Wednesday, February 5, 2014

gravitational Rydberg constant

CQW celebrates 100 posts (and looks back on almost 3e4 views)!

what's the equivalent of the Rydberg-constant for a gravitationally bound hydrogen-analogue in Newtonian gravity (i.e. the ground state energy of "Keplerium")? would a similar construction work for Schwarzschildium?

5 comments:

  1. setting up a bound system with the field generating mass $M$ and an orbiting mass $m$ where gravity $GmM/r^2$ provides the centripetal force $m\upsilon^2/r$ needed and introducing angular momentum quantisation $L=n\hbar$ with an integer number $n$ for the angular momentum $L=m\upsilon r$ yields
    \begin{equation}
    r \times r_s = 2\lambda_c^2\times n^2
    \end{equation}
    for the orbit radius as a function of the Compton wavelength $\lambda_c=\hbar/(mc)$ and
    \begin{equation}
    E = \left(\frac{r_s}{\lambda_c}\right)^2\frac{mc^2}{4}\times\frac{1}{n^2}
    \end{equation}
    for the potential energy, with the Schwarzschild radius $r_s=2GM/c^2$.

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  2. we at CQW like this solution a lot, actually, because after all everything's given in terms of the two length scales of the system: $\lambda_c$ for quantum mechanics and $r_s$ for gravity, as well as the rest mass energy $mc^2$ of the orbiting body.

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  3. Another way of thinking about it is via the fine-structure constant and its gravitational analogon. The original hydrogen Hamiltonian is
    $$ H = \frac{p^2}{2m} - \frac{e^2}{r} = \frac{p^2}{2m} - \frac{\alpha \hbar c}{r} $$
    Then the ground state energy is
    $$ E = -\frac{1}{2} \alpha^2 m c^2 $$
    For a gravitationally bound system we have
    $$ \tilde{H} = \frac{p^2}{2m} - \frac{G M m}{r} = \frac{p^2}{2m} - \frac{ \tilde{\alpha} \hbar c}{r} $$
    with $\tilde{\alpha} = \frac{G M m}{\hbar c}$ (which is not the standard definition of the gravitational fine structure constant btw.)
    The ground state energy is then
    $$ E = -\frac{1}{2} \tilde{\alpha}^2 m c^2 $$
    Whether the same would work in GR depends of course on the masses. If $M=m_p$ and $m=m_e$, we really have a gravitationally bound "hydrogen". In that case the "gravitational Bohr radius" will be so large that GR would exactly look like Newtonian gravity. For larger masses, there will be corrections.

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    Replies
    1. we at CQW thank you, Bjoern, a lot for this alternative description! of course it makes a lot of sense to identify $\tilde{\alpha}\simeq r_s/\lambda_c$, and it's true that the gravitationally bound system seems more intuitive than the electrically bound system!

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  4. CQW would like to add one further comment on Schwarzschildium: the $\ell$-degeneracy is a consequence of the Newtonian potential and this of course would be violated by relativistic corrections close to the central body.

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