Time derivative of Noether charge

Question

I understand that the Noether charge can be written as $$ Q= \int~\mathrm d^3 x ~J^0$$ and the time derivative of the Noether charge is zero $$ \dot Q=0 $$

but how would you explicitly calculate it?

Answer 1

While Sebastian nailed it in his comment, I'll post the same as a proper answer.

The four-vector current $J^{\mu} = (J^0, J^i) \equiv (J^0, {\vec J})$ obeys the continuity equation $$\partial_{\mu} J^{\mu} = 0,$$ or in other words: $$\frac{\partial J^0}{\partial t} + {\vec \nabla}\cdot{\vec J} = 0$$ This represents the statement of local conservation of the Noether charge (whatever that may be: electric charge, baryon number, or simply any Noether charge).

The next bit is global conservation. The same follows from the above by integrating both sides over entire space, which yields: $$\frac{\partial}{\partial t} \left(\displaystyle\int~\mathrm d^3 x ~J^0 \right)+ \left(\int ~\mathrm d^3 x \ {\vec \nabla}\cdot{\vec J} \right)= 0$$ The second term on the right reduces to a surface integral with the use of the divergence theorem, and the first term in simply $\partial Q/\partial t$. Now, since the surface integral is on the surface enclosing the volume, and we have taken the integration over entire space, we are talking about a surface integral at $r \to \infty$. Thus, the behavior of the surface integral $\displaystyle\int \left({\vec J}\cdot {\hat n}~ \mathrm dS \right)$ depends on the behavior of ${\vec J}$ as a function of $r$. In fact, by substituting for the area element, $\mathrm dS = r ~\mathrm dr ~\mathrm d\theta$, it may be verified that this surface term would $\to 0$, as $r \to \infty$, provided $J$ falls faster than $1/r^2$. In this case, and only in this case, the global conservation of the Noether charge holds: $$\frac{\partial Q}{\partial t} = 0\,.$$