eigenmodes:
$\phi = \hat{\phi} exp(i \tilde{\phi}_u x^u)$
$\tilde{\phi} =$ phase vector.
$\phi = \hat{\phi} exp(i (\tilde{\phi}_t t + \tilde{\phi}_i x^i))$
Let $\tilde{\phi}_t = -\omega$, for $\omega =$ the angular frequency
Therefore $\tilde{\phi}^t \approx \omega$
Let $\tilde{\phi}_i = k_i =$ the wave vector.
Therefore $|\tilde{\phi}_i| = \frac{2 \pi}{\lambda}$, for $\lambda =$ the wavelength.
Therefore $\phi = \hat{\phi} exp(i (k_i x^i - \omega t))$

eigenmodes of the metric perturbation approximation, with the inverse metric approximated by flat metric
This is the linearization method used by (that one chapter in) MTW, and (page 50 or so in) Keifer, "Quantum Gravity":
$g_{ab} = \eta_{ab} + f_{ab}$
$g^{ab} \approx \eta^{ab}$
$f_{ab} = \hat{f}_{ab} exp(i \tilde{f}_u x^u)$
$g_{ab,c} = f_{ab,c} = i f_{ab} \tilde{f}_c$
$\Gamma_{abc} = \frac{i}{2}(g_{ab,c} + g_{ac,b} - g_{bc,a})$ $= \frac{1}{2}( f_{ab} \tilde{f}_c + f_{ac} \tilde{f}_b - f_{bc} \tilde{f}_a)$
${\Gamma^a}_{bc} = \frac{i}{2}( {f^a}_b \tilde{f}_c + {f^a}_c \tilde{f}_b - \tilde{f}^a f_{bc})$

gravitational acceleration:
$\ddot{x}^i + {\Gamma^i}_{ab} \dot{x}^a \dot{x}^b = 0$
$\ddot{x}^i + \frac{i}{2}( {f^i}_a \tilde{f}_b + {f^i}_b \tilde{f}_a - \tilde{f}^i f_{ab}) \dot{x}^a \dot{x}^b = 0$
$\ddot{x}^i = -i ({f^i}_a \tilde{f}_b \dot{x}^a \dot{x}^b - \frac{1}{2} \tilde{f}^i f_{ab} \dot{x}^a \dot{x}^b)$
...and that's as simple as it gets, right?

eigenmodes of the full metric
Let's try again, perturbing the metric as a whole. And forget that stupid "raise by $\eta^{ab}$ rule" -- who came up with that?
$g_{ab} = \hat{g}_{ab} exp(i \tilde{g}_u x^u)$
so $g_{ab,c} = i g_{ab} \tilde{g}_c$
Likewise $g^{ab} = \hat{g}^{ab} exp(-i \tilde{g}_u x^u)$
so $g_{ac} g^{cb} = \hat{g}_{ac} exp(i \tilde{g}_u x^u) \hat{g}^{cb} exp(-i \tilde{g}_u x^u) = \hat{g}_{ac} \hat{g}^{cb}$
Therefore define $\hat{g}^{ab}$ to be the inverse of $\hat{g}_{ab}$, so $\hat{g}_{ac} \hat{g}^{cb} = \delta_a^b$

Representing the wave 4-vector in terms of the derivatives:
$\tilde{g}_a = \frac{n}{n} \tilde{g}_a$
$= -i \frac{1}{n} g^{uv} g_{uv} \cdot i \tilde{g}_a$
$= -i \frac{1}{n} g^{uv} g_{uv,a}$

