Wednesday, June 4, 2014

Dirac's large numbers

can you express the ratio between the size of the Universe and the size of the electron with fundamental constants and quote a numerical value?

2 comments:

  1. the classical electron radius is estimated by equating the rest mass to the Coulomb self-energy
    \begin{equation}
    \frac{e^2}{4\pi r_e} = m_ec^2
    \end{equation}
    and can be expressed in terms of the Compton-wavelength $\lambda_e=h/(m_ec)$ and the fine structure constant $\alpha$: \begin{equation}
    r_e = \alpha\lambda_e
    \end{equation}
    putting this quantity in relation with the size of the Universe $\chi_H = c/H_0$ yields
    \begin{equation}
    \frac{r_e}{\chi_H} = \alpha\frac{\lambda_e}{\chi_H} = \alpha\frac{hH_0}{m_ec^2} \simeq 10^{-40}
    \end{equation}

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  2. CQW thinks that there's a funny follow up calculation... you could compare the Rydberg energy $E_\mathrm{Ry}=hcR$ to the mass contained in the Hubble radius and get the following ratio:
    \begin{equation}
    \frac{E_\mathrm{Ry}}{Mc^2} = \alpha\frac{l_p^2}{\lambda_e\chi_H}
    \simeq 10^{-40}
    \end{equation}
    with the Planck length $l_p=\sqrt{hG/c^3}$, the Rydberg constant $R=\alpha^2/2/\lambda_e$ and the mass $M=4\pi/3\rho_\mathrm{crit}\chi_H^3$ inside the Hubble volume, with $\rho_\mathrm{crit}=3H_0^2/(8\pi G)$.

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