Finding Noether Charges from Action

[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.

 

[Hao]

Have you found the answer to this question?

 

[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:
<br /> \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)<br />
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:
<br /> \frac{\partial\mathcal{L}}{\partial\phi_i}=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\right)<br />
We can see that our first equation's first term can be replaced with the RHS of the E-L giving:
<br /> \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)<br />
Which we can bring into one term by the product rule:
<br /> \delta\mathcal{L}=\partial_\mu\left(\frac{\partial_\mu\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta_{\phi_i}\right)<br />
Since as we stated \delta\mathcal{L}=0 we can make the identification:
<br /> \partial_\mu\underbrace{\left(\frac{\partial\mathcal{L}}{\partial\left(\partial_\mu\phi_i\right)}\delta_{\phi_i}\right)}_{J^\mu}<br />
As this fulfils \partial_\mu J^\mu=0

We can then look at the \delta\phi_i terms under an infinitesimal change:
<br /> \delta\phi_i&amp;=\phi_i-\phi&#039;_i=\phi_i-\left(\phi_i+\epsilon^a\left(T_a\right)^i_j\phi^j\right)\\<br /> &amp;=-\epsilon^a\left(T_a\right)^i_j\phi^j<br />