Notes for AKT-140117/0:26:27

Just some generalization for the least action principle and Euler-Lagrange equation for the classical cases. In the calculus of variation, we have developed a tool for describing various physics situation. In general, let $q$ be the coordinate of a particular configuration space and ${\dot {q}}$ be its time derivative. Then, the action is described as

${\mathcal {L}}=\int _{t_{i}}^{t_{f}}dtL\left(q,{\dot {q}}\right),$

where $t_{i},t_{f}$ are the initial time and final time, respectively. The integrand $L$ is known as the Lagrangian and is assumed to be time-independent for convenience. The idea here is to find the path that minimize the action ${\mathcal {L}}$. Now, we introduction the idea of variation, which can be viewed as an infinitesimal shift from the original path; however, it does not change the terminal points. Since the path we are interested is the path that minimizes the action, then the variation of the action should be 0 and that is

$0=\delta {\mathcal {L}}=\delta \int _{t_{i}}^{t_{f}}dtL\left(q,{\dot {q}}\right)=\int _{t_{i}}^{t_{f}}dt\delta L\left(q,{\dot {q}}\right)$ $=\int _{t_{i}}^{t_{f}}dt\left({\frac {\partial L}{\partial q}}\delta q+{\frac {\partial L}{\partial {\dot {q}}}}\delta {\dot {q}}\right)=\int _{t_{i}}^{t_{f}}dt\left({\frac {\partial L}{\partial q}}\delta q+{\frac {\partial L}{\partial {\dot {q}}}}{\frac {d}{dt}}\left(\delta q\right)\right).$

Then, by the integration by parts, we have that

$0=\int _{t_{i}}^{t_{f}}dt\left({\frac {\partial L}{\partial q}}-{\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}}}\right)\right)\delta q+\int _{t_{i}}^{t_{f}}{\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}}}\delta q\right)=\int _{t_{i}}^{t_{f}}dt\left({\frac {\partial L}{\partial q}}-{\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}}}\right)\right)\delta q+{\frac {\partial L}{\partial {\dot {q}}}}\delta q{\big |}_{t_{i}}^{t_{f}}=\int _{t_{i}}^{t_{f}}dt\left({\frac {\partial L}{\partial q}}-{\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}}}\right)\right)\delta q,$

since the boundary term does not vary so that $\delta q\left(t_{i}\right)=\delta q\left(t_{f}\right)=0$. Thus, we arrive at the point where the classical particle must obey the path where the equation

${\frac {d}{dt}}\left({\frac {\partial L}{\partial {\dot {q}}}}\right)-{\frac {\partial L}{\partial q}}=0.$

This equation is known as the Euler-Lagrange equation.