Why is dependence on derivatives not a problem in the definition of canonical energy-momentum tensor?

Question

Let $\mathcal L(\phi,\partial\phi)$ be a Lagrangian for a field $\phi$. It is known that the Lagrangian $\mathcal L$ and the Lagrangian $\mathcal L+\partial_\mu K^\mu$ produces the same physics, provided that $K$ depends on spacetime points through $\phi$ and $x$ only (eg. not through $\partial\phi$).

This can be seen because if the divergence term is varied, we get $$ \int_{\partial\mathcal D}d\sigma_\mu\delta K^\mu=\int_{\partial\mathcal D}d\sigma_\mu\left(\frac{\partial K^\mu}{\partial\phi}\delta\phi+\frac{\partial K^\mu}{\partial\phi_{,\mu}}\delta\phi_{,\mu}\right), $$ where I have assumed that $K$ also depends on $\phi_{,\mu}$. If only the first term was present in the brackets, then because $\delta\phi|_{\partial\mathcal D}=0$, the variation of this term would vanish. However because of the term proportional to $\delta\phi_{,\mu}$, this is no longer true.

On the other hand, if we consider invariance of a Lagrangian under spacetime-translations, because the Lagrangian is a scalar (at least in SR), under the transformation $x^\mu\mapsto x^\mu+\epsilon a^\mu$ it gets varied to $$ \delta \mathcal L=a^\mu\partial_\mu\mathcal L=\partial_\mu(a^\mu\mathcal L). $$ So in this case, $$ K^\mu=a^\mu\mathcal L, $$ but $\mathcal L$ depends on $\partial\phi$, so by the things said in the first part of this post, this is not a good $K^\mu$.

How to resolve this?

EDIT: Upon rereading my post, I realize I have been a bit too brief. To contextualize this better, a variation is a symmetry of the action if the Lagrangian gets varied to $\delta\mathcal L=\partial_\mu K^\mu$. This is a symmetry because of what I said in the first part.

This is utilized in the derivation of the canonical SEM tensor, as translations provide a symmetry of the Lagrangian because it gets varied to $\partial_\mu (a^\mu\mathcal L)$.

Answer 1

I restrict attention on the proof of the fact that $\phi(x)$ and $\phi(x+a)$ are simultaneously solutions of E.L. equations of ${\cal L}(\phi, \partial \phi)$. I.e., spacetime displacements are (dynamical) symmetries for ${\cal L}(\phi, \partial \phi)$. Otherwise the question is too vague.

I think this is not the right way to tackle the problem. You don't use in your attempt of proof the crucial hypothesis:

$\qquad \qquad \qquad \qquad \qquad$ ${\cal L}$ does not depend explicitly on $x$.

Without this fact it is false that the solutions of E-L equations (i.e. the stationary points of the action with standard boundary conditions) are preserved under spacetime translations.

The condition $${\cal L}'(x,\phi(x), \partial \phi(x))= {\cal L}(x,\phi(x), \partial \phi(x))+ \partial_\mu K^{\mu}(x,\phi(x))\tag{1}$$ is just sufficient to produce the same field equations for ${\cal L}$ and ${\cal L}'$. But it is not necessary.

Furthermore it works also if an explicit dependence on $x$ shows up, whereas the absence of $x$ is crucial here.

All that suggests that using (1) is not a good idea to prove that $\phi(x+ a)$ satisfies the same E.L. equations generated by $${\cal L}(\phi(x), \partial \phi(x))\quad \mbox{ (I stress that no explicit dependence on $x$ appears)}$$ if $\phi(x)$ does.

Instead, a proof of this fact entirely relies on

(i) $\frac{\partial}{\partial x^\mu} = \frac{\partial}{\partial (x+ a)^\mu}$

and

(ii) ${\cal L}$ does not explicitly depend on $x$.

Using them, it is easy to prove that $$\left.\left(\frac{\partial {\cal L}}{\partial \phi} - \frac{\partial}{\partial x^\mu}\frac{\partial {\cal L}}{\partial \partial_\mu \phi}\right)\right|_{\phi(x+a)} = \left.\left[\left.\left(\frac{\partial {\cal L}}{\partial \phi} - \frac{\partial}{\partial x^\mu}\frac{\partial {\cal L}}{\partial \partial_\mu \phi}\right)\right|_{\phi(z)}\right]\right|_{z=x+a}$$ The right-hand side vanishes for every value of $z$ by hypothesis so that $$\left.\left(\frac{\partial {\cal L}}{\partial \phi} - \frac{\partial}{\partial x^\mu}\frac{\partial {\cal L}}{\partial \partial_\mu \phi}\right)\right|_{\phi(x+a)} = 0\:.$$

Answer 2

1. At the heart of OP's question (v2) seems to be the fact that for a quasisymmetry
   $$\delta S ~\sim~ \int_V \mathrm{d}^nx~d_{\mu}k^{\mu}, $$ the $k^{\mu}$ functions are allowed to (and typically do) depend on derivatives of the fields $\phi$ without spoiling the conclusions of Noether's (first) theorem.

2. The Lagrangian density could in principle depend on higher derivatives, although there could be a price to pay, cf. e.g. this and this Phys.SE posts.

3. Adding total divergence terms to the Lagrangian density is also discussed in this Phys.SE post and links therein.

4. Specifically OP considers spacetime translation symmetry, i.e. energy-momentum conservation. This is e.g. discussed in this Phys.SE post and links therein.

Answer 3

I think @Qmechanic 's answer is great because it brings a lot of interesting topics to the table. Let me just add this.

Take this example: $L=\partial_\mu \phi \partial^{\mu} \phi $. Let's apply a finite translation: $x\rightarrow x-a$. The Lagrangian becomes $L \rightarrow L'= \partial_\mu \phi' \partial^{\mu} \phi' $ with $\phi'=\phi'(z)=\phi(x+a)$, $z=x+a$. You are free to write this as $L'= \partial'_\mu \phi' \partial'^{\mu} \phi'$ with $\partial'_\mu=\frac{\partial}{\partial z^\mu}$. Then, if we variate with respect to $\phi'$, we will reproduce the same equations of motion. But notice that from the point of view of the original field $\phi(x)$, this transformed Lagrangian $L'$ depends on all of its derivatives, since $\phi'=\phi(x+a)=\sum^{\infty}_{k=0} \frac{1}{k!} a^{k} \partial_{k} \phi(x)$.

As you mention, an infinitesimal transformation will generically insert higher order derivatives. Generically we can't expect this transformed quantity to vanish on-shell without the need for additional artifices/boundary conditions, as it was pointed out in the other answer and this should not surprise us. And, as this example points out, once we perform a symmetry transformation, there is no point in using the original fields (evaluated in the same position) to variate the action.