# Conserved charges given conserved current via Noether's theorem

Let $j^{\mu}_{a}$ be the conserved current associated with an infinitesimal symmetry transformation, cf. Noether's theorem. The conserved charge associated with $j^{\mu}_{a}$ is $$Q_a = \int d^{d-1}x j^{0}_a,\,\,\,\,\,\,\,(1)$$ where $j^0_a$ is the time component of $j^{\mu}_a$ and $d^{d-1}x$ is the spatial integration measure.

The time derivative is $$\dot{Q_a} = \int d^{d-1}x \partial_{0} j^{0}_a = -\int d^{d-1}x \partial_{i}j^i\,\,\,\,\,\,(2)$$ which vanishes by the generalized divergence theorem.

I was wondering if someone could explain how equations (1) and (2) are arrived at. I am wondering if in eqn (1), the author skipped a step and by integrating over the timelike component of $j$ reduced the integral to an integral purely over space and by setting $\mu = 0$ as a result of already integrating over the time component.

I guess that $\dot{Q_a} = \frac{d}{dt}Q_a$ and since $j$ is a function of coordinates, when that derivative is brought into the integral, it turns to a partial and the subscript $0$ denotes a time derivative?

Many thanks for any clarity on the above.

A $j^\mu_a$ current is conserved if it satisfies the equations $$\partial_\mu j^\mu_a = 0 \implies \partial_t j^0_a + \partial_i j^i_a = 0 ~~~~~~~~~~~...... (1)$$ The first equation in your question is simply a definition. We can define an object called "charge" as $$Q_a = \int d^{d-1} x j_a^0$$ Conservation of the current then implies that the charge does not change with time, i.e. $$\frac{dQ_a}{dt} = \frac{d}{dt} \int d^{d-1} x j_a^0 = \int d^{d-1}x \partial_t j_a^0 = -\int d^{d-1}x \partial_i j_a^i ~~~~~~~~ (2)$$ The last equality above was obtained from (1). Now, the last quantity is a boundary integral (by Stokes theorem). It is assumed in field theory that the currents vanish at the boundaries and therefore it is zero. We therefore find $$\frac{dQ_a}{dt} = 0$$ Whenever this is true, we say that the charge $Q_a$ is conserved.
• I assumed (maybe incorrectly) that we were working in Minkowski space where the metric is $g_{\mu\nu} = \eta_{\mu\nu} = diag(-1,1,1,1)$. If you are working in Euclidean space then $g_{\mu\nu} = \delta_{\mu\nu} = diag(1,1,1,1)$. In this case, there is no minus sign in (1) and there is an extra minus sign in the last equality in (2). – Prahar Jun 13 '14 at 15:33
• How is the metric manifest in the equation $\partial_{\mu}j^{\mu}_a = 0$ though? You assumed correctly, but I don't see how it fits into the equation. Thanks! – CAF Jun 13 '14 at 15:37