Comparison between formulations of Noether's theorem

Question

Version 1:

An infinitesimal variation on the fields $\phi\mapsto\phi'$ is said to be a symmetry if $\delta \mathcal{L}:=\mathcal{L}(\phi',\partial\phi')-\mathcal{L}(\phi,\partial\phi)$ is a total derivative $\partial_\mu F^\mu$. If this is the case, let $\phi'(x)=\phi(x)+\delta\phi(x)$. Then $$0=\delta\mathcal{L}-\partial_\mu F^\mu=\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\delta\phi-\partial_\mu F^\mu=\delta\phi\left(\frac{\partial\mathcal{L}}{\partial\phi}-\partial_\mu\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\right)+\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\delta\phi-F^\mu\right).\tag{1}$$ Therefore, on-shell, the current $$j^\mu:=\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\delta\phi-F^\mu\tag{2}$$ is conserved.

Energy-momentum tensor: $\delta\phi(x)=\epsilon^\mu\partial_\mu\phi$ and $\delta\mathcal{L}=\epsilon^\mu\partial_\mu\mathcal{L}=\partial_\mu(\epsilon^\mu\mathcal{L})$, which yields $j^\mu=\epsilon^\nu T_\nu^\mu$, with $$T_\nu^\mu=\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\nu\phi-\delta^\mu_\nu\mathcal{L}.\tag{3}$$

Advantages:

1. There is no mention to horizontal changes ($x\mapsto x'=x+\delta x$). In particular, the only data required is a vector field in the space of field configurations.
2. It includes the possibility that the action is modified by a boundary term.
3. The derivation is simple.

Disadvantages: It doesn't yield a method of computing $F^\mu$. I always get confused when I try to compute $F^\mu$ because what I end up doing is $\frac{\partial\mathcal{L}}{\partial\phi}\delta\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\delta\phi.$

Version 2: An infinitesimal variation $x\mapsto x'=x+\delta x$ and $\phi\mapsto\phi'$ with $\phi'(x')=\phi(x)+\delta\phi(x)$ is a symmetry if $\delta S_\Omega(\phi):=S_{\Omega'}(\phi')-S_\Omega(\phi)=0$. After a computation one gets that for a general transformation (not necessarily a symmetry) $$\delta S_\Omega(\phi)=\int d^D x\left(-\partial_\mu\delta x^\nu T^\mu_\nu+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\delta\phi+\delta\phi\frac{\partial\mathcal{L}}{\partial\phi}\right).\tag{3}$$ Assuming $\delta x=0$ and that $\delta\phi$ vanishes on $\partial\Omega$, one gets to the Euler-Lagrange equations. On the other hand, if $\delta x^\mu=\omega^a X_a^\mu(x)$ and $\delta\phi(x)=\omega^a \mathcal{F}_a(\phi(x))$, one obtains $$\delta S_\Omega(\phi)=\int d^D x\omega^a\left(-\partial_\mu X_a^\nu T^\mu_\nu+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\mathcal{F}_a(\phi)+\mathcal{F}_a(\phi)\frac{\partial\mathcal{L}}{\partial\phi}\right)-\int d^D x\partial_\mu\omega^a j^\mu_a,\tag{4}$$ with $$j^\mu_a=T^\mu_\nu X^\nu_a-\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\mathcal{F}_a(\phi).\tag{5}$$ Now, assume that the transformation is a symmetry when $\omega^a$ is constant. Then, by definition of symmetry the stuff in the first parenthesis vanishes (this is in fact what one needs to check to ensure this is a symmetry). Then, returning to the case $\omega^a$ may vary with $x$, we obtain $$\delta S_\omega(\phi)=-\int d^Dx\partial_\mu\omega^aj^\mu_a.\tag{6}$$ Then, by definition of the equations of motion (I have a comment on this below), this variation vanishes on shell. Choosing $\omega^a$ to vanish on $\partial\Omega$ one can integrate by parts and conclude that $j^\mu_a$ is conserved.

