Consider any differentiable transformation $q_{j}\rightarrow q_{j}^{\lambda} = f_{j}^{\lambda}(q_{1}, \ldots, q_{N})$ parametrized by $\lambda$ (I'll place it as a superscript). For any such transformation, we can write $$ L(q_{j}^{\lambda}, \dot{q}_{j}^{\lambda}, t) - L(q_{j}^{0}, \dot{q}_{j}^{0}, t) = G^{\lambda} $$ where $G^{\lambda} = G^{\lambda}(q_{j}, \dot{q}_{j}, t)$.

Now trivially, for any given solution $q_{j} = q_{j}(t)$ to the Euler-Lagrange equation, we can write $G^{\lambda} = \frac{dF^{\lambda}}{dt}$. Does this mean that any transformation changes the Lagrangian up to a total time-derivative? If so, doesn't that mean any transformation is a quasi-symmetry?

The Noether charge for this should be $$ Q = p_{j}\frac{dq_{j}^{\lambda}}{d\lambda}\Big|_{\lambda=0} - \frac{dF^{\lambda}}{d\lambda}\Big|_{\lambda=0}, $$ and it seems like we my write $$ Q = p_{j}\frac{dq_{j}^{\lambda}}{d\lambda}\Big|_{\lambda=0} - \frac{d}{d\lambda}\Big|_{\lambda=0}\int_{0}^{t} G^{\lambda} \, dt. $$

I've tried out some examples, and it seemed to have given me consistent results for physically meaningless transformations. I was able to generate various expressions that indeed resulted in invariant quantities when plugging in solutions to the Euler-Lagrange equations, but it's not clear what significance any of them had, and all of this seems to trivialize Noether's theorem in my mind. Can anyone clarify if I'm doing something wrong?

I think what would be most helpful is, what would be an explicit example of a differentiable transformation that's not a quasi-symmetry? I would like to see exactly why such an example would fail.

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    $\begingroup$ If you have to impose that the Euler-Lagrange equations are satisfied, it's not a symmetry. $\endgroup$ Commented Jun 7, 2023 at 17:22

2 Answers 2

  1. Not all transformations of the variables are a quasi-symmetry of the action. For a counterexample, see e.g. this related Phys.SE post.

  2. It is perhaps helpful to stress that one of the requirements of being a quasi-symmetry is that it holds off-shell, i.e. without assuming the Euler-Lagrange (EL) eqs. of motion. If we assume the EL eqs. of motion to hold, then any infinitesimal variation of the Lagrangian is trivially a total derivative.

  3. A related question is whether any Lagrangian is a total derivative, see this Phys.SE post.

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    $\begingroup$ Thank you, this post was helpful! I tried to write why I was wrong in my own words (borrowing from other posts). Despite my reasoning errors, it turns out the quantity $$ Q = p_{j}\frac{dq_{j}^{\lambda}}{d\lambda}\Big|_{\lambda=0} - \frac{d}{d\lambda}\Big|_{\lambda=0}\int_{0}^{t} G^{\lambda} \, dt $$ really is a conserved quantity for any solution to the Euler-Lagrange equation. What intrigues me is that this works even if we don't have any quasi-symmetry. Do happen to know any significance of this? This manages to generate all sorts of contorted expressions that end up being constant. $\endgroup$ Commented Jun 8, 2023 at 1:25

I'll try to explain my reasoning errors in my own words. Actually, I think it is actually interesting where I went wrong.

Now trivially, for any given solution $q_{j} = q_{j}(t)$ to the Euler-Lagrange equation, we can write $G^{\lambda} = \frac{dF^{\lambda}}{dt}$.

This is false!

Not only is this false, but it is misleading in a way that lends itself to multiple wrong ways of thinking. I'll explain three ways in which someone might think my assertion is true, and explain why each one is actually false. Only interpretation (2) is the correct one and our statement is still not true there.

