Deducing Archimedes's Principle from the momentum equation for incompressible fluids

Question

Consider an incompressible fluid, with constant density $\rho$, subject to a conservative body force, so that $g = −∇\chi$ for some potential function $\chi$.

$$\frac{ \partial{\textbf{u}}}{\partial{t} } +(\nabla \times \textbf{u} ) \times \textbf{u} = -\nabla \Big( \frac{p}{\rho}+\frac{1}{2} |\textbf{u}|^2 +\chi \Big)$$

Where $p$ is the pressure

How can you deduce Archimedes’ Principle from this, that an oject immersed in a fluid at rest experiences a buoyancy force equal to the weight of fluid displaced.

Answer 1

Half a year had past and it's ripe to answer the question :-)

It is fairly easy. First, since Archimedes’ Principle is essentially about hydrostatics, then $\boldsymbol u = 0$, i.e. nothing moves. Thus your equation turns into

$$\nabla p = - \rho \boldsymbol g$$

Then you have to integrate pressure over the whole body's surface to find the force exerted on the body by the fluid:

$$\boldsymbol F = \int_S p \cdot d \boldsymbol s = \int_V \nabla p \, dV = - \rho \boldsymbol g \int_V dV = - \rho V \boldsymbol g$$

$V$ is the volume of the body, $S$ is its surface. That's all. I've used Gauss theorem here:

$$\int_V \left(\nabla\cdot\mathbf{F}\right)dV=\int_S(\mathbf{F}\cdot\mathbf{n}) \, dS$$

However as you see I've integrated $\nabla p$ inside the body, just like there was fluid and no body. If you are a mathematician, I think you may say that given $p(\boldsymbol x)$ satisfying the hydrostatic equation outside the body, there is a unique smooth continuation of $p$ inside the body, satisfying the same equation.

But I'm rather a physician and I would say, that we can substitute the body by a bubble full of fluid and the forces acting on that bubble from the outside fluid won't change. That's just because these forces depend only on the shape of the body.

UPD Oops, I've missed the minus sign, I've corrected it --- the force exerted on the body is opposite to $\boldsymbol g$.