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.

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