Why is the symmetry variation $\delta_s q$ different from the ordinary variation $\delta q$?

Question

I was reading about symmetry of action when I came before the symmetry variation in Particles and Quantum Fields by H. Kleinert; there he wrote:

Symmetry variations must not be confused with ordinary variations $\delta q(t)$ used to derive the Euler-Lagrange equations. While the ordinary variations $\delta q(t)$ vanish at initial and final times, $\delta q(t_b) = \delta q(t_a) = 0,$ the symmetry variations $\delta_s q(t)$ are usually non-zero at the ends.

So, isn't $\delta_s q$ a virtual variation? For, if it would be, it should have vanished in the fixed initial and final time, $t_a$ and $t_b$, isn't it?

Could anyone explain me why the symmetry variation is different from the ordinary variation?

Answer 1

What Kleinert calls symmetry variations and ordinary variations are used in 2 different contexts. Both are off-shell variations.

1. Symmetry variations should leave the action invariant up to boundary terms. (In the affirmative case, one can then apply Noether's theorem to deduce a conservation law.) They have typically a specific prescribed form with possibly both horizontal and vertical components, i.e. components in $t$- and $q$-space, respectively.

2. Ordinary variations are performed to find Euler-Lagrange equations. They are general vertical transformations that satisfy pertinent boundary conditions. Boundary conditions must be imposed to get rid of boundary terms.