Do an action and its Euler-Lagrange equations have the same symmetries?

Question

Assume a certain action $S$ with certain symmetries, from which according to the Lagrangian formalism, the equations of motion (EOM) of the system are the corresponding Euler-Lagrange equations.

Can it happen that the equations of motion derived by this procedure have different kinds and/or numbers of symmetries than the action one has started with? And if yes, are there underlying principles that state why which kind of symmetries the action does not have can emerge in corresponding EOMs or which kind of symmetries of the action can potentially disappear in the EOMs derived from the Euler-Lagrange equations?

Answer 1

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

1. The action $S$.

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

3. 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, the EOM (2.) must have a symmetry (wrt. the same transformation).

Examples:

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

2. Another example is a non-relativistic free point particle where the Lagrangian $$\tag{3}L~=~\frac{1}{2}m\dot{q}^2$$ is not invariant under the Galilean symmetry $$\tag{4}\dot{q}\quad \longrightarrow \quad\dot{q}+v,$$ nor the dilation/scale symmetry $$\tag{5} q \quad \longrightarrow \quad \lambda q,$$ but the EOM $$\tag{6}\ddot{q}~=~0$$ is. In the case of the Galilean symmetry (4), the Lagrangian changes by a total time derivative $$\tag{7} L \quad \longrightarrow \quad L +mv\frac{d}{dt}\left( q +\frac{vt}{2}\right).$$ Thus (4) is actually an example of a quasi-symmetry of the action. See also this Phys.SE post.

3. The simple harmonic oscillator (SHO) $$\tag{8} m\ddot{q}~=~-kq $$ is not invariant under the temporal symmetry $$\tag{9} t \quad \longrightarrow \quad \lambda t,$$ but the trivial solution $q=0$ is.

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$^1$ 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.