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

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.

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 $$t$$.

• I'm sorry, I'm still not familiar with field theory (I have not reached such lessons yet). Could you please explain what this implies for my situation? Thank you so much!!!
– Oski
Dec 17, 2021 at 19:16

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 $$d/dt$$...)

• Could you please specify what you mean by $R$? I'm not familiar with this symmetry definition! Thank you :)
– Oski
Dec 17, 2021 at 19:46
• @Oski It means some arbitrary differential function of $t$; $dL/d\epsilon$ is allowed to be any such function's derivative.
– J.G.
Dec 17, 2021 at 20:23

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,

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

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

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

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

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

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

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

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.

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