Does adding a total derivative to a Lagrangian change its symmetry and/or associated constants of motion?

Question

I'm learning about symmetries and Noether's theorem and I'm stuck on this issue:

Suppose you have a system described by a Lagrangian $L(q,\dot q,t)$, and an infinitesimal transformation $T$ which is a symmetry of the system. Let $Q$ be the constant of motion associated with this symmetry.

Let's now consider the Lagrangian $L'=L+\frac{d}{dt}F(q,t)$, that is, we add a total derivative to $L$. I know that $L'$ satisfies the same Euler-Lagrange equations as $L$, but how about symmetries? Is $T$ a symmetry for $L'$ as well and is $Q$ a constant of motion?

I know that a transformation $T$ can be shown to be a symmetry through the symmetry test, $$ \frac{\partial L}{\partial q}\delta q + \frac{\partial L}{\partial \dot q}\delta \dot q + \frac{\partial L}{\partial t}\delta t + L \frac{d (\delta t)}{d t} + \frac{d}{dt}\delta G=0$$ and an associated constant of motion may be found by using Noether's theorem, $$ \frac{\partial L}{\partial \dot q}\delta q - \left[ \frac{\partial L}{\partial \dot q} \dot q - L \right]\delta t + \delta G = 0 $$

I suspect that this may be related to the $\delta G$ term on the symmetry test and Noether's theorem – in class we often assume $\delta G = 0$ – but I cannot seem to figure it out.

Answer 1

Let us consider field theory (rather than point mechanics$^1$) to be as general as possible. Then assume that the Lagrangian density is changed by a total spacetime derivative $$ \tilde{\cal L} - {\cal L}~=~\Delta{\cal L}~=~d_{\mu}F^{\mu}. \tag{A}$$ The infinitesimal transformations are of the form $$\begin{align} \delta x^{\mu}~=~& x^{\prime \mu} - x^{\mu} ~=~\epsilon X^{\mu}\qquad \text{(horizontal variation)}\cr \delta_0\phi^{\alpha}(x)~=~& \phi^{\prime\alpha}(x) - \phi^{\alpha}(x)~=~\epsilon Y_0^{\alpha}\qquad \text{(vertical variation)}\cr \delta\phi^{\alpha}(x)~=~& \phi^{\prime\alpha}(x^{\prime}) - \phi^{\alpha}(x)~=~\epsilon Y^{\alpha}\qquad \text{(full variation)}. \end{align} \tag{B}$$ Technically the calculations are a bit cumbersome since only the vertical transformation commutes with the total spacetime derivative $$[\delta_0, d_{\mu}]=0.\tag{C}$$ However, using the standard Noether formulas, one may show that

• The transformation (B) is a quasisymmetry for the action $\tilde{S}$ iff it is a quasisymmetry for the action $S$.

• In the affirmative case, the Noether current
  $$ \tilde{J}^{\mu}~=~J^{\mu} \tag{D}$$ and the Noether charge $$ \tilde{Q} ~=~Q\tag{E}$$ are unchanged.

--

$^1$ Point mechanics is just field theory in 0+1D, i.e. $x^{\mu}$ is just time $t$.

Answer 2

I think, as long as Schwartz Theorem holds, and partial derivatives commute, your $T$ is a symmetry of both. What is meant by saying that some Lagrangian $L$ is symmetric under a transformation mapping $q \mapsto q_{\epsilon} = q + \epsilon \eta $ etc., is that $ \delta L := \frac{dL(q_{\epsilon}, \dot{q}_{\epsilon}, t_{\epsilon})}{d\epsilon} |_{\epsilon =0} = \frac{dR}{dt} $ (it really is a statement about how only in first order the Lagrangian changes). Now given that for your Lagrangian $\frac{dL(q_{\epsilon}, \dot{q}_{\epsilon}, t_{\epsilon})}{d\epsilon} |_{\epsilon =0} = \frac{dR}{dt}$, the variation for $L' = L + \frac{dF}{dt}$ is then $\frac{dL'(q_{\epsilon}, \dot{q}_{\epsilon}, t_{\epsilon})}{d\epsilon} |_{\epsilon =0} =\frac{dL(q_{\epsilon}, \dot{q}_{\epsilon}, t_{\epsilon})}{d\epsilon} |_{\epsilon =0} + \frac{d}{d\epsilon}\frac{dF}{dt}|_{\epsilon =0} =^{?} \frac{d}{dt} ( R + \frac{dF}{d\epsilon}|_{\epsilon = 0}) $. And as long as the last equality holds, you can consider $T$ being also a symmetry for $L'$, albeit having a different conserved quantity. (changing the differentiation order should not be an issue, the only worry is the evaluation at $\epsilon =0$ that is done, before $d/dt$...)

Answer 3

Adding a total derivative to a Lagrangian will change the conserved quantities found from Noether's (first) theorem, such as energy and momentum. Therefore even though the Euler-Lagrange equation does not change, the constants of motion will. For example, if you have the Klein Gordon scalar field Lagrangian,

\begin{equation} \mathcal{L}_{KG} = - \frac{1}{2} \partial_\mu \phi \partial^\mu \phi \end{equation}

From Noether's theorem you will get for the energy-momentum tensor,

\begin{equation} T_{KG}^{\gamma\rho} = - \frac{1}{2} \eta^{\gamma\rho} \partial_\mu \phi \partial^\mu \phi + \partial^\gamma \phi \partial^\rho \phi \end{equation}

Now we can add a total derivative to the Lagrangian, if we follow the total derivative added by Kuzmin and McKeon [1],

\begin{equation} \mathcal{L} = -\frac{1}{2} \partial_\mu \phi \partial^\mu \phi + \partial_\mu (\frac{1}{3} \phi \partial^\mu \phi) \end{equation}

Using this Lagrangian we get the so-called new improved energy momentum tensor of Callan-Coleman-Jackiw [2] directly from Noether's theorem,

\begin{equation} T^{\mu\nu}_{CCJ} = - \frac{1}{6} \eta^{\mu\nu} \partial_\alpha \phi \partial^\alpha \phi + \frac{1}{3} \eta^{\mu\nu} \phi \partial_\alpha \partial^\alpha \phi + \frac{2}{3} \partial^\mu \phi \partial^\nu \phi - \frac{1}{3} \phi \partial^\mu \partial^\nu \phi \end{equation}

which is an energy-momentum tensor that was obtained by Callan-Coleman-Jackiw by a improvement method not related to Noether's theorem. For both Lagrangians above, the same Euler-Lagrange equation of motion $EL = \square \phi$ is obtained.

This example is a good one for a couple of reasons; it shows that adding a total divergence to the Lagrangian will change the resulting conserved quantities obtained using Noether's theorem, and it shows that a total divergence added to a Lagrangian can be used to bypass improvement methods which are not related to Noether's theorem if one wishes to obtain specific conserved quantities directly from Noether's theorem.

References:

[1] https://journals.aps.org/prd/abstract/10.1103/PhysRevD.64.085009

[2] https://www.sciencedirect.com/science/article/abs/pii/0003491670903945