Gauge transformations

[PreposterousUniverse]

I have introduced the Lorentz gauge on my perturbed metric $\gamma_{\alpha\beta}$ given by $\partial^{a}\gamma_{\alpha\beta}$. However, there remains the freedom to make further gauge transformations $$\gamma_{\alpha\beta} \rightarrow \gamma_{\alpha\beta} + \partial_{\alpha}\xi_{\beta} + \partial_{\beta}\xi_{\alpha}$$ provided that $$\partial^{\beta}\partial_{\beta}\xi^{a}=0$$

But I what I don't understand is that after taking the divergence on $\gamma_{\alpha\beta} + \partial_{\alpha}\xi_{\beta} + \partial_{\beta}\xi_{\alpha}$ and using the condition $\partial^{\beta}\partial_{\beta}\xi^{\alpha}=0$ we are still left with the term $$\partial^{\beta}\partial_{\alpha}\xi_{\beta}$$ But this term must be zero in order for the lorentz condition to hold. But how can this term be zero?

 

[haushofer]

Isn't the Lorentz gauge given by the divergence of the traceless-metric $\overline{\gamma}$? See e.g. eqn.11 and onwards of

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwj-lZC_3JrsAhUBNOwKHQ82CsIQFjAAegQIBRAC&url=http://www.tapir.caltech.edu/~chirata/ph236/lec08.pdf&usg=AOvVaw0TqLVNZvAwJTMyQSqpZG03

I.e. you're right in your question: for your version of the Lorentz gauge, you also need divergence-free vector fields defining your residual coordinate transformations.