Consider the Lagrangian $L(q_i,\dot{q_i},t)$ for $i=1,2, ...n$. Transform (invertibly) $q_i$ to another set of generalized coordinates $s_i=s_i(q_j,t)$. Now, in a different scenario, consider transformation of $q_i$ under some group, so that $q_i \rightarrow q_i'=f(q_j,\epsilon_k)$ where $f(q_j,\epsilon_k)$ is some function of $q_j$'s and parameters $\epsilon_k$. My doubts are

  1. Does the the Lagrangian transform as $L\rightarrow L^{'}(s_i,\dot{s_i},t)=L(q_i(s_j,t),\dot{q_i}(s_j,\dot{s_j},t),t)$

or as $L\rightarrow L^{'}(s_i,\dot{s_i},t)=L(s_i,\dot{s_i},t)$ ?

  1. Is there a difference in the way $L$ (or the Lagrangian density $\mathcal{L}$) transform under (1) above mentioned coordinate transformation and (2) Group transformation of coordinates $q_i$ (or fields $\phi_i$)

  2. Does the magnitude of $L$ change in either of the transformations of kind (1) and (2)?

  3. Does form invariance of Lagrangian imply invariance (magnitude wise) of the corresponding action?

  • $\begingroup$ What do you mean by the "magnitude" of the lagrangian? $\endgroup$
    – A. Bordg
    Jun 28, 2020 at 16:03
  • $\begingroup$ @A.Bordg I mean its numerical value (at any time) $\endgroup$ Jun 28, 2020 at 16:15
  • $\begingroup$ See also Eqs. (1)-(12) here. $\endgroup$
    – J.G.
    Jun 29, 2020 at 6:48
  • $\begingroup$ @aneetkumar I completed my answer with more details. $\endgroup$
    – A. Bordg
    Jul 2, 2020 at 14:53
  • $\begingroup$ @aneetkumar Are you ok with the distinction between passive and active transformations? $\endgroup$
    – A. Bordg
    Jul 3, 2020 at 14:54

1 Answer 1


Regarding your first question, the Lagrangian transforms as a scalar. It means that $$L'({\bf s},{\bf \dot{s}},t) = L({\bf q}({\bf s},t),{\bf \dot{q}}({\bf s},{\bf \dot{s}},t),t)\,.$$ Transformations ${\bf q} \rightarrow {\bf s}({\bf q},t)$ are called point transformations. Actually, Lagrange's equations are invariant under point transformations, but not under the much larger class of canonical transformations, i.e. transformations of the form \begin{align*} {\bf q}\rightarrow {\bf s}({\bf q},{\bf p},t) & & {\bf p}\rightarrow {\bf P}({\bf q},{\bf p},t)\,. \end{align*} On the other hand, Hamilton's equations are invariant under canonical transformations. This is one of the advantages of the Hamiltonian formalism.

Regarding your third question, it may be useful to distinguish between passive and active transformations. Passive transformations consist in erasing all the labels on the points of the configuration space and replacing them with new ones, leaving unchanged the system. Clearly, passive transformations do not change the value of the Lagrangian. However, active transformations, i.e. instructions to change the system in some way, usually change the value of the Lagrangian, the few exceptions being called symmetries.

  • $\begingroup$ Thanks for your answer. Can you give an example of a generic active transformation? Would the group transformation, as stated in the question, be an active transformation? $\endgroup$ Jul 5, 2020 at 15:57
  • $\begingroup$ @aneetkumar The same transformation can be seen etiher as passive or active. For instance x' = x + 1 can be understood as relabelling the points such that x=0 is now x'=1 (passive transformation), or as an instruction to move the particle one unit to the right (active transformation). $\endgroup$
    – A. Bordg
    Jul 5, 2020 at 19:00
  • $\begingroup$ Then I can't reconcile the fact that active transformations can change L if both active and passive are in fact same thing $\endgroup$ Jul 6, 2020 at 4:27

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