6
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ They are not, but $\delta L=0$ is very common special case. $\endgroup$ – Blazej Sep 26 '17 at 13:26
6
$\begingroup$

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}

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ I don't understand why would we want a new name for it? $\endgroup$ – Blazej Sep 26 '17 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 '17 at 13:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.