For example, this symmetry:

$$\delta q^{i}=\epsilon(q^{i}-2\dot{q}^{i}t)$$

it's derivative is:

$$\delta\dot{q}^{i}=-\epsilon(\dot{q}^i +2\ddot{q}^i t)$$

There appears $\ddot{q}^{i}$ in this expression, so I am tempted to replace the equations of motion there, but don't know if that is valid. I know you can't do that kind of stuff sometimes.

What I am trying to show is that the EOM of the Lagrangian $L=\frac{1}{2}\dot{q}_{i}\dot{q}^{i}-V(q_{i}q^{i})$ are invariant (or maybe not) under this symmetry.

  • 1
    $\begingroup$ On the level of the Lagrangian, the e.o.m. are not valid. Are you trying to show the invariance of the Lagrangian or of the equations of motion? $\endgroup$
    – ACuriousMind
    Sep 23, 2016 at 0:36
  • $\begingroup$ The invariance of the equations of motion, to investigate if that is a Noether's symmetry or not. $\endgroup$
    – rsaavedra
    Sep 23, 2016 at 0:38
  • 2
    $\begingroup$ Noether's theorem pertains to symmetries of the action/quasi-symmetries of the Lagrangian, not to symmetries of the equation of motion. I'm not sure what you're trying to do. $\endgroup$
    – ACuriousMind
    Sep 23, 2016 at 0:44
  • $\begingroup$ Isn't taking $\delta (e.o.m) = (e.o.m)$ a way of proving that a symmetry is Noetherian? Taking the variation of the equations of motion and proving that the symmetry leaves the equations invariant? Sorry, I am new in this subject. $\endgroup$
    – rsaavedra
    Sep 23, 2016 at 0:47
  • $\begingroup$ In arxiv.org/abs/1207.5001 is a nice example of a symmetry that leads to the Runge-Lenz vector. It is a symmetry only after using the equations of motion. But it is non-trivial because it doesnt need all of them $\endgroup$ Feb 3 at 4:53

3 Answers 3


Yes, to show that an infinitesimal transformation $\delta$ is a symmetry of the equations of motion $\text{EOM}=0$, one may use the equations of motion $\text{EOM}=0$ (and consequences thereof) to reduce the expression $\delta(\text{EOM})$ to zero.

Example: The equation $x=0$ is invariant under scale symmetry. Let $\text{EOM}\equiv x$ and $\delta=\varepsilon x\frac{\partial}{\partial x}$, where $\varepsilon$ is an infinitesimal parameter.

It seems relevant to stress the following points, partly mentioned by user ACuriousMind in comments above:

  1. A symmetry of the Euler-Lagrange (EL) equations does not imply a quasi-symmetry of the action, cf. e.g. this Phys.SE post.

  2. A quasi-symmetry of an action is by definition assumed to be valid off-shell. In particular, one is not allowed to use the EL eqs. when proving that a transformation is a quasi-symmetry of the action. In fact, if one assumes EL eqs. to hold, then any infinitesimal variation of the action is trivially a boundary integral.

  3. Noether's theorem relies on an action formulation and a quasi-symmetry thereof.


Yes it is valid. The problem can be viewed this way. There is a manifold, a fiber bundle (a jet bundle actually) $A$, with basis $\mathbb R$ where the coordinate $t$ ranges and fibers $Q_t$ locally covered by coordinates $q^1,\ldots, q^n, \dot{q}^1, \ldots, \dot{q}^n$.

The solutions of E.-L. equations arising from a Lagrangian $L=L(t, q, \dot{q})$ are nothing but the integral curves $$\mathbb R \ni t \mapsto \gamma(t) = (t, q(t), \dot{q}(t)) \in A$$ of the dynamical vector field with local form

$$Z = \frac{\partial}{\partial t} + \sum_k \dot{q}^k \frac{\partial}{\partial q^k} + \sum_k A^k(q, \dot{q})\frac{\partial}{\partial \dot{q}^k}$$

Imposing that $\gamma$ is an integral curve of $Z$ we have

$$1=1, \quad \frac{dq^k}{dt} = \dot{q}^k(t)\:, \quad \frac{d\dot{q}^k}{dt} = A^k(t, q(t), \dot{q}(t))$$

the last requirement is nothing but the Euler-Lagrange differential equation system written into its normal form, separating on the left-hand side the derivatives of highest order from the other derivatives.

A one-parameter group of dynamical symmetries first of all admits a generator $X$. This is a vector field on $A$ whose integral lines are nothing but the evolution of the points of $A$ subjected to the symmetry.

A dynamical symmetry, by definition, moves solutions of the E.-L. solutions into solutions of E.-L- solutions. It is possible to prove that this is equivalent to $$[X,Z]=0 \tag{1}$$ where the bracket is the standard Lie bracket of vector fields. What is the natural structure of $X$ used in classical mechanics? I mean the one leading to the standard formulation of Noether theorem. Here is

$$X = 0\frac{\partial}{\partial t} + \sum_k X^k(t,q, \dot{q}) \frac{\partial}{\partial q^k} + \sum_k Z(X^k)(t, q, \dot{q})\frac{\partial}{\partial \dot{q}^k}\tag{2}$$

The factor $0$ means that the fixed time $t$ fibers are fixed under the symmetry. The functions $X^k(t,q)$ are arbitrary and $$Z(X^j) := \frac{\partial X^j}{\partial t} + \sum_k \dot{q}^k \frac{\partial X^j}{\partial q^k} + \sum_k A^k(q, \dot{q})\frac{\partial X^j}{\partial \dot{q}^k}\tag{3}$$ This is exactly you are doing written into another language.

The term $A^k$ disappears if $X$ is only function of $t$ and $q$. In this case one says that the symmetry is geometric.

To be complete I conclude saying that, if the Lagrangian of the system is $L$ and $Z$ is constructed out of it, and $X$ has the form (2), the invariance condition
$$X(L)=0 \tag{4}$$
implies both (1), so that we have a dynamical symmetry, and $$Z(N)=0\tag{5}$$ where $$N(t,q,\dot{q}) = \sum_{k=1}^n X^k \frac{\partial L}{\partial \dot{q}^k}\:.$$ The identity (5) just says that $N$ is a conserved quantity along the motion of the system.


Yes, in classical field theory and classical Lagrangian systems the Noether Theorem is valid on shell, i.e., taking the eom's into account. The equivalent quantum field theory version for expectation values is the Ward-Takahashi identities and they can work off shell, and remain valid after renormalization.

See the description of Noethers applicability and the on shell use at Wikipedia at https://en.wikipedia.org/wiki/Noether%27s_theorem

Noether's theorem can also be stated as the off shell Noether charge is conserved on shell. You do need the eom's to get to conservation in time.


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