It is a well known fact that the gravitational field (in General Relativity and direct generalizations of it) has no local energy-momentum density.

Usually there are two reasons stated, one is heuristic, one is mathematical.

  • Heuristic reasoning: The gravitational field can be "turned off" at any one $x\in M$ by a proper choice of reference frame, which implies there are problems defining a local density.

  • Mathematical reasoning: The (Einstein-Hilbert) stress-energy (SEM) tensor of a matter field $\psi$ is defined as $$ T_{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g^{\mu\nu}} $$ where $S_m$ is the matter action. It is, in some sense, the response of the matter field action to changes in the metric. On the other hand, the response of the gravitational action to a change in the metric is $$ \frac{1}{16\pi G}G_{\mu\nu}=\frac{\delta S_{EH}}{\delta g^{\mu\nu}}, $$ which is basically the vacuum EoMs of GR, clearly this is not to be interpreted as a SEM tensor. So the gravitational field has no SEM tensor.

Moreover, the covariant conservation law $$ \nabla_\mu T^{\mu\nu}=0 $$ is usually interpreted not as a genuine conservation law, but rather the exchange of energy-momentum between the matter field and the gravitational field.


Let $\psi$ be a matter field, whose target space carries a representation of a Lie group $G$, with action $$ S_m [\psi]=\int d^4x\ \mathcal L_m(\psi,\partial\psi)$$ such that the Lie group $G$ is a group of symmetries of $S_m$. From Noether's theorem, one may obtain $k$ (we have $h=\dim G$) conserved currents satisfying $$ \partial_\mu\mathcal J^\mu _a=0. $$

Let us turn the global symmetry into a gauge symmetry, introduce the gauge connection/gauge field $\mathcal A_\mu=A^a_\mu T_a$ ($T_a$ are a set of generators for $\mathfrak g$) through the covariant derivative $$ D_\mu\psi=\partial_\mu \psi + \mathcal A_\mu\psi, $$ and introduce a self-action $$ S_{GF}[A]=\int d^4 x\ \mathcal L_{GF}(A,F) $$ for the gauge field.

One may show the following:

  • The current $$ j^\mu_a=\frac{\delta S_m}{\delta A^a_\mu} $$ is an equivalent current to the Noether current $\mathcal J^\mu _a$ provided that $A\rightarrow 0$. This includes the conservation law $\partial_\mu j^\mu_a=0$.

  • With $A_\mu^a\neq 0$, the current satisfies the identity $$ D_\mu j^\mu_a=\partial_\mu j^\mu_a-A_\mu^cC_{c\ a}^{b}j^\mu_b=0, $$ where $ D_\mu $ is the covariant derivative in the coadjoint bundle.


The situation here is clearly analogous to the situation with GR but with $T^{\mu\nu}$ replaced with $j^\mu_a$. In particular:

  • The conservation law $D_\mu j^\mu_a=0$ cannot be interpreted as a real conservation law, it expresses the fact that the matter field can exchange charge with the gauge field.

  • It does not make sense to define a gauge current for the gauge field $A$, only for the matter field $\psi$.

  • Does that mean that in a nonabelian gauge theory, the local charge current density of the gauge field is ill-defined?

If so, I would find it odd, since it is not something that's usually stated, plus unlike gravity, where you can "turn off" the "field strength" $\Gamma^\rho_{\mu\nu}$ by a choice of reference frame at a point, the gauge curvature $\mathcal F=d\mathcal A+\frac{1}{2}[\mathcal A,\mathcal A]$ cannot be eliminated at points via gauge transformations.

  • $\begingroup$ Related: physics.stackexchange.com/q/137061/2451 $\endgroup$
    – Qmechanic
    Sep 20, 2018 at 14:32
  • $\begingroup$ @Qmechanic I do not understand point 5) in your answer to the question you linked. In particular, it can be shown that the "full" current $j^\mu_a=\delta S_m/\delta A^a_\mu$ transforms under the coadjoint representation of $G$, so it is gauge-covariant, and it does satisfy a conservation law $D_\mu j^\mu_a$, albeit one with covariant derivatives instead of ordinary derivatives. Or do I misunderstand something? $\endgroup$ Sep 20, 2018 at 15:42


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