We briefly introduced some terms at the start of my Fluids class, and one of them was pressure. I'm looking for a rigorous definition of pressure.

Wikipedia gives this definition for pressure: $p = \frac{dF_n}{dA}$ ($p$ is the pressure, $F_n$ is the normal force, and $A$ is the area of the surface on contact). Pressure is the rate at which normal force increases as the area of surface on contact increases.

However, the rate of which the normal force increases on the surface considered depends on how $A$ is increased. Perhaps we extend the surface a little into a region where a great deal of force acts on it, or maybe we extend it into a region where no force acts on it.

$p = \frac{dF_n}{dA}$ is an ambiguous definition because it does not specify how A must be increased.

I'm looking for an unambiguous, rigorous definition. Thanks!


To elaborate: $p = \frac{dF_n}{dA} = \lim_{\Delta A \to 0} \frac{F_n(A + \Delta A) - F_n(A)}{\Delta A}$. The problem is that $F_n(A + \Delta A)$ is ambiguous, since it depends on how $\Delta A$ is selected.

  • 3
    $\begingroup$ It's not ambiguous because the $d$'s in there indicate that you're supposed to consider an infinitely small area $A$ over which to check the force. If you complain that the area is too big and is feeling different amounts of force in different places, you didn't make $dA$ small enough :) $\endgroup$
    – DanielSank
    Commented Sep 30, 2014 at 22:59
  • $\begingroup$ The unambiguous, rigorous definition is that of normal force per area. It's not clear what else you have in mind. $\endgroup$
    – CuriousOne
    Commented Sep 30, 2014 at 23:00
  • $\begingroup$ @CuriousOne: OP is confused because he/she realizes that $A$ might feel different forces in different places. OP has to grok "infinitesimal". $\endgroup$
    – DanielSank
    Commented Sep 30, 2014 at 23:01
  • $\begingroup$ @CuriousOne: Actually "normal force per area" is kind of ambiguous. If I don't tell you how big to make $A$ you could get different results. The definition is only well-formed if I specify that you should take a limit, i.e. $dF/dA$. $\endgroup$
    – DanielSank
    Commented Sep 30, 2014 at 23:04
  • 1
    $\begingroup$ Related questions and answers that would probably suffice: physics.stackexchange.com/questions/107824/… physics.stackexchange.com/questions/100820/… $\endgroup$
    – tpg2114
    Commented Oct 1, 2014 at 3:15

1 Answer 1


Sorry for the enormous delay, I was caught with more work than I thought, then, here it is.

@DanielSank, relativity is not necessary, it would help with what you said, since it would pinpoint exactly you are calling momentum in your system.

My answer would be an extension of DanielSank comment. When there is the conservation of a continuous quantity, the usual way to treat it is through an continuity equation:

$\frac{\partial A}{\partial t}+\nabla \cdot \vec J_A = S_A$

Where one would read that the quantity '$A$' is conserved if the source '$S_A$' is equal to zero. $\vec J_A$ is called 'current' associated with the quantity $A$. Also if $\int |A| dV < \infty$, the value $Q =\int A\ dV$ is constant in time.

Now, to discuss pressure, one would try to write this kind of continuity equation for a fluid. In this setting, the usual approach for normal fluids is to fix $A=\rho \vec v$, which is interpreted as the momentum density of the fluid. One would get:

$\partial_t (\rho \vec v) + \nabla \cdot \sigma = \vec f$

$\vec f$ is the 'volumar force', i.e., the force for unit of volume throughout the fluid. One good example is the newtonian gravity force, which would result in $\vec f = \rho \vec g = - \rho g \hat z$. The definition of pressure comes when decomposing $\sigma$.

Since $A=\rho \vec v$ is a vector field, $\sigma$ is a given point is a tensor (let's restrict to 3x3), called the stress tensor, thus, we would write a 3x3 matrix for it. Using angular momentum conservation, one can show that $\sigma$ is a symmetric matrix. The decomposition goes as follows:

$\sigma_{ij} = (p+\Pi) \delta_{ij}+ \pi_{ij}$

Where $p = p(\rho)$ would be a function of just the density (and other thermodynamic quantities of the fluid), $\Pi$ and $\pi_{ij}$ could depend also on the velocity field of the fluid, and $\pi_{ij}$ is traceless: $\sum_i \pi_{ii} = 0$. What is usually called the pressure is the term $p$.

If there is no velocity field, $\Pi$ and $\pi_{ij}$ zero, thus one can read the pressure as one third of the trace of stress tensor:

$p = \frac{1}{3} \sum_i \sigma_{ii}$

In this context, and also setting $\vec f=0$, one identifies the integrated momentum flow with the pressure, because one gets, for a surface $\Sigma=\partial \Omega$, boundary of a region $\Omega$:

$\int_\Omega\ (\nabla \cdot \sigma )\ dV = \int_{\partial \Omega}\ \sigma \cdot d\vec S = \int_{\partial \Omega}\ p\ d\vec S$

Thus interpreting the change of the momentum in a given region as the integrated pressure over the boundary surface. One can directly see that if either $\Pi$ or $\pi_{ij}$ is non-zero, there are other contributions to the momentum flow.

It is still possible to use the trace as a definition of pressure, where we would get:

$p + \Pi = \frac{1}{3} \sum_i \sigma_{ii}$

So, $\Pi$, associated with the bulk viscosity of the fluid, acts as a second 'pressure' contribution. The only way to separate the two is through an equation of state, which can't be extract just looking to the conservation laws of the system. Since most liquid systems are treated as 'incompressible' ($\nabla \cdot \vec v =0$), people usually consider $\Pi=0$.

On newtonian fluids, these expressions reduce to the more usual ones:

$\pi_{ij} = \mu\left[\partial_i v_j + \partial_j v_i - \frac{2}{3} (\nabla \cdot \vec v)\delta_{ij}\right]$

$\Pi = - \zeta (\nabla \cdot \vec v)$

This is a purely macroscopical way of approaching hydrodynamics and the definition of pressure, without resorting to any microscopic/kinectic argument, thus, it's not clear where the information about the microscopic x macroscopic scale enters, but this will take another post.

The conection with lagriangians and relativity goes as follows:

When people deal with 'ideal'/non-dissipative systems, it's normal to approach them using an action $I=\int dt\ dV\ \mathcal L$, where $\mathcal L$ is called the 'lagrangian density' of the system. If the action have symmetries, for example, time or space translation, it's possible to write continuity equations associated with those symmetries, this result is called 'Noether Theorem'.

The continuity equation associated with the space translation invariance is the conservation of momentum equation. Also, the Noether Theorem explicitly defines what is the momentum density of your system, without the need to any complicated convention.

When one has Lorentz relativity, there is a more tight connection between time and space, which induce a tighter conection between energy momentum, and what happens is that it identifies the energy current with the momentum density for an isolated system.

edit: I saw @tpg2114 comment just after I completed my answer, thus I'm posting it anyway also because there is a comment here that isn't on DumpsterDoofus answer on the other thread.

  • $\begingroup$ It's better to have posted the answer than just rely on people going to the other questions I linked. This is more complete anyway. $\endgroup$
    – tpg2114
    Commented Oct 1, 2014 at 16:27
  • $\begingroup$ A related answer that conects with Kinectic Theory: http:physics.stackexchange.com/questions/67966/fluids-in-thermodynamic-equlibrium/67971#67971 $\endgroup$
    – Hydro Guy
    Commented Oct 1, 2014 at 22:26

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