Energy-momentum tensor: $X^\mu_\nu=\delta^\mu_\nu$ and $\mathcal{F}_\nu(\phi(x))=0$.

Advantages:

1. It shows explicitly how to check if a transformation is a symmetry.
2. It yields an efficient method to compute the current (namely the coefficients of $\partial_\mu\omega^a$).
3. It gives an explicit formula for the conserved current.

Disadvantages:

1. The proof is long.
2. It requires horizontal transformations
3. At least in its present form it does not allow for variations of the action by boundary terms (I imagine this can be corrected easily).

Question: What is the relationship between these two formulations of Noether's theorem. I am particularly interested on why the first only requires the data of a vector field on the space of field configurations.

Side question: In version 2 there seems to be a loop hole. The vanishing of the variation of the action uses the on-shell condition. However, the Euler-Lagrange equations do not contemplate horizontal transformations. Then why can we guarantee that $\delta_\Omega S(\phi)=0$ on-shell?

Answer 1

It turns out to be easier to compare both versions if one uses the functional changes $\bar{\delta}\phi(x):=\phi'(x)-\phi(x)=\phi(x-\delta x)+\delta\phi(x-\delta x)-\phi(x)=-\delta x^\mu\partial_\mu\phi(x)+\delta\phi(x)$. It is precisely in terms of this functional changes that the first version of Noether's theorem is written in. The variation in the second version is $$\delta S_\Omega(\phi)=\int_\Omega d^Dx\left(\partial_\mu(\delta x^\mu\mathcal{L})+\frac{\partial\mathcal{L}}{\partial\phi}\bar{\delta}\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu\bar{\delta}\phi\right),$$ as can be quickly checked with from the formula in the question using the relation between $\delta$ and $\bar{\delta}$ (this is all found e.g. in Ramond's "Field Theory: A Modern Primer").

As a first remark, notice that the principle of stationary action remains valid even when including horizontal transformations, as long as these vanish on $\partial\Omega$. Indeed, in the equation above these transformations only appear through the total derivative $\partial_\mu(\delta x^\mu\mathcal{L})$. Moreover, in this case $\delta=\bar{\delta}$ on $\partial\Omega$ so that there is no ambiguity whether one should ask for $\bar{\delta}\phi|_{\partial\Omega}=0$ or $\delta\phi|_{\partial\Omega}=0$.

As a second remark, one can now include the possibility that the action varies through boundary terms. Namely, the theorem now goes like this. Consider varitions $\delta x^\mu=\epsilon X^\mu$ and $\bar{\delta}\phi=\epsilon G\phi$ where $G$ is some differential operator (unlike $\mathcal{F}$ in the statement of the question above which was in general a matrix). We then have $$\delta S_\Omega(\phi)=\int_\Omega d^Dx\epsilon\left(\partial_\mu(X^\mu\mathcal{L})+\frac{\partial\mathcal{L}}{\partial\phi}G\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu G\phi\right)+\int_\Omega d^Dx\partial_\mu\epsilon\left(X^\mu\mathcal{L}+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi\right).$$ Now, assume that whenever $\epsilon$ is constant we have $\delta S_\Omega(\phi)=\epsilon\int_\Omega d^Dx\partial_\mu F^\mu$. Then $$\partial_\mu F^\mu=\partial_\mu(X^\mu\mathcal{L})+\frac{\partial\mathcal{L}}{\partial\phi}G\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu G\phi.$$ (Side remark: Notice that the last two terms of this equation are simply $\bar{\delta}\mathcal{L}$ of the first version of Noether's theorem. Thus including horizontal changes has modified the boundary term. We will say more about this at the end.) We conclude that under arbitrary $\epsilon$ $$\delta S_\Omega(\phi)=\int_\Omega d^Dx\epsilon\partial_\mu F^\mu+\int_\Omega d^Dx\partial_\mu\epsilon\left(X^\mu\mathcal{L}+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi\right).$$ To conclude let us restrict to $\epsilon$ vanishing at the origin. Then we can integrate by parts and get $$\delta S_\Omega(\phi)=\int_\Omega d^Dx\epsilon\partial_\mu \left(F^\mu-X^\mu\mathcal{L}-\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi\right).$$ The argument is now finished by restricting to $\phi$ on-shell. Indeed, in this case the variation has to vanish for all $\epsilon$ vanishing at the boundary. As we remarked above, this is not spoiled by the presence of horizontal variations. Then by the fundamental theorem of calculus of variations we have $\partial_\mu j^\mu=0$, where, explicitely, $$j^\mu=F^\mu-X^\mu\mathcal{L}-\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi.$$

