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  Can Einstein's coefficient explain aspects of gravity?

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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 Be 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 Bk 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 Bk to Be,

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

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

asked Jul 13 in Theoretical Physics by Hyperthought (5 points) [ no revision ]

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