= Solution
First take the spacetime trace of the generalized field equation. Since $n_\mu Z^\mu=-\Theta$ and spacetime has dimension four, this gives
$$
{}^{(4)}R+2\nabla_\mu Z^\mu-2\Theta=-8\pi T.
$$
Next contract twice with the unit normal. The result is
$$
R_{\mu\nu}n^\mu n^\nu
+2n^\mu n^\nu\nabla_\mu Z_\nu-\Theta
=8\pi\rho+4\pi T.
$$
Adding the trace equation to twice this normal projection cancels $T$. The <Scalar Gauss equation> then converts the curvature terms to
$$
{}^{(3)}R+K^2-K_{\mu\nu}K^{\mu\nu}.
$$
For the derivative terms, part (iii) gives
$$
2\nabla_\mu Z^\mu+4n^\mu n^\nu\nabla_\mu Z_\nu
=2D^\mu Z_\mu+2n^\mu n^\nu\nabla_\mu Z_\nu.
$$
Differentiating $n^\nu Z_\nu=-\Theta$ along $n^\mu$ yields
$$
n^\mu n^\nu\nabla_\mu Z_\nu
=-n^\mu\nabla_\mu\Theta-Z_\nu a^\nu.
$$
Combining these identities produces
$$
\boxed{
{}^{(3)}R+K^2-K_{\mu\nu}K^{\mu\nu}
-2n^\mu\nabla_\mu\Theta-2Z_\mu a^\mu
-4\Theta+2D^\mu Z_\mu=16\pi\rho.
}
$$
Because $\Theta$ is a <scalar field> and $n^\mu=\alpha^{-1}(1,-\beta^i)$,
$$
n^\mu\nabla_\mu\Theta
=\frac1\alpha(\partial_t-\beta^m\partial_m)\Theta.
$$
The <normal acceleration> is spatial, so $Z_\mu a^\mu=\Theta_\mu a^\mu$. Part (iv) also gives $D^\mu Z_\mu=D^\mu\Theta_\mu-K\Theta$. Solving the preceding constraint for $\partial_t\Theta$ gives
$$
\partial_t\Theta=\beta^m\partial_m\Theta+\frac\alpha2
\left[
{}^{(3)}R+K(K-2\Theta)-K_{\mu\nu}K^{\mu\nu}
-2\Theta_\mu a^\mu-4\Theta
+2D^\mu\Theta_\mu-16\pi\rho
\right].
$$
Thus the constants in this <Z4 formulation> evolution equation are
$$
\boxed{d_1=-2,\qquad d_2=-2,\qquad d_3=2,\qquad d_4=-16\pi.}
$$
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