Chapter 7 Covariance of the Lagrangian formalism

7.1 Change of dependent variables

Consider a non-singular change of variables of the generalized Lagrange coordinates,

\begin{equation} q^\alpha =q^\alpha (q^{\prime \beta },t),\quad {\rm det}\ (\frac {\partial q^\alpha }{\partial q^{\prime \beta }})\neq 0,\label {chov} \end{equation}

with inverse transformation denoted by \(q^{\prime \beta }=q^{\prime \beta }(q^\alpha ,t)\). The Lagrangian in the new coordinates is

\begin{equation} \label {eq:1a} L'(q',\dot q',t)=L(q(q',t),\frac {d}{dt} q(q',t),t):=L| \end{equation}

where the bar means that one has done the change of variables inside the function, with \(\frac {d}{dt} q^\alpha (q^{\prime \beta },t)=\frac {\partial q^\alpha }{\partial q^{\prime \gamma }}\dot q^{\prime \gamma }+\frac {\partial q^\alpha }{\partial t}\).

By direct computation one finds

\begin{multline} \label {eq:2e} \frac {\delta L'}{\delta q^{\prime \beta }}=\frac {\partial L'}{\partial q^{\prime \beta }}-\frac {d}{dt}\frac {\partial L'}{\partial \dot q^{\prime \beta }}\\=\frac {\partial L}{\partial q^\alpha }|\frac {\partial q^\alpha }{\partial q^{\prime \beta }}+\frac {\partial L}{\partial \dot q^\alpha }|\big (\frac {\partial ^2 q^\alpha }{\partial q^{\prime \beta }\partial q^{\prime \gamma }} \dot q^{\gamma '}+\frac {\partial ^2 q^\alpha }{\partial q^{\prime \beta }\partial t}\big )-\frac {d}{dt}\big (\frac {\partial L}{\partial \dot q^\alpha }|\frac {\partial q^\alpha }{\partial q^{\prime \beta }}\big )\\=\frac {\partial q^\alpha }{\partial q^{\prime \beta }} \Big [\frac {\partial L}{\partial q^\alpha }|-\frac {d}{dt}\big (\frac {\partial L}{\partial \dot q^\alpha }|\big )\Big ]. \end{multline} Indeed, when \(\frac {d}{dt}\) in the last term of the first line hits \(\frac {\partial q^\alpha }{\partial q^{\prime \beta }}\) the previous term in the first line is cancelled. As a consequence, out of the second and third term of the first line, one only remains with \(\frac {d}{dt}\) that hits \(\frac {\partial L}{\partial \dot q^\alpha }|\).

Furthermore,

\begin{equation} \label {eq:3a} \frac {d}{dt}\big (\frac {\partial L}{\partial \dot q^\alpha }|\big )=\frac {\partial ^2 L}{\partial q^{\beta }\partial \dot q^\alpha }|\frac {d}{dt}q^\beta (q^{\prime \gamma },t)+ \frac {\partial ^2 L}{\partial \dot q^{\beta }\partial \dot q^\alpha }|\frac {d^2}{dt^2}q^\beta (q^{\prime \gamma },t) +\frac {\partial ^2 L}{\partial t\partial \dot q^\alpha }|=\big (\frac {d}{dt}\frac {\partial L}{\partial \dot q^\alpha }\big )|, \end{equation}

where \(f(q,\dot q,\ddot q,t)|=f\big (q(q',t),\frac {d}{dt}q(q',t),\frac {d^2}{dt^2}q(q',t),t\big )\). Hence,

\begin{equation} \label {eq:4a} \boxed {\frac {\delta L'}{\delta q^{\prime \beta }}=\frac {\partial q^\alpha }{\partial q^{\prime \beta }} \big (\frac {\delta L}{\delta q^{\alpha }}\big )|} . \end{equation}

This equation captures the covariance of Euler-Lagrange equations under changes of Lagrange coordinates. The meaning is as follows. Suppose that \(\bar q^\alpha (t)\) is a natural trajectory, i.e., a solution to the Euler-Lagrange equations of motion,

\begin{equation} \label {eq:5} \frac {\delta L}{\delta q^{\alpha }}(\bar q^\alpha (t),\frac {d}{dt} \bar q^\alpha (t), t)=0. \end{equation}

If one changes variables, \(q^{\prime \beta }=q^{\prime \beta }(q^\alpha ,t)\), with inverse given in (7.1), it follows that \(\bar q^{\prime \beta }(t)=q^{\prime \beta }(\bar q^\alpha (t),t)\) is a solution for the variational pinciple defined by \(L'(q',\dot q',t)\). Indeed, (7.5) is equivalent to

\begin{equation} \label {eq:6} \frac {\partial q^{\prime \beta }}{\partial q^\alpha }\frac {\delta L'}{\delta q^{\prime \beta }}\big (q'(q,t), \frac {d}{dt}q'(q,t),\frac {d^2}{dt^2}q'(q,t),t\big )=\frac {\delta L}{\delta q^\alpha }(q,\dot q,\ddot q,t). \end{equation}

When evaluating at \(\bar q^\alpha (t)\), the RHS vanishes. The result that \(\frac {\delta L'}{\delta q^{\prime \beta }}\) evaluated at \(\bar q^{\prime \beta }(t)\) vanishes then follows from invertibility of \(\frac {\partial q^{\prime \beta }}{\partial q^\alpha }\).

