1
$\begingroup$

We are doing Lorentz-invariant Lagrangian field theory in Minkowski spacetime, and I'm now considering only the form of Noether's theorem where the fields are varied.

Assume that $\delta \phi$ is a symmetry of the Lagrangian. The variation is of the form $\phi_\epsilon(x)=\phi(x)+\epsilon a\delta\phi(x)$, where $\epsilon$ is the continuous parameter and $a$ is just a constant separated (because I dislike the "infinitesimal" formulation, so I'll make $a$ a function later, instead of $\epsilon$).

It is known that one way of obtaining the Noether current is to consider $a=a(x)$ a function of spacetime. If $\psi_\epsilon$ is a symmetry with constant $a$, then it no longer will be when we make it a function. The variation in the action is then of the form $$ \delta S=\int d^4x\ \mathfrak J^\mu\partial_\mu a(x),$$ since if $a$ is a constant, then the variation should vanish. One may then show that $\mathfrak J$ is the same current as one'd obtain via the normal means (with that said, I don't think I have ever seen an explicit derivation of this anywhere).

On the other hand, the conservation of this current comes from the fact that on-shell, $\delta S$ should vanish for all variations, so partially integrating, we get $$ \delta S=0=-\int d^4x\ \partial_\mu\mathfrak J^\mu a, $$ and since this is true for all $a$s, we get $\partial_\mu\mathfrak J^\mu=0$.

Preliminary question: Partial integration implies that $a$ is a function which vanishes on the boundary - possibly at infinity. However we have arrived at the expression by considering that the variation for constant $a$ is a symmetry. However the need for $a$ to vanish on the boundary, and the need for $a$ to have a well defined "constant limit" is conflicting. If both is to be kept, the constant $a$ has to zero, however then there is no content in variation with respect to constant $a$ being a symmetry, since if $a$ is zero, it is trivially a symmetry. How to resolve this?$\square$

Many sources also use this procedure for the Stress-Energy-Momentum tensor (SEM tensor from now on), even though

  • the original translation symmetry is a quasisymmetry, not an exact symmetry;

  • the SEM tensor arises from changes in the coordinates, not changes in the fields.

I don't know how quasisymmetries can be incorporated into this scheme, but I do know that for constant-parameter translations, eg. $x^\mu_\epsilon=x^\mu+\epsilon a^\mu$ ($a$ const), the fields do not transform "internally", and the volume element is unaffected, so a constant parameter translation is effectively equivalent to the field transformation $$ \phi_\epsilon(x)=\phi(x)+\epsilon a^\nu\partial_\nu\phi(x). $$

So my logic is that the Lagrangian formulation shouldn't care about whether this variation actually comes from the variation of the coordinates, or variations of the form $\delta\phi=a^\mu\partial_\mu\phi$. This'll be a point I'll come back later.

So dazed and confused, I try to explicitly apply this "local parameter" scheme to the SEM tensor. The variation of the derivative when we consider $a^\mu$ to be a function is then $$ \delta\partial_\mu\psi=a^\nu\partial_\nu\partial_\mu\phi=\partial_\mu(a^\nu\partial_\nu\phi)-\partial_\mu a^\nu\partial_\nu\phi. $$

Now, the variation of the Lagrangian is equivalent to $$ \delta\mathcal L=a^\nu\partial_\mu\mathcal L=\partial_\nu(a^\nu\mathcal L)-\partial_\mu a^\nu\delta^\mu_\nu\mathcal L, $$ but calculating it piece by piece gives $$ \delta\mathcal L\approx\partial_\mu\left(\frac{\partial\mathcal L}{\partial\partial_\mu\phi}a^\nu\partial_\nu\phi\right)-\frac{\partial\mathcal L}{\partial(\partial_\mu\phi)}\partial_\nu\phi\partial_\mu a^\nu, $$ where $\approx$ means I have thrown out the Euler-Lagrange terms, which vanish on-shell.

Several things of note:

  • If we order the total divergences to one side, it is precisely the divergence of the SEM tensor. So it can be assumed to vanish on-shell. But if we use this scheme to derive the SEM tensor, we cannot know that.

  • Because of this, I'll simply throw them out as boundary terms (I have only considered the variation of the Lagrangian, but we can now assume the action integral around it is there), which implies that $a^\mu(x)$ vanishes at infinity/the boundary.

Throwing away the boundary terms we have $$ \delta S=-\int d^4x \partial_\mu a ^\nu \delta^\mu_\nu \mathcal L, $$ but also $$ \delta S=-\int d^4x\ \partial_\mu a ^\nu \frac{\partial\mathcal L}{\partial\partial_\mu\phi}\partial_\nu\phi. $$

The problem is that if I now assume that since we are perturbing around a $\phi$ that satisfies the EOMs, the variation must vanish identically, therefore we get that both terms involved in the SEM tensor must vanish.

On the other hand, if I order them to one side, we get $$ 0=\int d^4x\ \partial_\mu a^\nu\Theta^\mu_{\ \nu}. $$ Since $a^\mu$ is arbitrary, it's derivative is resonably arbitrary, so it seems to me this implies that the SEM tensor vanishes, rather than its divergence (EDIT: In a flash of insight I realized this reasoning of mine is incorrect, so this is a nonissue).

Once again, on the other hand, partially integrating, we get $$ 0=-\int d^4x \partial_\mu\Theta^\mu_{\ \nu}a^\nu, $$ from which the conservation of $\Theta$ follows as it should.

Question: Essentially, is this derivation correct? I am asking, because though I have gotten the intended solutions, I feel there have been a rather large amount of handwave-y steps along the way. I feel the throwing out of the boundary terms is unjustified, and I could have also gotten the conclusion that $\Theta$ itself vanishes.

Furthermore, we never actually used the fact that $\delta S$ vanishes because we are varying a stationary point, but all of literature that discusses this "trick" ("trick" = making the constant symmetry parameter a function) mentions it is necessary.

Is this correct? Is there a better way to obtain this? What about the disposability of boundary terms as per "Question 1"?

Bonus question: I know I shouldn't be asking too many questions in one post, but I feel this strongly connects here.

I know that there are formulations of Noether's theorem, in which the coordinates are also varied. Apparantly "the trick" works there too (see di Francesco - Conformal Field Theory - though without much explicit derivations sadly), however if I considered the variation $x^\mu_\epsilon=x^\mu+\epsilon a^\mu(x)$ fundemental, I would have had to include the variation of the Jacobian determinant, and took into account the "internal" transformation of the field $\phi$. In a Lorentz-invariant theory, what we mean under a field's transformation by a general coordinate transform (which $x^\mu_\epsilon$ is, in infinitesimal form) is quite frankly, up to debate.

So I find it incredulous that both ways lead to the same result. Is there any quick/succint answer to why does both work, even though we vary a lot more things in the second approach?

$\endgroup$

closed as too broad by Qmechanic Jan 24 '18 at 12:24

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The title (v1) seems to be a misnormer: The post seems to be about the Noether $x$-dependence trick rather than actually gauging a global symmetry into a local symmetry. It is still only about Noether's 1st theorem; not Noether's 2nd theorem. Related: physics.stackexchange.com/q/99853/2451 and links therein. $\endgroup$ – Qmechanic Jan 24 '18 at 10:44
  • $\begingroup$ I'm closing this post (v1) as too broad, cf. this meta post. To reopen it, for starters, consider to ask the preliminary subquestion separately. $\endgroup$ – Qmechanic Jan 24 '18 at 12:24