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:

$$\mathcal{L}(\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi,\partial_\sigma(\phi-\omega^\mu_\nu x^\nu\partial_\mu\phi))$$

The variation is:

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


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...

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

 

[paranormal]

Yes, you are on the right track. The last term in the sum can be made zero by using the Euler-Lagrange equations. In the Euler-Lagrange equations, the term involving the variation of the Lagrangian with respect to the field itself, which in this case is $\frac{\partial \mathcal{L}}{\partial \phi}$, is always multiplied by the variation of the field, which in this case is $\delta \phi$. Since the variation of the field is zero, this term will also be zero. Therefore, the last term in the sum will be zero and you will arrive at the desired result:

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