Noether charge in multisymplectic geometryby ygor.geurts@gmail.com Tags: charge, geometry, multisymplectic, noether 
#1
Jun808, 05:00 AM

P: n/a

Hi,
I'm currently looking for the mathematical foundation behind the claim, often found in field theory/string theory books that the noether charge associated to a symmetry of the lagrangian is the generator of that symmetry, ie. its poisson bracket with a field from the lagrangian, generates the change in the field. Recently I came across the article 'Covariant Momentum maps & field theory part I/II' by Gotay and Marsden. In there I've found that the concept of a Noether current is in fact given by the covariant momentum map in the multisymplectic classical field theory formulation. In particular, this map satisfies the mathematical analog of noether's theorem as explained in the article. In part II of the article they define an instantaneous momentum map, loosely defined by the integral of the covariant momentum map over a cauchy hypersurface. My question is now, whether this is the mathematical object that corresponds to the Noether charge (which in physics is defined as the integral of the time component of the current over a spacelike hypersurface)? Moreover, if that is so, somehow I still can't find an explicit demonstration (in the above article or any other mathematical article) that this charge then generates the transformation, ie is somehow related to the generators of the Lie algebra of the Lie group that encodes the symmetry. Could anyone help me out on clarifying these matters? Thanks in advance, Ygor 


#2
Jun1008, 05:00 AM

P: n/a

The Noether charge is not the integral of the timelike component of a
density. It is usually presented that way because you're in flatspace and the Cauchy surfaces normally used there are orthogonal with respect to a fixed timelike direction. On the contrary, it's the integral of a *3form* density. If Q is the charge, then the current density would be J^m, putting J in script to emphasize that it's a tensor density. It's integrated against the 3 form basis (d^3x)_{m}. For surfaces normal to the timelike direction futureoriented, the 3form corresponding to the surface would be (dx ^ dy ^ dz) = (d^3x)_0, in Minkowski coordinates. More generally, you'd have (d^3x)_{m} = 1/6 epsilon_{mnrs} (dx^n ^ dx^r ^ dx^s). Moreover, the charge is not a functional of Cauchy surfaces. It's merely presented that way, since the primary context of the concept is the Noether theorem and all matters related to it. It's a functional of 3submanifolds (and more generally, 3chains). So, if S is a 3 manifold (or 3chain) then the charge associated with Q(S) would be the integral of the 3form J = J^m (d^3x)_m over S. The *actual* content of the Noether theorem merely states that Q(S) = Q(S') for any two 3surfaces that comprise the 2 parts of the boundary of a 4region; dO = S  S'. They can even be compact. In fact, I think that's the only place where it's truly rigorous. If you were to take O to be the noncompact region wedged between two noncompact 3surfaces S and S', then you'd have to add in various technical assumptions about how everything behaves in the neighborhood of infinity. 


#3
Jun1108, 05:00 AM

P: n/a

On Jun 7, 3:49 pm, ygor.geu...@gmail.com wrote:
> Hi, > > I'm currently looking for the mathematical foundation behind the > claim, often found in field theory/string theory books that the > noether charge associated to a symmetry of the lagrangian is the > generator of that symmetry, ie. its poisson bracket with a field from > the lagrangian, generates the change in the field. This theorem goes back to Hamiltonian mechanics. It generalizes to field theory in a straightforward way. Unfortunately, I haven't found an explicit demonstration of it in any of my classical mechanics books. The proof is not particularly complicated, so here it is. Consider a Lagrangian function L(v,x) defined on a configuration space with coordinates x^i. Let X^i be the coordinate components of the vector field that generates a symmetry of the Lagrangian. Then, there exists a conserved quantity I(v,x) = L,i X^i. The notation L,ij,klm will represent derivatives, with this example representing derivatives of L with respect to v^i, v^j, x^k, x^l and x^m. The velocity varies under the symmetry transformation as (' is time derivative): (x^i + eps X^i + ...)' = v^i + eps X^i,k v^k + ... . Invariance of the Lagrangian dictates L,,i X^i + L,i X^i,k v^k = 0 . (*) The EulerLagrange equations are (L,i)' = L,,i . It is easy to show that I(v,x) is a constant of motion: I' = (L,i)' X^i + L,i X^i,k v^k (**) = L,,i X^i + L,i X^i,k v^k = 0 by (*) . That's the first half of the theorem you wanted. The other half of the theorem states that the conserved quantity I, as a function on the phase space, will generate the phase space extension of the configuration space vector field X^i. The phase space is coordinatized by x^i and p_j = L,j. In terms of the the x^i, v^j coordinates, the phase space extension of X^i is X = X^i @x^i + X^i,k v^k @v^i , where @x^i and @v^i are the coordinate basis for vector fields. Using the chain rule, we express this basis in terms of the @x^i, @p_j basis: [ @v^i ] [ L,ji 0 ] [ @p_j ] [ ] = [ ] [ ] . [ @x^i ] [ L,j,i 1 ] [ @x^i ] Therefore X = X^i ( L,j,i @p_j + @x^i) + X^i,k v^k L,ji @p_j = X^i @x^i + ( L,j,i X^i + L,ji X^i,k v^k ) @p_j = X^i @x^i + [ (L,,i X^i v^k + L,i X^i,k v^k),j  L,i X^i,j ] @p_j = X^i @x^i + [ 0  L,i X^i,j ] @p_j by (**) = X^i @x^i  p_i X^i,j @p_j . Using the symplectic form, we can transform this vector field into a 1form: w = X^i dp_i  p_i X^i,j (dx^j) = X^i dp_i + d(X^i) p_i = d(p_i X^i) = d(L,i X^i) = dI . In other words, the above demonstrates that X is precisely the Hamiltonian vector field generated by the phase space function I(p,x) = p_i X^i, which is presicely the same quantity as the conserved quantity I(v,x) obtained in the first half of the calculation. Hope this helps. Igor 


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