# Does a constant factor matter in the definition of the Noether current?

This is a very basic Lagrangian Field Theory question, it is about a definition convention. It takes much more time to typeset it than answering, but here it is:

Consider a field Lagrangian with only a kinetic term,

$$L = \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi$$

Consider the very simple transformation $\phi \rightarrow \phi + \alpha$ ($\alpha$ constant), and so I understand here that $\alpha$ plays the role of $\delta\phi$. I determine the Noether current as $$\frac{\partial L}{\partial[\partial_{\mu}\phi]}\delta\phi$$

and the result is $$\alpha\partial_\mu\phi$$

But in Peskin & Schroeder (just above eq 2.14), the result they give is:

$$\partial_\mu\phi$$

And it doesn't seem to be an erratum. I don't care that "localized" Lagrangian very much (hey, wait before closing, please), but a very general question arises:

Is $\alpha$ dropped simply because $\partial_\mu\phi$ is too a conserved quantity (and so under "conserved current" one understands the general concept, momentum, energy or whatever, regardless of its value), or am I missing some other very basic detail that is assumed to be known by the reader?

Later edit: I have eventually understood this question and more, by reading the beginning of chapter 22 of Srednicki. I am finding that book (well, the free preprint for the moment) crystal clear, it seems excellent.

-

I) Let us for simplicity address OP's question in the context of point mechanics where $q^i$ are generalized position coordinates on some manifold $M$ [instead of considering field theory with fields $\phi^{\alpha}(x)$]. OP's question is rooted in the difference between

1. on one hand, an infinitesimal variation $$\tag{1} q^i~\rightarrow \widetilde{q}^i~=~ q^i+\delta q^i$$ of the generalized position coordinates, or equivalently, $$\tag{2} \delta q^ ~:=~ \widetilde{q}^i-q^i;$$

2. and on the other hand, that of a generator/Lie algebra element/vector field $$\tag{3} Y~=~Y^i\frac{\partial}{\partial q^i},\qquad Y^i~=~Y^i(q),$$ which is not infinitesimal (although $Y$ is sometimes confusingly referred to as an 'infinitesimal generator' in the literature).

Both concepts $\delta$ and $Y$ are linear derivations that satisfy Leibniz rule, and the interrelation between the two is given by

$$\tag{4} \delta q^i~=~\epsilon Y^i,$$

where $\epsilon$ in eq. (4) is an infinitesimal parameter. The mathematical concept of a vector field $Y$ is tied in a bijective manner to the concept of a flow$^1$

$$\tag{5} \sigma:~]\!-\!c,c[ ~\times~ M~\to~ M, \qquad ]\!-\!c,c[ ~\subseteq~ \mathbb{R},$$

where

$$\tag{6} \frac{d}{d\epsilon}\sigma^i(\epsilon,q)~=~Y^i(\sigma(\epsilon,q)), \qquad\sigma^i(\epsilon=0,q)~=~q^i.$$ A flow $\sigma$ satisfies $$\tag{7} \sigma^i(\epsilon,\sigma(\epsilon^{\prime},q))~=~\sigma^i(\epsilon+\epsilon^{\prime},q).$$ Note that in eq. (7), it is understood that $\epsilon$ and $\epsilon^{\prime}$ are real numbers in the interval $]\!-\!\frac{c}{2},\frac{c}{2}[ \subseteq \mathbb{R}$, and not infinitesimal.

II) The (bare) Noether charge

$$\tag{8} Q ~=~p_i Y^i$$

is (in this case) momentum

$$\tag{9} p_i ~:= ~\frac{\partial L}{\partial \dot{q}^i}$$

times generator $Y^i$. In particular, the definition (5) of the Noether charge $Q$ does not depend on the $\epsilon$ parameter.

--

$^1$ We ignore the possibility that the domain $]\!-\!c,c[$ could depend on the initial position $q\in M$.

