through a sphere, but we will come back to these idea’s on a later date. It is possible to associate the combination of charge to some ‘’normal’’ charge case such as: [math]E_{tot} = \frac{e^2}{R} = \frac{Gm^2}{R} + \frac{\hbar c}{R}$

Edited by Dubbelosix, 03 February 2019 - 05:18 AM.

### #27 Dubbelosix

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Posted 03 February 2019 - 06:52 AM

If $\phi$ depends only on the radius then it is known that:

$\nabla^2 \phi = \frac{1}{r} \frac{\partial^2}{\partial r^2} (r \phi)$

and this is equal to

$\frac{\partial^2}{\partial r^2} (r \phi) = \frac{m^2c^2}{\hbar^2} (r \phi)$

Thinking of $(r \phi)$ as the independent variable, it's solution takes

$r\phi = K^{- \mu r}$

Or

$\phi = K^{\frac{- \mu r}{r}}$

This is known as the Yukawa potential. So why are we going through it? Well one such term does feature in one of the equations we have derived:

$\Box \phi = (\frac{\partial^2}{\partial \tau^2} + \alpha \nabla^2 + \frac{m^2c^2}{\hbar^2})\phi$

and so would be a general solution to the diffusion equation as well.

### #28 Dubbelosix

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Posted 04 February 2019 - 01:36 PM

$- (|\mathbf{k}|^2 + \frac{\omega^2}{c^2})\frac{m}{R} = \frac{m^2c^2}{\hbar^2}\phi = \frac{m^2c^2}{\hbar^2}\frac{c^2}{4 \pi G}$

divide by c^2 and distribute an acceleration term:

$- (|\mathbf{k}|^2 + \frac{\omega^2}{c^2})\frac{m}{c^2R}\frac{\partial v}{\partial t} = \frac{m^2c^2}{\hbar^2}\frac{\phi}{c^2}\frac{\partial v}{\partial t} = \frac{m^2c^2}{\hbar^2}\frac{a}{4 \pi G}$

This implements Sciama's gravielectric field:

$\mathbf{E} = \frac{m}{r^2} + \frac{\phi}{c^2}\vec{a}$

From these equations we can define the gravielectric field as:

$\frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2 \frac{m}{r^2} + \frac{\omega^2}{c^2}\frac{\phi}{c^2}\vec{a})$

and so we tidy up:

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = - (|k|^2 \frac{mc^2}{r^2} + \omega^2\frac{\phi}{c^2}\vec{a})= |k|^2 \frac{mc^2}{r^2} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a}$

In which the gravimagnetic field (torsion field) is:

$\omega = -\frac{\Omega}{2}$

The equation is really nice and interpreting it will be fun. I'll need to rewrite this into the gravielectromagnetic study that I did. For a homogeneous medium, the group velocity of a wave is related to the phase

$v^2_g = \frac{c^2}{n-\lambda(\frac{dn}{d\lambda})} = \frac{v^2}{1 - \frac{\lambda}{n}(\frac{dn}{d\lambda})}$

A slight modification to this takes:

$\frac{v^2_g}{\lambda^2} = \frac{c^2}{n-\lambda^3(\frac{dn}{d\lambda})} = \frac{v^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})}$

Plugging this into the gravielectric equation gives:

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2\ \frac{mc^2}{n-\lambda^3(\frac{dn}{d\lambda})} + \omega^2\frac{\phi}{c^2}\vec{a}) = (|k|^2\frac{mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a})$

Edited by Dubbelosix, 05 February 2019 - 08:16 AM.

### #29 Dubbelosix

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Posted 04 February 2019 - 01:51 PM

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2\ \frac{mc^2}{n-\lambda^3(\frac{dn}{d\lambda})} + \omega^2\frac{\phi}{c^2})\vec{a} = (|k|^2\frac{mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a)}$

In fact

$\omega = -\frac{\omega}{2} = \frac{G\hbar}{2c^2r^3}$

Edited by Dubbelosix, 04 February 2019 - 04:32 PM.

