# Momentum Equations for Micropolar Fluid

I am looking for the derivation of Momentum Equations for Micropolar Fluid $$\rho\frac{D\vec V}{Dt}=-\nabla p+(\mu+k_1^*)\nabla^2\vec V+k_1^*(\nabla\times\vec N^*)+\vec J\times\vec B ,\\ \rho j^*\frac{D\vec N^*}{Dt}=\gamma^*\nabla^2\vec N^*+k_1^*(-2\vec N^*+\nabla \times \vec V),$$

that how we can derive these equations for momentum and micro-rotation from the momentum equation for Navier-Stokes equation $$\rho \left[ \frac{\partial\vec V}{\partial t}+\vec V\cdot\nabla\vec V\right]=\text{div}\vec T+\vec F.$$

Also what can be the stress tensor and couple stress tensor for micro-polar fluid and stress tensor for Navier-Stokes equation.

• The equation with $\mathrm{div}\,\mathbf{T}$ is not the Navier-Stokes equation (which is a special case of it) but is more general one, and is (usually) called Cauchy's equation. Also, could you write your equations in TeX (or at least, crop the images to remove whitespace). – user23660 Oct 10 '13 at 7:10

Micropolar fluids are fluids with microstructures. They belong to a class of fluids with a nonsymmetric stress tensor. Micropolar fluids consist of rigid, randomly oriented or spherical particles with their own spins and microrotations, suspended in a viscous medium.

Physical examples of micropolar fluids can be seen in ferrofluids, blood flows, bubbly liquids, liquid crystals, and so on, all of them containing intrinsic polarities.

The following (including notations) is based on textbook G. Lukaszewicz, Micropolar Fluids: Theory and Applications, Birkhauser, Boston, 1999 http://books.google.ru/books?id=T3l9cGfR9o8C

We start with Cauchy momentum equation $$\rho \frac{D\vec{v}}{Dt} = \rho \vec{f} + \nabla \cdot \hat{T},$$ where $\vec{f}$ is body force and $\hat{T}$ is stress tensor.

If we assume that fluid is polar, that is it has its own internal angular momentum (independent of the motion of fluid as a whole) then we need an additional equation expressing conservation of angular momentum (for nonpolar fluid conservation of angular momentum is a consequence of Cauchy equation): $$\rho \frac{D}{Dt}(\vec{l} + \vec{x}\times\vec{v}) = \rho \vec{x} \times \vec{f} + \rho\vec{g} + \nabla \cdot (\vec{x} \times \hat{T} + \hat{C}).$$ Here $\vec{l}$ is an intrinsic (internal) angular momentum per unit mass, $\vec{g}$ is body torque and $\hat{C}$ is a new object called couple stress tensor (this equation could be seen as its definition).

Now in order to close this system of equation we need to express the stress tensor and couple stress tensor through the characteristics of fluid dynamics. For this we need to make certain assumptions: the absence of preferred direction and position, reduction to hydrostatic pressure in the absence of deformations and linear dependence on the velocity spatial derivatives $v_{i,j}$ (deformation). For polar fluid we also define the vector field $\vec{\omega}$ -- microrotation which represents the angular velocity of rotation of particles of the fluid. We further assume that the fluid is isotropic and $\vec{l} = I \,\vec{\omega}$ with $I$ a scalar called the microinertia coefficient. Couple stress tensor should be a linear function of the spatial derivatives of the microrotation field: $\omega_{i,j}$. All this assumptions allow us to specify the general form for stress tensor: $$T_{ij}=(- p +\lambda v_{k,k})\,\delta_{ij}+\mu\,(v_{i,j}+v_{j,i})+\mu_r\,(v_{j,i}-v_{i,j}) - 2 \mu_r\, \epsilon_{mij}\omega_m,$$ and couple stress tensor: $$C_{ij} = c_0\, \omega_{k,k} \delta_{ij} + c_d\, (\omega_{i,j}+\omega_{j,i}) + c_a \,(\omega_{j,i}-\omega_{i,j}).$$ Note: we have three parameters $c_0$, $c_d$ and $c_a$ (called coefficients of angular viscosities) because there are three irreducible representations for the action of $SO(3)$ group on rank 2 tensor: scalar (times $\delta_{ij}$), traceless symmetric tensor, antisymmetric tensor. The same for the triplet $(\lambda,\mu, \mu_r)$ (which are called second viscosity coefficient, dynamic Newtonian viscosity and dynamic microrotation viscosity).

Substituting this expressions for $T_{ij}$ and $C_{ij}$ into Cauchy equation and angular momentum equation we obtain the equations from the question. The notations correspondence is: $$\vec{\omega} \to \vec{N}^{*}, \qquad \mu_r \to k_1^{*}, \qquad I\to j^{*} \qquad c_a+c_d \to \gamma^{*}.$$ Additionally we see that the equations in the question assume $\mathrm{div} \vec{v} =0$ (incompressible flow) and $\mathrm{div} \vec{\omega} =0$ -- this one is generally not true, but symmetries of the system could make it so. Also we see that there are no body torque and the body force term corresponds to Lorentz force.