$\tilde{g}_a \tilde{g}_b = \frac{n}{n} \tilde{g}_a \tilde{g}_b$
$= -\frac{1}{n} g^{uv} g_{uv} \cdot (i^2) \tilde{g}_a \tilde{g}_b$
$= -\frac{1}{n} g^{uv} g_{uv,ab}$

connection coefficients:
$\Gamma_{abc} = \frac{i}{2}(g_{ab} \tilde{g}_c + g_{ac} \tilde{g}_b - g_{bc} \tilde{g}_a)$
${\Gamma^a}_{bc} = g^{au} \frac{i}{2}(g_{ub} \tilde{g}_c + g_{uc} \tilde{g}_b - g_{bc} \tilde{g}_u)$
$= g^{au} \frac{i}{2}(g_{ub} \tilde{g}_c + g_{uc} \tilde{g}_b - g_{bc} \tilde{g}_u)$
$= \frac{i}{2}(\delta^a_b \tilde{g}_c + \delta^a_c \tilde{g}_b - g_{bc} \tilde{g}^a)$

gravitational acceleration:
$\ddot{x}^i + {\Gamma^i}_{ab} \dot{x}^a \dot{x}^b = 0$
$\ddot{x}^i + \frac{i}{2}(\delta^i_a \tilde{g}_b + \delta^i_b \tilde{g}_a - g_{ab} \tilde{g}^i) \dot{x}^a \dot{x}^b = 0$
$\ddot{x}^i + i (\dot{x}^i \tilde{g}_u \dot{x}^u - \frac{1}{2} \tilde{g}^i g_{ab} \dot{x}^a \dot{x}^b) = 0$
Now we can use $g_{ab} \dot{x}^a \dot{x}^b = (\dot{x})^2 = -1$
$\ddot{x}^i = -\frac{i}{2} \tilde{g}^i - i \dot{x}^i \tilde{g}_u \dot{x}^u$
Now we are left with:
(particle acceleration) = $-\frac{i}{2} \times$ (the geometry wave 4-vector) $+$ (particle velocity) $\times$ (the geometry phase in the particle's direction)

Let's consider a basis decomposition of the metric:
$g_{ab} = {e_a}^I {e_b}^J \eta_{IJ}$
And let's look at modes of the basis:
${e_a}^I = {\hat{e}_a}^I exp(i \tilde{e}_u x^u)$
So ${{e_a}^I}_{,b} = i {e_a}^I \tilde{e}_b$
Then $g_{ab,c} = ({e_a}^I {e_b}^J \eta_{IJ})_{,c} = i ({e_a}^I \tilde{e}_c {e_b}^J \eta_{IJ} + {e_a}^I {e_b}^J \tilde{e}_c \eta_{IJ})$
$ = i \tilde{e}_c (2 {e_a}^I {e_b}^J \eta_{IJ})$
$ = 2 i \tilde{e}_c g_{ab}$
So from this, we can say $\tilde{e}_u = \frac{1}{2} \tilde{g}_u$, or the wave 4-vector of ${e_a}^I$ is half the wave 4-vector of $g_{ab}$

Einstein Field Equations
Now let's continue on to look at the Einstein Field Equations
${\Gamma^a}_{bc,d} = \frac{i}{2}(\delta^a_b \tilde{g}_c + \delta^a_c \tilde{g}_b - g_{bc} \tilde{g}^a)_{,d}$
$= -\frac{i}{2} g_{bc,d} \tilde{g}^a$
$= \frac{1}{2} \tilde{g}^a g_{bc} \tilde{g}_d$

