**Summery**

Three equations of interest to me derived coherently was:

[math]F = \frac{1}{2} \rho v^2\ f\ A = m\frac{2 \pi c\ k_BT}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G \frac{Mm}{r^2}[/math]

[math]\frac{1}{f}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - \frac{A_b}{A_f} \cdot \rho v^2 = - \frac{A_b}{A_f} \cdot T^{00} [/math]

[math] - \frac{1}{f}\frac{A_f}{A_b}(\frac{mv^2}{n^2x^3} + \frac{mv^2}{n^2y^3} + \frac{mv^2}{n^2z^3}) = - g^{00}\ \frac{\Lambda c^4}{8 \pi G} = T^{00} [/math]

The first was a drag equation unified with thermodynamic gravity in two dimensions. The second equation was transcended into a three dimensional format and the third explained the cosmological constant as a thrust. The thrust in this sense, had to be greater than the drag of gravity:

[math]Thrust >\ Drag[/math]

or much greater than the drag

[math]Thrust >>\ Drag[/math]

We also deducted the law of gravitational mass and inertial mass was also equivalent to the drag of the mass, extending the Einstein equivalence:

[math]m_d = m_g = m_i[/math]

Instead of a negative mass, the thrust appears to be thermodynamically-related to a pressure difference as found from previous independent speculations from Bernoulli's equations when using a correction factor to Einstein's equivalence principle.

**Extending the Investigation for the Compton Wavelength**

We will use the master equation linking drag to the thermodynamic entropy:

[math]F = \frac{1}{2} \rho v^2\ f\ A = m\frac{2 \pi c\ k_BT}{\hbar} = m\frac{4 \pi c}{\hbar} \frac{E}{N} = m\frac{4 \pi c^2}{\hbar} \frac{M}{N} = \frac{4 \pi GMm}{A} = G \frac{Mm}{r^2}[/math]

And remind ourselves of the corrections we introduced, one by Lorentz (which included the dispersion of an acoutsic gravitational wave in the modified theory of mine) and the other I derived for the extension into three dimensions in which the realization was that the index of refraction had to be quantized, revealing that the principal quantum number plays the identical role:

[math] 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}[/math]

[math]\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[/math]

And that

[math]f = \frac{2F}{\rho v^2 A} = \frac{A_b}{A_f} \frac{B}{Re^2_L}[/math]

Moving forward, the pressure was defined from a Planck unit investigation which led to a derivation involving the wavelength:

[math]P = \frac{1}{2} \rho v^2\ f = 2 \pi \rho_Ac^2\ (\frac{mc}{\hbar}) = 2 \pi \rho_Ac^2\ \lambda^{-1}[/math]

Solving for the wavelength we have, and obtain the Doppler shift:

[math]P = \frac{\rho}{4 \pi \rho_A} (\frac{v^2}{c^2})\ f = \frac{\rho}{4 \pi \rho_A}\ (1 - \frac{1}{\gamma^2}) = \lambda^{-1}[/math]

And the inverse of the force equation is:

[math]F^{-1} = \frac{2}{\rho v^2\ f\ A} = \frac{\hbar}{2 \pi mc}(\frac{1}{k_BT}) = \frac{r^2}{GMm} = \frac{r^2}{\hbar c}[/math]

If either the force or the density increases, likewise the wavelength becomes larger! Since the force is proportionally inverse to the kinetic energy, This reserves a specific rule in physics in which short wavelegths carry more energy than longer wavelengths. Suppose we take one factor of [math]r^2[/math] for the Schwarzchild radius, we would have

[math]r^2_s = \frac{G^2M^2}{c^4}[/math]

we obtain a simple equation

[math]F^{-1} = \frac{G}{c^4}(\frac{M_e}{M_P})(\frac{m_e}{m_P}) = \frac{G}{c^4}\ \frac{Gm^2}{\hbar c} = \alpha_G \ (\frac{G}{c^4})[/math]

In which [math]\frac{G}{c^4}[/math] has units of the inverse of the upper limit of the gravitational force.

**Edited by Dubbelosix, 01 July 2019 - 06:27 PM.**