# Relation between fixed constraints and time derivative of the Lagrangian

I have had some trouble interpreting and proving the following statement from Fasano, Marmi's "Analytical Mechanics" (page 139):

"... $${\partial L}/{\partial t} \neq 0$$ (1) only if the constraints are in motion (2)."

I do not understand wether this is an equivalence or an implication and, in that case, in which direction it goes.

I think that (1) $$\implies$$ (2) is not true for the contrapositive would be "if the constraints are fixed then $${\partial L}/{\partial t} = 0$$" which is nonsense since the system could still be in motion. Viceversa, I am not sure on how to go about proving (2) $$\implies$$ (1).

As always any comment or answer is much appreciated and let me know if I can explain myself clearer!

1. A holonomic constraint $$f({\bf X},t)=0$$ is by definition in motion/moving iff $$\partial f/\partial t \not \equiv 0$$, cf. Definition 1.35 on p. 53.

2. Ref. 1 means that $$(1)\Rightarrow (2)$$. For the implication to be true, Ref. 1 is apparently only considering a Lagrangian of the form $$L=T-U$$ with a standard kinetic term $$T=\sum_{i=1}^N \frac{1}{2}m_i\dot{\bf X}_i^2$$ a conservative potential $$U({\bf X})$$, and only holonomic constraints.

References:

1. A. Fasano & S. Marmi, Analytic Mechanics, 2006; p. 139.
• Yeah you are right, what I have written in the question is ultimately wrong (I forgot that when the constraints are fixed we can always choose a parametrization that makes the explicit dependence on time vanish). I just wanted to ask another thing: is there any clear and easy counterexample of an holonomic and conservative system such that (2) does not imply (1)? Commented Oct 6, 2022 at 22:58