# Noether current for SO(3) rotation

1. Mar 20, 2012

### ClaraOxford

This is a problem from my theoretical physics course. We were given a solution sheet, but it doesn't go into a lot of detail, so I was hoping for some clarification on how some of the answers are derived.

1. The problem statement, all variables and given/known data

For the Lagrangian L=1/2(∂μTμ∅-m2T∅) derive the Noether currents and charges.

2. Relevant equations

jμa=∂L/(∂(∂μa))*$\Phi$ - $\Lambda$$\mu$α

Here, the lambda term is zero, because the Lagrangian is invariant under SO(3).

a → ∅a + $\Phi$εα

3. The attempt at a solution

We were first told to show that the above Lagrangian satisfies SO(3) symmetry (this was fine). The solution sheet then states that infintessimal transformations can be written as ∅a → ∅a-itc(Tc)abb, where (Tc)ab=-iεcab

I could not work out how to derive this though.

Using the above info, I can see that $\Phi$ac = -i(Tc)abb, taking εα = tc

Then I just need to calculate ∂L/∂(∂μa)
Is this just ∂μa?? I'm not sure how to calculate this when there's 2 derivatives, one with a superscript and one with a subscript. And does the transpose affect things?

2. Mar 20, 2012

### fzero

You probably want ${j^\mu}_\alpha$ on the left side to match indices.

You could either take this as a definition of the Lie algebra of the orthogonal group or else show that the finite matrices $\exp (i t^a T_a)$ are orthogonal. This is a generalization of the way we can parametrize 2d rotations by an angle.

You can write

$$\partial_\mu \phi^a \partial^\mu \phi_a = \delta_{ab} \eta^{\mu\nu} \partial_\mu \phi^a \partial_\nu \phi^b,$$

then use

$$\frac{\partial}{\partial(\partial_\nu \phi^b) } \partial_\mu\phi^a = \delta^a_b \delta^\nu_\mu.$$

Since there are 2 factors of $\partial\phi$, you'll need to use the product rule which brings in a factor of 2 in the answer.

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