# Is there a higher dimension analogue of Noether's theorem?

So I have recently read the proof of Noether's theorem from the book variation calculus by Gelfand. Basically, what I have already seen is that for any single integral functional, if we have a transformation that keeps the functional invariant, we can derive a quantity that doesn't change along any solution of the Euler equations of the functional.

My question: Is there an analogue that work for multiple integral functional? That is, the corresponding system of Euler Lagrange equations are not ODEs, but rather PDEs. Can we define a quantity that is invariant on the whole solutions of these PDEs? The same argument used in Gelfand's proof for single integral functional clearly doesn't work. Can we have something that doesn't change not only with respect with one variable t, but is unchanged everywhere on the whole space like $$R^n$$ as long as we have a killing vector field for the functional?

• The argument goes through exactly the same with more dimensions. Noether's theorem is routinely used in field theory, which takes place in 3+1 dimensions. – Javier Mar 16 '19 at 3:12
• Another way of generalizing is to consider, instead of point charges arising from the Noether theorem, some spatially extended objects. Look up "generalized global symmetries". – mavzolej Mar 16 '19 at 7:39

In field theory, you often consider "Lagrangian densities" which are to be integrated over space-time instead of just over time.

For example, where as in the one dimensional case you would write

$$S = \int dt L$$ in field theory you would write $$S = \int d^4 x \mathcal{L}.$$ The equation of motion will be a PDE.

Noether's theorem, instead of giving you a conserved quantity $$Q$$ which satisfies $$\dot Q = 0$$, would now give you a conserved current $$J^\mu$$ (where $$\mu = 0, 1, 2, 3$$ and $$\mu = 0$$ is the time component and $$\mu = 1,2,3$$ are the space components) which satisfies $$\sum_\mu \frac{d}{d x^\mu} J^\mu = 0$$. You can still also find the a conserved quantity $$Q$$, which satisfies $$\dot Q = 0$$, defined by

$$Q = \int d^3 x J^0$$ and integrating over any fixed time.

• Thank you for your helpful answer. According to your answer, Noether's theorem doesn't give a quantity Q such that $dQ/dx^𝜇$ = 0 for all 𝜇 = 0,1,2,3,? . But, Noether's theorem only give 4 quantities $𝐽^𝜇$ that when taken derivative with respect to $x^𝜇$, the sum of the 4 derivatives is 0? I can understand your last argument about defining a quantity that 𝑄˙=0. – zhongyuan chen Mar 16 '19 at 16:01

Yes, already Noether herself in her seminal 1918 paper generalized her 1st & 2nd theorem to field theories in $$n$$ dimensions.