# Homework Help: Finding Noether Charges from Action

1. Nov 20, 2008

### malawi_glenn

Giving an action, a general one:

$$S = \int dt L(q^i,\dot{q}^i,t)$$

now assume this action is invariant under a coordinate transformation:

$$q^i \rightarrow q^i + \epsilon ^a (T_a)^i_jq^j$$

Where T_a is a generator of a matrix Lie group.

Now one should be able to find the consvered quantities, the "Noether Charges", and how those relate to the matrix lie group.

BUT HOW?

I have never done so much in school about actions, just lagrangians and hamolitonians.

For instance, if one only considered translation: $$q^i \rightarrow q^i + \epsilon q^i$$, and if the Lagrangian/hamiltonian is invariant under translations -> we know that the linear momentum is conserved. But how do we show it with the action and noether currents/charges?

Now this is a quite general question, I have never quite understood this, and is related to what I asked a couple of days ago in the math section about Lie Subgroups. I am trying to appreciate Group Theory, in perticular Lie Groups. This is a "general example" on its application to classical mechanics which I found somewhere, but then I found out that I am totally lost when it comes to performing the "searches" for noether charges.

2. May 29, 2009

### Hao

Have you found the answer to this question?

3. May 31, 2009

### samr

I'm going to use $\mathcal{L}\left(\phi,\partial_\mu\phi,t\right)$ as my lagrangian as this is the notation I'm used to typing :) But this doesn't change the process.

Ok since you know that the action is invariant under this transformation we have that $\delta\mathcal{L}=0$, so that:
$$\delta\mathcal{L}=\frac{\partial\mathcal{L}}{\partial\phi_i}\delta_{\phi_i}+\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta\left(\partial_\mu\phi_i\right)$$
And we can then change $\delta\left(\partial_\mu\phi_i\right)$ to $\partial_\mu\left(\delta\phi_i\right)$.
Using the Euler-Lagrange equation:
$$\frac{\partial\mathcal{L}}{\partial\phi_i}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\right)$$
We can see that our first equation's first term can be replaced with the RHS of the E-L giving:
$$\delta\mathcal{L}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\right)\delta_{\phi_i}+\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\partial_\mu\left(\delta\phi_i\right)$$
Which we can bring into one term by the product rule:
$$\delta\mathcal{L}=\partial_\mu\left(\frac{\partial_\mu\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta_{\phi_i}\right)$$
Since as we stated $\delta\mathcal{L}=0$ we can make the identification:
$$\partial_\mu\underbrace{\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta_{\phi_i}\right)}_{J^\mu}$$
As this fulfils $\partial_\mu J^\mu=0$

We can then look at the $\delta\phi_i$ terms under an infinitesimal change:
$$\delta\phi_i&=\phi_i-\phi'_i=\phi_i-\left(\phi_i+\epsilon^a\left(T_a\right)^i_j\phi^j\right)\\ &=-\epsilon^a\left(T_a\right)^i_j\phi^j$$