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From the book Introduction to Classical Mechanics With Problems and Solutions by David Morin, page 236 states:

Noether's Theorem: For each symmetry of the Lagrangian, there is a conserved quantity.

Whereas the Wikipedia page states:

Noether's (first) theorem states that any differentiable symmetry of the action of a physical system has a corresponding conservation law

Does the action and Lagrangian have identical symmetries and conserved quantities?

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up vote 6 down vote accepted

Yes, provided one uses the correct notions of symmetry for the action and the lagrangian.

The setup.

We assume throughout that the action can be written as the integral of a local Lagrangian. Namely, let $\mathcal C$ be the configuration space of the system, then for any admissible path $q:[t_a, t_b]\to \mathcal C$, there exists a local function $L$ of paths such that \begin{align} S[q] = \int_{t_a}^{t_b} dt \,L_q(t). \end{align} Let a smooth, $\epsilon$-deformation $q(t) \to \hat q(t, \epsilon)$ of paths be given. We will use the $\delta$ notation for first order changes in quantities under such a deformation.

Symmetry defined.

We say that this deformation is a symmetry of the action $S$ provided there exists a local function of paths $B_q$ such that \begin{align} \delta S[q] = B_q(t_b) - B_q(t_a) \end{align} for all admissible paths $q:[t_a, t_b]\to \mathcal C$. In other words, the action only changes to first order by a boundary term. We say that this deformation is a symmetry (or what Qmechanic calls a quasisymmetry in his response) of the Lagrangian $L$ provided there exists a local function $\Lambda_q$ of paths such that \begin{align} \delta L_q(t) = \frac{d\Lambda_q}{dt}(t) \end{align} for all admissible paths $q:[t_a, t_b]\to \mathcal C$. In other words, the lagrangian only changes to first order up to a total derivative.

Equivalence of notions of symmetry.

Using these definitions, one can show that a given deformation is a symmetry of $S$ if and only if it is a symmetry of $L$.

Notice that for any deformation, and for any admissible path $q:[t_a, t_b]\to \mathcal C$, one has \begin{align} \delta S[q] = \int_{t_a}^{t_b} dt\,\delta L_q(t) \end{align} Suppose now, that a given deformation is a symmetry of $S$, and let a path $q:[t_a, t_b]\to\mathcal C$ be given. For each $t\in [t_a, t_b]$ we have \begin{align} \int_{t_a}^{t} dt'\,\delta L_{q}(t') = B_{q}(t) -B_{q}(t_a), \end{align} Since the deformation is a symmetry of $S$. Taking the derivative of both sides with respect to $t$, and using the fundamental theorem of calculus on the left, we obtain \begin{align} \delta L_q(t) = \dot B_q(t) \end{align} for all $t\in[t_a, t_b]$. Identifying $B$ with $\Lambda$, we find that the deformation is a symmetry of the lagrangian.

I'll leave the converse to you.

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First some terminology:

  1. In general an infinitesimal transformation of a field theory consists of a so-called horizontal infinitesimal transformation $$ \delta x^i ~=~x^{\prime i}- x^i$$ of the base manifold, and a so-called vertical infinitesimal transformation $$ \delta_0\phi^{\alpha}(x)~=~\phi^{\prime \alpha}(x)-\phi^{\alpha}(x) $$ of the fields. The full infinitesimal transformation of the fields reads $$ \delta\phi^{\alpha}(x)~=~\phi^{\prime \alpha}(x^{\prime})-\phi^{\alpha}(x). $$

  2. A quasisymmetry of a local action $S=\int_V d^dx ~{\cal L}$ means that the infinitesimal change $\delta S$ is a boundary term under the quasisymmetry transformation. A symmetry of an action is the special case $\delta S=0$.

  3. A quasisymmetry of a Lagrangian (density) ${\cal L}$ means that the infinitesimal change $\delta {\cal L}$ is a total divergence under the quasisymmetry transformation, cf. this Phys.SE answer. A symmetry of a Lagrangian (density) is the special case $\delta {\cal L}=0$.

