This is a delayed answer at a slightly lower level than @Bence Racsko’s great answer. Here, we shall work only with first order Lagrangians, and deal with a theory for sections of vector bundles (a lot of this is a surface level treatment of what can be found in Christodoulou’s book The Action Principle and Partial Differential Equations).
The ‘discrepancy’ with the comment made by @Prahar/the link to Banados’ paper vs Greiner is that in the former they ascribe everything to field variations whereas in the latter he considers both field variations and flows of vector fields (i.e Lie-derivative effects). Why the discrepancy?
- Greiner uses both so that he can stay within the realm of symmetries of the action rather than quasisymmetries. To be clear: I’m not Greiner, so I can’t speak to his motivations, but what I’m saying is that in his presentation he stays fully within the realm of symmetries, and so to obtain some of the conserved currents, he needs to consider flows of vector fields as well.
- If you want to ascribe everything to field deformations then (barring some special cases, e.g when we have a Lagrangian which is covariant in a certain sense (I’ll get back to this later)) you’ll always end up working with quasisymmetries of the action.
My preference is to use both field variations and flows in the domain, because I find it less artificial (i.e we’re not artificially lumping a flow in the domain with the field variation). Having said all of this, when one does the full calculation, the Noether current is the same (just as @Bence Racsko’s answer and his linked answer say).
Below is as complete yet concise of a description of everything I could come up with. I have split it into definitions, ‘basic’ computational lemmas, and finally the statement and proof of various versions of Noether’s theorem. This assumes you’re able to follow along most of my link in the comments.
Notation.
Let $(E,\pi,M)$ be a vector bundle and let’s denote $n:=\dim M$ and $k:=\text{rank}(E)$ (i.e $\dim E-\dim M$). Let $J^1(M,E)$ be first jet bundle of maps $M\to E$.
- $M$ is thought of as space/spacetime. Here we denote local coordinates as $x\equiv (x^{\mu})=(x^1,\dots, x^n)$.
- the vector bundle $E$ is where the field takes values; adapted local coordinates here are denoted $(x,u)\equiv (x^{\mu},u^i)=(x^1,\dots, x^n, u^1,\dots, u^k)$. So, given a field $\phi$ (i.e a smooth section $\phi:M\to E$) we can consider its components $\phi^i:=u^i\circ \phi$. If we take $E=M\times\Bbb{R}^n$, we’re dealing with a simple vector-valued theory; one can also consider more complicated bundles (e.g tensor bundles, spinor bundles etc).
- The space $J^1(M,E)$, if you’re not familiar with it, is just to be thought of as a ‘big enough manifold’ in order to capture information about fields and their first derivatives and ‘treat these as independent variables’. We shall denote local coordinates here as $(x,u,v)=(x^{\mu},u^i, v^{i}_{\mu})$; also for each smooth map $\phi:M\to E$ there’s something called the first jet prolongation, denoted $j^1\phi$, which just means capture all information about first derivatives. The local form of $j^1\phi$ in these coordinates is just
\begin{align}
(x^1,\dots, x^n)\mapsto
\left(x^1,\dots, x^n;
\phi^1(x),\dots, \phi^k(x);
\left(\frac{\partial\phi^i}{\partial x^{\mu}}(x)\right)_{\substack{i\in \{1,\dots, k\}\\
\mu\in\{1,\dots, n\}}}\right)
\end{align}
or more concisely, it’s just $x\mapsto (x,\phi(x),\partial\phi(x))$.
Henceforth, we shall assume the manifold $M$ is (orientable and) oriented.
Definitions.
Now, we give some basic definitions so we’re all on the same page. We keep the notation $(E,\pi, M)$ from above.
Definition (nice open set)
For convenience, let’s call an open set $\Omega\subset M$ nice if Stokes’ theorem is valid on $\Omega$. This is always the case if $\Omega$ has smooth boundary and compact closure.
Definition (Lagrangian $n$-form, Action functional).
A Lagrangian $n$-form on $(E,\pi,M)$ is a smooth fiber-bundle morphism $\mathcal{L}:J^1(M,E)\to \bigwedge^n(T^*M)$. Given any section $\phi:M\to E$, we shall denote $\mathcal{L}_{\phi}:=\mathcal{L}\circ (j^1\phi)$; this is is an $n$-form on $M$, hence can be integrated on nice open sets.
