Put into one sentence, Noether's first Theorem states that a continuous, global, off-shell symmetry of an action S implies a local on-shell conservation law. By the words on-shell and off-shell are meant whether Euler-Lagrange equations of motion are satisfied or not.
Now the question asks if continuous can be replace by discrete?
It should immediately be stressed that Noether Theorem is a machine that for each input in form of an appropriate symmetry produces an output in form of a conservation law. To claim that a Noether Theorem is behind, it is not enough to just list a couple of pairs (symmetry, conservation law).
Now, where could a discrete version of Noether's Theorem live? A good bet is in a discrete lattice world, if one uses finite differences instead of differentiation. Let us investigate the situation.
Our intuitive idea is that finite symmetries, e.g., time reversal symmetry, etc, can not be used in a Noether Theorem in a lattice world because they don't work in a continuous world. Instead we pin our hopes to that discrete infinite symmetries that become continuous symmetries when the lattice spacings go to zero, can be used.
Imagine for simplicity a 1D point particle that can only be at discrete positions qt∈Za on a 1D lattice Za with lattice spacing a, and that time t∈Z is discrete as well. (This was, e.g., studied in J.C. Baez and J.M. Gilliam, Lett. Math. Phys. 31 (1994) 205; hat tip: Edward.) The velocity is the finite difference
vt+12:=qt+1−qt∈Za,
and is discrete as well. The action S is
S[q]=∑tLt
with Lagrangian Lt on the form
Lt=Lt(qt,vt+12).
Define momentum pt+12 as
pt+12:=∂Lt∂vt+12.
Naively, the action S should be extremized wrt. neighboring virtual discrete paths q:Z→Za to find the equation of motion. However, it does not seem feasible to extract a discrete Euler-Lagrange equation in this way, basically because it is not enough to Taylor expand to the first order in the variation Δq when the variation Δq∈Za is not infinitesimal. At this point, we throw our hands in the air, and declare that the virtual path q+Δq (as opposed to the stationary path q) does not have to lie in the lattice, but that it is free to take continuous values in R. We can now perform an infinitesimal variation without worrying about higher order contributions,
0=δS:=S[q+δq]−S[q]=∑t[∂Lt∂qtδqt+pt+12δvt+12]
=∑t[∂Lt∂qtδqt+pt+12(δqt+1−δqt)]
=∑t[∂Lt∂qt−pt+12+pt−12]δqt+∑t[pt+12δqt+1−pt−12δqt].
Note that the last sum is telescopic. This implies (with suitable boundary conditions) the discrete Euler-Lagrange equation
∂Lt∂qt=pt+12−pt−12.
This is the evolution equation. At this point it is not clear whether a solution for q:Z→R will remain on the lattice Za if we specify two initial values on the lattice. We shall from now on restrict our considerations to such systems for consistency.
As an example, one may imagine that qt is a cyclic variable, i.e., that Lt does not depend on qt. We therefore have a discrete global translation symmetry Δqt=a. The Noether current is the momentum pt+12, and the Noether conservation law is that momentum pt+12 is conserved. This is certainly a nice observation. But this does not necessarily mean that a Noether Theorem is behind.
Imagine that the enemy has given us a global vertical symmetry Δqt=Y(qt)∈Za, where Y is an arbitrary function. (The words vertical and horizontal refer to translation in the q direction and the t direction, respectively. We will for simplicity not discuss symmetries with horizontal components.) The obvious candidate for the bare Noether current is
jt=pt−12Y(qt).
But it is unlikely that we would be able to prove that jt is conserved merely from the symmetry 0=S[q+Δq]−S[q], which would now unavoidably involve higher order contributions. So while we stop short of declaring a no-go theorem, it certainly does not look promising.
Perhaps, we would be more successful if we only discretize time, and leave the coordinate space continuous? I might return with an update about this in the future.
An example from the continuous world that may be good to keep in mind: Consider a simple gravity pendulum with Lagrangian
L(φ,˙φ)=m2ℓ2˙φ2+mgℓcos(φ).
It has a global discrete periodic symmetry φ→φ+2π, but the (angular) momentum pφ:=∂L∂˙φ=mℓ2˙φ is not conserved if g≠0.
This post imported from StackExchange Physics at 2015-10-04 21:40 (UTC), posted by SE-user Qmechanic