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?


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.