Symmetries of Lagrangians

The definition of a symmetry of a theory is quite clear at the level of a Lagrangian. We say a Lagrangian $\mathcal{L}(\phi,\partial_\mu \phi)$ is symmetric under the transformation $\phi \mapsto \phi + \epsilon \delta \phi_s$ if, when evaluated off-shell, the Lagrangian varies by a total derivative as

$$ \delta \mathcal{L}_\text{off-shell} = \epsilon \partial_\mu F^\mu(\phi), \quad (1)$$

where the symmetry transformation $\delta \phi_s$ is not arbitrary but depends upon the Lagrangian. However, we also know that if $\bar{\phi}$ is a solution to the equations of motion, then, if we vary this field as $\bar{\phi} \mapsto \bar{\phi} + \delta \phi$, the Lagrangian varies as

$$ \delta \mathcal{L}_\text{on-shell} = \epsilon \partial_\mu \left( \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \delta \phi\right) \quad (2)$$

where the variation $\delta \phi$ is completely arbitrary this time. Evaluating both of these variations on-shell by setting $\phi = \bar{\phi}$ and $\delta \phi_s = \delta \phi$, one equates the two variations to yield Noether's theorem $\partial_\mu j^\mu =0$ where $j^\mu = \frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \delta \phi - F^\mu$.

Symmetries of Actions

However, at the level of the action the definition of a symmetry does not seem to be as clear to me. I have seen this answer which states the action is totally invariant under a symmetry transformation, i.e. when $\phi \mapsto \phi'$ then $S[\phi] = S[\phi']$. However, if I was to use (1) above, my definition of a symmetry of the action would actually be

$$ \delta S = \int \mathrm{d}^n x \epsilon \partial_\mu F^\mu $$

which will integrate to a boundary term which is not zero in general. This is a the definition of a symmetry that equation (2.62) of this review uses.


Consider non-relativistic point particle mechanics for now. The equations (1) and (2) still apply in this case. Consider the Lagrangian $L(x,\dot{x}) = \frac{1}{2}m\dot{x}^2$ and action $ S[x] = \int_{t_i}^{t_f} \mathrm{d}t L(x(t),\dot{x}(t))$. If I performed a time-translation of my trajectories $x(t)$, I would say

$$ x(t) \mapsto y(t) = x(t-\epsilon) = x(t) - \epsilon \dot{x}(t) + O(\epsilon^2)$$

in which case my variation is given by $\delta x(t) = -\epsilon \dot{x}(t)$. Substituting this into my Lagrangian, we see that it transforms as

$$ L \mapsto L - \epsilon \frac{d}{dt} \left( \frac{1}{2} m \dot{x}^2 \right) + O(\epsilon^2)$$

which is a total derivative so by equation (1) I would say this is a symmetry. However, my action under this transformation is not invariant:

$$ S[x] \mapsto S[y] = S[x] - \epsilon \int_{t_i}^{t_f} \mathrm{d} t \frac{d}{dt}\left( \frac{1}{2}m \dot{x}^2 \right) = S[x] - \epsilon \left[ \frac{1}{2}m \dot{x}^2 \right]_{t_i}^{t_f}. $$

Now in field theory it seems we would circumnavigate these boundary contributions by noting that the limits of our action integral are at infinity and we assume the field dies off as $|x| \rightarrow \infty$ so the boundary terms vanish. But it seems that these boundary terms are paramount in deriving Noether's theorem. I should hope that they aren't zero or surely (1) and (2) above would both integrate to zero so Noether's theorem, as proven on pages 17 and 18 of this review, would be trivially stating that $0=0$?.

My question

Is a symmetry of the action a transformation $\phi \mapsto \phi'$ such that $S[\phi] = S[\phi]$?


1 Answer 1


The notion that is needed for Noether's theorem is not strict symmetry (where the action is strictly preserved under the transformation), but rather quasi-symmetry (where the action is only preserved up to boundary terms).

Concerning terminology, see also this related Phys.SE post.


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