### #30 Dubbelosix

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Posted 04 February 2019 - 04:24 PM

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2\ \frac{mc^2}{n-\lambda^3(\frac{dn}{d\lambda})} + \omega^2\frac{\phi}{c^2}\vec{a})= |k|^2 \frac{mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a}$

Expanding to first order we get:

$= (|k|^2\ \frac{ mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\Omega^2}{2c^2}\frac{m \vec{a}}{\lambda_1 \lambda_2}) [\lambda_2 - \lambda_1]\theta_{GR}$

$\theta_{GR} = \frac{(1 + \gamma)}{2}$ is the angle of deflection.

Expanding the potential is trivial stuff:

$\Delta \phi = \phi_1 - \phi_2 = \frac{m}{\lambda_1 \lambda_2}(\lambda_2 - \lambda_1)$

The refractive index for gravitational radiation is proportional to $\sqrt{\epsilon_G \mu_G}$ (permittivity and permeaility) and is represented as:

$n = \sqrt{\frac{\epsilon_G \mu_G}{\epsilon_0 \mu_0}}$

A high refractive index for the equation $\frac{1}{\sqrt{\epsilon_G \mu_G}}$ causes a low speed of light (such as found round strong gravitational fields of black holes). To have some insight into the wavenumber it is related to the following terms:

$k = \frac{2 \pi}{\lambda} = \frac{2 \pi f}{v} = \frac{\omega}{v} = -\frac{\Omega}{2v}$

or for the wavelength

$\lambda = \frac{2 \pi}{k} = -\frac{4 \pi v}{\Omega} = \frac{v}{f}$

Edited by Dubbelosix, 04 February 2019 - 04:54 PM.

### #31 Dubbelosix

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Posted 05 February 2019 - 08:59 AM

From these equations we can define the gravielectric field as:

$\frac{1}{c^2}\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2 \frac{m}{r^2} + \frac{\omega^2}{c^2}\frac{\phi}{c^2}\vec{a})$

and so we tidy up:

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = - (|k|^2 \frac{mc^2}{r^2} + \omega^2\frac{\phi}{c^2}\vec{a})= |k|^2 \frac{mc^2}{r^2} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a}$

In which the gravimagnetic field (torsion field) is:

$\omega = -\frac{\Omega}{2}$

The equation is really nice and interpreting it will be fun. I'll need to rewrite this into the gravielectromagnetic study that I did. For a homogeneous medium, the group velocity of a wave is related to the phase

$v^2_g = \frac{c^2}{n-\lambda(\frac{dn}{d\lambda})} = \frac{v^2}{1 - \frac{\lambda}{n}(\frac{dn}{d\lambda})}$

A slight modification to this takes:

$\frac{v^2_g}{\lambda^2} = \frac{c^2}{n-\lambda^3(\frac{dn}{d\lambda})} = \frac{v^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})}$

Plugging this into the gravielectric equation gives:

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2\ \frac{mc^2}{n-\lambda^3(\frac{dn}{d\lambda})} + \omega^2\frac{\phi}{c^2}\vec{a}) = (|k|^2\frac{mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a})$

Torsion is defined through the Heaviside equations:

$\Omega = \frac{G}{2c^2}\frac{\mathbf{L} - 3 (\mathbf{L} \cdot \frac{\mathbf{r}}{r})\frac{\mathbf{r}}{r}}{r^3}$

The gravitational field strength defined as an acceleration is not far from the definition of the torsion field.

