= Solution
Let $g=\dot\gamma$ and use $G_{yx}=g$ for the source's <velocity gradient> convention. Seek the homogeneous stationary <stress> reached under steady <simple shear flow>. The <material derivative> of that <stress tensor> vanishes. In the $xy$ plane,
$$
G=\begin{pmatrix}0&0\\g&0\end{pmatrix},\qquad
E=\frac12\begin{pmatrix}0&g\\g&0\end{pmatrix}.
$$
The $xx$, $yy$ and $zz$ components of the <constitutive equation> give, respectively,
$$
T_{xx}-(2\tau-\alpha)gT_{xy}=0,\qquad
T_{yy}+\alpha gT_{xy}=0,\qquad T_{zz}=0.
$$
Thus, putting $D=1+\alpha(2\tau-\alpha)g^2$ and using the <shear stress> calculated below,
$$
\boxed{T_{xx}=\frac{\mu(2\tau-\alpha)g^2}{D},\qquad T_{yy}=-\frac{\mu\alpha g^2}{D},\qquad T_{zz}=0.}
$$
The out-of-plane shear components vanish on the homogeneous branch. The two <normal-stress differences> are
$$
N_1=2\tau gT_{xy}=\frac{2\mu\tau g^2}{D},\qquad
N_2=-\alpha gT_{xy}=-\frac{\mu\alpha g^2}{D},
\qquad \boxed{\frac{N_2}{N_1}=-\frac{\alpha}{2\tau}.}
$$
The ratio is defined at nonzero shear when $N_1\ne0$; at $g=0$ both differences vanish, and the displayed constant is their limiting ratio. An arbitrary isotropic <pressure> in the total <Cauchy stress tensor> cancels from both differences; it is not an arbitrary addition to the constitutively defined extra <stress> $\mathsf T$ itself.
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