When we calculate the spin-orbit interaction in a Hydrogen atom we just work in the electron's frame of reference: the proton is moving and produces a magnetic field which the electron's spin interacts with.

We can show here that the answer is $$ \Delta H = \frac{2\mu_B}{\hbar m_e e c^2}\frac{1}{r}\frac{\partial U(r)}{\partial r} \mathbf{L} \cdot\mathbf{S}$$ where $U(r)$ is the potential energy = $eV(r)$ with $V(r) = \frac{1}{4\pi\epsilon_0}\frac{e}{r}$ for a proton.

NOW: I want to get the same answer from the reference frame of the proton, where the proton is stationary and the electron is moving. Since Physics must be the same in all frames of reference, we should get the same answer.

I guess that the only this can happen is if the electron's magnetic field (due to its motion, i.e. charged particle moving around) interacts with the electron's own spin.

We can calculate the current density $\mathbf{j}$ of the electron in Hydrogen, and it is given by: $$ j_\phi=-e\frac{\hbar m}{\mu r\sin\theta}\left|\psi_{nlm}\left(r,\theta,\phi\right)\right|^2 $$ (derivation found here on page 6)

I could use the Biot-Savart law to calculate the magnetic field due to this current density: $$\textbf{B} = \frac{\mu_0}{4\pi} \frac{1}{r^2} \int \textbf{J}d^3\textbf{r}$$ where the integration should be (at least classicaly) the along the current loop.

Here, I get stuck.

Does anyone know how to get the $\textbf{L}\cdot\textbf{S}$ factor from this approach?

When we calculate the spin-orbit interaction in a Hydrogen atom we just work in the electron's frame of reference: the proton is moving and produces a magnetic field which the electron's spin interacts with.

We can show here that the answer is $$ \Delta H = \frac{2\mu_B}{\hbar m_e e c^2}\frac{1}{r}\frac{\partial U(r)}{\partial r} \mathbf{L} \cdot\mathbf{S}$$ where $U(r)$ is the potential energy = $eV(r)$ with $V(r) = \frac{1}{4\pi\epsilon_0}\frac{e}{r}$ for a proton.

NOW: I want to get the same answer from the reference frame of the proton, where the proton is stationary and the electron is moving. Since Physics must be the same in all frames of reference, we should get the same answer.

I guess that the only way this can happen is if the electron's magnetic field (due to its motion, i.e. charged particle moving around) interacts with the electron's own spin.

We can calculate the current density $\mathbf{j}$ of the electron in Hydrogen, and it is given by: $$ j_\phi=-e\frac{\hbar m}{\mu r\sin\theta}\left|\psi_{nlm}\left(r,\theta,\phi\right)\right|^2 $$ (derivation found here on page 6)

I could use the Biot-Savart law to calculate the magnetic field due to this current density: $$\textbf{B} = \frac{\mu_0}{4\pi} \frac{1}{r^2} \int \textbf{J}d^3\textbf{r}$$ where the integration should be (at least classicaly) along the current loop.

Here, I get stuck.

Does anyone know how to get the $\textbf{L}\cdot\textbf{S}$ factor from this approach?