Noether Current Derivation for SO(3) Rotation?

ClaraOxford
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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.

Homework Statement



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


Homework Equations



jμa=∂L/(∂(∂μa))*\Phi - \Lambda\muα

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

a → ∅a + \Phiεα


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 \Phiac = -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?
 
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ClaraOxford said:
jμa=∂L/(∂(∂μa))*\Phi - \Lambda\muα

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

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.

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.

Using the above info, I can see that \Phiac = -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?


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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