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

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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

\(\bar{\upsilon}_C\;\;{then}\;\;\;f_C=\frac{1}{\alpha}\)

Electron Be Frequency

\(\upsilon_{B_e}\;\;{then}\;\;\;f_{B_e}={\sqrt{\alpha}}\)

Planck Frequency

\(\upsilon_P\;\;{then}\;\;\;f_P=\frac{\sqrt{\alpha_G}}{\alpha}\)

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.

Note

\(B_{e}=\dfrac{r_{e}}{m_{e}}=\dfrac{c^{3}}{h\;\nu^2_e}\)

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

2 Answers

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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

\(\upsilon_{\Lambda}\;\;{then}\;\;\;f_{\Lambda}=\frac{{\alpha^2}}{\alpha_G}=f_P^{-2}\)

Applying this to dark energy density 

\(u_{DE}=\dfrac{\Lambda}{\kappa}=\frac{A^2_{\Lambda}}{8\pi\;B_P}=\frac{u_P}{8\;\pi}\dfrac{\nu_{\Lambda}^{2}}{\nu_p^{2}}=\frac{u_P}{8\;\pi}\;6.8*10^{-122}=5.3*10^{-10}\;\frac{J}{m^3}\)
------

answered Aug 3, 2022 by Hyperthought (5 points) [ no revision ]

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