Difference between revisions of "CM Course Notes"

Since we expect oscillatory motion of a normal mode (where ω is the same for both masses), we try: \[ x_1(t) = A_1 e^{i \omega t} \\ x_2(t) = A_2 e^{i \omega t} \]

\[\frac{\mathrm{d}}{\mathrm{d}t} \left ( \frac {\partial L}{\partial \dot{q}_j} \right ) = \frac {\partial L}{\partial q_j} \]

\[p = \left( p - \frac{e}{c} A \right), \text{ where p inside the brackets is the quantum mechanical momentum } p = -i \hbar \nabla \text{ and A is the gauge invariant vector potential.}\] Canonical momentum is not the momentum fired out of a cannon, but rather the momentum that satisfies $p = \frac{\partial \mathcal{L}}{ \partial \dot{q}}$, where $\mathcal{L}$ is the Lagrangian of the system. \[H=\frac{1}{2m} \left(p-\frac{qA}{c}\right)^2 +q\phi\]

\[H = \dot{\mathbf{q}} \frac{\partial \mathcal{L}}{\partial \dot{\mathbf{q}}} - \mathcal{L} \]

\[ \vec{v}' = \vec{v} - \vec{\omega} \times \vec{r} \]

\[\gamma \beta = \sqrt{\frac{\beta^2}{1- \beta^2}}= \sqrt{\frac{A}{1-\beta^2} + B} \\ \text{In the words of the notorious J.D. Jackson, ''we see'' that }A = 1 \text{ and }B = -1 \text{ , such that } \\ \gamma \beta = \sqrt{\frac{1}{1-\beta^2} - 1}=\sqrt{\gamma^2 - 1} \] E.g.: Lim#3021