# Navier-Stokes equations for rotating falling object

I am considering the problem of a rigid body moving in a fluid, free to translate and rotate (say, for example, a falling body in a fluid). Some of the approaches I found in some articles involve a form of the NS equations such as:

$$\frac{\partial\mathbf{v}}{\partial t} + \left[\left( \mathbf{v}-\mathbf{u}-\Omega \times \mathbf{r} \right) \cdot \nabla \right] \mathbf{v} + \Omega \times \mathbf{v} = -\nabla p + \nu \Delta \mathbf{v} \\ \nabla \cdot \mathbf{v}=0$$

Here, the body moves with rigid motion $$\mathbf{u} + \Omega \times \mathbf{r}$$ while $$\mathbf{v}$$ is the fluid velocity. In particular, when some authors present this equation they specify that $$\mathbf{v}$$ is the velocity field with respect to a fixed frame of reference "projected" onto a system of axes which rotate with the body (but the origin of these axes is fixed). Furthermore, these equations hold on a control volume that moves with the body. However, it is not clear to me how to rigorously derive these equations, and what this "projection" actually is.

I know that Arbitrary Eulerian Lagrangian formulations involve using as reference an arbitrary frame, that is, if at time $$t$$ the fluid domain around the moving body is $$\mathcal{F}(t)$$, having defined a sufficiently smooth mapping $$\hat{T}_f (t): \hat{\mathcal{F}} \rightarrow \mathcal{F}(t)$$, I can pull back the equations on a fixed domain $$\hat{\mathcal{F}}$$. In particular starting from $$\rho_f (\frac{\partial\mathbf{v}}{\partial t} + \mathbf{v} \cdot \nabla \mathbf{v}) = \text{div } \boldsymbol{\sigma}_f$$

where $$\boldsymbol{\sigma}_f$$ is the stress tensor and the equations are defined on $$\mathcal{F}(t)$$, if one defines $$\hat{\mathbf{v}}(\hat{x},t) := \mathbf{v} \left( \hat{T}_f(\hat{x},t),t \right), \qquad \hat{p}(\hat{x},t) := p \left( \hat{T}_f(\hat{x},t),t \right) \qquad \forall \hat{x} \in \hat{\mathcal{F}}$$

standard pullbacks and Piola transforms yield the NS equations for a moving domain on the reference $$\hat{\mathcal{F}}$$: $$\rho_f \hat{J}_f \left( \frac{\partial \hat{\mathbf{v}}}{\partial t} + \hat{\mathbf{F}}_f^{-1} (\hat{\mathbf{v}} - \partial_t \hat{T}_f ) \cdot \nabla \hat{\mathbf{v}} \right) = \hat{\text{div }} \left( \hat{J}_f \hat{\boldsymbol{\sigma}_f} \hat{\mathbf{F}}_f^{-T} \right)$$

where the Cauchy stress tensor in the reference system is

$$\hat{\boldsymbol{\sigma}_f} := -\hat{p} \mathcal{I} + \rho_f \nu_f \left( \hat{\nabla} \hat{\mathbf{v}} \hat{\mathbf{F}}_f^{-1} + \hat{\mathbf{F}}_f^{-T} \hat{\nabla} \hat{\mathbf{v}}^T \right)$$ and $$\hat{\mathbf{F}}_f := \hat{\nabla} \hat{T}_f, \qquad \hat{J}_f := \text{det} \left( \hat{\mathbf{F}}_f \right) >0$$

Now my hope is that if $$\partial_t \hat{T}_f = \mathbf{u} + \Omega \times \mathbf{r}$$, where $$\Omega$$ is the axial vector of the skew matrix $$\dot{Q}Q^T$$, and $$\hat{\mathbf{F}}_f = \hat{\nabla} \hat{T}_f = Q$$, with $$Q$$ orthogonal rotation matrix, then somehow the equations simplify to the one I am trying to obtain, but I cannot work out the details. The final goal as I understand it is that if the control volume follows the body, it can be seen as fixed and the numerical solution of the problem is simplified since the mesh discretization will not move or be deformed. Is it possible to obtain the starting equation with a pullback on a reference fixed domain, when the transformation is a rotation?

If I understand this correctly, the first equation is written in a possibly non-inertial reference frame (the rigid body), whose center of mass has coordinate $$\widehat{\mathbf{x}}_G$$ in the absolute reference frame, and $$\mathbf{r} = \mathbf{x}-\mathbf{x}_G = \widehat{\mathbf{x}} - \widehat{\mathbf{x}}_G=\widehat{\mathbf{r}}$$.
Then $$\mathbf{u}$$ would be just the time derivative of the center of mass of the rigid body in the absolute reference frame, $$\mathbf{u} = \frac{d}{dt}\widehat{\mathbf{x}}_G.$$
$$\widehat{\mathbf{x}} = \widehat{\mathbf{x}}_G + \widehat{\mathbf{r}} = \int_0^t \mathbf{u}dt + \int_0^t \Omega\times \mathbf{r} dt,$$ where the cross product $$\Omega\times\mathbf{r}$$ appears just to keep track of a possible rotation of the two reference frames.