Gauge fixing conditions in general relativity Is there a limit to gauge fixing conditions we can impose in gravity ? I have seen two gauge fixing conditions. The DeDonder gauge $\partial_\mu g^{\mu\nu}$ and then in 3+1 formalism the gauge fixing condition $\nabla^2 t = 0$ is imposed where $t$ is the time coordinate. What if I imposed $\nabla^2 x_i = 0$ where $x_i$ is some spatial coordinate. Maybe a combination $\nabla^2(t^2+x^2) = 0$. How is the gauge fixing condition decided upon?
 A: In the case of linearised GR, one of the main reasons for choosing our gauge fixing condition is just sheer convenience. If we're doing a metric perturbation of $g_{\mu \nu}  = \eta_{\mu \nu} + h _{\mu \nu} $, then our corresponding action that we get from perturbing our Ricci scalar is the Fierz-Pauli action which is second order in $h $ (since the first and zeroth orders vanish), the precise form of which you can look up in David Tong's notes on linearised gravity. 
$$ S_{FP} = \int d ^ 4 x \, L (h) .$$
Now, this yields a very difficult and verbose equation of motion to solve for $h_{\mu \nu } $ (which you can look up yourself in the same chapter), so to make things a bit easier we employ the fact that metric diffeomorphisms should leave our action unchanged, since they're just changes in coordinates. So, our action should be invariant under the change 
$$ h _{\mu \nu } \rightarrow h_{\mu \nu } + \partial _ \mu \xi _ \nu + \partial _\nu \xi _ \mu. $$ This freedom allows us 4 free parameters to fix, since $\xi$ has four components, and we choose constraining the four degrees of freedom via the de-Donder gauge because it simplifies our equation of motion in a vacuum to the simple 
$$ \Box  h _{\mu \nu} - \frac{1}{2} \Box h \eta_{\mu \nu }  = 0 $$ Now, this is great because we can now set $\bar {h}_{ \mu \nu } = h _{\mu \nu } - \frac{1}{2} h \eta _{\mu \nu } $ which leaves us with $\Box \bar{ h}_{\mu \nu } = 0 $, which is just a wave equation which we can solve easily. 