  1. We know that on its own, $G^{\lambda}(a_{1}, \ldots, a_{N}, b_{1}, \ldots, b_{N}, t)$ is a function $\mathbb{R}^{N}\times\mathbb{R}^{N}\times\mathbb{R}\rightarrow\mathbb{R}$ (let's ignore the role of $\lambda$ for now by fixing it). We can define $$ F^{\lambda}(a_{j}, b_{j}, t) = \int_{0}^{t} G^{\lambda}(a_{j}, b_{j}, \tau)\, d\tau. $$ If we assume all the $a_{j}$'s and $b_{j}$'s are independent scalar variables, then $\frac{dF^{\lambda}}{dt} = \frac{\partial F^{\lambda}}{\partial t} = G^{\lambda}$. The problem is, when we are discussing Lagrangians we are plugging in paths $q_{j}(t)$'s for $a_{j}$ and velocities $\dot{q}_{j}(t)$'s for $b_{j}$, so the total time-derivative does not coincide with $\frac{\partial F^{\lambda}}{\partial t}$. Instead, we have to use the chain rule and we would obtain $\frac{dF^{\lambda}}{dt} \ne G^{\lambda}$.
  2. Let's suppose $G^{\lambda} = \frac{dF^{\lambda}}{dt}$ was written with generic paths plugged-in in mind. By generic, I mean the relationship holds for any path $q_{j}(t)$ plugged-in. In that case, whatever expression we have for $L(q_{j}^{\lambda}, \dot{q}_{j}^{\lambda}, t) - L(q_{j}^{0}, \dot{q}_{j}^{0}, t)$, it needs to be of the form $$ \frac{\partial F^{\lambda}}{\partial q_{j}}\dot{q}_{j} + \frac{\partial F^{\lambda}}{\partial \dot{q}_{j}}\ddot{q}_{j} + \frac{\partial F^{\lambda}}{\partial t} $$ without any specific form of $q_{j}(t)$ being exploited. Note that $L(q_{j}^{\lambda}, \dot{q}_{j}^{\lambda}, t) - L(q_{j}^{0}, \dot{q}_{j}^{0}, t)$ has no dependence on a generic $\ddot{q}_{j}(t)$, so the second term in the above displayed expression is going to be zero. In any case, in this context, I still turn out to be wrong, because there is an explicit counterexample. See Counterexample 1 below.
  3. There is another sense in which we might be temped to assert $G^{\lambda} = \frac{dF^{\lambda}}{dt}$. Maybe we are asserting this when we are plugging in a specific, particular path $q_{j}(t)$. In that case, for a given fixed path $q_{j}(t)$, we can take $$ F^{\lambda}_{q}(t) = \int_{0}^{t} G^{\lambda}(q_{j}(\tau), \dot{q}_{j}(\tau), \tau)\, d\tau. $$ Notice that the function "$F^{\lambda}_{q}$" defined is specific to the given $q_{j}(t)$, so I labeled it accordingly. This has the problem that our function is now dependent on $t$ and the entire path $q$ itself. For each different path, we need to invoke a different function, so our relationship is not universal with respect to any differentiable transformation and we lose generality.

(2) is the only correct interpretation by what is meant when people say, "the Lagrangians differ by a total time-derivative," and in that interpretation it is completely false to say that we are guaranteed to have quasi-symmetries. Below I give a specific example where we fail to have a quasi-symmetry under this interpretation.

Counterexample 1. I'll use the example that Qmechanic has in this post. Let $$ L(q, \dot{q}, t) = \frac{\dot{q}^{2}}{2} - \frac{q^{2}}{2} $$ and consider the transformation $q\rightarrow \lambda q$. Then $L(q^{\lambda}, \dot{q}^{\lambda}, t) - L(q^{1}, \dot{q}^{1}, t) = (\lambda^{2}-1)(\dot{q}^{2}-q^{2})/2$. Suppose there exists a function $F^{\lambda}(q, \dot{q}, t)$ for which $$ \frac{1}{2}(\lambda^{2}-1)(\dot{q}^{2}-q^{2}) = \frac{\partial F^{\lambda}}{\partial q}\dot{q} + \frac{\partial F^{\lambda}}{\partial \dot{q}}\ddot{q} + \frac{\partial F^{\lambda}}{\partial t}. $$ Now remember that this should hold for all possible paths $q(t)$ (and accompanying $\dot{q}(t), \ddot{q}(t)$). Since the LHS has no $\ddot{q}$-dependence, the second term on the RHS has to be zero, so $\frac{\partial F^{\lambda}}{\partial\dot{q}} = 0$. Rearrange terms to get $$ \frac{1}{2}(\lambda^{2}-1)\dot{q}^{2} - \frac{\partial F^{\lambda}}{\partial q}\dot{q} = \frac{1}{2}(\lambda^{2}-1)q^{2} + \frac{\partial F^{\lambda}}{\partial t}. $$ The RHS has no $\dot{q}$-dependence, so $\frac{1}{2}(\lambda^{2}-1)\dot{q} - \frac{\partial F^{\lambda}}{\partial q} = 0$, so $F^{\lambda} = \frac{1}{2}(\lambda^{2}-1)\dot{q}q + C$. However we had $\frac{\partial F^{\lambda}}{\partial\dot{q}} = 0$, which forces $\lambda = \pm 1$. This contradicts the fact that $\lambda$ varies continuously, so $F^{\lambda}$ does not exist. $$\tag*{$\blacksquare$}$$


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