Direction of forces due to pressure

Question

If I understood correctly then if you have a 'pressure field' and you have a pressure difference between two points in it, then there is a force acting. (This is the explanation of Archimedes principle , the upthrust is due to pressure difference at top and bottom of immersed object)

So, I've started to think of pressure as a 'potential' of sorts sort of like 'voltage'. Now the problem is you define pressure using the equation

$${p} * \vec{dA} = \vec{dF}$$

But on the other hand, $$\vec{F} = -q \nabla V$$

What is weird to mean is that voltage is related to force (indirectly) by an integral while pressure is a potential defined using a derivative of sorts.

Particularly speaking can 'potential functions' be both defined by derivatives and integrals? I was of the opinion that you can only get them using integrals

Now coming to the main question, clearly force points in the direction of steepest descent of potential in the electrodynamics case. However how would one figure out the direction due to pressure?

Answer 1

In a material that cannot support shear stresses, i.e. in the inviscid approximation, the equations of static equilibrium can be written as:

$$-\nabla p + {\bf b} = 0$$

where ${\bf b}$ is the external force per unit volume required to equilibrate the pressure gradient. Hence the force per unit volume due to the pressure gradient itself is $-\nabla p$. Let's call that ${\bf b}_p$. Then we can integrate ${\bf b}_p$ over the entire volume of the sample to get the net force due to the pressure on the sample as:

$${\bf F}_{net}=\int_V {\bf b}_p dV=\int_V -\nabla p dV$$

Applying the divergence theorem to the far right-hand side gives:

$${\bf F}_{net}=-\int_S p{\bf n} dS$$

This is exactly what you should expect, and the integral of the external body force over the volume will then equilibrate this net force due to the pressure.

Answer 2

Taking a small cube of liquid in equilibrium, oriented according an arbitrary cartesian axis: $F_z(z + \Delta z) + \delta B_z - F_z(z) = 0$, where $\delta B_z$ is the component of a body force (the weight for example) in $z$ direction.

For a generic material, $F_z$ would produce normal and shear stresses. For a liquid with negligible viscosity, only normal stress is meaningful: $(\Delta x \Delta y)(\sigma_z(z + \Delta z) - \sigma_z(z)) + \delta B_z = 0$

The same for the other axis, resulting in:

$$\Delta x \Delta y\Delta \sigma_z + \delta B_z = 0$$ $$\Delta x \Delta z\Delta \sigma_y + \delta B_y = 0$$ $$\Delta y \Delta z\Delta \sigma_x + \delta B_x = 0$$

Dividing by $\Delta V = \Delta x \Delta y \Delta z$ and taking the limit when $\Delta V$ tends to zero:

$$\frac{\partial \sigma_z}{\partial z} + b_z = 0$$ $$\frac{\partial \sigma_y}{\partial y} + b_y = 0$$ $$\frac{\partial \sigma_x}{\partial x} + b_x = 0$$

where $b_i$ are the components of the body force per unit of volume.

Besides the absence of shear stresses, all normal stresses are equal at a given point of a liquid, so we can call them by the same name: $-p(x,y,z)$. The minus sign is because they are compressive stresses in the volume element.

We can then write, (and now taking $\mathbf b = \mu \mathbf g$):

$$\nabla p - \mu \mathbf g = 0$$