We are considering a transformation, which may transform the field variables $\phi^{\alpha}(x)$ and which may transform the space-time points $x^{\mu}$. The transformation in turn apply to

The action $S$.

The Euler-Lagrange equations = the equations of motion (EOM).

A solution of EOM.

If any of the items 1-3 are invariant under the transformation, we speak of a symmetry of the corresponding item.

If a solution (3) doesn't have a symmetry that the EOM (2) have, we speak of spontaneous symmetry breaking.

Next let us recall the definition of an (off-shell$^1$) *quasi-symmetry* of the action. It means that the action changes by a boundary integral under the transformation.

In general, if an action (1) has a quasi-symmetry, then the EOM (2) must have a symmetry (wrt. the same transformation).

*Examples:*

One example is the Maxwell Lagrangian density (in vacuum without the $J^{\mu}A_{\mu}$ source term)
$$\tag{1.1}{\cal L} ~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\frac{1}{2}(\vec{E}^2-\vec{B}^2), $$
which doesn't have electromagnetic $SO(2,\mathbb{R})$ duality symmetry
$$\tag{1.2}(\vec{E}, \vec{B})\quad \longrightarrow \quad(\vec{E}\cos\theta - \vec{B}\sin\theta, \vec{B}\cos\theta + \vec{E}\sin\theta),$$
while the Euler-Lagrange equations (the Maxwell's equations in vacuum) are symmetric under electromagnetic duality.

Another example is a non-relativistic free point particle where the Lagrangian
$$\tag{2.1}L~=~\frac{1}{2}m\dot{q}^2$$
is not invariant
under the Galilean symmetry
$$\tag{2.2}\dot{q}\quad \longrightarrow \quad\dot{q}+v,$$
nor the dilation/scale symmetry
$$\tag{2.3} q \quad \longrightarrow \quad \lambda q,$$
but the EOM
$$\tag{2.4}\ddot{q}~=~0$$
is invariant. In the case of the Galilean symmetry (2.2), the Lagrangian changes by a total
time derivative
$$\tag{2.5} L \quad \longrightarrow \quad L +mv\frac{d}{dt}\left( q +\frac{vt}{2}\right).$$
See also this Phys.SE post. Thus (2.2) is actually an example of a quasi-symmetry of the action. [It is an instructive exercise to derive the corresponding Noether charge $Q$. At the infinitesimal level, the Galilean transformation (2.2) reads
$$ \tag{2.6}\delta \dot{q}~=~\delta v~=~\varepsilon, \qquad \delta q~=~\varepsilon t,\qquad \delta L ~=~ \varepsilon\frac{df}{dt}, \qquad f ~:=~mq. $$
The bare Noether charge is
$$ \tag{2.7} Q^0~=~t \frac{\partial L}{\partial \dot{q}}~=~t m\dot{q}, $$
while the full Noether charge is
$$ \tag{2.8}Q~=~Q^0-f~=~m(\dot{q}t-q),$$
which is conserved on-shell, cf. Noether's Theorem. The (non-relativistic) Galilean boosts generator (2.8) should be compared to the (relativistic) Lorentz boosts generators $tP-xE$ in relativistic theories, cf. e.g. this Phys.SE post.]

The simple harmonic oscillator (SHO)
$$\tag{3.1} m\ddot{q}~=~-kq $$
is not invariant under the temporal symmetry
$$\tag{3.2} t \quad \longrightarrow \quad \lambda t,$$
but the trivial solution $q=0$ is.

--

$^1$ Here the word *off-shell* indicates that the EOM are not assumed to hold under the specific transformation. In case of continuous transformations, if we assume the EOM to hold, then *any* infinitesimal variation of the action is trivially a boundary integral.

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Qmechanic