Let $R = \beta^{1/2} \text{exp}(\vec E/2)$ where $\beta \in \mathbb R$ and $\vec E$ is a timelike bivector, $\hat \rho = P + \vec S$, and $R^\dagger R = \beta (L_0 + \vec L)$. Take $\vec L (\vec S_\parallel + \vec S_\perp) = (\vec S_\parallel - \vec S_\perp) \vec L$. Then
$
\begin{aligned}
R \hat \rho R^\dagger &= \beta (PL_0 + P \vec L + \vec S_\parallel L_0 + \vec S_\parallel \cdot \vec L + \vec S_\perp)
\end{aligned}
$
and
$
\begin{aligned}
\frac{1}{2}\big\{R^\dagger R, \hat \rho\big\} &= \beta \big( P L_0 + P \vec L + \vec S L_0 + \vec S \cdot \vec L \big) \\
&= \big(R \hat \rho R^\dagger - \beta \vec S_\perp\big) + \beta \vec S_\perp L_0 \\
&= R\hat \rho R^\dagger + \beta (L_0 -1) \vec S_\perp
\end{aligned}
$
which imply
$
\begin{aligned}
R \hat \rho R^\dagger - \frac{1}{2}\big\{R^\dagger R, \hat \rho\big\} &= \beta (1-L_0) \vec S_\perp.
\end{aligned}
$
Note that
$R \hat \rho R^\dagger - \langle R\hat \rho R^\dagger\rangle = \langle R \hat \rho R^\dagger\rangle_2 = (R\rho \widetilde R)\wedge \gamma_0= \beta(P \vec L + \vec S_\parallel L_0 + \vec S_\perp)$
carries significantly more boost information than the Lindblad term, and is also trace-free.
*Why does the Lindblad throw away so much of the boost?*