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
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