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The de Donder gauge is often used to simplify the linearised equations of motion of general relativity. If the metric is linearised as $g_{ab} = \bar g_{ab} + \gamma_{ab}$, then the de Donder gauge reads
$\nabla^a(\gamma_{ab} - \frac{1}{2}\bar g_{ab}\gamma) = 0$.

The partial differential equation for the gauge transformation vector $v^a$ is $ \nabla^b\nabla_b v_a + R_a^b v_b = \nabla^a(\gamma_{ab} - \frac{1}{2}\bar g_{ab}\gamma)$.

In chapter 7.5 of Wald, I read that this equation can always be solved because it is of the form $g^{ab}\nabla_a\nabla_b \phi_i + \sum_j (A_{ij})^a\nabla_a \phi_j + \sum_j B_{ij}\phi_j + C_i$.
Theorem 10.1.2 of Wald says that in a globally hyperbolic spacetime this equation has a well posed initial value formulation on any spacelike Cauchy surface.

In stead of de Donder gauge, I want to use a similar gauge:
$\nabla^a(\gamma_{ab} - n \bar g_{ab}\gamma) = 0$.
The partial differential equation changes to
$ \nabla^b\nabla_b v_a + (1 -2n)\nabla_a\nabla_b v^b + R_a^b v_b = \nabla^a(\gamma_{ab} - n\bar g_{ab}\gamma)$.

This equation is not covered by theorem 10.1.2 of Wald. My question is: is the existence of a solution for this equation guaranteed in an AdS background when $n=1$?

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I am pretty sure that the term $\nabla_a (\nabla_b v^b)$ which is a gradient of the divergence $\nabla\cdot v$ that you don't cancel for $n\neq 1/2$ is bad for the solvability. When the term cancels, the equations for individual components $v_a$ are pretty much independent, but they get mixed up if the term is there which probably damages the existence or uniqueness of solutions. There is a good reason why only the de Donder gauge with the right coefficient is being used but I don't have the answer in my head clearly enough to post it as a full answer. – Luboš Motl Jan 19 '12 at 16:59
@LubošMotl Thanks. Do you have any suggestion where to look for the answer? – sjdh Jan 19 '12 at 17:15
up vote 1 down vote accepted

It may be far too late for this to be of any value to you, but in this paper by Mora et al:, it is shown that such a gauge choice does exist (for any $n$ in your notation).

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Great, I did not find an answer yet. Thank you! – sjdh Oct 1 '14 at 2:08

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