Theorem behind Archimedes principle of buoyancy?

[Hiero]

I was thinking about why the buoyant force on an object should depend solely on it's volume and not shape. It seems loosely like the divergence theorem in that an integral over the surface is determined by the volume. There is a big difference though; in the divergence theorem we integrate scalars (flux/divergence) but to find the buoyant force we must integrate a vector.

By making the vector-analogous arguments behind the divergence theorem, I am led to the following conclusion. Suppose you have a scalar field P, and a volume V with surface S(V), then I believe:
_S(V)∫∫ P dA = _V∫∫∫ ∇P dV (where dA is the outward pointing area-element)

In the example of buoyancy, the scalar field P is the pressure P = C-ρgz with C being constant, ρ being the fluid density, g being the gravitational field strength, and +z being vertically upwards. Then ∇P = -ρge_z.
The net force due to fluid pressure is F = ∫∫ P (-dA) = -∫∫∫ ∇S dV = -∫∫∫(-ρge_z) dV = ρge_z∫∫∫ dV = (ρV)ge_z; Which is Archimedes principle.
(The -dA is because force comes from integrating P over the inward pointing normal.)I am wondering what is the name of this theorem I have stated? Is it somehow just a restatement of the divergence theorem? (If so; how to get between the two?) I cannot find anything about it.

 

[Orodruin]

It is a generalisation of the divergence theorem. It can be easily derived from the divergence theorem by multiplying it with each of the Cartesian basis vectors in turn or derived using the same argumentation that leads to the divergence theorem. In fact, the divergence theorem is just a collection of a number of integrals of this form.

 

[sophiecentaur]

I was thinking about why the buoyant force on an object should depend solely on it's volume and not shape.

I think this is counter intuitive, in the same way that it doesn't depend on orientation, either. The reason for this unsatisfactory nagging in the back of one's mind is that it is easy to confuse Force with Potential. Potential depends both on shape and orientation and I think the intuitive abjection is based on this ( and subjective experience, of course).

 

[Hiero]

Thank you @Orodruin, I now see the connection... a vector field is just a collection of 3 (in 3-Dim.) scalar fields. Let us choose Cartesian (orthonormal & coordinate independent) basis vectors e_x, e_y, e_z. The divergence of a vector field P = P_xe_x + P_ye_y + P_ze_z can be written as ∇⋅P = e_x⋅∇P_x + e_y⋅∇P_y + e_z⋅∇P_z ... So to get the 'original' flux/divergence theorem from my version, we do like this:

∫∫∫ ∇⋅P dV = ∫∫∫ (e_x⋅∇P_x + e_y⋅∇P_y + e_z⋅∇P_z) dV = e_x⋅∫∫∫ ∇P_x dV + e_y⋅∫∫∫ ∇P_y dV + e_z⋅∫∫∫ ∇P_z dV =*(by my version of the theorem)*= e_x⋅∫∫ P_x dA + e_y⋅∫∫ P_y dA + e_z⋅∫∫ P_z dA = ∫∫ (e_xP_x+e_yP_y+e_zP_z)⋅dA = ∫∫ P⋅dA

The other direction (showing my version results from the original) is eluding me. :\@sophiecentaur Sorry, I did not catch your drift :\ (what potential are we speaking of?)

 

[Orodruin]

The other direction (showing my version results from the original) is eluding me. :\

It is not much more difficult. Given the divergence theorem, you have that
$$
\vec e_i \cdot \oint_S p\, d\vec S = \oint_S p\vec e_i \cdot d\vec S = \{\mbox{divergence theorem}\} = \int_V \nabla \cdot p\vec e_i \, dV = \vec e_i \cdot \int_V \nabla p \, dV,
$$
since $\nabla\cdot \vec e_i = 0$. Thus, all components of $\oint_S p\, d\vec S$ and $\int_V \nabla p \, dV$ are the same and the integrals must therefore be the same.

