2
$\begingroup$

Noether's theorem for continuous symmetries gives rise to a conserved charge. Mathematiccally, what does it mean for a charge to be (i) locally and (ii) globally conserved? I think, a locally conserved charge needs not arise only from a local gauge symmetry. For example, conservation of energy is a local conservation law which arises from translational invariance in time.

$\endgroup$
  • $\begingroup$ Are you suggesting that time translation invariance is a global symmetry? It's not. That's why energy isn't globally conserved $\endgroup$ – Jim Oct 12 '16 at 15:22
  • $\begingroup$ @Jim- Poincare symmetries are not gauge symmetries. If the parameters used to represent the transformations of a group are independent of spacetime, then that transformation is global. Right? $\endgroup$ – SRS Oct 12 '16 at 15:28
6
$\begingroup$

A quantity $Q$ with a density $\rho(t,x)$ (such that $Q(V) = \int_V \rho(t,x)\mathrm{d}x$ is the amount of $Q$ inside $V$) is globally conserved if the global amount of $Q$ is constant, i.e. $\int\rho(x)$ does not depend on time if we integrate over the entire space $M$: $$\frac{\mathrm{d}}{\mathrm{d}t}\int_M \rho(t,x)\mathrm{d}x=0\tag{1}$$ Such a quantity is locally conserved if for every volume $V$ we have that $$\frac{\mathrm{d}}{\mathrm{d}t}\rho = \nabla j,\tag{2}$$ where $j$ is the current density of $Q$, i.e. the function says how much of $Q$ crosses a given unit area in a given unit time.

Noether's theorem says that to every quasi-symmetry of the action there exists a locally conserved quantity. The notion of "local conservation" is unrelated to the notion of "local symmetries" (i.e. gauge symmetries).

There is a slightly different notion of local vs. global conservation in the context of e.g. general relativity (The following is paraphrased from this article by Baez, for more discussion on global conservation of energy in GR, see this question and its linked questions): Here, you'll often find people stating that something (like energy or momentum) is locally but not globally conserved, and what it means is that while the differential statement like $\dot{\rho} = \nabla j$ holds, its classically equivalent integral formulation $$\int_V \rho(t_1,x)\mathrm{d}x - \int_V \rho(t_2,x)\mathrm{d}x = \int_{\partial V}\int_{t_0}^{t_1} \rho(t,x)\mathrm{d}t\mathrm{d}S\tag{3}$$ doesn't hold if $\rho,Q$ are not scalars, but instead vectors/tensors as in the case of the energy-momentum tensor. In this case, the differential equation is a "local" statement, and the integral formulation is a "global" statement.

$\endgroup$

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.