As a final remark, let us comment on whether horizontal changes are necessary or not. Well, definitely the second version, in our present version were we have allowed for boundary terms, is at least as powerful as the first. The first is in fact recovered by setting $X^\mu=0$. In particular, the energy-momentum tensor can be recovered by setting $X^\mu=0$ and $G=-\partial_\nu$, as in the first version, or setting $X^\mu=\delta^\mu_\nu$ and $G=-\partial_\nu$, as in the perspective of the second version. Perhaps more surprisingly, it turns out that the first version is as powerful as the second. Indeed, assume that the conditions for the second are satisfied. In particular we have $$\partial_\mu F^\mu=\partial_\mu(X^\mu\mathcal{L})+\frac{\partial\mathcal{L}}{\partial\phi}G\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu G\phi,$$ for some $F^\mu$. Then define $\tilde{F}^\mu:=F^\mu-X^\mu\mathcal{L}$. We then have $$\partial_\mu \tilde{F}^\mu=\frac{\partial\mathcal{L}}{\partial\phi}G\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu G\phi,$$ Moreover, we have $$j^\mu=F^\mu-X^\mu\mathcal{L}-\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi=\tilde{F}^\mu-\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi.$$ Thus, we could have recovered the same Noether current if we set $X^\mu=0$. We conclude that horizontal variations are not necessary to obtain Noether currents as long as we are willing to have variations of the action by boundary terms. On the other hand, although I don't have any examples in mind at the moment, presumably one cannot in general hide any boundary variation as a space variation (setting $X^\mu=-F^\mu/\mathcal{L}$ seems like a weird thing to do in general.

To sum up:

Consider an infinitesimal variation $\phi\mapsto\phi'=\phi+\epsilon G\phi$. We say this is an infinitesimal symmetry of our system if for constant $\epsilon$ we have that $$\delta S_\Omega(\phi):=S_\Omega(\phi')-S_\Omega(\phi)=\epsilon\int_\Omega\partial_\mu F^\mu$$ for some $F^\mu$. It is important to note that in general $F^\mu$ will depend on $\phi$ and this must be true for any $\phi$ disregardless of whether it is on-shell or not. The first non-trivial statement is that an $F^\mu$ satisfies the condition above if and only if $$\partial_\mu F^\mu=\frac{\partial\mathcal{L}}{\partial\phi}G\phi+\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}\partial_\mu G\phi.$$ (We leave as an interesting side remark that whenever the transformation $\phi\mapsto\phi'$ comes from a horizontal transformation $x\mapsto x'=x+\epsilon X^\mu$, one can usually take $F^\mu=-X^\mu\mathcal{L}$. But that is the whole role that horizontal variations play.)

Now, assume that we have an infinitesimal symmetry as above. For any $F^\mu$ that witnesses that $\phi\mapsto\phi'$ is a symmetry, the current $$j^\mu=F^\mu-\frac{\partial\mathcal{L}}{\partial\partial_\mu\phi}G\phi$$ is conserved.

Finally, it is usually a good idea to compute this current by computing $\delta S_\Omega(\phi):=S_\Omega(\phi')-S_\Omega(\phi)$ for a arbitrary varying $\epsilon$. One can read of $F^\mu$ (and in the meantime check whether this is indeed a symmetry) and $j^\mu$ from the formula $$\delta S_\Omega(\phi)=\int d^D x\epsilon\partial_\mu F^\mu+\int_\Omega d^D x\partial_\mu\epsilon (F^\mu-j^\mu).$$