I have recently started studying classical field theory. Noether's theorem states that every differentiable symmetry of the action of a physical system has a corresponding conservation law. But I find that often when solving for conserved currents, it is assumed that they are due to invariance of Lagrangian i.e; $\delta L = 0$. Are both the statements same always?

  • 3
    $\begingroup$ They are not, but $\delta L=0$ is very common special case. $\endgroup$
    – Blazej
    Sep 26, 2017 at 13:26

2 Answers 2


No, they are not the same. To see why, even in classical mechanics, suppose we have symmetry transformation $q \rightarrow q + \epsilon K$ that leaves the Lagrangian invariant. This means that we must have \begin{equation} \lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}\left(L(q+\epsilon K, \dot{q}+\epsilon\dot{K},t) - L(q,\dot{q},t)\right) = \frac{\partial L}{\partial q}K +\frac{\partial L}{\partial \dot{q}}\dot{K} = 0 \end{equation} $Then$ you use the fact that the equations of motion are satisfied to write $\frac{\partial L}{\partial q}=\frac{d}{dt}\frac{\partial L }{\partial \dot{q}}$ and this implies \begin{equation} \frac{d}{dt}\frac{\partial L}{\partial \dot{q}}K+\frac{\partial L}{\partial \dot{q}}\dot{K}=\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}K\right)=0 \end{equation} i.e. the quantity $\frac{\partial L}{\partial \dot{q}}K$ is conserved.

A symmetry of the action is a transformation that leaves the action invariant whether or not the equations of motion are satisfied. In this case the same procedure yields the condition \begin{equation} \frac{\partial L}{\partial q}K + \frac{\partial L}{\partial \dot{q}}\dot{K} = \frac{dM}{dt} \end{equation} where $M$ is a function of $q, \dot{q}, t$. If such an M exists, we say that the action is invariant under the symmetry transformation.

It's very easy to see that when we impose the equations of motion LHS becomes $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}K\right)$ and we can derive a conserved quantity: \begin{equation} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}K - M\right)=0. \end{equation}

The simplest possible example of a symmetry transformation which is a symmetry of the action but not of the Lagrangian is time translation in systems where the Lagrangian has no explicit time dependence. When we shift the time by an arbitrary small $\epsilon$, the generalized coordinates $q$ change according to $q(t) \rightarrow q(t) + \epsilon \dot{q}(t)$, therefore $K=\dot{q}$. But \begin{equation} \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{q}}\dot{q}\right)=\frac{\partial L }{\partial q}\dot{q} + \frac{\partial L}{\partial \dot{q}}\ddot{q}=\frac{dL}{dt}\neq 0 \end{equation} In this case $M =L$ and the conserved quantity is \begin{equation} H = \frac{\partial L}{\partial \dot{q}}\dot{q} - L. \end{equation}

  • $\begingroup$ I may be a little bit late to the party, but what is the function K that you describe here? Is it a function of q(t) and t? $\endgroup$
    – Tachyon209
    Oct 17, 2020 at 8:54
  • $\begingroup$ Also, I guess M should be a function of $q$ and $t$, and not $q$, $\dot{q}$ and $t$ because once you include velocities too, then you would have to add an additional constraint that velocity along the path vanishes at the endpoints. $\endgroup$
    – Tachyon209
    Oct 17, 2020 at 9:06
  • $\begingroup$ As shown in the case of time translation invariance, $K$ can be a function of $\dot{q}$ also. $\endgroup$ Oct 18, 2020 at 10:16
  • $\begingroup$ In the same example, $M$ is the Lagrangian which is a function of $\dot{q}$.The Lagrangian is a function of $q$ and $\dot{q}$ treated as independent variables. The fact that in the Lagrangian formalism you derive the eqs of motions varying the action keeping the coordinates fixed at the endpoints (without requiring velocities to vanish) is another story. It means that for the trajectories that satisfy the eqns of motions as a function of time, $q(t)$ and $\dot{q}(t)$ are not independent. See tinyurl.com/yxfpqfg7 . Indeed, I never assumed the velocities to vanish at the endpoints. $\endgroup$ Oct 18, 2020 at 10:19
  • $\begingroup$ Why do you say it is a symmetry of the action if the change in L is dM/dt? The action changes by M evaluated at the boundaries. $\endgroup$
    – Sam Gralla
    Dec 21, 2023 at 18:39

First of all, a notion of strict symmetry should be relaxed to a notion of quasisymmetry to be as general as possible. Both versions of Noether's theorem remain true, but the action version is more general.

  • $\begingroup$ I don't understand why would we want a new name for it? $\endgroup$
    – Blazej
    Sep 26, 2017 at 13:29
  • $\begingroup$ Different authors use different terminologies. It is often useful to use the word quasisymmetry to emphasize the more general notion. $\endgroup$
    – Qmechanic
    Sep 26, 2017 at 13:44

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