# Interaction between body and fluid

I'm trying to understand locomotion of bodies in fluids with low Reynolds number and I'm having a hard time to understand the interactions between the body and the fluid.

Suppose $D\subset \mathbb{R}^n$ is a region filled with a fluid (on the plane if $n = 2$ and on the space if $n = 3$). If the fluid is incompressible and homogenous, then the velocity field $\mathbf{u}$ satisfies

$$\dfrac{D\mathbf{u}}{Dt} = -\nabla \tilde{p} + \nu \nabla^2\mathbf{u}$$

Where $\tilde{p} = p/\rho_0$ and $\nu = \mu/\rho_0$. This equation tell us about the evolution of the flow itself and we use together with it the no-slip boundary condition, that is, $\mathbf{u} = 0$ on $\partial D$.

Now, suppose a body is inside the fluid. We can think of the body as a subset $B\subset D$ and to "tell the equation" the body is there we also apply the boundary condition $\mathbf{u} = 0$ on $\partial B$ if this body is fixed (like a column or something like that). This aims just to change the geometry of the situation.

A different thing happen, however, if the body moves. In that case, we have not $B\subset D$ fixed, but a sequence of subsets, $B_t\subset D$, one for each $t$. We can then let $\sigma_t : A\subset \mathbb{R}^2\to D$ parametrize $\partial B_t$ and we gain a map $\sigma : A\times \mathbb{R} \to D$.

In that case, at time $t$, the velocity of the point on the boundary initialy labeled by $(a,b)\in \mathbb{R}^2$ will be simply

$$\mathbf{v}_B(a,b,t) = \dfrac{\partial \sigma}{\partial t}(a,b,t)$$

And then we change the no-slip condition to $\mathbf{u}(\sigma(a,b,t),t) = \mathbf{v}_B(a,b,t)$.

That is all fine, but now comes the part I don't understand: how do we model the interaction between the body and the fluid? I mean, there are things like drag force, buyoancy force and certainly the motion in a fluid with low or high Reynolds number should be different. I even read things like: "in low Reynolds number, the forces and torques on any body inside the fluid are zero".

My question is exactly how those interactions between the fluid and the body are modeled in the most general setting. I thought on using the stress tensor $T$ to compute forces per unit area on the body surface, but I'm not sure this is the way things are done.

Can someone give an outline about this or give a reference where I can read more?

• Where did you read "in low Reynolds number, the forces and torques on any body inside the fluid are zero"? It does not make sense - there should always be some viscous force, surely? – Floris Nov 19 '14 at 0:41
• @Floris, I've found some articles saying that the net forces and torques on the body should be zero, not each force. I forgot to write this down. The viscous force on the body is then obtained with the stress tensor as I imagined? – Gold Nov 19 '14 at 0:55

## 2 Answers

In your approach, drag force is a result of the calculation rather than additional effects: it is the surface integral of the normal stress vector $T n$ where $n$ is the normal to $B$.

Buoyancy force will arise if you add gravity to your fluid.

Of course in order to close the system you need to prescribe how the body moves: if you're making it move (imposed displacements) then what you get is the resisting force you feel, else you need to write a force balance equation for your body.

Can someone give an outline about this or give a reference where I can read more?

Low reynolds number (Laminar) flows are reversible. There is a really good old Video about the issue: https://www.youtube.com/watch?v=51-6QCJTAjU&list=PL0EC6527BE871ABA3&index=7

The most interersting stuff to your question is shown at 21:00 -> The precise answer at 27:00 ->

Short, because the laminar flow is reversible the normal propusion through turbulence is simply not possible. The forces and Torgues inside are indeed zero. Thus the only way to create a viscous propulsion is to move in a way which overcomes this reversibility. All this is shown in this video very understandable way.