I A proof of Archimedes' Principle

[DaTario]

TL;DR

Hi All, is there a proof to Arquimedes Principle and the expression for the buoyancy force?

Hi All, is there a proof to Arquimedes Principle and the expression for the buoyancy force? In case there is a proof, may we refer to this as Arquimedes theorem? (Buoyancy force = density of the fluid x acceleration of gravity x submerged volume)

Best Regards,
DaTario

 

[jbriggs444]

Hi All, is there a proof to Arquimedes Principle and the expression for the buoyancy force? In case there is a proof, may we refer to this as Arquimedes theorem? (Buoyancy force = density of the fluid x acceleration of gravity x submerged volume)

It follow more or less automatically from Pascal's principle that the pressure in a stationary fluid at equilibrium and under no net force is the same everywhere.

You layer onto this the idea that a uniform gravity field on a fluid of uniform density results in a fixed pressure at any given depth given by $P=\rho g h$.

Then you realize that "buoyancy" is nothing more than the difference in the net force along the bottom of an object compared to the force along its top.

From there you realize that the net force difference on any given column within the object is equal to the volume of that column times $\rho$ times $g$. Add up all the columns and you get that $\sum F=\rho g V$.

If one is going to be careful about the proof, one needs that the object is completely surrounded by the fluid and that the fluid is all connected -- free to flow so as to equalize pressure.

 

[ergospherical]

It's fairly straightforward to transform the buoyant force integral $-\int_{\partial V} p d\mathbf{S}$ into a volume integral over $V$ by pure vector calculus.

A more intuitive explanation is that said buoyant force integral depends only on the shape of the boundary $\partial V$. Therefore, for the purpose of determining the buoyant force, one can replace the body with fluid. The parcel of fluid is obviously in equilibrium along with the rest of the fluid, which implies that the buoyant force is nothing but minus the weight of the parcel of fluid.

 

[vanhees71]

Just to work that out: For a static fluid the total force density must be 0, i.e.,
$$-\vec{\nabla} P +\rho \vec{g}=0.$$
This implies that the buoyancy force of a body with volume $V$ and boundary surface $\partial V$ is
$$\vec{F}_{\text{buo}}=-\int_{\partial V} \mathrm{d}^2 \vec{f} P=\int_V \mathrm{d}^3 x (-\vec{\nabla} P)=-\vec{g} \int_V \mathrm{d}^3 x \rho=-\vec{g} m_V,$$
where $m_V$ is the mass of the displaced fluid, which is Archimedes's principle.

 

[DaTario]

Just to work that out: For a static fluid the total force density must be 0, i.e.,
$$-\vec{\nabla} P +\rho \vec{g}=0.$$
This implies that the buoyancy force of a body with volume $V$ and boundary surface $\partial V$ is
$$\vec{F}_{\text{buo}}=-\int_{\partial V} \mathrm{d}^2 \vec{f} P=\int_V \mathrm{d}^3 x (-\vec{\nabla} P)=-\vec{g} \int_V \mathrm{d}^3 x \rho=-\vec{g} m_V,$$
where $m_V$ is the mass of the displaced fluid, which is Archimedes's principle.

So we use Gauss Law, basically. Is there any obstacle to including every possible solid shape in this proof? My guess is that this is the reason why Arquimedes Principle is not named as a law or a theorem (as in Stevin's).

 

[vanhees71]

Yes, it's Gauss's law, and the proof shows that it's completely general, i.e., independent of the shape of the solid. Of course it rests on Pascal's Law that the static stress tensor for a fluid is isotrophic, i.e., $\sigma_{jk}=-P\delta_{jk}$. It's a good question, why it's named Archimendes's Principle and not Archimendes's Law or Theorem. I guess it's simply for historical reasons.