Question about statement of Noether's theorem

dEdt

In a lecture on Classical Mechanics by Susskind, he says that for Noether's theorem to hold, we have to have a differential transformation of the coordinates which does not depend on time explicitly ie from [itex]\vec{q}\rightarrow \vec{q}'(\varepsilon,\vec{q})[/itex], where s is some parameter. I don't see why it's necessary that the transformation doesn't contain a time term -- as far as I can tell, his proof didn't require that assumption, but perhaps it crept in somewhere.

Here's a link to the lecture: http://www.youtube.com/watch?v=FZDy_...feature=relmfu (start at around 31 min).

vanhees71

This is only the most simple case. More generally there can be symmetries with explicit time dependence in the symmetry transformation. The most prominent example is invariance under Galilei (Newtonian physics) or Lorentz (special relativistic physics) boosts.

There has been another thread on this topic recently in this forum. Just search for it!