$\mathbf{H} = \frac{G}{2c}\frac{\mathbf{L} - 3 (\mathbf{L} \cdot \frac{\mathbf{r}}{r})\frac{\mathbf{r}}{r}}{r^3}$

knowing this, we can see that the product $\Omega^2\ \vec{a}$ reveals a simplification in:

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2\ \frac{mc^2}{n-\lambda^3(\frac{dn}{d\lambda})} + \omega^2\frac{\phi}{c^2})\vec{a} = |k|^2\frac{mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\Omega^2}{2}\frac{\phi}{c^2}\vec{a}$

In fact, keep in mind

$\omega = -\frac{\Omega}{2} = \frac{G\hbar}{2c^2r^3}$

As the last term is the general solution. So we need two factors of the torsion field and one factor of gravitational acceleration field strength, in doing this I get:

$\frac{G^3}{6c^5}\frac{\mathbf{L}^3 - 9 (\mathbf{L}^3 \cdot \frac{\mathbf{r}}{r})\frac{\mathbf{r}}{r}}{r^9} = \frac{G^3\hbar^3}{6c^5r^9}$

Edited by Dubbelosix, 05 February 2019 - 05:26 PM.

### #32 Dubbelosix

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Posted 05 February 2019 - 09:27 AM

And so written out is given as ~

$\frac{\partial^2 \mathbf{E}}{\partial t^2} = -(|k|^2\ \frac{mc^2}{n-\lambda^3(\frac{dn}{d\lambda})} + \omega^2\frac{\phi}{c^2}\vec{a}) = |k|^2\frac{mv^2}{1 - \frac{\lambda^3}{n}(\frac{dn}{d\lambda})} + \frac{\phi}{2c^2}\frac{G^3}{6c^5}\frac{\mathbf{L}^3 - 9 (\mathbf{L}^3 \cdot \frac{\mathbf{r}}{r})\frac{\mathbf{r}}{r}}{r^9}$

Edited by Dubbelosix, 05 February 2019 - 09:38 AM.

### #33 Dubbelosix

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Posted 06 February 2019 - 04:08 PM

Planck suggested an alteration for the energy-mass equivalence as:

$E = mc^2 + PV$

In which he linked the last term to the dynamic pressure which would involve internal kinetic energy also. Due to the equpartition laws, and related virial theorems, the pressure is related to the kinetic energy and is related lastly with the thermal contribution of the motion of $N$-particles ~

$\frac{3}{2}PV = \sum_{N=i} \frac{1}{2}m_i v^2 = \frac{3N}{2}k_BT$

We'll come back to this... There was dispute in literature about what transformation law is acceptable in relativity... Einstein eventually said Ott's case was correct... but I don't know enough if he also held his case was correct as well.

In which the Ott covariant transformation is

$T = T_0 \sqrt{1 - \frac{v^2}{c^2}} = \gamma T$

And the suggestion by Einstein and Planck was

$T = \frac{2T_0}{\sqrt{1 - \frac{v^2}{c^2}}} = \frac{2T}{\gamma}$

par a factor of two that we have picked up through this derivation. The equation will be universal in understanding the relationship between the two transformation laws:

$T’ = \gamma T$

$T’ = \frac{T}{\gamma}$

I don't know if Ott or Einstein was aware, but both terms are relative in entropy:

$\Delta \mathbf{S} = \frac{\Delta S}{k_B} = 2T \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2T}{\sqrt{1 - \frac{v^2}{c^2}}}$

This means that a body in motion may appear both hotter and cooler, and this makes sense when you notice the problem is also direction-dependent which entails the redshift. Now, the extension to the rest energy would I assume take the form the kinetic energy $E_k$:

$E_k = \sum_{N=i} \frac{1}{2}m_i v^2 + \frac{3}{2}PV + \frac{3}{2}Nk_BT$

We know how energy transforms under the Lorentz contraction, but what of temperature? For that matter, what about the middle term, the pressure and volume? Well, I came to realize that the volume and the temperature had to transform in exactly the same way! If temperature is covariant, then the next question is entropy covariant?

