# Internal Symmetries (Prove Lagrangian is invariant)

1. Oct 19, 2014

### Breo

1. The problem statement, all variables and given/known data
Note: There is an undertilde under every $$\phi$$

Imagine $$\phi ^t M \phi$$ . M is a symmetric, real and positive matrix. Prove L is invariant:

$$\mathcal{L} = \phi ^t M \phi + \frac{1}{2} \partial_\mu \phi ^t \partial ^\mu \phi$$

Trick: Counting parameters.

2. Relevant equations

Internal symmetries:

$$x \longrightarrow x' = x$$

$$\phi ^{(\alpha)} \longrightarrow \phi '^{(\alpha)} = \sum_{\substack{\beta = 1}} \Omega_{\alpha \beta} \phi ^{(\beta)}$$

Being:

$$\Omega \in O)M)$$
$$\Omega \Omega ^t = \mathbb {1}$$

3. The attempt at a solution

Ok, the second term in the right hand of the Lagrangian is easy to prove is invariant (I have the proof from classes), so we can obvious it.

About the trick, there are $$\frac{M(M-1)}{2}$$ parameters, which should be the same that dimension, right? ( Or is M parameters?) because the O(M) group has the same equation for dimension. The issue comes from the reason M matrix must have dimension 2 ( the fields have 1 dimension each one in a 4-spacetime) but this makes the equation above to gives a non-natural number of M.... so I'm lost here.

Going further:

I think we need to prove: $$\delta \mathcal {L} = 0$$. I tried this:

We know as M is symmetric and positive:

$$M = M^t$$ and $$A^t M A \geq 0$$

So:

$$\phi^t \Omega^t \Omega M^t \Omega^t \Omega \phi = \phi '^t \Omega M \Omega^t \phi '$$

But this is nothing....

Help?

EDIT: I just did a new calculation way using conjugate momentums in the current calculation and reached that L would be invariant if $$\phi = - \phi^t$$ haha I forgot something this time...

Last edited: Oct 19, 2014
2. Oct 24, 2014

### Staff: Admin

Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

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