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?

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

ReplyDelete\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 $r_s=2GM/c^2$.

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.

ReplyDeleteAnother way of thinking about it is via the fine-structure constant and its gravitational analogon. The original hydrogen Hamiltonian is

ReplyDelete$$ H = \frac{p^2}{2m} - \frac{e^2}{r} = \frac{p^2}{2m} - \frac{\alpha \hbar c}{r} $$

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.

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!

DeleteCQW 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.

ReplyDelete