So I have the following curiosity: Consider for example, in QED, the quantity

$$ j^\mu\equiv\partial_\nu (\lambda(x) F^{\mu \nu}) $$

where $\lambda(x)$ is an arbitrary scalar function of spacetime, constructed from elements of the theory, e.g.

$$ \lambda(x)=A_\mu A^\mu F_{\rho \sigma} F^{\rho \sigma} $$

or anything else you can think of. Then, since $F^{\mu \nu}$ is antisymmetric, $\partial_\mu j^\mu=0$ identically.

In fact this can be generalized to any theory; construct from various elements an antisymmetric two-rank tensor, and a scalar, multiply them together, and there is a corresponding conserved current for any choice you make. If these currents are not trivial (e.g. giving only vanishing charges) then it appears that all theories give an infinite landscape of conserved currents. Is this so? Am I missing something? How is this justified logically?

  • Maybe this obvious, but why is $\partial_\mu j^\mu=(\partial_\mu\lambda)(\partial_\nu F^{\mu\nu}) = 0$ necessarily? – WAH Oct 16 at 19:26
  • It is $\partial_\mu \partial_\nu (\lambda F^{\mu \nu})=-\partial_\nu \partial_\mu (\lambda F^{\nu \mu})=-\partial_\mu \partial_\nu (\lambda F^{\mu \nu})$ where in the last equality I just renamed the dummy indices. – Valentina Oct 16 at 19:31
  • I agree that $(\partial_\mu\partial_\nu \lambda)F^{\mu\nu}=0$ and $\lambda \partial_\mu\partial_\nu F^{\mu\nu}=0$. So then $\partial_\mu\partial_\nu(\lambda F^{\mu\nu})=(\partial_\mu\lambda)(\partial_\nu F^{\mu\nu})$. It's not clear to me that the last quantity is always identically 0. – WAH Oct 16 at 19:36
up vote 5 down vote accepted

OP wrote (v2):

If these currents are not trivial [...]

In fact most$^1$ of OP's currents are trivial$^2$. The corresponding charges $$Q ~:=~ \int_V d^3x~ j^0~=~\int_V d^3x~\sum_{i=1}^3d_i(\lambda F^{i0})~=~\int_{\partial V} d^2x~(\ldots) ~=~0$$ vanish if the components $\lambda F^{i0}$ fall off fast enough $o(r^{-2})$ at spatial infinity $\partial V$.

--

$^1$ The constant case $\lambda=1$ corresponds to the conserved electric current, cf. Maxwell's equations. A non-constant term in $\lambda$ typically falls off too fast at spatial infinity $\partial V$ to produce a non-trivial conservation law.

$^2$ Nevertheless, in a general gauge theory, the second Noether identity does in fact lead to the existence of a superpotential with an infinite hierarchy of conserved surface charges at the spatial boundary, cf. e.g. this Phys.SE post.

  • So it seems to be, by the same token, that any current created by using a derivative has trivial charges. Anyway, I think this clarifies it, thanks. – Valentina Oct 16 at 19:49
  • However, how big or small is "most", anyways? Is it "most" as in "all but a finite number", or "most" as in some mathematician's sense like "of measure zero" which could still leave an uncountably infinite number of possible such non-trivial quantities? – The_Sympathizer Oct 17 at 2:15
  • I updated the answer. – Qmechanic Oct 17 at 5:45

If we have $\partial_\nu F^{\mu\nu}=0$ on-shell, as happens e.g. in EM without a source current, $F^{\mu\nu}X_\nu$ is conserved provided $\partial_\mu X_\nu$ is symmetric, as happens e.g. with $X_\nu=\partial_\nu\lambda$. Not all these conserved currents are trivial. See Sec. 2.1.2 of my thesis, which generalises this to curved spacetime.

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