You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
$\newcommand{\bm}{\mathbf}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\pp}{\partial}$
E-L is deduced from the Hamilton’s principle
$$\delta S=\delta\int L(\bm q, \dot{\bm q}, t)dt=0$$
It is easy to find that, for $L'=L+\dd f(\bm q, t)/\dd t$ or change of
variables $\bm q\to\bm Q$, the min of $\delta S$ will not change. Here
we want to prove it the hard way—using E-L equations.
The original E-L Equations are:
$$\left(\frac{\dd}{\dd t}\frac{\pp}{\pp \dot q_i}-\frac{\pp}{\pp q_i}\right)L=0$$
Commutator $\left [\dfrac{\mathrm{d}}{\mathrm{d} t}, \dfrac{\partial}{\partial q_i}\right ]f(\mathbf q, t)=0$
Proof
$$\left[\frac{\dd}{\dd t},\frac{\pp}{\pp q_i}\right]f(\bm q, t) = \left[\dot q_i\frac{\pp}{\pp q_i}+\frac{\pp}{\pp t},\frac{\pp}{\pp q_i}\right]f(\bm q, t)=0$$
Condition $L'=L+\dfrac{\mathrm{d} f(\mathbf q, t)}{\mathrm{d} t}$
We we change generalized coordinates $\bm q\to\bm Q$, the Lagrangian:
$$L(\bm q, \dot{\bm q}, t)\to L'(\bm Q, \dot{\bm Q}, t)=L\big[\bm q(\bm Q, t), \dot{\bm q}(\bm Q, \dot{\bm Q}, t), t\big]$$
We want to prove:
$$\left(\frac{\dd}{\dd t}\frac{\pp}{\pp \dot Q_i}-\frac{\pp}{\pp Q_i}\right)L'=0$$