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Noether charge in multisymplectic geometry

by ygor.geurts@gmail.com
Tags: charge, geometry, multisymplectic, noether
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ygor.geurts@gmail.com
#1
Jun8-08, 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

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Rock Brentwood
#2
Jun10-08, 05:00 AM
P: n/a
The Noether charge is not the integral of the time-like component of a
density. It is usually presented that way because you're in flat-space
and the Cauchy surfaces normally used there are orthogonal with
respect to a fixed time-like direction.

On the contrary, it's the integral of a *3-form* 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
future-oriented, the 3-form 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 3-submanifolds (and more generally, 3-chains). So, if S is a 3-
manifold (or 3-chain) then the charge associated with Q(S) would be
the integral of the 3-form
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 3-surfaces that comprise the 2 parts of the boundary
of a 4-region; 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 non-compact region wedged between two non-compact 3-surfaces
S and S', then you'd have to add in various technical assumptions
about how everything behaves in the neighborhood of infinity.

Igor Khavkine
#3
Jun11-08, 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 Euler-Lagrange 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
1-form:

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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