# On bohr- sommerfeld theory

by Kennalj
Tags: bohr, sommerfeld, theory
 P: 6 So when relativity is applied to an orbiting electron, you get: $$p_{\varphi}=mr^2 \dot{\varphi}, \quad m=\frac{m_{0}}{\sqrt{1-\beta^2}}, \quad \beta=\frac{v}{c}$$ Change the rectangular coodinates into the polar coordinates, $$x = r cos \varphi, \qquad y = r sin \varphi$$ The nucleus is at the origin. The equation of motion is (the Coulomb force condition), $$\frac{d}{dt}m\dot{x}= - \frac{kZe^2}{r^2}cos \varphi, \quad \frac{d}{dt}m\dot{y}= - \frac{kZe^2}{r^2}sin \varphi$$ Using the next condition (the angular momentum $$p_{\varphi}$$ is the constant), $$\frac{d}{dt}= \frac{d\varphi}{dt} \frac{d}{d\varphi}= \frac{p_{\varphi}}{mr^2} \frac{d}{d\varphi}$$ So the equation of the motion is ($$u= 1/r$$), $$\frac{d}{dt}m\dot{x}= - \frac{p_{\varphi}^2}{mr^2}(u+\frac{d^2 u}{d\varphi^2}) cos \varphi$$ In the case of y, change the upper cos into sin. Combine this with the Coulomb force condition, $$\frac{d^2 u}{d \varphi^2}+u = \frac{kZe^2 m_{0}}{p_{\varphi}^2} \frac{1}{\sqrt{1-\beta^2}}$$ Using the energy $$W$$ (of #4) and erase the $$\beta$$, the solution can be expressed as, $$u = \frac{1}{r} = C (1+ \epsilon cos \gamma \varphi)$$ And, the condition of the quantization is, (using the partial integration) $$\oint p_{r}dr= p_{\varphi} \epsilon^2 \gamma \oint \frac{sin^2 \varphi d \varphi}{(1+\epsilon cos \varphi)^2} = p_{\varphi} \gamma \oint (\frac{1}{1+\epsilon cos \varphi}-1) d\varphi=n_{r} h$$ And, we should use the following mathematical formula, too, $$\frac{1}{2\pi} \oint \frac{d \varphi}{1+ \epsilon cos \varphi} = \frac{1}{\sqrt{1-\epsilon^2}}$$ my question is, assuming r= radius, how does momentum = mr and furthermore p(lorentz)=mr^2(lorentz)