Micropolar equations model In most of the articles about the 3D micropolar equations, these equations are written as
\begin{equation}
\begin{cases}
\partial_t \textbf{u}-(\nu+\kappa)\Delta \textbf{u} -2\kappa\nabla\times \textbf{w}+\textbf{u}\cdot\nabla \textbf{u}+\nabla\Pi=0,\quad x \in \mathbb{R}^3,\; t\geq 0,\\
\partial_t \textbf{w}-\gamma\Delta \textbf{w}+4\kappa \textbf{w}-\mu\nabla\nabla\cdot \textbf{w} -2\kappa\nabla\times \textbf{u}+\textbf{u}\cdot\nabla \textbf{w}=0, \\
\nabla \cdot \textbf{u}=0,
\end{cases}
\end{equation}
where
$\textbf{w} =\textbf{w}(x,t)=(w_1(x,t),w_2(x,t),w_3(x,t))$ the field of microrotation representing the angular velocity of the rotation of the fluid particles, $\textbf{u} =\textbf{u}(x,t)=(u_1(x,t),u_2(x,t),u_3(x,t))$ denotes the fluid velocity,  $\Pi(x,t)$ the scalar pressure, $\kappa$ denotes the micro-rotation viscosity, $\nu$ the Newtonian kinematic viscosity, and $\gamma$ and $\mu$ the angular viscosities.
I want to know if the above written equations are considered in the general form.
That is, for example, if we consider the following ones (we change the signe in front of $2\kappa\nabla\times \textbf{u}$)
\begin{equation}
\begin{cases}
\partial_t \textbf{u}-(\nu+\kappa)\Delta \textbf{u} -2\kappa\nabla\times \textbf{w}+\textbf{u}\cdot\nabla \textbf{u}+\nabla\Pi=0,\quad x \in \mathbb{R}^3,\; t\geq 0,\\
\partial_t \textbf{w}-\gamma\Delta \textbf{w}+4\kappa \textbf{w}-\mu\nabla\nabla\cdot \textbf{w} +2\kappa\nabla\times \textbf{u}+\textbf{u}\cdot\nabla \textbf{w}=0, \\
\nabla \cdot \textbf{u}=0,
\end{cases}
\end{equation}
do the last equations are also considered as micropolar equations?
Thanks in advance.
 A: No, you cannot simply flip the sign of $\kappa$ in the second equation.  The proper physical formulation of the (isotropic, incompressible) micropolar equations is
\begin{equation}
\begin{cases}
\rho(\partial_t u + u \cdot \nabla u) = \nabla \cdot \mathbb{T} + f \\
\operatorname*{div} u =0 \\
j(\partial_t \omega + u \cdot \nabla \omega) = \nabla \cdot \mathbb{M} +2 \mathbb{T}^\ast + g
\end{cases}
\end{equation}
where $\rho >0$ is the constant fluid density, $j >0$ is the constant microinertia, $f$ is the external force, $g$ is the external microtorque,  $\mathbb{T}$ is the stress tensor, $\mathbb{M}$ is the couple stress tensor, and the vector $\mathbb{T}^\ast$ is given by
\begin{equation}
\mathbb{T}^\ast_i = \sum_{a,b} \epsilon_{aib} \mathbb{T}_{ab}
\end{equation}
where $\epsilon$ is the Kronecker symbol.   The stress tensor is given by
\begin{equation}
\mathbb{T} = \mu(\nabla u + (\nabla u)^T) - p I + \kappa \left(\frac{1}{2} \nabla \times u - \omega \right)_\ast
\end{equation}
where for a vector $v$ the tensor $v_\ast$ is defined via
\begin{equation}
(v_\ast)^\ast = v,
\end{equation}
i.e. the lower star inverts the upper star mapping defined above.  The couple stress tensor is given by
\begin{equation}
\mathbb{M} = \alpha (\nabla \cdot \omega) I + \beta (\nabla \omega + (\nabla \omega)^T) + \gamma (\nabla \times \omega)_\ast.
\end{equation}
There are then various constraints placed on the coefficients $\mu,\kappa,\alpha,\beta,\gamma$ from the thermodynamic requirement that the Clausius-Duhem inequality holds:
$$
\mathbb{T} : (\nabla u - \omega_\ast) + \mathbb{M} : \nabla \omega \ge 0.
$$
All of this can be found in the old papers deriving the micropolar model, say those of Eringen.  The essential point for your question is that the parameter $\kappa$ couples to both $u$ and $\omega$ via $\mathbb{T}$ and so you can't flip the sign in one equation without breaking the structure in terms of the tensors.  Also, for the  physical reason mentioned above, as written we must have $\kappa >0$ in the micropolar model.