7.2 Change of the independent variable

The action is a functional, it produces a number if one evaluates it on a trajectory,

\begin{equation} \label {eq:7} S[q]=\int ^{t_f}_{t_i}dt\, L\Big (q(t),\frac {d}{dt} q(t),t\Big )\in \mathbb {R}. \end{equation}

Under an arbitrary reparametrisation of time, \(t=t(t')\), with inverse denoted by \(t'=t'(t)\), one finds the same number provided that

\begin{equation} \label {eq:8} S[q]=\int ^{t'=t'(t_f)}_{t'=t'(t_i)}dt'\, \frac {dt}{dt'} L\Big (q(t(t')),\frac {dt'}{dt}\frac {d}{dt'} q(t(t')),t(t')\Big ). \end{equation}

This suggests that the Euler-Lagrange equations are covariant under \(t=t(t')\) and \(q^{\prime \alpha }(t')=q^\alpha (t(t'))\) provided that

\begin{equation} \label {eq:9} \boxed {L'\Big (q',\frac {d q'}{dt'},t')=\frac {dt}{dt'} L\Big (q(t(t')),\frac {dt'}{dt}\frac {d}{dt'} q(t(t')),t(t')\Big )}. \end{equation}

with

\begin{equation} \frac {d q^{\prime \alpha }}{dt'}=\frac {dq^\alpha }{dt}|_{t=t(t')}\frac {dt}{dt'}=\dot q^{\alpha }|_{t=t(t')}\frac {dt}{dt'}. \label {eq:10} \end{equation}

Consider first a simple example. If \(L=\frac 12 m \dot q^2\), \(\frac {\delta L}{\delta q}=0\iff \ddot q=0 \iff q(t)=vt + c\) with \(v,c\in \mathbb R\). Let \(t,t'>0\) and take \(t=\sqrt {t'} \iff t'=t^2\).

It follows that \(\frac {dt'}{dt}=2t=2\sqrt {t'}\), and the solution becomes \(q'(t')=v\sqrt {t'}+c\). The new Lagrangian is

\begin{equation} L'=\frac {dt}{dt'}\frac 12 m (\frac {dt'}{dt}\frac {dq'}{dt'})^2 =m\sqrt {t'}(\frac {dq'}{dt'})^2\label {eq:12}. \end{equation}

The new equations of motion are

\begin{equation} \label {eq:11} \frac {\delta L'}{\delta q'}=\frac {\partial L'}{\partial q'}-\frac {d}{dt'}\frac {\partial L'}{\partial \frac {dq'}{dt'}}= -\frac {d}{dt'}\Big (2m\sqrt {t'}\frac {dq'}{dt'}\Big )= -m\frac {1}{\sqrt {t'}}\frac {dq'}{dt'}-2m\sqrt {t'}\frac {d^2q'}{dt'^2}=0, \end{equation}

which is equivalent to

\begin{equation} \frac {d^2q'}{dt'^2}=-\frac {1}{2t'}\frac {dq'}{dt'}\label {eq:13}. \end{equation}

This equation holds since

\begin{equation} \label {eq:14} \frac {dq'}{dt'}=\frac 12 \frac {v}{\sqrt {t'}},\quad \frac {d^2q'}{dt'^2}=-\frac {1}{4}\frac {v}{(\sqrt {t'})^3}. \end{equation}

The proof in the general case proceeds as follows.

\begin{equation} \label {eq:15} \frac {\delta L'}{\delta q^{\prime \alpha }}=\frac {\partial L'}{\partial q^{\prime \alpha }} -\frac {d}{dt'}\frac {\partial L'}{\partial \frac {dq^{\prime \alpha }}{dt'}}= \frac {dt}{dt'}\frac {\partial L}{\partial q^{\alpha }}|-\frac {d}{dt'}\Big (\cancel {\frac {dt}{dt'}} \frac {\partial L}{\partial \dot q^{\alpha }}|\cancel {\frac {dt'}{dt}}\Big ) =\frac {dt}{dt'}\Big [\frac {\partial L}{\partial q^{\alpha }}|-\frac {d}{dt}\big (\frac {\partial L}{\partial \dot q^\alpha }|\big )\Big ] =\frac {dt}{dt'}\Big [\frac {\delta L}{\delta q^\alpha }\Big ]|, \end{equation}

where the last equality holds provided one can show that

\begin{equation} \label {eq:16} \frac {d}{dt}\big (\frac {\partial L}{\partial \dot q^\alpha }|\big )=\big [\frac {d}{dt}\frac {\partial L}{\partial \dot q^\alpha }\Big ]|. \end{equation}

As in the preceding section, \(\frac {\delta L'}{\delta q^{\prime \alpha }}=\frac {dt}{dt'}(\frac {\delta L}{\delta q^\alpha })|\) means that, if \(q^\alpha (t)\) is a natural trajectory for \(L(q,\dot q,t)\), then so is \(q^{\prime \alpha }(t')=q^{\alpha }(t(t'))\) for \(L'(q',\frac {dq'}{dt'},t')\).

