Physics with Calculus/Thermodynamics/Pressure and Temperature

In its most basic and most conceptual form, pressure is force per unit area. Of course, we all have a basic intuition about what pressure is. However, it's a little more slippery to define rigorously because we could in principle have a different pressure at every point, then it's not really force per unit area on a macroscopic scale. Furthermore, what force and what area are we even talking about?

What we really mean is that if we draw a small closed surface (like a sphere) around some point in a fluid, then the pressure is the outward force from the fluid inside the surface divided by the area or the surface. Of course, the surface must be orientable (you can't draw a Klein bottle around it), but more importantly, as the surface you draw gets smaller (say you choose a sphere, then shrink the radius), the pressure calculated must get closer to a definite value. Of course, these are all mathematical subtleties, and there are many more, all of which may be ignored.

One thing you should realize is that for any given surface, the total force on it is the surface integral of the pressure. Equivalently, for a small surface (infinitesimal), the force is merely the pressure times the area.

Notice also that pressure cannot produce a shear force, only a normal force.

Now, imagine a small rectangular prism of fluid of dimensions dx, dy, dz. Let's look at the two opposite sides normal to the x axis, let's call them A, and A'. Now, we want to find out the total force on this cube from the fluid surrounding it. The total force in the x direction is the force on A' minus the force on A. The force on A is the pressure at A times the surface area, let's call it dy dz, so P(x) dy dz. The force on A' is not quite the same; if we take the linear approximation, which becomes exact as dx, dy, and dz become smaller, then it is

$P(x)\,dy\,dz+{\frac {\partial P}{\partial x}}\,dx\,dy\,dz$. All the higher order terms can be neglected because when they are divided by dx dy dz, they will become zero.

The force in the x direction is then

${\frac {\partial P}{\partial x}}\,dx\,dy\,dz$.

Repeating the process for the other faces gives us

$d{\vec {F}}=\left({\frac {\partial P}{\partial x}},{\frac {\partial P}{\partial y}},{\frac {\partial P}{\partial z}}\right)\,dV$

Dividing through by dV (strictly speaking, we should be using delta's instead of d's, but you get the same result), and changing notation, we get

${\vec {F}}=\nabla P$

where ${\vec {F}}$ is force density.