Given a Lagrangian $n$-form, we can define the associated action functional $S$ by defining for each smooth section $\phi$ of $(E,\pi,M)$ and each nice open set $\Omega$, $S[\phi,\Omega]:=\int_{\Omega}\mathcal{L}_{\phi}$.
Definition (Current).
A current (in the sense of Noether) on $(E,\pi,M)$ is a smooth fiber bundle morphism $K:J^1(M,E)\to \bigwedge^{n-1}(T^*M)$.
So, for each section $\phi$ of $(E,\pi,M)$, we get a smooth $(n-1)$-form on $M$ defined by $K_{\phi}:=K\circ (j^1\phi)$.
The word ‘current’ has several different meanings (electrical current, and also a current in the sense of de Rham/ Schwartz), so I’m just specifying what a current in this Noether-context is. Notice that $K_{\phi}$ is an $(n-1)$-form, meaning it can be integrated over nice hypersurfaces (in particular, over the boundary of a nice open set $\Omega$).
Definition (Conserved current)
Let $\mathcal{S}$ be a set of smooth sections of $(E,\pi,M)$, and let $K$ be a current on $(E,\pi,M)$. We say that $K$ is a conserved current relative to the collection $\mathcal{S}$ if for every $\phi\in \mathcal{S}$, we have that $K_{\phi}$ is a closed form, i.e $d(K_{\phi})=0$.
In particular, suppose we have a Lagrangian $n$-form $\mathcal{L}$; let’s define $\mathcal{S}_{\mathcal{L}}$ to be the set of smooth sections which satisfy the Euler-Lagrange equations of $\mathcal{L}$. If $K$ is conserved relative to $\mathcal{S_L}$, we shall also say that $K$ is conserved (relative to $\mathcal{L}$).
You may be wondering why I make this definition rather than requiring (for simplicity) $\mathcal{S}$ to be all sections. Well, it turns out in the proof of Noether’s theorem later on, we only consider solutions of the equations of motion (the ELE).
Definition (Field deformations.)
Let $\mathcal{S}$ be a set of smooth sections of $(E,\pi,M)$. By a 1-parameter deformation of fields in $\mathcal{S}$ we mean a collection of maps $T_{\epsilon}$ indexed by a small real number $\epsilon$ such that for each $\phi\in\mathcal{S}$, we have a section $T_{\epsilon}\phi$ (the ‘transformed field’); furthermore we require $T_0\phi=\phi$ and that the mapping $(\epsilon,p)\mapsto (T_{\epsilon}\phi)(p)$ to be smooth from a suitable open subset of $\Bbb{R}\times M\to E$.
We denote $\delta\phi:=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}(T_{\epsilon}\phi)$. This is a smooth section of $(E,\pi,M)$, called the (field) variation of $\phi$ relative to the deformation $T_{\epsilon}$.
Finally, we define symmetries and quasisymetries:
Definition ((Quasi-)Symmetries)
Let $\mathcal{L}$ be a Lagrangian $n$-form on $(E,\pi,M)$ with action functional $S$. Also, let $\mathcal{S_L}$ be the set of solutions of the Euler-Lagrange equations. Let $T_{\epsilon}$ be a 1-parameter deformation of $\mathcal{S_L}$ and let $X$ be a vector field on $M$ with flow $\Phi_{\epsilon}$.
We say that the pair $(X,T_{\epsilon})$ is a symmetry of the action functional if for all $\phi\in\mathcal{S_L}$ and all nice open sets $\Omega\subset M$, we have
\begin{align}
\frac{d}{d\epsilon}\bigg|_{\epsilon=0}S[T_{\epsilon}\phi,\Phi_{\epsilon}(\Omega)]:=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\int_{\Phi_{\epsilon}(\Omega)}\mathcal{L}_{T_{\epsilon}\phi}=0.
