(\[inta\]). The latter equation shows that $\lambda_{0}^{b(1)}=0$.

Any term in $\lambda_{\mu\nu}^{(2)}$ which is not strictly required by the integrability conditions can be absorbed into $\lambda_{\mu\nu}^{(1)}$, so there is no loss of generality if we take $\lambda_{00}^{(2)}= 0$. Moreover, from Eq. (\[s2\]) we see that $\lambda_{0j}^{(2)}=0$ and $\lambda_{j}^{a(1)}=0$ is a solution to the integrability condition. Therefore, we can write \_[ij]{}\^[(2)]{}=-\_[ij]{} where we have put =

The quadratic equations then read R\[FR\[\^[(2)]{}\]\] = -N \^2 + (\^[(1)]{})\^2 F \[sec1\] and R\[FR\[\^[b(2)]{}\]\] = -\^2 \^[b(1)]{} \[secb\]

Note that the left hand side of Eq. (\[sec1\]) vanishes when integrated against $\omega$ (which means that the equation has a solution) but does not vanish when integrated agaist $\omega/F$. Therefore, the solution to Eq. (\[sec1\]) is \^[(2)]{} = (\^[(1)]{})\^2 - N - A \[zeta20\] where A= \^[-1]{}

Similarly, we obtain from Eq. (\[secb\]) \^[b(2)]{} = - \^[b(1)]{} + \[zeta2b\] where B\^b = \^[b(1)]{}\^2 Note that $\Lambda \propto (\zeta^{(1)})^2$ is already quadratic in $\tau$, which means that $\zeta^{b(2)}$ is actually third order and therefore can be neglected.

We are ready to compute $\Pi_2^{\mu\nu}$ and $J_2^{a\mu}$; see Eqs. (\[shearfull\]) and (\[jdfull\]). We have \_2\^ = p\^p\^Note that $\Pi_2^{0i}=0$, but $\Pi_2^{00}\neq 0$, so $\Pi_2^{\mu\nu}$ is not the true correction 