If we consider a transformation of a field $\Phi \rightarrow \Phi + \alpha \frac{\partial \Phi}{\partial \alpha}$ which is not a symmetry of a lagrangian then one can show that the Noether current is not conserved but that instead $\partial_{\mu}J^{\mu} = \frac{\partial L}{\partial \alpha}$.

I think the way this is derived is as follows $$\delta S = \int d^4 x \, \delta L = \int d^4 x \left( \left( \frac{\partial L}{\partial \Phi} \delta \Phi - \partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \right) \delta \Phi + \partial_{\mu} \left(\frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta \Phi \right) \right)$$ Then the first term is zero due to the equations of motion and so we are left with the second total divergence term with $$J^{\mu} = \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \frac{\partial \Phi}{\partial \alpha}$$ so we are left with $$\delta S = \int d^4 x \, \alpha \partial_{\mu} J^{\mu}.$$

Writing out $$\delta L = \frac{\partial L}{\partial \Phi}\delta \Phi + \frac{\partial L}{\partial (\partial_{\mu} \Phi)} \delta (\partial_{\mu} \Phi),$$ inserting $\delta \Phi = \alpha \, \partial \Phi/\partial \alpha$ we see that $$\delta L = \alpha \frac{\partial L}{\partial \alpha}.$$ Then we can compare with the above and deduce the result.

My questions are:

What permits the use of the equations of motion here? If the equations of motion hold then $\delta S = 0$ identically in that the solutions to such equations minimise the action. Using the equations of motion gives me $\int \partial_{\mu} J^{\mu} d^4 x = \delta S$ in the end as shown above but since I used the equations of motion isn't this just equal to zero? And also since we are always left with an integral of a total divergence isn't this always zero on the physical assumption that the field variations vanish at infinity/boundary of experiment?

I've seen the nice questions and answers posted e.g here https://physics.stackexchange.com/question/327999/ and the answer here by Qmechanic Which transformations *aren't* symmetries of a Lagrangian?

Basically I'd like to understand what was said in that answer and see it in practice with the above non symmetry of the lagrangian.


3 Answers 3


Let me present for you the derivation of Noether's theorem that I really like and that makes all assumptions of Noether's theorem clear. Let $\varphi$ be the set of fields in the theory so that $$ S[\varphi] = \int d^4 x {\cal L}( \varphi(x), \cdots ) $$ Now, suppose we have a connected, continuous symmetry of the action $\varphi(x) \to \varphi_\alpha'(x)$ parameterized by a real number $\alpha \in {\mathbb R}$ so that $\alpha = 0$ corresponds to the identity transformation $\varphi'_{\alpha=0}(x) = \varphi(x)$. The claim that the transformation above is a symmetry of the action (see my answer to What constitutes a symmetry for Noether's Theorem?) is $$ S[\varphi'_\alpha] = S[\varphi] $$ Note that this symmetry is off-shell.

Now, take $\alpha$ to be small and write $$ \varphi'_{\alpha}(x) = \varphi(x) + \alpha \psi(x) + {\cal O}(\alpha^2) ~. $$ Then, $$ S[\varphi + \alpha \psi] = S[\varphi] + {\cal O}(\alpha^2) \qquad \qquad (1) $$

OK! So far, so good. Now, consider a different field redefinition $\varphi(x) \to \varphi'(x) = \varphi(x) + \alpha f(x)\psi(x)$. Note that, now since $f(x)$ is a function as opposed to a constant, the above field redefinition is not a symmetry of the action. This means that if I consider $S[\varphi + \alpha f \psi]$ there is in general an ${\cal O}(\alpha)$ unlike what's happening in (1). However, we know that this ${\cal O}(\alpha)$ term vanishes whenever $f(x)=$ constant since in that case, we would have a symmetry. This means that the ${\cal O}(\alpha)$ term must take the form $$ S[\varphi + \alpha f \psi] = S[\varphi] + \alpha \int d^4 x \partial_\mu f(x) j^\mu(x) + {\cal O}(\alpha^2) \qquad \qquad (2) $$ A derivative must act on $f(x)$ in the ${\cal O}(\alpha)$ term so that it vanishes when $f(x)=$ constant.

Now, here comes the kicker. In (2), let $\varphi=\varphi_0$ where $\varphi_0$ be a solution to the equations of motion. The, by the variational principle $$ S[\varphi_0+\alpha \delta \varphi] = S[\varphi_0] + {\cal O}(\alpha^2) $$ Note that this is true for any variation $\delta \varphi$ (there are caveats to this as well, but we will not discuss this). We then find $$ S[\varphi_0 + \alpha f \psi] = S[\varphi_0] + \alpha \int d^4 x \partial_\mu f(x) j^\mu(x) + {\cal O}(\alpha^2) = S[\varphi_0] + {\cal O}(\alpha^2) $$ which implies that if and only if $\varphi$ is taken to be on-shell, then $$ \int d^4 x \partial_\mu f(x) j^\mu(x) = 0~. $$ This is true for any function $f(x)$ (subject to the same caveats mentioned previously). So we can take $f(x) = \delta^4(x-y)$ and the equation above then reduces to $$ \partial_\mu j^\mu(y) = 0~. $$

Let us summarize all our assumptions. We assume that there exists a connected continuous global off-shell symmetry of the action that is generated by a real parameter. We have then proved that given this, there exists a current $j_\mu(x)$ which is conserved $\partial_\mu j^\mu(x)=0$ on-shell.