 

[Chestermiller]

I have a different version. The normal force exerted by the surrounding fluid on a differential section of the object surface is $-p\mathbf{n}dA$ where $\mathbf{n}$ is the local outward directed unit normal. The upward component of this force is $-p(\mathbf{n}\cdot \mathbf{e_z})dA=-(p\mathbf{e_z})\cdot \mathbf{n}dA$. So, applying the divergence theorem, we get:
$$F=\int_A{-(p\mathbf{e_z})\cdot \mathbf{n}dA}=-\int_V{\nabla \cdot }(p\mathbf{e_z})dV=\rho g V$$

 

[Orodruin]

I have a different version. The normal force exerted by the surrounding fluid on a differential section of the object surface is $-p\mathbf{n}dA$ where $\mathbf{n}$ is the local outward directed unit normal. The upward component of this force is $-p(\mathbf{n}\cdot \mathbf{e_z})dA=-(p\mathbf{e_z})\cdot \mathbf{n}dA$. So, applying the divergence theorem, we get:
$$F=\int_A{-(p\mathbf{e_z})\cdot \mathbf{n}dA}=-\int_V{\nabla \cdot }(p\mathbf{e_z})dV=\rho g V$$

This is actually nothing else than the $i = 3$ case in post #5.

 

[Charles Link]

I think I could add something to this=perhaps it was already mentioned: There is a force per unit volume in the fluid given by $\vec{f}_V=-\nabla p$. If there is equilibrium, this force per unit volume due to the pressure gradient must be equal and opposite the gravitational force per unit volume $\vec{f}_g=-\delta g \hat{z}$ where $\delta$ is the density of the fluid. Thereby, we have $\nabla p=-\delta g \hat{z}$. This is used in evaluating the integrals of post #5 by @Orodruin to compute the total buoyant force .

 

[Orodruin]

This is used in evaluating the integrals of post #5 by @Orodruin .

No it isn't. All that was used was the divergence theorem. The scalar field $p$ does not need to be the pressure, the theorem is general.

 

[Charles Link]

No it isn't. All that was used was the divergence theorem. The scalar field $p$ does not need to be the pressure, the theorem is general.

I agree, but you didn't evaluate your last integral. In @Chestermiller version, (post #6), he wrote the final result (he evaluated the integral), but didn't show where it comes from.

 

[Orodruin]

I agree, but you didn't evaluate your last integral. In @Chestermiller version, he wrote the final result, but didn't show where it comes from.

The integral was already evaluated in the OP. The question was about the relation to the divergence theorem.

 

[Hiero]

It is not much more difficult. Given the divergence theorem, you have that
$$
\vec e_i \cdot \oint_S p\, d\vec S = \oint_S p\vec e_i \cdot d\vec S = \{\mbox{divergence theorem}\} = \int_V \nabla \cdot p\vec e_i \, dV = \vec e_i \cdot \int_V \nabla p \, dV,
$$
since $\nabla\cdot \vec e_i = 0$. Thus, all components of $\oint_S p\, d\vec S$ and $\int_V \nabla p \, dV$ are the same and the integrals must therefore be the same.

Brilliant! Thank you Orodruin, you are consistently insightful when I come to PF, so thank you for your time invested.

@Chestermiller I like and dislike your method. Firstly why I dislike it; because you used the symmetry of that particular field to claim that only the z component (of force) will survive. I prefer the form in the OP which works regardless of simplifying symmetries. But I like your presentation because it makes very clear the connection between the two forms of the theorem... (If Orodruin had not yet posted#5, your presentation would have inspired me towards it.) For the general case we just do what you did, in each component.

 

[Chestermiller]

This is actually nothing else than the $i = 3$ case in post #5.

Oops. I reinvented the wheel...again.

 

[sophiecentaur]

@sophiecentaur Sorry, I did not catch your drift :\ (what potential are we speaking of?)

