# On bohr- sommerfeld theory

1. Jan 15, 2010

### Kennalj

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)