It seems the heart of OP's question is the following.

It is well-known that boundary conditions (BCs) are necessary in order to derived Euler-Lagrange (EL) equations. Does the infinitesimal transformations in Noether's theorem satisfy the pertinent BCs?

Answer: This is typically not the case. So boundary terms (BTs) will not necessarily vanish. See also this related Phys.SE post.

Example 1: Consider a free particle $$L~=~\frac{1}{2}m\dot{q}^2.$$ Infinitesimal off-shell translational symmetry $$\delta q~=~\epsilon$$ (which leads to on-shell momentum conservation via Noether's theorem) does not satisfy Dirichlet BCs $$ q(t_i)~=~q_i\quad\text{and}\quad q(t_f)~=~q_f. $$

Example 2: Consider a harmonic oscillator $$L~=~\frac{1}{2}m\dot{q}^2-\frac{1}{2}kq^2.$$ Infinitesimal translation transformation $$\delta q~=~\epsilon$$ (which is not an off-shell quasi-symmetry) leads to the following variation of the Lagrangian $$ \delta L~=~\frac{\partial L}{\partial q}\delta q+\frac{\partial L}{\partial\dot{q}}\frac{d}{dt}\delta q ~=~-(m\ddot{q}+kq)\delta q +\frac{d}{dt}(p\delta q)~\stackrel{\text{on-shell}}{\approx}~ \epsilon \frac{dp}{dt}, $$ which is not necessarily zero but guaranteed to be total time derivative on-shell. This corresponds to the well-known fact that momentum $$ p~:=~\frac{\partial L}{\partial\dot{q}}~=~m\dot{q} $$ is not conserved on-shell in the harmonic oscillator.

  • $\begingroup$ Ah ok I see, but I liked it :P If we use the equations of motion then $\delta S = 0$ so we get $\delta S = 0 = \int d^4 x \alpha \partial_{\mu} J^{\mu} \sim J^{\mu}|_{\text{boundary}}$ Doesn't the boundary term have to vanish to satisfy this equality? $\endgroup$
    – CAF
    Apr 23, 2017 at 12:05
  • $\begingroup$ @Qmechanic I just wondered if what I say is in contradiction to what you wrote about non vanishing boundary conditions? $\endgroup$
    – CAF
    Apr 23, 2017 at 18:02
  • $\begingroup$ @CAF: It seems so. It is only true the EL eqs. imply that $\delta S$ is a BT, not that it is zero. To additionally conclude that $\delta S=0$, one needs more information, e.g. obtained by the use of BCs. $\endgroup$
    – Qmechanic
    Apr 23, 2017 at 18:13
  • $\begingroup$ what bothers me is that the EL equations are derived with $\delta S =0$ in place for extremisation of action so any use of them in a subsequent analysis will implicitly use also $\delta S =0 $? $\endgroup$
    – CAF
    Apr 24, 2017 at 9:18
  • $\begingroup$ It is true that the EL eqs. were originally derived under assumption of BCs, but imposing the EL eqs. do not by themselves enforce the BCs. Whether EL eqs. are satisfied or not is an independent statement from whether BCs are satisfied or not. $\endgroup$
    – Qmechanic
    Apr 24, 2017 at 9:43

What's confusing you is that you've jumped the gun in your first equation, by racing ahead with the derivation of the Euler-Lagrange equations. Start instead with $$\delta S = \int d^4x \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_\mu \phi)}\delta(\partial_\mu \phi) $$ In the derivation of the Euler-Lagrange equations, we impose the principle of least action that $\delta S =0 $, then and integrate by parts to obtain your first equation. The boundary term can be assumed to vanish, but as $\delta \phi$ is arbitrary, we demand that the coefficient of $\delta \phi$ must vanish to satisfy the least action principle, which implies our equation of motion.

For a symmetry transformation $\delta \phi$ is not arbitrary; it is a symmetry of the action. This means that it leaves the action unchanged. As an example, a Lagrangian for a complex field $$L =\frac{1}{2}\partial_\mu \phi \partial^\mu\phi^* + \frac{1}{2}m^2 \phi^* \phi$$ has a symmetry $\phi \rightarrow e^{i\alpha} \phi$, corresponding to an infinitesmal variation $\delta\phi = i\alpha \phi$. This transformation leaves the Lagrangian $L$ unchanged; we don't need to use the equations of motion to enforce this.