\end{align}
We say that the pair $(X,T_{\epsilon})$ is a quasisymmetry of the action functional if there exists a current $K$ on $(E,\pi,M)$ such that for all $\phi\in\mathcal{S_L}$ and all nice open sets $\Omega\subset M$, we have
\begin{align}
\frac{d}{d\epsilon}\bigg|_{\epsilon=0}S[T_{\epsilon}\phi,\Phi_{\epsilon}(\Omega)]:=\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\int_{\Phi_{\epsilon}(\Omega)}\mathcal{L}_{T_{\epsilon}\phi}=\int_{\partial\Omega}K_{\phi}.
\end{align}
We shall call $K$ a surface current (relative to the action $S$ and the deformations induced by $(X,T_{\epsilon})$).
Note that the $K$ in the definition of quasisymmetries is not unique; we can (by Stokes’ theorem) add to $K$ any conserved current, and conversely if $K_1,K_2$ both satisfy the definition of quasisymmetry, then $K_1-K_2$ is a conserved current.
’Basic’ Computations.
Here, we present just purely computational lemmas, a-priori devoid of any relation to Noether’s theorem. I want to make this separation so that later on, the statement and logic of the proof of Noether’s theorem are distilled to the bare essentials, and so that we can analyze various terms separately.
Lemma 1.
Let $\mathcal{L}$ be a Lagrangian $n$-form on $(E,\pi,M)$ and $f$ a smooth section of $E$. Then,
There’s a current $J_{\mathcal{L},f}$ on $(E,\pi,M)$ such that in terms of any adapted coordinate chart $(x,u,v)=(x^{\mu},u^i,v^i_{\mu})$ on the jet bundle $J^1(M,E)$, we have that
\begin{align}
J_{\mathcal{L},f}=\sum_{\mu=1}^n(-1)^{\mu-1}\frac{\partial \lambda}{\partial v^i_{\mu}}\bigg|_{(x,u,v)}f^i(x)\cdot dx^1\wedge\cdots\wedge \widehat{dx^{\mu}}\wedge\cdots\wedge dx^n,
\end{align}
where $\lambda$ is the function (on the domain of the adapted coordinates on the jet bundle) such that $\mathcal{L}=\lambda(x,u,v)\,dx^1\wedge\cdots\wedge dx^n$.
For any smooth section $\phi$ of $(E,\pi,M)$, let us denote $J_{\mathcal{L},f,\phi}:=J_{\mathcal{L},f}\circ (j^1\phi)$. The exterior derivative of the $(n-1)$-form $J_{\mathcal{L},f,\phi}$ in local coordinates is
\begin{align}
d(J_{\mathcal{L},f,\phi})&=\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial \lambda}{\partial v^i_{\mu}}\bigg|_{(x,\phi(x),\partial\phi(x))}\cdot f^i(x)\right),
\end{align}
or in more traditional notation, we have
\begin{align}
d(J_{\mathcal{L},f,\phi})&=\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi^i)}\bigg|_{(x,\phi(x),\partial\phi(x))}\cdot f^i(x)\right).
\end{align}
The proof of the first part is a routine exercise in verifying that the definitions in two different adapted coordinate charts are actually equal (this boils down to an application of the chain rule). The second part is an easy computation. I’ll leave the details to you.
In particular, if we’re given a a 1-parameter deformation $T_{\epsilon}$ of the set of solutions $\mathcal{S}_{\mathcal{L}}$ of the Euler-Lagrange equations, then for each $\phi\in \mathcal{S_L}$, we can consider the field variation $\delta\phi$ and use this as our $f$ above. And in this light, I’m sure you can recognize that $d(J_{\mathcal{L},\delta \phi,\phi})=\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi^i)}\delta\phi^i\right)$ is precisely one of the terms we encounter in Noether’s theorem.
Lemma 2.
Let $\mathcal{L},X,\Phi_{\epsilon},T_{\epsilon},\mathcal{S_L}$ have the same meanings as above. Then, for each $\phi\in\mathcal{S_L}$ and nice open set $\Omega$, we have
\begin{align}
\frac{d}{d\epsilon}\bigg|_{\epsilon=0}S[T_{\epsilon}\phi,\Phi_{\epsilon}(\Omega)]&=\int_{\Omega}d\left(J_{\mathcal{L},\delta\phi,\phi}+ X\lrcorner \mathcal{L}_{\phi}\right).
