Transformation of Lagrangian and action

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?

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

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.