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 \(L'=L+\dfrac{\mathrm{d} f(\mathbf q, t)}{\mathrm{d} t}\)
$$\begin{aligned}
\left(\frac{\dd}{\dd t}\frac{\pp}{\pp \dot q_i}-\frac{\pp}{\pp q_i}\right)L'
&=\left(\frac{\dd}{\dd t}\frac{\pp}{\pp \dot q_i}-\frac{\pp}{\pp q_i}\right)\frac{\dd f}{\dd t}\\
&=\frac{\dd}{\dd t}\frac{\pp}{\pp q_i}f-\frac{\pp}{\pp q_i}\frac{\dd}{\dd t}f\\
&=\left[\frac{\dd}{\dd t},\frac{\pp}{\pp q_i}\right]f\\
&=0\end{aligned}$$
Condition \(\mathbf q\to\mathbf Q\)
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:
$$\left(\frac{\dd}{\dd t}\frac{\pp}{\pp \dot Q_i}-\frac{\pp}{\pp Q_i}\right)L'=0$$
$$\begin{aligned}
\mathrm{LHS}
&=\left(\frac{\dd}{\dd t}\frac{\pp \dot q_j}{\pp \dot Q_i}\frac{\pp}{\pp \dot q_j}-\frac{\pp q_j}{\pp Q_i}\frac{\pp }{\pp q_j}-\frac{\pp \dot q_j}{\pp Q_i}\frac{\pp }{\pp\dot q_j}\right)L\\
&=\left(\frac{\dd}{\dd t}\frac{\pp q_j}{\pp Q_i}\frac{\pp}{\pp\dot q_j}-\frac{\pp q_j}{\pp Q_i}\frac{\pp }{\pp q_j}-\frac{\pp \dot q_j}{\pp Q_i}\frac{\pp }{\pp\dot q_j}\right)L\\
&=\left[\left(\frac{\dd}{\dd t}\frac{\pp}{\pp Q_i}q_j\right)\frac{\pp}{\pp \dot q_j}+\frac{\pp q_j}{\pp Q_i}\frac{\dd}{\dd t}\frac{\pp}{\pp \dot q_j}-\frac{\pp q_j}{\pp Q_i}\frac{\pp }{\pp q_j}-\left(\frac{\pp}{\pp Q_i}\frac{\dd}{\dd t}q_j\right)\frac{\pp }{\pp\dot q_j}\right]L\\
&=\left(\left[\frac{\dd}{\dd t}, \frac{\pp}{\pp Q_i}\right]q_j\right)\frac{\pp}{\pp \dot q_j}L+\frac{\pp q_j}{\pp Q_i}\left(\frac{\dd}{\dd t}\frac{\pp}{\pp \dot q_j}-\frac{\pp }{\pp q_j}\right)L\\
&=0\end{aligned}$$
Comments
comments powered by Disqus