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Let ${\gamma }_{\alpha \beta }={\delta }_{\alpha \beta }$ be the positive Euclidean spatial background metric. Write a tensor-sector metric perturbation as $h_{00}=h_{0\alpha }=0$, $h_{\alpha \beta }=-2E_{\alpha \beta }$, with ${\gamma }^{\alpha \beta }{E}_{\alpha \beta }=0$ and ${\partial }^{\alpha }{E}_{\alpha \beta }=0$. Spatial indices here are raised with $\gamma$, and $\square ={\partial }_t^2-\Delta$.

Substitute the linearized Levi-Civita connection into the cover's curvature definition. To first order,In the transverse-traceless tensor sector $h=0$ and ${\partial }^c{h}_{ac}=0$, so $\delta C=0$ andThe tensor part of the Linearized Einstein equations therefore givesIn a calculation strictly confined to the tensor sector, the source on this equation means its tensor projection; it is already trace-free there.

For a generic compact source, trace-free is not the same as transverse-traceless. For example, the static compactly supported spatial stress $T^{ij}={\partial }^i{\partial }^j\chi -{\gamma }^{ij}\Delta \chi$, with smooth compactly supported $\chi$, is divergence-free, but its trace-free part has divergence $2{\partial }^j\Delta \chi /3$, generally nonzero. There is a precise way to reproduce the unprojected trace-free formula used in the question while retaining this distinction. In Lorenz gauge in linearized gravity, the trace-reversed metric perturbation ${\bar{h}}_{ab}=h_{ab}-{\eta }_{ab}h/2$ obeysDefine its spatial trace-free half-strain byThen its wave equation has exactly the displayed trace-free source $\delta {\hat{T}}^{\alpha \beta }=\delta T^{\alpha \beta }-{\gamma }^{\alpha \beta }\delta T^{\mu }{}_{\mu ,\mathrm{spatial}}/3$. The physical tensor radiation is obtained afterwards with the transverse-traceless projector. Without that projection, this $E$ is a Lorenz-gauge trace-free coefficient, not necessarily a transverse tensor mode. At leading far-zone order, the same directional projector is applied to both sides of the quadrupole field formula below.

The flat contracted Bianchi identity implies stress-energy conservation, ${\partial }_b\delta T^{ab}=0$. LetCompact support removes boundary terms. Differentiate and integrate by parts once:A second time derivative, another integration by parts and symmetry of the stress-energy tensor giveTake the spatial trace-free part. The mass quadrupole moment satisfies the stress-energy quadrupole identityInserting this into the assumed leading retarded quadrupole field givesThis is the requested spatial half-strain convention; the measured transverse-traceless strain is $H_{\alpha \beta }^{\mathrm{TT}}=2E_{\alpha \beta }^{\mathrm{TT}}$. The formula selects the retarded source field and does not include an independently added homogeneous gravitational wave. Dropping the source-dependent part of retardation also needs the long-wavelength source approximation, besides $R/r\ll 1$.

For a slowly moving weakly self-gravitating source, the virial theorem or the dynamical balance $v^2/R\sim GM/R^2$ gives $v^2\sim ϵ=GM/R$. A varying nonspherical mass quadrupole moment has characteristic size $M{R}^2$ and dynamical time $\tau \sim R/v$, so $\ddot{Q}\sim M{R}^2/{\tau }^2\sim M{v}^2$. Consequently the quadrupole strain scaling of a weak self-gravitating source isThis is a characteristic scale, not a nonzero lower bound: a stationary or exactly spherical source does not radiate. Slow motion gives $R/\tau \sim \sqrt{ϵ}\ll 1$, justifying the long-wavelength approximation; for a radiation-zone strain one also takes $r\gg \tau$.