Can Einstein's coefficient explain aspects of gravity?

Question

Are the below observation valid?

Shortly after developing his theory of General Relativity, Einstein developed his Quantum Theory of Radiation. If we converge these two theories we can develop Einstein’s A and B Gravitational Coefficient. For the moment we will focus on the B Gravitational Coefficient, which blends the classical and quantum relationships.

\(B=\frac{l}{m}=\frac{c^3}{h\;v^2}\)

As such, Einstein’s B_e coefficient for the electron is;

\(B_e=\frac{r_e}{m_e}=\frac{2.8\times 10^{-15}\;m}{9.1\times 10^{-31}\;kg}= 3.1\times 10^{15} \frac{m}{kg}\)

In Einstein's field equations, Einstein’s gravitational constant kappa is proportional to Einstein's B_k gravitational coefficient, (at the Planck scale).

\(B_\kappa=\frac{\kappa}{8 \pi}=\frac{G}{c^2}=B_P= \frac{l_P}{m_P}=7.4\times 10^{-28} \frac{m}{kg}\)

Comparing the ratio of B_k to B_e,

\(\frac{B_\kappa}{B_e}=2.4\times 10^{-43} \)

This ratio is exactly equal to the relative strength of the gravitational to the electrostatic force of two electrons. All values of B_x are frequency squared scaled versions of Einstein's B_k gravitational coefficient;

\(B_x=B_{\kappa} \times \frac{v^2_{\kappa} }{v^2_x}\)

As the B coefficient is proportional to the inverse frequency squared the following holds:

\(\frac{G\;{m^2_e}}{k_e\;q_e^2}=\frac{B_\kappa}{B_e}=\frac{\upsilon_{Be}^2}{\upsilon_{B\kappa}^2}=2.4\times 10^{-43} \)

From the above we can also conclude the weakness of the gravitational force is related to a frequency squared and the probability of stimulated emission/absorption of energy due to curvature.

Comments

This is very impressive. I got these formulas in a different way