PlanetPhysics/Archimedes Principle

Archimedes' Principle states that

When a floating body of \htmladdnormallink{mass {http://planetphysics.us/encyclopedia/CosmologicalConstant.html} $$M$$ is in equilibrium with a fluid of constant density, then it displaces a mass of fluid $$M_d$$ equal to its own mass; $$M_d = M$$.}

Archimedes' principle can be justified via arguments using some elementary classical mechanics. We use a Cartesian coordinate system oriented such that the $$z$$-axis is normal to the surface of the fluid.

Let $$\mathbf{g}$$ be The Gravitational Field (taken to be a constant) and let $$\Omega$$ denote the submerged region of the body. To obtain the net force of buoyancy $$\mathbf{F}_B$$ acting on the object, we integrate the pressure $$p$$ over the boundary of this region $$ \mathbf{F}_B = \int_{\partial\Omega}{-p\mathbf{n}\,dS} $$Where $$\mathbf{n}$$ is the outward pointing normal to the boundary of $$\Omega$$. The negative sign is there because pressure points in the direction of the inward normal. It is a consequence of Stokes' theorem that for a differentiable scalar field $$f$$ and for any $$\Omega\subset\mathbb{R}^3$$ a compact three-manifold with boundary, we have $$ \int_{\partial\Omega}{f\mathbf{n}\,dS} = \int_\Omega{\nabla f\,dV} $$therefore we can write $$ \mathbf{F}_B = -\int_\Omega{\nabla p\, dV} $$Now, it turns out that $$\nabla p = \rho_f\mathbf{g}$$ where $$\rho_f$$ is the volume density of the fluid. Here is why. Imagine a cubical element of fluid whose height is $$\Delta z$$, whose top and bottom surface area is $$\Delta A$$ (in the $$x-y$$ plane), and whose mass is $$\Delta m$$. Let us consider the forces acting on the bottom surface of this fluid element. Let the z-coordinate of its bottom surface be $$z$$. Then, there is an upward force equal to p(z)\Delta A\mathbf{e}_z$$ on its bottom surface and a downward force of $$-p(z + \Delta z)\Delta A\mathbf{e}_z + \Delta m\mathbf{g}. These forces must balance so that we have $$ p(z)\Delta A = p(z + \Delta z)\Delta A - \Delta m|\mathbf{g}| $$a simple manipulation of this equation along with dividing by $$\Delta z$$ gives $$ \frac{p(z + \Delta z) - p(z)}{\Delta z} = \frac{\Delta m}{\Delta A\Delta z}|\mathbf{g}| = \frac{\rho_f \Delta A\Delta z}{\Delta A\Delta z}|\mathbf{g}| = \rho_f |\mathbf{g}| $$taking the limit $$\Delta z \to 0$$ gives $$ \frac{\partial p}{\partial z} = \rho_f|\mathbf{g}| $$Similar arguments for the $$x$$ and $$y$$ directions yield $$ \frac{\partial p}{\partial x} = \frac{\partial p}{\partial y} = 0 $$putting this all together we obtain $$\nabla p = \rho_f\mathbf{g}$$ as desired. Substituting this into the integral expression for the buoyant force obtained above using Stokes' theorem, we have $$ \mathbf{F}_B = -\int_\Omega{\rho_f\mathbf{g}\, dV} = -\rho_f\mathbf{g}\int_\Omega{dV} = -\rho_f\mathbf{g}=Vol= (\Omega) $$where we can pull $$\rho_f$$ and $$\mathbf{g}$$ outside of the integral since they are assumed to be constant. But notice that $$\rho_f=Vol= (\Omega)$$ is equal to $$M_d$$, the mass of the displaced fluid so that $$ \mathbf{F}_B = -M_d\mathbf{g} $$But by Newton's second law, the buoyant force must balance the weight of the object which is given by $$M \mathbf{g}$$. It follows from the above expression for the buoyant force that $$ M_d = M $$which is precisely the statement of Archimedes' Principle.