I don't think so, the term doesn't have to be a function of entropy at all. We may see how the change in entropy becomes dependent on the only covariant object which can lead to a variation in the temperature due to Lorentz contractions:

$\Delta \mathbf{S} = \log_2(\frac{V_2}{V_1}) = \log_2(2) = 1$

Or as a more compact argument,

$\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})$

Which reminds me a four-dimensional entropy equation which measured the temperature through a similar ratio. In such cases, a ratio like

$\frac{V_0}{V} = \frac{T_0}{T}$

is something I suspect can be formed. The equation I suggested not long ago for the transformation of the volume lead to a change in the entropy:

$\Delta S = Nk_B\ \log_2(\frac{V_2}{V_1})$

On my blog a while back I derived the following (with some adjustments for clarity here):

$\Delta S = Nk_B \log_2(\frac{T_2}{T_1})$

So it stands to reason the identity may hold to satisfy a covariant temperature following the same transformation laws as the four volume.

$\Delta S = Nk_B(\log_2(\frac{V_2}{V_1}) = \log_2 (\frac{T_2}{T_1}))$

The only two invariant quantities in this equation is entropy and the Boltzmann constant. The covariant temperature transforms in exactly the same way as the volume and should be approximated under the kinetic energy of the equipartition theorem. It is probably also a case that the four volume would transform as:

$S = \sqrt{(\frac{1}{(nk_BT)^2} - \frac{1}{(pV_x)^2} - \frac{1}{(pV_y)^2} - \frac{1}{(pV_z)^2)}}$

In any case, it seems we have collected all the information we need to construct the final energy-mass equivalence through using the common and ideal gas laws.

$E_k = \sum_{N=i} \frac{1}{2}m_i v^2 + \frac{3}{2}PV + \frac{3}{2}Nk_BT$

and you know, just for a pure stab in the dark, since the transformation laws apply to the last two terms, it will mathematically also to the first term, the kinetic energy ensemble. It's not so much in the dark, when you realize the kinetic, pressure and temperature are all fundamentally related that they have their own definition of a gas law, so there was some pressure there that pushed me.

$E_k = \sum_{N=i} \frac{1}{2}m_i v^2 + \frac{3}{2}PV + \frac{3}{2}Nk_BT$

All we need now is the entropy equation which shows the truthfulness behind both Ott's statement and Einstein's:

$\Delta \mathbf{S} = 2T \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = \frac{2T}{\sqrt{1 - \frac{v^2}{c^2}}}$

It is interesting to note, the factor two in this equation is a historical relic of their derivation - some uses for it some with normalizing a kinetic energy and this would bring us back to our energy form, in short version:

$E = mc^2 + PV + everything\ else$

and we want this because my intention originally was to find the general solution to the rest mass, but I had to resort to the kinetic energy as how it was defined through the ideal gas law variables. Let us also remind ourselves of the law in which the transformation of the volume was the same with temperature and it looks like:

$\Delta S = Nk_B(\log_2(\frac{V_2}{V_1}) = \log_2 (\frac{T_2}{T_1}))$

Now, plugging the Lorentz transform into the whole system, gives ~

$E_k = (\sum_{N=i} \frac{1}{2}m_i v^2 + \frac{3}{2}PV + 3Nk_BT) \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = (\sum_{N=i} \frac{1}{2}m_i v^2 + \frac{3}{2}PV + \frac{3}{2}Nk_BT ) \frac{2}{\sqrt{1 - \frac{v^2}{c^2}}}$

Now this is starting to get interesting, since all we have done to obtain this equation, including the transformation laws, is from stipulations of relativity. As noted, the factor of two removes the kinetic interpretation behind the terms in the paranthesis, so we correct this now, but in doing so, we find a very unique interpretation of how the kinetic energy is related to the rest energy by a transformation law!! This means, in some strange way, the kinetic energy can be equal to the rest mass, through a transformation law:

$E_k = (\sum_{N=i} \frac{1}{2}m_i v^2 + \frac{3}{2}PV + 3Nk_BT) \sqrt{1 - \frac{v^2}{c^2}} \frac{1}{1 - \frac{v^2}{c^2}} = (\sum_{N=i} m_ic^2 + 3PV + 3Nk_BT ) \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$

But totally feasible since the extra Lorentz factor demonstrates a boost in the velocity of the system it speaks of.... but being able to equate it in a fashion like this, is really unusual.

