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)
<br /> \Rightarrow\partial_\alpha\partial_\beta\phi^\beta (x)=\partial_\alpha\partial_\beta\phi^\beta(x)+\mu ^2g_{\alpha\beta}\phi^\beta(x)<br />

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