In order to show (7.17), we have

\begin{multline} \label {eq:17} \frac {d}{dt}\Big [\frac {\partial L}{\partial \dot q^\alpha }\big (q'(t'),\frac {dt'}{dt}\frac {dq'}{dt'},t(t')\big )\Big ]\\=\frac {\partial ^2L}{\partial q^\beta \partial \dot q^\alpha }|\frac {dt'}{dt} \frac {dq^{\prime \beta }}{dt'}+\frac {\partial ^2L}{\partial \dot q^\beta \partial \dot q^\alpha }|\big (\frac {d^2t'}{dt^2}\frac {dq^{\prime \beta }}{dt'} +(\frac {dt'}{dt})^2\frac {d^2q^{\prime \beta }}{dt'^2}\big )+\frac {\partial ^2L}{\partial t\partial \dot q^\alpha }|\cancel {\frac {dt'}{dt}}\cancel {\frac {dt}{dt'}}, \end{multline} while

\begin{equation} \label {eq:18} \Big [\frac {d}{dt}\frac {\partial L}{\partial \dot q^\alpha }\Big ]|=\frac {\partial ^2L}{\partial q^\beta \partial \dot q^\alpha }|\dot q^\beta |+ \frac {\partial ^2L}{\partial \dot q^\beta \partial \dot q^\alpha }|\ddot q^\beta |+\frac {\partial ^2L}{\partial t\partial \dot q^\alpha }|. \end{equation}

When taking into account (7.11), the first terms on the RHS do agree, since so do the last terms, we only have to show that middle terms do as well, which is the case if

\begin{equation} \label {eq:19} \ddot q^\beta |=\frac {d^2t'}{dt^2}\frac {dq^{\prime \beta }}{dt'} +(\frac {dt'}{dt})^2\frac {d^2q^{\prime \beta }}{dt'^2}. \end{equation}

Differentiating (7.11) with respect to \(t'\) gives

\begin{equation} \label {eq:20} \frac {d^2 q^{\prime \beta }}{dt'^2} =\frac {d}{dt'}( \frac {dt}{dt'} \dot q^\beta |)=\frac {d^2t}{dt'^2}\dot q^\beta |+(\frac {dt}{dt'})^2\ddot q^\beta |, \end{equation}

which implies that

\begin{equation} \label {eq:21} \ddot q^\beta |=(\frac {dt'}{dt})^2 \frac {d^2 q^{\prime \beta }}{dt'^2}-(\frac {dt'}{dt})^2\frac {d^2t}{dt'^2}\dot q^\beta |. \end{equation}

This gives the result when using (7.11) and

\begin{equation} \label {eq:22} 0=\frac {d}{dt}(\frac {dt'}{dt}\frac {dt}{dt'})=\frac {d^2t'}{dt^2}\frac {dt}{dt'} +(\frac {dt'}{dt})^2\frac {d^2t}{dt'^2}\Longrightarrow \frac {d^2t'}{dt^2}=-(\frac {dt'}{dt})^3\frac {d^2t}{dt'^2}. \end{equation}

7.3 Combined change of variables

When combining the results of the two previous sections, Euler-Lagrange equations are covariant under

\begin{equation} \label {eq:23} \left \{\begin{array}{l} q^\alpha =q^\alpha (q^{\prime \beta },t') \\ t=t(t') \end {array}\right . \end{equation}

with inverse

\begin{equation} \label {eq:23a} \left \{\begin{array}{l} q^{\alpha }=q^{\prime \alpha }(q^{\beta },t) \\ t'=t'(t) \end {array}\right . \end{equation}

provided that

\begin{equation} \label {eq:24} \boxed {L'\Big (q^{\prime \beta }(t'),\frac {d}{dt'}q^{\prime \beta }(t'),t'\Big ) =\frac {dt}{dt'} L\Big (q^\alpha (q^{\prime \beta }(t'),t'),\frac {dt'}{dt}\frac {d}{dt'} q^\alpha (q^{\prime \beta }(t'),t'),t(t')\Big )}, \end{equation}

with \(\frac {d}{dt'} q^\alpha (q^{\prime \beta }(t'),t') =\frac {\partial q^\alpha }{\partial q^{\prime \gamma }}\frac {d q^{\prime \gamma }}{dt'}+\frac {\partial q^\alpha }{\partial t'}\). Indeed, showing that

\begin{equation} \label {eq:25} \frac {\delta L'}{\delta q^{\prime \beta }}=\frac {dt}{dt'}\big (\frac {\delta L}{\delta q^\alpha }\big )|\frac {\partial q^\alpha }{\partial q^{\prime \beta }}. \end{equation}

can now be done in a straightforward way by combining the intermediate steps of the proofs of the previous two sections.