Speaking with generality, whilst the above transformation left $L$ itself invariant, $\delta \phi$ will be a symmetry of the action if it only modifies the Lagrangian by a divergence term $L \rightarrow L + \partial _ \mu j^{\mu} (x)$ which is assumed to vanish at infinity, so as to leave the action invariant. So $$\delta S = \int d^4x \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_\mu \phi)}\delta(\partial_\mu \phi) = \int d^4x \partial _ \mu j^{\mu} (x) =0$$ - to repeat, not by the principle of least action, but by the definition of a symmetry. We can write, at this stage trivially, $$ \int d^4x\left( \frac{\partial L}{\partial \phi}\delta\phi + \frac{\partial L}{\partial(\partial_\mu \phi)}\delta(\partial_\mu \phi) - \partial _ \mu j^{\mu} (x)\right) =0$$ In fact, by our definition of $\partial _ \mu j^\mu$ as the local change in the Lagrangian $L$, we don't even need that integral sign there, but let's agree to integrate the functional derivatives by parts before we get rid of it. We thus find $$\left( \frac{\partial L}{\partial \phi} - \partial_\mu \frac{\partial L}{\partial(\partial_\mu \phi)}\right)\delta \phi + \partial_\mu\left(\frac{\partial L}{\partial(\partial_\mu \phi)}\delta \phi -j^{\mu} (x)\right) =0$$ Now, let's go "on-shell" and think about what happens when the equations of motion are satisfied. Then the first term vanishes, so we infer that the current $$J^{\mu} =\frac{\partial L}{\partial(\partial_\mu \phi)}\delta \phi -j^{\mu} (x)$$ is conserved.

Typically speaking, this current is not trivial, and you can check that it is conserved when $\phi$ satisfies its equation of motion. For the example above, you should find something like $$J^{\mu} = \phi^* \partial^\mu \phi -\phi \partial^\mu \phi^*$$ which vanishes if and only if $\phi$ satisfies the Klein-Gordon equation. (In the quantum theory, $\phi$ is a charged field, and $J^{\mu}$ is the current density.)

To be clear about the relevance of the equations of motion in the conservation law, consider @Qmechanic 's example of a free particle above. Translation symmetry is a symmetry of the action; if you have some mad path from A to B that zig-zags every which way, its action will be equally ridiculous if you shift all the points on that path to the left by 5cm. But along that hypothetical zig-zag path momentum is not conserved; it does not obey the equation of motion. Instead, along the path that does satisfy the equation of motion (i.e. a straight line), momentum is conserved. I hope that helps clear things up.

  • $\begingroup$ Thanks for your answer. I feel I am closer to understanding the issue but few other comments: In the case of a transformation that is not a symmetry, then we have $$\delta S \neq 0 = \int d^4 x \, \left(\left( \frac{\partial L}{\partial \phi} - \partial_{\mu} \frac{\partial L}{\partial (\partial_{\mu} \phi)}\right) \delta \phi + \partial_{\mu} \left( \frac{\partial L}{\partial (\partial_{\mu} \phi)} \delta \phi\right)\right)$$ Going on shell we then have $$\delta S \neq 0 = \int d^4 x \partial_{\mu} J^{\mu}.$$ Now, are we still allowed to use the EL equations here (cont. in next comment) $\endgroup$
    – CAF
    Apr 28, 2017 at 8:36
  • $\begingroup$ .....even though $\delta S \neq 0$ because this only means the transformation was not a symmetry. We can still perturb around the classical EL trajectories but the fact it is not a symmetry transformation just means the solutions to the EL equations will change (ie not invariant under the transformation)? $\endgroup$
    – CAF
    Apr 28, 2017 at 8:40
  • $\begingroup$ So this means that the boundary term $J^{\mu}|_{\text{boundary}} \neq 0 $ which I am guessing is what Prahar and Qmechanic perhaps alluded to in their posts. I suppose this means for a non symmetry transformation, the current does not vanish at the boundaries - but then how can one subsequently show that the EL equations still hold given this non vanishing boundary? $\endgroup$
    – CAF
    Apr 28, 2017 at 8:45
  • $\begingroup$ First of all, be careful to distinguish between the variation of the field $\delta\phi$ And that of the action $\delta S$. In the example relating to conservation of momentum, the variation of the "field" $q(t)$ was a constant shift $a$. This does not vanish on the boundary. However, this transformation leaves the lagrangian unchanged, so $\delta L$=0, which does vanish on the boundary! $\endgroup$
    – rwold
    Apr 28, 2017 at 18:05
  • $\begingroup$ Now consider the transformation $ \phi\rightarrow \phi + a$. Try that out with our complex scalar lagrangian and you'll find $$\delta S = \int m^2(a^* \phi + c.c)$$. The fields will vanish at the boundary, but this integral won't. But, if $\phi$ solves the Klein-Gordon equation, then you can re-write this as $\delta S =a \int \partial_\mu (\partial^\mu \phi) =0$ $\endgroup$
    – rwold
    Apr 28, 2017 at 18:18

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