DR. Nobody
New member
- Joined
- Dec 8, 2014
- Messages
- 15
\(\displaystyle \int \partial \mathcal{L} dt = \int \frac{m(\dot{x} \cdot \dot{x})}{\sqrt{1 - \beta}} dt = \hbar\) [1]
I wrote this not long ago... classical physics can admit other terms. We shall get to those in a minute -
\(\displaystyle \mathcal{L}=\frac{m}{2}\dot{r}\cdot \dot{r}\)
(which is essentially the same as [1] without the time integral) -
The correct full form of the energy is a classical given equation as
\(\displaystyle \mathcal{L}=\frac{m}{2}\dot{r}\cdot\dot{r}+qA\cdot \dot{r}-q\phi\)
\(\displaystyle \int \partial \mathcal{L} dt = \frac{1}{2}\int \frac{\frac{m\dot{r}\cdot \dot{r}+qA\cdot \dot{r} -q\phi}{{\sqrt{1 - \beta}}\ dt\)
which dimensionally gives the angular quantization \(\displaystyle \hbar\).
The additions allow us to talk about the particle experiencing a potential energy contribution, plus an extra contribution which is quantum mechanically an extra phase which any charged particle gets when it is moving along a vector potential. In other words, the angular momentum component is really produced by a particle moving with an extra phase in a potential, with the additional potential energy contribution from classical physics.
I wrote this not long ago... classical physics can admit other terms. We shall get to those in a minute -
\(\displaystyle \mathcal{L}=\frac{m}{2}\dot{r}\cdot \dot{r}\)
(which is essentially the same as [1] without the time integral) -
The correct full form of the energy is a classical given equation as
\(\displaystyle \mathcal{L}=\frac{m}{2}\dot{r}\cdot\dot{r}+qA\cdot \dot{r}-q\phi\)
\(\displaystyle \int \partial \mathcal{L} dt = \frac{1}{2}\int \frac{\frac{m\dot{r}\cdot \dot{r}+qA\cdot \dot{r} -q\phi}{{\sqrt{1 - \beta}}\ dt\)
which dimensionally gives the angular quantization \(\displaystyle \hbar\).
The additions allow us to talk about the particle experiencing a potential energy contribution, plus an extra contribution which is quantum mechanically an extra phase which any charged particle gets when it is moving along a vector potential. In other words, the angular momentum component is really produced by a particle moving with an extra phase in a potential, with the additional potential energy contribution from classical physics.
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