Derivative of d'Alembertian (KG Equation proof) I'm trying to obtain the Klein-Gordon equation using a specific lagrangian:
$$\mathcal{L} = -\frac{1}{2} \phi(\Box+\mu^2)\phi$$
and the generalized Euler-Lagrange equation:
$$\frac{\partial L}{\partial \phi} - \partial_\alpha \left(\frac{\partial L}{\partial(\partial_\alpha \phi)}\right)  -  \partial_\alpha  \partial_\beta \left(\frac{\partial L}{\partial(\partial_\alpha \partial_\beta \phi)}\right) + ... = 0$$
I'm almost done, but there is one term that I'm having trouble calculating:
$$-\frac{1}{2} \partial_\alpha  \partial_\beta \frac{\partial (\phi \Box \phi)}{\partial(\partial_\alpha \partial_\beta \phi)} = -\frac{1}{2} \partial_\alpha  \partial_\beta \frac{\partial (\phi \partial_\gamma \partial^\gamma \phi)}{\partial(\partial_\alpha \partial_\beta \phi)}$$
But this is far as I have gone with this term. Any help is welcome!!!
 A: Try changing the D'Alembertian term of your Lagrangian density
$$\phi\square\phi=\phi\partial_\mu(\partial^\mu\phi)=\partial_\mu\big(\phi\partial^\mu\phi\big)-\partial_\mu\phi\partial^\mu\phi,$$
the first term is a total divergence, so it won't contribute to equations of motion and can be dropped from the Lagrangian density. Now you have a LD with only first derivatives.
In case you still want to calculate that term,
$$\frac{\partial(\phi\partial_\gamma\partial^\gamma\phi)}{\partial(\partial_\alpha\partial_\beta\phi)}=\frac{\partial(g^{\gamma\nu}\phi\partial_\gamma\partial_\nu\phi)}{\partial(\partial_\alpha\partial_\beta\phi)}=g^{\gamma\nu}\phi\delta^\alpha_\gamma\delta^{\beta}_\nu=g^{\alpha\beta}\phi$$
but note that your third term in Euler-Lagrange equations must be with a $+$ sign, i.e.
$$\frac{\partial L}{\partial \phi} - \partial_\alpha \left(\frac{\partial L}{\partial(\partial_\alpha \phi)}\right)  +  \partial_\alpha  \partial_\beta \left(\frac{\partial L}{\partial(\partial_\alpha \partial_\beta \phi)}\right)  = 0$$
