Variation of Lagrangian under Lorentz transformations

[jinawee]

Homework Statement

Prove that under an infinitesimal Lorentz transformation: x^\mu \to x^\mu+\omega^\mu_\nu x^\nu so: \phi\to\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi the Lagrangian varies as:

\delta \mathcal{L}=-\partial_\mu(\omega^\mu_\nu x^\nu \mathcal{L})

The Attempt at a Solution

The new Lagrangian will be:

<br /> \mathcal{L}(\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi,\partial_\sigma(\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi))

The variation is:

<br /> \delta\mathcal{L}=\frac{\partial \mathcal{L}}{\partial \phi}\delta\phi+\frac{\partial \mathcal{L}}{\partial (\partial_\sigma\phi)}\partial_\sigma \delta \phi=-\omega^\mu_\nu \left[x^\nu\partial_\mu \phi \frac{\partial \mathcal{L}}{\partial \phi}+\partial_\sigma(x^\nu\partial_\mu\phi)\frac{\partial \mathcal{L}}{\partial (\partial_\sigma\phi)}\right]


If the last term in the sum vanished somehow, I would arrive to the solution, but I can't see how it is zero. Or have I made a mistake before?

I think I've done some progress:

-\omega^\mu_\nu \left[\partial_\sigma(x^\nu\partial_\mu\phi)\frac{\partial \mathcal{L}}{\partial (\partial_\sigma\phi)}\right]=-\omega^\mu_\nu \left[ \partial_\mu\phi\frac{\partial \mathcal{L}}{\partial (\partial_\nu\phi)}+x^\nu \partial_\mu\phi\mathcal{L}\right]=

Could I use Euler-Lagrange equations to make this zero?

 

[bloby]

Seems to me that there is a simple answer: what kind of field is the lagrangian density? How do these fields transform? And then with a property of $\omega$ it seems ok.

In your calculations I think there is something wrong with $\partial_{\sigma}$: it's not a scalar.

 

[bloby]

Could you develop

\delta \mathcal{L}=-\partial_\mu(\omega^\mu_\nu x^\nu \mathcal{L})

 

[bloby]

I'm going to bed...

$$ \partial_{\sigma}\phi \to \partial_{\sigma}\phi-\omega^\mu_\nu x^\nu\partial_\mu\partial_\sigma\phi+\omega^\mu_\sigma\partial_\mu\phi $$