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Noether's theorem gives rise to quantities that are conserved over time. But does it also give rise to quantities that are conserved over space?

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In basic Lagrangian mechanics (of the sort that is covered in a sophomore-level classical mechanics class), no it doesn't. The reason is that time plays a special role in the basic Lagrangian theory: it's the only independent parameter, which everything else is expressed as a function of. This is related to the fact that the action is the integral of the Lagrangian over time, not space, and that in turn means that the "classical" version of Noether's theorem only works for conservation in time.

However, when you generalize to field theory, the situation changes: in field theory, both the time and space coordinates are considered to be independent parameters, so that everything else is expressed as a function of both time and space. In particular, instead of the classical Lagrangian, you have a Lagrangian density $\mathcal{L}$, which lets you express the action as a spacetime integral,

$$S = \int \mathcal{L}\mathrm{d}^4x$$

So the field theory version of Noether's theorem doesn't assign any special status to time. Instead of temporal conservation laws ($\frac{\mathrm{d}Q}{\mathrm{d}t} = 0$), it gives you spacetime conservation laws of the form

$$\frac{\partial j^\mu}{\partial x^\mu} = 0$$

You can turn this into a temporal conservation law by integrating the current $j^\mu$ over a spacelike volume (i.e. all of space at a single moment in time):

$$Q = \int_{t=\text{const}} g_{\mu\nu} j^\mu\mathrm{d}x^\nu\quad\to\quad\frac{\mathrm{d}Q}{\mathrm{d}t} = 0$$

but you can just as well turn it into a spatial conservation law by integrating over a spatiotemporal volume:

$$Q = \int_{x=\text{const}} g_{\mu\nu} j^\mu\mathrm{d}x^\nu\quad\to\quad\frac{\mathrm{d}Q}{\mathrm{d}x} = 0$$

So in this way, yes it is possible to create a spatial conservation law using Noether's theorem.

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You say: "In nonrelativistic classical mechanics, no it doesn't". Why can't you have a Lagrangian density over Galilean space-time with a Euclidean metric and from this derive conserved quantities over space? – John McVirgo Jul 10 '11 at 0:04
@John: because classical mechanics doesn't use the concept of the Lagrangian density. That's an idea from field theory. Then again, the distinction is pretty artificial... I guess what I wanted to say was that it isn't done that way in classical physics, though you're right that it could be. I'll think of some way to rephrase that. – David Z Jul 10 '11 at 0:08
The Lagrangian density is used in classical mechanics, such as for an elastic material. You can use v << c for non-relativistic mechanics to simplify the relativistic expressions for the conserved quantities which shows they're equivalently all independent of time. – John McVirgo Jul 10 '11 at 0:47
OK, well maybe "classical" isn't the right word then. "Particle mechanics" perhaps. – David Z Jul 10 '11 at 2:49

Yes, in a field-theoretic setting. Consider a $3+1$ dimensional flat space-time for simplicity. Noether's theorem gives rise to a conservation law of the form

$$d_{\mu} J^{\mu}(x)=0,$$

i.e. the Noether current $J^{\mu}(x)$ is a divergence-less four-current. [We use the symbol $d_{\mu}$ (rather than $\partial_{\mu}$) to stress the fact that the derivative $d_{\mu}$ is a total derivative, which involves both implicit differentiation through the field variables $\phi^{\alpha}(x)$, and explicit differentiation wrt. $x^{\mu}$.]

Say we want to consider a quantity $Q$ that is independence of the $x^1$-coordinate. Define the so-called 'charge' $Q=Q(x^1)$ by integrating $J^1(x)$ over the $2+1$ dimensional plane with fixed $x^1$-coordinate $x^1$. Then it is easy to show using a $2+1$ dimensional divergence theorem that

$$d_{1} Q(x^1)=0$$

by imposing pertinent boundary conditions.

The above construction may be vastly generalized to a geometrically covariant framework, where the preferred direction is given by a non-vanishing vector field, which can be time-like, space-like, light-like, or some combination.

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