Let's write each term for what it is:

$E_k = \frac{1}{2}\sum_{N=i} \sqrt{1 - \frac{v^2}{c^2}} \frac{m_i v^2}{1 - \frac{v^2}{c^2}} + \frac{3}{2}\sqrt{1 - \frac{v^2}{c^2}} \frac{PV}{1 - \frac{v^2}{c^2}} + \frac{3}{2}\sqrt{1 - \frac{v^2}{c^2}} \frac{Nk_BT}{1 - \frac{v^2}{c^2}} = \frac{\sum_{N=i} m_ic^2}{\sqrt{1 - \frac{v^2}{c^2}}} + \frac{3PV}{\sqrt{1 - \frac{v^2}{c^2}}} + \frac{3Nk_BT}{\sqrt{1 - \frac{v^2}{c^2}}}$

Since we approved the identity, at least in theory to support the transformation,

$\Delta S = Nk_B(\log_2(\frac{V_2}{V_1}) = \log_2 (\frac{T_2}{T_1}))$

But! This would be an entropy related to three different terms, the mass, the pressure and the gas constant and/or the Boltzmann constant

I suspect we can write the stipulations in more manageable ways, but first, I'll demonstrate what I mean. For instance, the Einstein equation can be summed up as (in theory), via the entropy itself:

$\mathbf{S} = m[(\frac{v_2}{v_1})^2 + P(\frac{V_2}{V_1}) + N k_B(\frac{T_2}{T_1})]\log_2$

There are three dimensionless numbers here, $(\frac{v_2}{v_1}\ squared, \frac{V_2}{V_1}, \frac{T_2}{T_1})$... and here is the rational way to deal with this... the three remaining variables are the mass, pressure and temperature. Systemically from law, the mass is attached to the definition of pressure, defined through a volume term. The pressure is related to temperature through a gas constant. This isn't a coincidence that we constructed this equation with these linear relationships, though, energy still does, here at least, have the dimensions of energy, which isn't uncharted territories. It's been known for a while, you can construct a definition of entropy from literally, anything within logical reason.

But as I pointed out, the left hand side, the enropy $\mathbf{S}$ has prosthetic units for now as an energy, but we have both a mass, pressure and a Boltzmann constant to deal with, meaning the right hand side is not technically accurate. The mathematical way to deal with, is make it entirely dimensionless!!! I will show you why, such an equation would look like:

$\mathbf{S} =[ \frac{m_2}{m_1}(\frac{v_2}{v_1})^2 + \frac{P_2}{P_1}(\frac{V_2}{V_1}) + N \frac{k_B}{\mathbf{R}}(\frac{T_2}{T_1})]\log_2$

It would have been very easy, say to construct a ratio of

$\frac{m_e}{m_P} = \frac{\hbar c}{Gm^2} = \alpha_G$

(the gravitational fine structure)

But this is far too fundamental for the theory, also, the Planck mass is equally too fundamental for any standard particle that we have experimentally measured - but the ratio has importance no doubt in hierarchy. . This is why, we make it still, a ratio simply numbered: The last term turns out to be cooler than even I initially expected, one that doesn't need to be numbered, because you canbot make a ratio to find differences in a constant. In this case, we have used the ratio of the Boltzmann constant over the gas constant $\frac{k_B}{\mathbf{R}}$.

Edited by Dubbelosix, 06 February 2019 - 04:43 PM.

### #34 Dubbelosix

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Posted 06 February 2019 - 04:16 PM