Invariance of the Lagrange equation under point transformations

This post is a derivation requested by some students in a course on classical mechanics at my university. Consider the Lagrangian ${L(q_i,\dot q_i,t)}$ which satisfies the Euler-Lagrange equation

$\displaystyle {\partial L \over \partial q_i} - \frac {d}{dt} {\partial L \over \partial \dot q_i}=0 ~.\ \ \ \ \ (1)$

Let ${s_j}$ be new coordinates such that we can express the ${q}$ as functions of the ${s}$ as

$\displaystyle q_i=q_i(s_j,t) ; \quad i=1,2,\ldots, n; \quad j=1,2,\ldots, n ~.\ \ \ \ \ (2)$

Such changes of variables are known as point transformations. It is relatively easy to show, by variational calculus, that the Euler-Lagrange equation is invariant under point transformations. Here we show this invariance explicitly via the chain rule.

First note that

$\displaystyle {\partial L \over \partial s_j}=\sum_i \left[ {\partial L \over \partial q_i}{\partial q_i \over \partial s_j} + {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} + {\partial L \over \partial t}{\partial t \over \partial s_j} \right] ~. \ \ \ \ \ (3)$

Since ${t}$ is an independent variable that depends on nothing else, and since the partial derivative is taken with other coordinates fixed, the last term is thus zero so that

$\displaystyle {\partial L \over \partial s_j}=\sum_i \left[ {\partial L \over \partial q_i}{\partial q_i \over \partial s_j} + {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} \right]~. \ \ \ \ \ (4)$

Next, consider that

$\displaystyle \dot q_i = \sum_j {\partial q_i \over \partial s_j}{\dot s_j} + {\partial q_i \over \partial t} \ \ \ \ \ (5)$

so that

$\displaystyle {\partial \dot q_i \over \partial \dot s_j}= {\partial q_i \over \partial s_j} ~. \ \ \ \ \ (6)$

Now consider that

$\displaystyle {\partial L \over \partial \dot s_j}=\sum_i \left[ {\partial L \over \partial q_i}{\partial q_i \over \partial \dot s_j} + {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial \dot s_j} \right]~. \ \ \ \ \ (7)$

The first term is zero from (2). Substituting (6) we get

$\displaystyle {\partial L \over \partial \dot s_j}=\sum_i \left[ {\partial L \over \partial \dot q_i}{\partial q_i \over \partial s_j} \right]~. \ \ \ \ \ (8)$

Hence

$\displaystyle \frac {d}{dt} {\partial L \over \partial \dot s_j}= \sum_i \left[ {\partial q_i \over \partial s_j}\frac {d}{dt}{\partial L \over \partial \dot q_i} + {\partial L \over \partial \dot q_i}\frac {d}{dt}{\partial q_i \over \partial s_j}\right]~. \ \ \ \ \ (9)$

Putting everything together we get

$\displaystyle \begin{array}{rcl} \displaystyle {\partial L \over \partial s_j} - \frac {d}{dt} {\partial L \over \partial \dot s_j} &=& \displaystyle \sum_i \left[ {\partial L \over \partial q_i}{\partial q_i \over \partial s_j} + {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} - {\partial q_i \over \partial s_j}\frac {d}{dt}{\partial L \over \partial \dot q_i} - {\partial L \over \partial \dot q_i}\frac {d}{dt}{\partial q_i \over \partial s_j} \right] \\ &=& \displaystyle \sum_i \left[ {\partial q_i \over \partial s_j} \left( {\partial L \over \partial q_i} -\frac {d}{dt}{\partial L \over \partial \dot q_i} \right) + {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} - {\partial L \over \partial \dot q_i}\frac {d}{dt}{\partial q_i \over \partial s_j} \right] \\ &=& \displaystyle \sum_i \left[ {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} - {\partial L \over \partial \dot q_i}\frac {d}{dt}{\partial q_i \over \partial s_j} \right] \\ &=& \displaystyle \sum_i \left[ {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} - {\partial L \over \partial \dot q_i}{\partial \dot q_i \over \partial s_j} \right] \\ &=& 0 ~. \end{array}$

Q.E.D.