${R^a}_{bcd} = 2 ( {\Gamma^a}_{b[d,c]} + {\Gamma^a}_{u[c} {\Gamma^u}_{d]b} )$
$= \tilde{g}^a g_{b[d} \tilde{g}_{c]} - \frac{1}{2} (\delta^a_u \tilde{g}_{[c} + \tilde{g}_u \delta^a_{[c} - \tilde{g}^a g_{u[c}) (\delta^u_{d]} \tilde{g}_b + \tilde{g}_{d]} \delta^u_b - g_{d]b} \tilde{g}^u)$
$= \tilde{g}^a g_{b[d} \tilde{g}_{c]} + \frac{1}{2} ( - \delta^a_u \tilde{g}_{[c} \delta^u_{d]} \tilde{g}_b - \delta^a_u \tilde{g}_{[c} \tilde{g}_{d]} \delta^u_b + \delta^a_u \tilde{g}_{[c} g_{d]b} \tilde{g}^u - \tilde{g}_u \delta^a_{[c} \delta^u_{d]} \tilde{g}_b - \tilde{g}_u \delta^a_{[c} \tilde{g}_{d]} \delta^u_b + \tilde{g}_u \delta^a_{[c} g_{d]b} \tilde{g}^u + \tilde{g}^a g_{u[c} \delta^u_{d]} \tilde{g}_b + \tilde{g}^a g_{u[c} \tilde{g}_{d]} \delta^u_b - \tilde{g}^a g_{u[c} g_{d]b} \tilde{g}^u ) $
$= \tilde{g}^a g_{b[d} \tilde{g}_{c]} + \frac{1}{2} ( - \delta^a_{[d} \tilde{g}_{c]} \tilde{g}_b - \delta^a_b \tilde{g}_{[c} \tilde{g}_{d]} + \tilde{g}^a \tilde{g}_{[c} g_{d]b} - \delta^a_{[c} \tilde{g}_{d]} \tilde{g}_b - \tilde{g}_b \delta^a_{[c} \tilde{g}_{d]} + \delta^a_{[c} g_{d]b} \tilde{g}^u \tilde{g}_u + \tilde{g}^a g_{[cd]} \tilde{g}_b + \tilde{g}^a g_{b[c} \tilde{g}_{d]} - \tilde{g}^a \tilde{g}_{[c} g_{d]b} ) $
$= \frac{1}{2} ( \tilde{g}^a g_{b[d} \tilde{g}_{c]} + \delta^a_{[c} g_{d]b} \tilde{g}^u \tilde{g}_u - \delta^a_{[c} \tilde{g}_{d]} \tilde{g}_b)$

$R_{ab} = {R^u}_{aub}$
$= \frac{1}{2} ( \tilde{g}^u g_{a[b} \tilde{g}_{u]} + \delta^u_{[u} g_{b]a} \tilde{g}^v \tilde{g}_v - \delta^u_{[u} \tilde{g}_{b]} \tilde{g}_a)$
$= \frac{1}{4} ( \tilde{g}^u g_{ab} \tilde{g}_u - \tilde{g}^u g_{au} \tilde{g}_b + \delta^u_u g_{ab} \tilde{g}^v \tilde{g}_v - \delta^u_b g_{ua} \tilde{g}^v \tilde{g}_v - \delta^u_u \tilde{g}_b \tilde{g}_a + \delta^u_b \tilde{g}_u \tilde{g}_a)$
$= \frac{1}{4} ( g_{ab} \tilde{g}_u \tilde{g}^u - \tilde{g}_a \tilde{g}_b + n g_{ab} \tilde{g}^u \tilde{g}_u - g_{ab} \tilde{g}^v \tilde{g}_v - n \tilde{g}_a \tilde{g}_b + \tilde{g}_a \tilde{g}_b )$
$= \frac{1}{4} n (g_{ab} (\tilde{g})^2 - \tilde{g}_a \tilde{g}_b)$
...for $n=4$ this gives $R_{ab} = g_{ab} (\tilde{g})^2 - \tilde{g}_a \tilde{g}_b$
...for $(\tilde{g})^2 = 0$ this gives $R_{ab} = -\frac{1}{4} n \tilde{g}_a \tilde{g}_b$

$R = g^{ab} R_{ab}$
$= \frac{n}{4} g^{ab} (g_{ab} (\tilde{g})^2 - \tilde{g}_a \tilde{g}_b)$
$= \frac{n}{4} (n (\tilde{g})^2 - (\tilde{g})^2)$
$= \frac{1}{4} n (n - 1) (\tilde{g})^2$
Therefore $(\tilde{g})^2 = \frac{4}{n (n-1)} R$
...for $n=4$ this gives $R = 3 (\tilde{g})^2$
...for $R = 0$ gives $(\tilde{g})^2 = 0$

Substitute the definition of $(\tilde{g})^2 = \frac{4}{n (n-1)} R$ into the definition of $R_{ab}$:
$R_{ab} = \frac{1}{4} n (\frac{4}{n (n-1)} R g_{ab} - \tilde{g}_a \tilde{g}_b)$
$= (\frac{1}{n-1} R g_{ab} - \frac{1}{4} n \tilde{g}_a \tilde{g}_b)$

