# 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!!!

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.

$$\frac{\partial L}{\partial \phi} - \partial_\alpha \left(\frac{\partial L}{\partial(\partial_\alpha \phi)}\right) + \partial_\alpha \partial_\beta \left(\frac{\partial L}{\partial(\partial_\alpha \partial_\beta \phi)}\right) = 0$$

• Hmmm, this seems good, but I cannot see how this will lead me to a squared term in order to vanish the 1/2 Mar 14 at 19:47
• Note that $\partial_\mu\phi\partial^\mu\phi=g^{\mu\nu} \partial_\mu \phi \partial_\nu \phi$, so when you take the derivative of that term with respect to $\partial_\alpha\phi$, there will be two contributions.
– AFG
Mar 14 at 19:51