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?

  • $\begingroup$ The charge isn't even a scalar under Lorentz transformations, so I don't see how it can be invariant under general coordinate transformations. $\endgroup$
    – Stratiev
    Jul 7 '20 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.