\end{align}
The proof is essentially given in the link in the comments (see section 3 of my answer there). But for completeness we give the argument here (albeit briefly; see the link for details on the computations… adapting it to the present vector bundle case is easy). We have
\begin{align}
\frac{d}{d\epsilon}\bigg|_{\epsilon=0}S[T_{\epsilon}\phi,\Phi_{\epsilon}(\Omega)]&:= \frac{d}{d\epsilon}\bigg|_{\epsilon=0}\int_{\Phi_{\epsilon}(\Omega)}\mathcal{L}_{T_{\epsilon}\phi}\\
&=\int_{\Omega}\frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Phi_{\epsilon}^*(\mathcal{L}_{T_{\epsilon}\phi})\\
&=\int_{\Omega}\frac{d}{d\epsilon}\bigg|_{\epsilon=0}(\mathcal{L}_{T_{\epsilon}\phi})+ \frac{d}{d\epsilon}\bigg|_{\epsilon=0}\Phi_{\epsilon}^*(\mathcal{L}_{\phi}).
\end{align}
For the second term, that’s precisely the definition of the Lie derivative $L_X(\mathcal{L}_{\phi})$; now use Cartan’s magic formula $L_X=d\iota_X+\iota_Xd$, and the fact that $\mathcal{L}_{\phi}$ is a top-degree form, so that $L_X(\mathcal{L}_{\phi})=d(X\,\lrcorner\,\mathcal{L}_{\phi})$. For the first term, we just compute in local coordinates (now on I’ll use the more common notation rather than the more precise one):
\begin{align}
\frac{d}{d\epsilon}\bigg|_{\epsilon=0}(\mathcal{L}_{T_{\epsilon}\phi})&=
\frac{\partial \mathcal{L}}{\partial\phi^i}\delta\phi^i+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi^i)}\frac{d}{d\epsilon=0}(\partial_{\mu}(T_{\epsilon}\phi)^i)\\
&= \frac{\partial \mathcal{L}}{\partial\phi^i}(\delta\phi)^i+\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi^i)}\partial_{\mu}((\delta\phi)^i)\tag{equality of mixed derivatives}\\
&=\left[\frac{\partial \mathcal{L}}{\partial\phi^i}-\frac{\partial}{\partial x^{\mu}}\left(\frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi^i)}\right)\right](\delta\phi)^i+ \frac{\partial}{\partial x^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial (\partial_{\mu}\phi^i)}(\delta\phi)^i\right)\\
&=0+ d(J_{\mathcal{L},\delta\phi,\phi}),
\end{align}
where we have invoked the Euler-Lagrange equations (recall we assumed $\phi$ was a solution) and Lemma 1 part (2). Using these results back above completes the proof.
Noether’s Theorem.
Finally, we come to a statement of Noether’s theorem. It is about as simple as can be given our preparation.
Noether’s Theorem.
Let $\mathcal{L}$ be a Lagrangian $n$-form on $(E,\pi,M)$, $S$ its action functional, $\mathcal{S_L}$ the set of solutions of the Euler-Lagrange equations. If $(X,T_{\epsilon})$ is a quasisymmetry of the action functional with surface current $K$, then for all $\phi\in\mathcal{S_L}$, we have that the $(n-1)$-form $J_{\text{Noether},\phi}:=J_{\mathcal{L},\delta\phi,\phi}+ X\,\lrcorner\,\mathcal{L}_{\phi}- K_{\phi}$ is closed.
By definition of a quasisymmetry, and Lemma 2, we have that for all solutions $\phi$ and all nice open sets $\Omega$,
\begin{align}
\int_{\Omega}d(J_{\mathcal{L},\delta\phi,\phi}+ X\,\lrcorner\,\mathcal{L}_{\phi})&=\int_{\partial\Omega}K_{\phi}.
\end{align}
Now, use Stokes’ theorem on the RHS, move it over to the LHS and use arbitrariness of $\Omega$ to deduce that $d(J_{\mathcal{L},\delta\phi,\phi}+ X\,\lrcorner\,\mathcal{L}_{\phi}-K_{\phi})$ vanishes identically on $M$. This proves the theorem.
