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

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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?

asked Jan 15, 2013 in Theoretical Physics by Dilaton (4,175 points) [ revision history ]
Related: physics.stackexchange.com/q/27500/2451

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

1 Answer

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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, then 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) $$\tag{1.1}{\cal L} ~=~ -\frac{1}{4}F_{\mu\nu}F^{\mu\nu}~=~\frac{1}{2}(\vec{E}^2-\vec{B}^2), $$ 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.

  2. 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.]

  3. 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
answered Jan 15, 2013 by Qmechanic (2,580 points) [ no revision ]
Thanks Qmechanic for these nice examples isslustrating both cases. Do you know the answer to my question concerning scale invariance too? And more generally, are there underlaying principles that state why which kind of symmetries the Lagrangian does not have can emerge in corresponding EOMs or which kind of symmetries of the Lagrangian can potentially disappear in the EOMs derieved from the Euler-Lagrange equations?

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Dilaton
Your Galilean symmetry example is slightly misleading, since two actions may differ by some constant and/or boundary terms yet still be the "same"! Consider: $\int (\dot{q}+v)^{2}\mathrm{d}t = \int\dot{q}^{2}\mathrm{d}t+2\int\dot{q}v\mathrm{d}t+(const.)$ and the second term, integrating by parts, contributes only boundary terms which don't affect action.

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Alex Nelson
@AlexNelson: I know. Note that the question has been updated with the Lagrangian (v1) replaced by the action (v2). I updated the answer.

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Qmechanic
Thanks for the nice update of your answer, I like and appreciate this.

This post imported from StackExchange Physics at 2014-03-12 15:25 (UCT), posted by SE-user Dilaton
@Qmechanic Interesting Set of examples, Regarding the first example, I don't think There is any way to rewrite the SO(2) transformation, in terms of the Vector Potential($A_\mu$). Reason It is relevant is because you cannot vary the action with respect to the E and B fields, atleast as far as I know.

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

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