$G_{ab} = R_{ab} - \frac{1}{2} g_{ab} R$
$ = \frac{1}{4} n (g_{ab} (\tilde{g})^2 - \tilde{g}_a \tilde{g}_b) - \frac{1}{2} g_{ab} \frac{1}{4} n (n - 1) (\tilde{g})^2 $
$ = \frac{1}{8} n ((3 - n) g_{ab} (\tilde{g})^2 - 2 \tilde{g}_a \tilde{g}_b)$
...for $n=4$ this gives $G_{ab} = -(\frac{1}{2} g_{ab} (\tilde{g})^2 + \tilde{g}_a \tilde{g}_b)$
...and for $R=0 \iff (\tilde{g})^2 = 0$ this gives $G_{ab} = -\frac{1}{2} \tilde{g}_a \tilde{g}_b$

Substitute the definition of $(\tilde{g})^2 = \frac{4}{n (n-1)} R$ into the definition of $G_{ab}$:
$G_{ab} = \frac{3 - n}{2 (n-1)} R g_{ab} - \frac{1}{4} n \tilde{g}_a \tilde{g}_b$

Zero Gaussian Curvature
Once again, $R = 0 \iff (\tilde{g})^2 = 0 \iff G_{ab} = -\frac{1}{2} \tilde{g}_a \tilde{g}_b$
$G_{ab} = 8 \pi T_{ab}$
So $-16 \pi T_{ab} = \tilde{g}_a \tilde{g}_b$

Matter stress-energy:
$T_{ab} = \rho h u_a u_b + p g_{ab}$, for $\rho =$ density, $p =$ pressure, $h =$ enthalpy, $u_a =$ 4-velocity.
$-16 \pi T_{tt} = -16 \pi (\rho h (u_t)^2 + p g_{tt}) = (\tilde{g}_t)^2$
$-16 \pi T_{ti} = -16 \pi (\rho h u_t u_i + p g_{ti}) = \tilde{g}_t \tilde{g}_i$
$-16 \pi T_{ij} = -16 \pi (\rho h u_i u_j + p g_{ij}) = \tilde{g}_i \tilde{g}_j$
For rest-frame $u_t = -1$, $u_i = 0$, and we find:
$-16 \pi T_{tt} = -16 \pi (\rho h + p g_{tt}) = (\tilde{g}_t)^2$
$-16 \pi T_{ti} = -16 \pi p g_{ti} = \tilde{g}_t \tilde{g}_i$
$-16 \pi T_{ij} = -16 \pi p g_{ij} = \tilde{g}_i \tilde{g}_j$

${T^a}_b = \rho h u^a u_b + \delta^a_b p$
$-16 \pi {T^a}_b = -16 \pi (\rho h u^a u_b + \delta^a_b P) = \tilde{u}^a \tilde{u}_b$
$-16 \pi {T^t}_t = -16 \pi (\rho h u^t u_t + P) = \tilde{u}^t \tilde{U}_t$
$-16 \pi {T^t}_i = -16 \pi (\rho h u^t u_i) = \tilde{u}^t \tilde{U}_i$
$-16 \pi {T^i}_j = -16 \pi (\rho h u^i u_j + \delta^i_j P) = \tilde{u}^i \tilde{u}_j$

Electromagnetism stress-energy:
$T_{ab} = \frac{1}{4\pi} (F_{au} {F_b}^u - \frac{1}{4} g_{ab} F_{uv} F^{uv})$
$-16 \pi T_{tt} = -2 (E^2 + B^2) = (\tilde{g}_t)^2$
$-16 \pi T_{ti} = 4 S_i = \tilde{g}_t \tilde{g}_i$
$-16 \pi T_{ij} = -2 g_{ij} (g^{kl} E_k E_l) + 4 (E_i E_j + B_i B_j) = \tilde{g}_i \tilde{g}_j$