The Potential Energy by virtue of its shape or orientation will change (which is why a stick floats horizontally and an iceberg rolls over occasionally etc. etc..). This is despite the displaced volume being the same. That's all.

 

[WignerE]

($\:h\:$ = depth of immersed horizontal surface from the rest open surface of the fluid)

(1) Firstly : Horizontal hydrostatic pressure force cancels out

Cut your body horizontally and take any section with infinitesimal height $\:\mathrm{d}h_{1}\:$ as in Figure. Then
\begin{align}
\mathbf{F}_{\text{horizontal}}&=\sum_{m=1}^{m=N}\left(- p\right)\Delta\mathbf{s}_{m}=\sum_{m=1}^{m=N}\left(- p\right)\left[\Delta\mathbf{r}_{m}\boldsymbol{\times}\left( \mathrm{d} h_{1}\mathbf{k}\right)\right]
\nonumber\\
&=\left(- p\right)\underbrace{\left(\sum_{m=1}^{m=N}\Delta\mathbf{r}_{m}\right)}_{=\mathbf{0}}\boldsymbol{\times}\left( \mathrm{d}h_{1}\mathbf{k}\right)=\mathbf{0}
\tag{01}
\end{align}
Don't worry if the perimeter of your cross section is a closed curve instead of a closed polygon. Then we have differentials $\:\mathrm{d}\:$ in place of Deltas $\:\Delta \:$ and integrals instead of sums
\begin{align}
\mathbf{F}_{\text{horizontal}}&=\oint\left(- p\right)\mathrm{d}\mathbf{s}=\oint\left(- p\right)\left[\mathrm{d}\mathbf{r}\boldsymbol{\times}\left(\mathrm{d}h_{1}\mathbf{k}\right)\right]
\nonumber\\
&=\left(- p\right)\underbrace{\left(\oint \mathrm{d}\mathbf{r}\right)}_{=\mathbf{0}}\boldsymbol{\times}\left( \mathrm{d}h_{1}\mathbf{k}\right)=\mathbf{0}
\tag{02}
\end{align}
(2) Secondly : Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.

Of course if this was a plate alone in a fluid then the upward buoyant force exerted by the fluid would be
\begin{equation}
\mathbf{B}_{\text{buoyant}}= p\left(h\right)\mathbf{S}_{A}-p\left(h-\mathrm{d} h_{1}\right)\mathbf{S}_{A}
=\rho g\underbrace{\mathrm{d} h_{1}S_{A}}_{V_{A}}\mathbf{k}=\left(\rho g V_{A}\right)\mathbf{k}
\tag{03}
\end{equation}
Now, on the first plate $\:A\:$of horizontal surface $\:S_{A}\:$ and infinitesimal height $\:\mathrm{d}h_{1}\:$ put the next plate$\:B\:$ of the body of horizontal surface $\:S_{B}\:$ and infinitesimal height $\:\mathrm{d}h_{2}\:$. Then
\begin{align}
\mathbf{B}_{\text{buoyant}}&= \underbrace{\left[-p\left(h\right)\left(-\mathbf{S}_{A}\right)\right]}_{A\: bottom}+\underbrace{\left[-p\left(h-\mathrm{d} h_{1}\right)\left(\mathbf{S}_{A}-\mathbf{S}_{B}\right)\right]}_{step}+\underbrace{\left[-p\left(h-\mathrm{d} h_{1}-\mathrm{d} h_{2}\right)\mathbf{S}_{B}\right]}_{B\: top}
\nonumber\\
&=\rho g\underbrace{\mathrm{d} h_{1}S_{A}}_{V_{A}}\mathbf{k}+\rho g\underbrace{\mathrm{d} h_{2}S_{B}}_{V_{B}}\mathbf{k}=\rho g \left(V_{A}+V_{B}\right)\mathbf{k}
\tag{04}
\end{align}
Any body could be cut in horizontal plates of finite surface area and infinitesimal height.