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:

[math]\hat{E} = i \hbar \frac{\partial}{\partial t}[/math]

And the momentum operator

[math]\hat{p} = -i \hbar \frac{\partial}{\partial x}[/math]

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

[math]E^2 = m^2c^4 + p^2c^2[/math]

identify the operators

[math]\hat{E}^2\psi = (c^2 \hat{p}^2 + m^2c^4)\psi[/math]

Which makes

[math]\hat{E}^2\psi = (c^2 \hat{p}^2 + m^2c^4)\psi[/math]

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

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

[math]- \hbar^2 \frac{\partial^2}{\partial t^2} \psi = \hbar^2 \frac{\partial^2}{\partial x^2}c^2\psi - m^2c^4\psi[/math]

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

[math]\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[/math]

Or more commonly encountered:

[math]\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[/math]

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

[math]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}[/math]

It is possible that the true form to write is:

[math]E = \sqrt{\frac{G^2m^4}{R^2} + \frac{\hbar^2 c^2}{R^2} + p^2c^2 } + PV[/math]

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 [math]\frac{\hbar c}{R}[/math]. 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

[math]E_0 = \frac{Gm^2}{R}[/math]

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 [math]mc^2[/math], 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:

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

then

[math](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}[/math]

And this produces:

[math] \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}[/math]

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

[math]\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[/math]

and simplify further

[math]\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[/math]

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