# 'Eulerian' description of a rigid body submerged in fluid

In this paper, equations of rigid body motion (eq 4 and 5 in the paper) are written in Eulerian form (eq 12 in the paper). The rigid body is submerged in a viscous incompressible fluid. $$m\frac{dw_G(t)}{dt}=\int_{\partial\Omega(t)}\sigma n\,ds+\int_{\Omega(t)}\rho f\,dx,$$

$$I\frac{dR(t)}{dt}=R\times(IR)+\int_{\partial\Omega(t)}r\times\sigma n \,ds+\int_{\Omega(t)}\rho r\times f\,dx,$$

are transformed into

$$\partial_t(\rho w)+\mathrm{div}(\rho w\otimes w)=\rho f+\frac{1}{\rho}\mathrm{div}(\rho\Sigma)-\frac{1}{\rho}\sigma\nabla\rho,$$ where $\rho$ is the density of the solid, $\sigma$ is the stress tensor of the fluid in which the solid is submerged, $\Sigma$ is being called an internal rigidity force and $f$ is an external force.

Could someone help me to understand how the 'Eulerian' equation has been obtained from the earlier equations?

• Looks like stokes theorem was used to get sine volume integrals. Are you familiar with the conversion from integral formulation to differential for the normal hydro equations? Commented Jun 9, 2015 at 12:47
• Commented Jun 9, 2015 at 14:36

Giving a full answer would take long, but here are a few steps to help you. The easiest is probably to start from the result, eq. (11). Multiply by a vector test function and integrate it over a given volume $V(t)$ within the solid phase, which is advected by the velocity $w$. Replace $w(x,t)$ by its value, defined in (3) -- note that $x_G$ has for derivative $w_G$.
Then taking the test function equal in turn to each direction of space $e_i$, or to $x \times e_i$, you will obtain the six scalar equations corresponding to (4) and (5).
Note that $\nabla \rho/\rho$ is a way to obtain the outward-pointing normal vector.