-
Wow! thanks! I have to decide tomorrow if I set the "answered" mark here instead, when I have some time to think about it (it is too late in Spain now, I was about to go to sleep). But thanks very much (+1) for the moment. – Eduardo Guerras Valera Feb 10 '13 at 6:22
Ouch! I realize now how stupid the question is... I studied something very similar (perhaps a particular case?) in elementary Lagrangian mechanics: the absence of a generalized coordinate in the Lagrangian leads to a conservation law on that coordinate, which was said to be "ignorable" (at least in spanish...). I guess this Noether-thing is sort of a more general statement... – Eduardo Guerras Valera Feb 10 '13 at 6:34
And, apart from the mathematical definition, it is also pretty intuitive that it has not much sense to consider "tens times the momentum". Momentum is conserved. Any. X( – Eduardo Guerras Valera Feb 10 '13 at 6:39
Just so that there's no confusion, Qmechanic's notation is slightly different, and he includes some (important and much appreciated) mathematical detail, but his response contains essentially the same content. Whether $\delta q$ is used to denote $q_\epsilon - q_0$ or $\frac{\partial}{\partial\epsilon} q_\epsilon|_{\epsilon=0}$ is a matter of notational taste. The point that the Noether charge does not depend on the infinitesimal parameter was present in our original discussion. – joshphysics Feb 10 '13 at 21:49
@Qmechanic, thanks for the answer and the many references. – Eduardo Guerras Valera Feb 14 '13 at 3:03

To an extent, the ambiguity is just a matter of notation. Some people would write a generic one-parameter transformation as $$\phi(x) \to \phi(x) + \alpha\delta\phi(x) + \mathcal O(\alpha^2)$$ so that $\delta\phi(x)$ is the coefficient of the small parameter $\alpha$ as with expressions in ordinary calculus like $$f(x+\alpha) = f(x) + \alpha f'(x) + \mathcal O(\alpha^2)$$ With this notation, the transformation $$\phi(x) \to \phi(x) + \alpha$$ would have $\delta\phi(x) = 1$, and in this case you would recover the Peskin result. The change in notation is really just a matter of taste when you are working with one-parameter families of transformations. As you point out, for any real number $\alpha$, both $\alpha\partial_\mu\phi$ and $\partial_\mu\phi$ are conserved, and their conservation equations are the same.

Addendum. When it comes down to it, in a physical context all that really matters is the equivalence class corresponding to all expressions for a conserved quantity that differ by a constant, nonzero multiplicative factor (since their corresponding conservation equations are the same), so the issue here is really just a matter of semantics.

Cheers!

-
Forgive me if I insist on something quite obvious (I am new in the field) but then, for instance, for a free particle Lagrangian $\frac{1}{2}mv^2$ the conserved current associated with $x \rightarrow x + c$ is $mv$ but NOT, say, ten times that, $10mv$ which, although it is also a conserved quantity, doesn't enter the definition of conserved current associated with that Lagrangian? (it is a matter of definition of the current, although both quantities are obviously conserved, I just want to know what is precisely understood by the words "conserved current") – Eduardo Guerras Valera Feb 9 '13 at 15:10
If you define the variation in the way that I did in the first equation, and if you define the conserved quantity via the expression in your second equation, then there is no ambiguity, and you obtain the standard result. In this classical mechanics case, this would correspond to defining $x \to x+ \alpha = x+\alpha\delta x$ in which case the conserved quantity would become $(\partial L/\partial \dot x) \delta x = m\dot x*1 = mv$ as desired. Or, for example, with angular momentum conservation in 3D, you would write $\mathbf x \to \mathbf x + \alpha \mathbf n\times \mathbf x$ so that – joshphysics Feb 9 '13 at 21:53
(contd.) $\delta\mathbf x = \mathbf n\times\mathbf x$ which would yield the standard normalization for angular momentum as the conserved current where $\mathbf n$ is a unit vector along the axis of rotation. – joshphysics Feb 9 '13 at 21:54
But if I define the variation and the conserved quantity as you say, I get the result, it is ok, and I can define whatever I want. The thing is how others understand and define things. Explicitely: If I tell you "the conserved current associated with a space translation for the Lagrangian of a free particle is ten times its linear momentum", would you say "No, the conserved current is the linear momentum, and not ten times it, although the latter quantity is also conserved, but it is not whatever is understood under the name Conserved Current associated with yak yak yak... ? That's my point – Eduardo Guerras Valera Feb 9 '13 at 22:04
I don't understand what you're trying to say. The notational convention I have espoused here is not an uncommon one, other people do use it too I assure you, and it is a consistent way to generate any conserved quantity you would like. Are you asking how the other notational convention would lead to the same expression for the conserved current (the one without an extra multiplicative factor)? – joshphysics Feb 9 '13 at 22:14