Equivalency of Gauge Conditions How is the Lorenz gauge condition $\partial_\mu \overline{h}^{\mu \nu}=0$ equivalent to the harmonic gauge condition $\Box x^\mu=0 $?
 A: The trick here is just to employ some tensor manipulations. When I first saw this condition too I thought it was a bit strange since one is a second order differential equation and the other is a first, right? Nevertheless, typically we write the harmonic or de-Donder expansion as $\bar{h}^{\alpha \beta} = -\sqrt{-g}g^{\alpha \beta} + \eta^{\alpha \beta}$. Now, by definition of the Laplace-Beltrami operator $\square$:
\begin{align}
0 &= \square x^{\mu} = g^{\rho \sigma} (\partial_{\rho} \partial_{\sigma} x^{\mu} - \Gamma^{\lambda}_{\rho \sigma} \partial_{\lambda} x^{\mu} )\\
&= g^{\rho \sigma} (\partial_{\rho} \delta^{\mu}_{\sigma} - \Gamma^{\lambda}_{\rho \sigma} \delta^{\mu}_{\lambda}) \\
&= - g^{\rho \sigma} \Gamma^{\mu}_{\rho \sigma} \tag{$\ast$}
\end{align}
Now, to even just borrow from Wikipedia (http://en.wikipedia.org/wiki/Harmonic_coordinate_condition): Consider the covariant derivative of the density $\sqrt{-g} g^{\alpha \beta}$, we have that:
$$0 = g^{\mu \nu}_{;\rho} = (g^{\mu \nu} \sqrt{-g})_{,\rho} + g^{\sigma \nu} \Gamma^{\mu}_{\sigma \rho} \sqrt{-g} + g^{\mu \sigma} \Gamma^{\nu}_{\sigma \rho} \sqrt{-g} - g^{\mu \nu} \Gamma^{\tau}_{\tau \rho} \sqrt{-g} \tag{$\ast\ast$}$$
Since by construction of the Levi-Cevita connection $\nabla$ (this is the definition) we have $\nabla_{\tau} g^{\alpha \beta} := g^{\alpha \beta}_{;\tau} = 0$ and hence $\sqrt{-g}_{;\alpha} = 0$ too (along with contracted index versions) we find, by contracting the indices $\rho$ and $\nu$ on $(**)$:
$$ 0 = (g^{\mu \nu} \sqrt{-g})_{,\nu} + g^{\sigma \nu} \Gamma^{\mu}_{\sigma \nu} \sqrt{-g} + g^{\mu \sigma} \Gamma^{\nu}_{\sigma \nu} \sqrt{-g} - g^{\mu \nu} \Gamma^{\sigma}_{\sigma \nu} \sqrt{-g}$$
Enforcing the harmonic condition $(*)$ now gives us that:
$$0 = (g^{\mu \nu} \sqrt{-g})_{,\nu} + g^{\mu \alpha} \Gamma^{\beta}_{\alpha \beta} \sqrt{-g} - g^{\mu \alpha} \Gamma^{\beta}_{\beta \alpha} \sqrt{-g}$$
Finally, since the metric tensor is symmetric in its indices (and therefore so are the structure constants (Christoffel symbols) $\Gamma$) we have:
$$0 = (g^{\mu \nu} \sqrt{-g})_{,\nu} = (g^{\mu \nu} \sqrt{-g})_{,\mu}$$
Which, recall by definition is:
$$\partial_{\mu} \bar{h}^{\mu \nu} = 0.$$
Since derivatives of the Minkowski metric vanish identically.
EDIT: Cleaned up some minor mistakes.
