# Noether's Theorem and the associated Noether Charge

1. Apr 3, 2009

### Kooklin

I've been trying to solve the problem of deriving the conserved "Noether Charge" associated with a transformation q(t) --> Q(s,t) under which the Lagrangian transforms in the following way:

L--> L + df(q,t,s)/dt (i.e. a full time derivative that doesn't depend on dq/dt)

I am guessing I need to take d/ds [ L + df(q,t,s)/dt ] = 0, which would mimic the derivation of the "Noether Charge" when the transformation leaves L invariant, but I am running into difficulties. Is this the right approach?? any help would be much appreciated.

Thanks =)

2. Apr 3, 2009

### turin

Do you know what Q and f are?

You must also consider the abstract variation of L in terms of the abstract variation of q, and use the Euler-Lagrange equation to eliminate all but total time derivatives. Then, you equate this abract variation to the explicit variation that you have shown (with f). The difference in time derivatives vanishes, and the time derivative is linear, so you will then have an abstract expression for a conserved quantity (i.e. constant in time). To relate it to the concrete transformation, you must assume that the transformation is continuous, and take it to first order for small values of the transformation parameter. I think this last concept may be what you are missing.

3. Apr 5, 2009

### Kooklin

Well I guess I'm not sure exactly how one would go about doing what you suggest.

I recognize the steps that I saw in the derivation for a Lagrangian that is invariant under the said transformation, soI think I understand the idea, but I'm definitely missing something...does anyone know where I might be able to find a reference that explains this process well?

If not, I guess I have the following question: should I be varying the total time derivative at all in the way I am doing?

d/ds [ L + df(q,t,s)/dt ] = 0

and basically try to rewrite this as a d/dt [expression] = 0?

Am I also right in thinking that the associated conserved quantity should look like the Hamiltonian?

4. Apr 5, 2009

### turin

OK, but I hope that you realize that the Lagrangian is not necessarily invariant under the transformation, and that's why df/dt is there. So, the principle is basically to cancel the variation in L w.r.t. δq with the variation in L w.r.t. δs.

I think it's in Goldstein's grad level CM text, but I don't have that book so I'm not sure how good it is. (Wow, I really should buy that one; it's a classic.)

That last part is basically what you need to do, but I've never seen the approach quite like what you are doing. Your approach may actually be better than the way that I do it, but it is hard to tell. Does your L in that expression depend on s (i.e. through q)? If everything is linear in s, then I think it turns out to be the same way as I do it. If not, then maybe you should set s in the resulting expression to s0, that is, whatever value of s is required to make the transformation trivial (like in the first coefficient of a Taylor series).

I don't think so; what gives you that idea? In general, no. Of course, it depends on what you mean by "look like the Hamiltonian". For example, the free-particle theory is invariant to Q(s,t) = q(t) + s.q0, and df/dt = 0 . The conserved charge is q0.p(t), where p(t) is the canonical momentum ( p(t) ≡ ∂L/∂(dq/dt) ). I don't think that q0.p(t) looks like the free-particle Hamiltonian.

Last edited: Apr 5, 2009