Does the Lorentz Condition Apply to the Given Vector Field Lagrangian?

[orentago]

Homework Statement

Given the Lagrangian density

$$L=-{1 \over 2}[\partial_\alpha\phi_\beta(x)][\partial^\alpha\phi^\beta(x)]+{1\over 2}[\partial_\alpha\phi^\alpha(x)][\partial_\beta\phi^\beta(x)]+{\mu^2\over 2}\phi_\alpha(x)\phi^\alpha(x)$$

for the real vector field $$\phi^\alpha(x)$$ with field equations:
$$[g_{\alpha\beta}(\square+\mu^2)-\partial_\alpha\partial_\beta]\phi^\beta(x)=0$$

Show that the field $$\phi^\alpha(x)$$ satisfies the Lorentz condition:
$$\partial_\alpha\phi^\alpha(x)=0$$

Homework Equations

See above.

The Attempt at a Solution

$$[g_{\alpha\beta}(\square+\mu^2)-\partial_\alpha\partial_\beta]\phi^\beta(x)=0$$
$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\square+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\partial^\beta\partial_\beta+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=\partial_\alpha\partial_\beta\phi^\beta(x)+\mu^2g_{\alpha\beta}\phi^\beta(x)$$
$$\Rightarrow\mu^2g_{\alpha\beta}\phi^\beta(x)=0$$
$$\Rightarrow\mu^2\phi^\beta(x)=0$$
$$\Rightarrow\mu^2\partial_\alpha\phi^\alpha(x)=0$$

I think I've done it, but I don't know if my method is correct. Would anyone be able to validate or refute this?

 

[fzero]

$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\square+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta(x)=g_{\alpha\beta}(\partial^\beta\partial_\beta+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta (x)=\partial_\alpha\partial_\beta\phi^\beta(x)+\mu ^2g_{\alpha\beta}\phi^\beta(x)$$

You made a mistake when you used the index $$\beta$$ for $$\square = \partial^\gamma\partial_\gamma$$. You have to use a dummy index and it's not the same as the one on $$\phi^\beta$$.

You can obtain the result by considering a particular derivative of the field equations.

 

[orentago]

Ok how about this:

$$\Rightarrow\partial_\alpha\partial_\beta\phi^\beta (x)=g_{\alpha\beta}(\partial^\gamma\partial_\gamma+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\partial^\alpha\partial_\alpha\partial_\beta\phi^\beta (x)=g_{\alpha\beta}\partial^\alpha(\partial^\gamma\partial_\gamma+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\square\partial_\beta\phi^\beta (x)=\partial_\beta(\square+\mu^2)\phi^\beta(x)$$
$$\Rightarrow\square\partial_\beta\phi^\beta (x)=\square\partial_\beta\phi^\beta(x)+\mu^2\partial_\beta\phi^\beta(x)$$
$$\Rightarrow\mu^2\partial_\beta\phi^\beta(x)=0$$
$$\Rightarrow\partial_\alpha\phi^\alpha(x)=0$$