As a remark, note that having a closed differential form $J_{\text{Noether},\phi}$ means that for every boundary $\partial\Omega$, we have (by Stokes’ theorem) $\int_{\partial\Omega}J_{\text{Noether},\phi}=0$, i.e the flux over every boundary vanishes; this means ‘what goes in goes out’, i.e it expresses a conservation law. Thus, Noether’s theorem can be succinctly stated as ‘every quasisymmetry of the action gives rise to a conservation law’ (by the above prescription).
Addressing the Discrepancies.
Let us observe the following trivial result.
Theorem (Where’s Waldo?)
Let $\mathcal{L}$ be a Lagrangian $n$-form, $X$ a vector field and $T_{\epsilon}$ a 1-parameter deformation of the fields in $\mathcal{S_L}$.
If $(X,T_{\epsilon})$ is a quasisymmetry of the action functional with surface current $K$, then $(0,T_{\epsilon})$ (i.e not considering any vector fields in the deformation) is a quasisymmetry with surface current $K^{(1)}=K- X\,\lrcorner\,\mathcal{L}$.
In particular, if $(X,T_{\epsilon})$ is a symmetry of the action functional, then $(0,T_{\epsilon})$ is a quasisymmetry with surface current $K^{(2)}=-X\,\lrcorner\,\mathcal{L}$.
This shows (as @Bence Racsko’s answers also mention), we can always consider the case of $X=0$ as long as we look for quasisymmetries. Then, the full Noether current is the same in both cases. The issue is really a matter of how you want to divide the various contributions, so it’s really more of a question of “where’s Waldo”, and the answer is it doesn’t matter.
However, if someone only considers field variations $(0,T_{\epsilon})$ and works only with symmetries of the action, then their Noether current is just $J_{\text{Noether},\phi}=J_{\mathcal{L},\delta\phi,\phi}$ without the other terms; hence they won’t be able to find more general conservation laws if they exist.
My Preference.
I personally prefer to allow myself the freedom to use the vector field $X$ and its flow, because in many concrete situations, it allows us to quickly spot symmetries of the action. Here’s a common situation.
Let $E$ is actually a some tensor bundle $T^r_s(TM)$ (I only make this assumption because here it’s clear that diffeomorphisms on the base manifold lift to diffeomorphisms of the total space $E$ and hence pullback of sections is well-defined; for general vector bundles this isn’t true) and let $\mathcal{L}$ be a Lagrangian $n$-form. Suppose $\mathcal{G}$ is a ‘good’ collection of diffeomorphisms of $M$, in the sense that for all smooth sections $\phi$ of the tensor bundle, and all diffeomorphisms $\Phi$, we have that
\begin{align}
\Phi^*(\mathcal{L}_{\phi})&=\mathcal{L}_{\Phi^*\phi}.
\end{align}
So, this is what I meant right at the beginning when I spoke of covariant Lagrangians. For such Lagrangians, we have that for all nice open sets $\Omega$, all $\Phi\in \mathcal{G}$ and all $\phi$,
\begin{align}
S[\Phi^*\phi,\Omega]&=S[\phi,\Phi[\Omega]].
\end{align}
This gives us an easy way to spot a symmetry of the action. If we have a vector field $X$ whose flow $\Phi_{\epsilon}$ always lies in $\mathcal{G}$, then we have for all $\epsilon$, all $\phi$ and all nice open sets $\Omega$,
\begin{align}
S[\Phi_{-\epsilon}^*\phi,\Phi_{\epsilon}(\Omega)]&=S[\Phi_{\epsilon}^*\Phi_{-\epsilon}^*\phi,\Omega]=S[\phi,\Omega],
\end{align}
so the $\epsilon$-derivative is $0$, and so $(X, T_{\epsilon}=(\Phi_{-\epsilon})^*)$ is a symmetry of the action. Notice how easy this was. This is the sort of thing we do in practice when looking out for translation/rotational invariances of the Lagrangian/action. This is why I like keeping around the effect of the vector field. At the end of the day, it of course doesn’t matter because the Noether current is the same, but still I like having my Waldo here.
