Symmetry of Euler-Lagrange equations and conservation laws

1. Continuous symmetry of the action implies a conservation law, but what if equations of motion have a continuous symmetry? Does it imply a conservation law?

2. Also is symmetry of equations of motion also a symmetry of the action?

I assume that the equations of motion I am talking about follow from the action, i.e. the equations of motion are Euler-Lagrange equations. Related examples could help and references also.

• To prove Noether's theorem you need a lagrangian. In general if you have equations of motion that don't follow from a lagrangian then you aren't guaranteed a conservation law. You can see a proof of noether's theorem in chapter 1 of these notes: damtp.cam.ac.uk/user/tong/qft.html. An example of an equation of motion (not following from a lagrangian) that has time translation invariance but no conserved current would be a damped harmonic oscillator. Aug 4, 2013 at 8:03
• I assume that the equation of motion i am talking about follows from the Lagrangian which defines the action Aug 4, 2013 at 8:11
• If the equation of motion comes from a lagrangian, and the lagrangian has a continuous symmetry, then Noether's theorem guarantees you a conserved current. Aug 4, 2013 at 8:15
• I am sorry but i am asking other way if you read the question carefully,Noether theorem is when action has the symmetry not the equation of motion. Aug 4, 2013 at 8:20
• Possible duplicate: physics.stackexchange.com/q/51327/2451 Aug 4, 2013 at 13:57

Different Lagrangian could give the same equation of motion. If you add non-dynamical variables, you could present a Lagrangian without symmetries, while the equation of motion has symmetries.

For instance, take the Lagrangian :

$$L(x,f) = \frac{f^2}{2}+\dot f ~x$$

Euler-Lagrange equations (applied to $x$ and $f$) give :

$$\dot f = 0, f = \dot x$$

That is :

$$\ddot x=0, f = \dot x$$

The Lagrangian does not respect explicitely translational $x$-invariance (because of the term $x$), but equations of movement respect translational $x$-invariance, and $\dot x$ is a conserved quantity.

$f$ is non-dynamical, because we can replace $f$ by its value due to the equation of movement and get a Lagrangian :

$$L(x) = \frac{\dot x^2}{2}$$

With applications of Euler-Lagrange equation, we obviously get the same equation of movement $\ddot x = 0$. This latter lagrangian obviously respects the translational $x$-invariance.

• The action for this lagrangian is invariant under translations. The lagrangian need only be invariant up to a total derivative for the action to be invariant, which this lagrangian is: $\delta L=\dot{f}$. Aug 4, 2013 at 13:27
• @Andrew : It is true that you may add a total derivative $-\frac {d (fx)}{d t}$ to get $L'(x,f) = \frac{f^2}{2} -f \dot x$ which is translational-invariant, but you have surface terms $[fx]_{t_1}^{t_2}$, and they are not explicitely translational invariant. Aug 4, 2013 at 14:19
• I've been staring at this for an hour, my first thought was that you are obviously wrong but you're not and this is a very interesting subtlety. It's strange since nothing physically happens at $t_1$ or $t_2$ to explicitly break any symmetries. My resolution would be to say the 'correct' lagrangian is $\frac{f^2}{2}+\dot{f} x - \frac{d(fx)}{dt}$, then if you shift $x\rightarrow x+a$ the action shifts by $\delta S = a ([f]_{t_1}^{t_2}-[f]_{t_1}^{t_2})=0$ and is translation invariant. But that's a very interesting point. I don't want to derail this question though so I'll check out. Aug 4, 2013 at 16:43
• @Andrew : Starting from the lagrangian $L(x,f) = (\frac{f^2}{2}+\dot f ~x)$, you could write this lagrangian $L(x,f) = L'(x,f) + \frac{d(xf)}{dt}$, where $L'(x,f) = (\frac{f^2}{2} - f ~\dot x)$ is translationnal invariant. But, anyway, in looking at the action $S_{12} = \int_{t_1}^{t_2}L$, with a translation $x \rightarrow x+a$, you get $S_{12}\rightarrow S_{12} + a(f(t_2) - f(t_1))$. So, there is no explicit translationnal invariance of the action. However, adding the equations of movement $\dot f=0$, we find that $S_{12}$ is invariant. So, we could speak of "hidden" symmetry of the action. Aug 5, 2013 at 8:51

You cannot prove the existence of a conserved current directly from the equations of motion in general just given the fact that the equations of motion have a continuous symmetry.

A counterexample to this is a damped harmonic oscillator, $\ddot{x}+\Gamma \dot{x}+\omega^2 x=0.$ This is invariant under $t\rightarrow t+\delta t$, but doesn't have a conserved quantity.

Now, given a specific equation of motion, you can look for constants of motion. For example, given $\ddot{x}=V'(x)$ you can prove that energy is conserved. For example see First integral of an equation of motion: $\mu\ddot r=-\frac{k}{r^2}$. But it's hard to see how this method relates to any particular symmetry of the eom, rather than the particular algebraic structure of the eoms.

To relate symmetries to conserved quantities, you should use the action formulation, as you know. Symmetries of the action are always symmetries of the equations of motion. However, not all equations of motion have a corresponding action (like the damped harmonic oscillator). There is no noether's theorem that operates directly at the level of the eoms, that's why I keep going back to the action.

It seems you want to know about the other way around, are symmetries of the eoms always symmetries of the action? I don't see how this could fail to be the case, but admittedly I can't prove it off the top of my head. Here's why I think that symmetries of the eoms should be symmetries of the action: Let's say the symmetry is translation invariance just for simplicity, you could generalize this argument to any symmetry easily. Let's say the eoms are translation invariant. Then I can set up my system at position $A$, let it evolve classically, and see what happens. Then I move my system to be position $B$ and set it up with identical initial conditions. Exactly the same motion must happen at position $B$ by translation invariance. But the action is just a function of the path taken--you would need to get a different number for the action evaluated on the paths taken at position $A$ then you would get if you evaluated the action on the paths at position $B$. But since the physics doesn't distinguish between $A$ and $B$ how could this be true: by simply calculating the action you could tell whether you were at $A$ or $B$. Admittedly it isn't a proof, but it's why I don't think it's true. Even if you could find a symmetry of the eoms that wasn't a symmetry of the action, you wouldn't be able to use Noether's theorem to show there was a conserved quantity, so there wouldn't be a conserved quantity associated with that symmetry.

Not all conservation laws follow from symmetries though.