# 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.

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 $\vec{F} = T \vec{A} = P \vec{A}$.

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).

Yes, area is a vector, which is the normal to the surface. ($\vec{A}=A\vec{n}$)

$\vec{F} = -P\vec{A}$

In this case P is simply the proportionality constant to the vectors F and A, which also mean that F and A has to be in the same direction (F is the normal force and not shear forces). The negative sign accounts for the fact that the force and normal vectors are pointing in the opposite direction.

Pressure is, by dimensional analysis, just a form of energy density or momentum flux density. In the most general terms, it is defined as a second rank tensor given by: $$\mathbb{P}_{s} = m_{s} \int d^{3}v \ \left( \textbf{v} - \textbf{U}_{s} \right) \left( \textbf{v} - \textbf{U}_{s} \right) \ f_{s}\left( \textbf{x}, \textbf{v}, t \right)$$ where $f_{s}\left( \textbf{x}, \textbf{v}, t \right)$ is the particle distribution function of species $s$ that defines a probability density in phase space (or the number of particles per unit spatial volume $d^{3}x$ per unit velocity volume $d^{3}v$, though a more careful expression would use relativistic momentum instead of velocity), $m_{s}$ is the mass of particle species $s$, and $\textbf{U}_{s}$ is an average velocity defined by: $$\textbf{U}_{s} = \frac{ 1 }{ n_{s} } \int d^{3}v \ \textbf{v} \ f_{s}\left( \textbf{x}, \textbf{v}, t \right)$$ In the pressure tensor definition, we can see there is a dyadic product of the same vector, which causes $\mathbb{P}_{s}$ to be a symmetric tensor, by definition. For instance, one can easily show that P$_{ij}$ = P$_{ji}$ for $i \ \neq \ j$. We also know that any real symmetric square matrix, $\mathbb{A}$, can be rewritten as a diagonal matrix, $\mathbb{D}$ = $\mathbb{Q}^{T} \cdot \mathbb{A} \cdot \mathbb{Q}$, where $\mathbb{Q}$ is a real orthogonal matrix.

We often separate the diagonal terms from the off-diagonal terms and call the latter the stress tensor, $\mathbb{S}$. Physically, the terms in $\mathbb{S}$ can be interpreted in the following way: S$_{ij}$ = $i^{th}$ component of the momentum flux density across the $x{\scriptstyle_{j}}$ = constant plane.

So to reduce $\mathbb{P}$ to a scalar is to say that $\mathbb{S}$ $\rightarrow$ 0 and each of the diagonal terms are degenerate such that $\mathbb{P}$ = P $\mathbb{I}$, where $\mathbb{I}$ is the unit matrix. This physically implies that the pressure has no preferred direction or orientation with respect to the basis used in the analysis.

If you want a really good review of pressure and stress tensors, with physically significant explanations, you should check out the paper by Longuet-Higgins and Stewart in 1964. They have some very good figures and explanations for the meaning of each term in the pressure tensor and why you can remove/approximate different parts of the tensor.

References
Longuet-Higgins, M.S., and R.W. Stewart "Radiation stresses in water waves; a physical discussion, with applications," Deep-Sea Research 11, pp. 529--562, 1964.

There are different mathematical way to define pressure (all equivalent), but perhaps the most common one is using the component of the force normal to the surface. That is why in the definition you are using you are actually dividing scalars. For this way to define pressure, you see http://en.wikipedia.org/wiki/Pressure.

You can also consider area as a vector, in such a case you will not be dividing vectors, but define pressure in the way already answered by t.c.

Because, Pressure=Force\Area pressure is a vector. quantity because the area is normal to the force, makes the area pseudo vector.Then by rule iii, Vector\Vector=Vector pressure is a scalar quantity

• I'm confused by the choice to say it's a vector but later revise it to a scalar. Can you add some clarifying text to explain this more? Mar 13, 2019 at 10:07

(bold typeface is used here to represent vectors)

You will remember the scalar product (dot product) of two vectors. It can e.g. be used in physics to define work as F.s , where F is the force vector and s is the distance over which the force applies. Two vectors are multiplied here to obtain a scalar (the work). Only the component of F in the direction of s is taken into acount by the scalar product.

Similarly, in the pressure equation, only the component of the force, perpendicular to the surface is needed. So, considering a plane surface, represented by vector A, and a force F, to me it sounds logical that we would define pressure as the division of two scalar products:

p = (F.A) / (A.A)

This equation is understandable on high school level physics, and shows the further use of vector quantities in physics.