A vertical quasisymmetry of a local action corresponds$^1$ to a vertical quasisymmetry of the Lagrangian (density). (However, a Jacobian factor from a horizontal transformation could complicate the correspondence.)

A vertical symmetry of an action is not necessarily a vertical symmetry of the Lagrangian (density), but the opposite holds.

It is more general to formula Noether's Theorem in terms of an action rather than a Lagrangian (density). This is also what Noether originally did in her 1918 paper.


$^1$ It is straightforward to deduce that a vertical quasisymmetry of the Lagrangian (density) leads to a vertical quasisymmetry of the action. If we know that an action has a vertical quasisymmetry for every integration region $V$, we can also easily deduce the other way by localization. However, if we only know the that the action has a vertical quasisymmetry for a single fixed integration region $V$, there could be possible topological obstructions in field configuration space that might invalidate that the Lagrangian (density) has a vertical quasisymmetry. Technically, the latter relies on an algebraic Poincare lemma of the so-called bi-variational complex, see e.g. Ref. 2.


  1. G. Barnich, F. Brandt and M. Henneaux, Local BRST cohomology in gauge theories, Phys. Rep. 338 (2000) 439, arXiv:hep-th/0002245.
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This is semantics, but it might be useful to mention that the term "symmetry" is also sometimes used to refer to what you call a "quasisymmetry," not that I think it's unhelpful to have terms that distinguish between invariance and invariance up to total derivative. – joshphysics Feb 11 '14 at 20:00
@joshphysics: Good point. – Qmechanic Feb 11 '14 at 20:06
Correction to the answer (v7): In the last line, Ref. 2 should be Ref. 1. – Qmechanic Feb 19 '14 at 17:46

The technical answer is $No$. Surprisingly I think Wikipedia gives the better definition, though I think both authors are trying to say the same thing. Let the action be defined as

$S[\varphi]=\int d^4x\ \mathcal L(\varphi(x),\partial_\mu\varphi(x))$

A differentiable symmetry is a symmetry of the functional that does not change the action


It is a differential symmetry because when this expression is interpreted as an action on the lagrange density $\mathcal L$ it does so by a differentiation. Assuming $\varphi'=\varphi+\delta\varphi$ where $\delta\varphi$ is an infinitesimal change in the function $\varphi(x)$ we get

$S[\varphi']-S[\varphi]=\int d^4x\ \left(\delta\varphi\frac{\partial}{\partial \varphi}\mathcal L+\partial_\mu\delta\varphi\frac{\partial}{\partial(\partial_\mu\varphi)}\mathcal L\right)=0$

To get Noether's theorem we have to integrate by parts, so it is necessary that the integral is included in the definition, and this is why it is more accurate to speak of the symmetry in terms of the action. Here is an example of an action which does not have the same symmetry of the Lagrangian depending only on boundary conditions of the integral:

$S_1=\int_0^\infty dr\int_0^{2\pi}rd\theta\ \mathcal L(\varphi,\partial_\mu\varphi)$

is spherically symmetric, so it conserves angular momentum by Noether's theorem, while

$S_2=\int_0^1dx\int_0^1dy\ \mathcal L(\varphi,\partial_\mu\varphi)$

does not. The integral breaks spherical symmetry that the lagrangian might otherwise preserve.

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OP, I am new to stackexchange (but a physics veteran) so I am not yet allowed to comment on the post itself, but you should know that the one you chose as the correct answer is only valid for one dimensional curves, and even there it is valid only for a special definition of symmetry that allows for boundary terms (called "quasi-symmetries," as QMechanic points out). I'm saying this because if you study anything more general than that, two-dimensional planes, QFT, string theory, for example, this result will not hold and the Lagrange symmetries may not be the same as the action symmetries because the measure (e.g. $\int d^4x$) can break the symmetry of $\mathcal L$.

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Thanks! I'll leave you to point this out as a comment to Josh's answer when you get your 50 rep, bearing in mind also that Noether only considered one dimensional curves in her paper. – Larry Harson Feb 13 '14 at 15:04
Welcome aboard. Maybe you could click on "tags" or "unanswered" at the top to answer the difficult questions and get your rep up. – Physiks lover Feb 13 '14 at 19:01

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