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  Dose Einstein's B coefficient determine the value of alpha?

+ 0 like - 1 dislike

Dose Einstein's B coefficient determine the value of alpha?

Einstein's B coefficient can be expressed as oscillator strength - \(f\) (a dimensionless value that expresses the probability transitions between energy levels).

\(B_x=\frac{c^3}{h\;\nu^2_x}= \frac{k_e\;e^2}{m_e\;h\;\upsilon_{x}}\;f\)

Solving for oscillator strength in terms of frequency we get;

\(f =\frac{m_e\;c^3}{k_e\;e^2}\frac{1}{\upsilon_x}=\left[\frac{\alpha_G^{0.5}}{\alpha} \nu_P\right]\frac{1}{\nu_x}=\frac{1.1\times 10^{23}}{\upsilon_x}\)

By oscillating the B coefficient's radiation field at specific frequencies we obtain;

Compton Frequency


Electron Be Frequency


Planck Frequency


From the above the value alpha is a resonant frequency of Einstein's B coefficient radiation field.

Another resonant frequency of Einstein's B coefficient radiation field generates a gravitational field.



asked Jul 21, 2022 in Theoretical Physics by Hyperthought (5 points) [ no revision ]

2 Answers

+ 0 like - 0 dislike

All dimensionless and dimensionful "numbers" must follow naturally from the corresponding equations. Then they have clear physical meaning. Simply combining the numbers as you like is a numerology, and thus it is void of physical meaning.

answered Jul 24, 2022 by Vladimir Kalitvianski (102 points) [ revision history ]
edited Jul 24, 2022 by Vladimir Kalitvianski
+ 0 like - 0 dislike

Applying the cosmological constant to the oscillator strength model gives us;

\(f =\frac{m_e\;c^3}{k_e\;e^2}\frac{1}{\upsilon_x}=\left[\frac{\alpha_G^{0.5}}{\alpha} \nu_P\right]\frac{1}{\nu_x}=\frac{1.1\times 10^{23}}{\upsilon_x}\)

Cosmological Constant


Applying this to dark energy density 


answered Aug 4, 2022 by Hyperthought (5 points) [ no revision ]

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