# Covariance of Noether's charge in GR

If we consider a theory of GR (the standard Einstein-Hilbert action) and a complex scalar field, we can easily see that we have a global $$U(1)$$ symmetry for the scalar field. Now, via Noether's theorem we can build a conserved current from which we obtain a conserved charge:

$$Q = \int d^3x \sqrt{-g}\, J^0.$$

The problem is that one cannot explicitly see whether this charge is actually a scalar under general coordinate transformations or not.

How could one prove it?

• The charge isn't even a scalar under Lorentz transformations, so I don't see how it can be invariant under general coordinate transformations. Jul 7, 2020 at 19:46

On a (pseudo-)riemannian manifold $$M$$, for a conservative vector field $$J^\mu , \ \nabla_\mu J^\mu = 0$$ (a "conserved current"), we have for its flux through the boundary of any submanifold $$S$$: $$\int_{\partial S} J^\mu n_\mu \mathrm{d}\mathbf{\sigma} = \int_S \nabla_\mu J^\mu \mathrm{d\mathbf{vol}} = 0$$ Where $$n^\mu$$ is the (outward-pointing) normal to the hypersurface $$\partial S$$ and $$\mathrm{d}\mathbf{\sigma}$$ is the volume form induced on it by the volume form $$\mathrm{d\mathbf{vol}}$$ of $$M$$ (which comes from the metric).
Then, by choosing a region $$S$$ whose boundary can be decomposed into 2 spacelike hypersurfaces $$\Sigma_1, \Sigma_2$$ (with $$\Sigma_2$$ being in the future light cone of $$\Sigma_1$$) joined by a timelike hypersurface $$T$$ (think a 4d cylinder with axis along the time direction), and taking the orientation of the normal to $$\Sigma_1$$ to be towards the future direction (thus the opposite orientation to the one it has as a piece of $$\partial S$$), we get: $$0 = \int_{\partial S} J^\mu n_\mu \mathrm{d}\mathbf{\sigma} = - \int_{\Sigma_1} J^\mu n_\mu \mathrm{d}\mathbf{\sigma} + \int_{\Sigma_2} J^\mu n_\mu \mathrm{d}\mathbf{\sigma} + \int_{T} J^\mu n_\mu \mathrm{d}\mathbf{\sigma}$$ Thus, calling $$Q_\Sigma$$ the flux of $$J^\mu$$ through the spacelike surface $$\Sigma$$: $$Q_{\Sigma_2} - Q_{\Sigma_1} = \int_{T} J^\mu n_\mu \mathrm{d}\mathbf{\sigma}$$ Then, if $$J^\mu$$ (or at least its flux) happens to be null on the timelike piece $$T$$ of the boundary, $$Q_{\Sigma_2} = Q_{\Sigma_1} = Q$$, and if we "glue together" more regions such that the past spacelike boundary of each is the future spacelike boundary of the previous one (and such that $$J_\mu$$ has zero flux on the timelike boundary of each), $$Q_{\Sigma_i} = Q$$ for all $$i$$. This is the conserved charge associated to the current $$J^\mu$$, and having obtained it in a coordinate-free way it's clearly invariant under coordinate transformations.
Computing $$Q_\Sigma$$ on a spacelike slice $$\Sigma_t$$ with constant time coordinate $$x_0 = t$$ gives the more familiar formula in your question, and if we have coordinates covering the whole spacetime we can take the family of surfaces $$\{\Sigma_t\}$$ and consider them connected by pieces of timelike surfaces "at spatial infinity" (where the current is usually taken to vanish), thus indeed for the charges $$Q_t$$ associated to this family of surfaces, $$\frac{\mathrm{d} Q_t}{\mathrm{d} t} = 0$$ holds.