forces - Mathematical Reasoning for Fluid Pressure as a Scalar

This question from a while ago and answers/comments to this question from earlier today both make heavy mention of the fact that fluid pressure is a scalar. Although this information was surprising to me, it makes sense intuitively, especially when considering Pascal's Principle and liquid depth pressure. However, I struggle when thinking about the mathematical reasons why pressure is a scalar, and the answers to the first question linked either appealed to intuition, seemed contentious, or unsatisfying.

My confusion comes from the fact that pressure is often defined at the high school/undergrad level as $$P\equiv \frac{F}{A}$$ where this not just a relationship, but often presented as the equation definition of pressure. Force is a vector, and I had assumed that area was a scalar. Therefore, I had always figured this was a scalar multiplication problem, and the product would also be a vector. Now that I know that pressure is a scalar, I just can't see how you would get a scalar for pressure out of this relationship. I see two possibilities:

1) For reasons I don't currently understand, area is treated as a vector in continuum physics. As a result, some type of vector algebra must be applied to divide a vector by another vector in order to yield a scalar. If this is the case, I would like to see at least a framework for this math in the answer.

2) If possibility 1 is not the case, then the other possibility in my mind is that $P=\frac{F}{A}$ is not truly the definition of pressure, and there exists some more basic definition for pressure that would explain its scalar nature.

Are either of these explanations correct, or is there a third possibility that I have not considered?

I've tried to make this a substantially different question than either of the other two questions that I've linked to, but if that does not appear to be the case, please let me know.

Answer

I would say pressure is better defined by $$ \vec{F} = P \vec{A}. $$ Yes, we are defining a quantity without having it all alone on the left-hand side. And yes, area is a vector. And as you guessed trying to divide one vector by another leads to trouble, so we won't do it.

Let me explain where this comes from and what it is shorthand for. In continuum mechanics we are often interested in flows of some quantity or another. In particular, we might want to consider flows of momentum, where these flows can go in any direction.

Let's work in Cartesian coordinates for simplicity. Then one can imagine $x$-, $y$-, or $z$-momentum flowing in the $x$-, $y$-, or $z$-direction, where the two coordinates are chosen independently. That is, there can be $x$-momentum moving in the $x$-direction, or it can be moving in the $y$-direction, or whatever. The rate of flow of $i$-momentum ($i$ being $x$, $y$, or $z$) per unit area in the $j$-direction is the $ij$-component of the stress tensor $T$: $T_{ij}$. And the time-rate-of-change of momentum is just another way of saying "force."¹

The way this tensor works is you choose a vector area $\vec{A}$. This is a vector with magnitude given by the same scalar everyone is familiar with, and also with direction given by the direction orthogonal to the surface. The application of $T$ to $\vec{A}$ gives another vector,² which is the force acting on the surface with area $\vec{A}$.

Now stress tensors appear in all sorts of places in physics, but they all behave similarly. In particular, their diagonal components are things like the $x$-force in the $x$-direction, which is your intuition for pressure: it directly pushes on things, as opposed to dragging them like shear. In fact, the off-diagonal elements are the shear forces per unit area.

In an isotropic medium, like a well-behaved fluid, the diagonal elements are in fact all the same: "$x$-pressure" is identical to "$y$-pressure." Call these values $P$. In the absence of shear, the stress tensor has a particularly simple representation: $$ T = \begin{pmatrix} P & 0 & 0 \\ 0 & P & 0 \\ 0 & 0 & P \end{pmatrix}. $$ Then clearly for any area $\vec{A}$ we have $\vec{F} = T \vec{A} = P \vec{A}$.

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¹Many treatments would just omit the whole talk of momentum I gave and express everything up to now as forces (per unit area). But the momentum way of looking at things works particularly well for generalizing the 3-dimensional stress tensor to the 4-dimensional stress-energy tensor used in relativity, and this is my field of study. Just thinking of forces rather than changes in momentum over time might work better in statics problem (there is still momentum flow, but it's hard to visualize since "nothing" is moving).

²If you are not familiar with tensor manipulations, it suffices here to think of the components $T_{ij}$ arranged as a matrix, and to think of the components $A_j$ arranged in a column vector. Then matrix multiplication tells you the components of the column vector $T \vec{A}$ are $F_i = T_{ij} A_j$ (summing over repeated indices).