# Define Pressure at A point. Why is it a Scalar?

I have a final exam tomorrow for fluid mechanics and I was just looking over the practice exam questions. They do not provide solutions. But pretty much I have to define pressure at a point and also say why pressure is scalar instead of a vector.

I am thinking pressure at a point is $P=\lim_{\delta A \to 0} \frac{\delta F}{\delta A}$. Please let me know if I am wrong.

But I do not know at all why pressure is a scalar instead of a vector. I know it has something to do with $d \mathbf{F}=-Pd \mathbf{A}$

• I don't why there are long answers. Its simply always true by definition. And yes, you are correct. $\delta F$ is the force projected on the normal, and $\delta A$ is obviously a scalar, so $\frac{\delta F}{\delta A}$ is a scalar. The pressure is not the stress state. – Mikael Öhman Dec 16 '11 at 18:36

The pressure is defined as the flow rate of x-momentum in the x-direction plus the flow-rate of the y-momentum in the y-direction plus the flow rate of the z-momentum in the z-direction divided by three. Each component of momentum is conserved, and flows locally from point to point in a fluid. Each component is like a charge, and has it's own current, which makes a tensor of stresses.

A fluid does not sustain shear, and this is true whether it is still or moving, by the principle of relativity. This means that if you put fluid between two plates, and squeeze, the force per-unit-area with which you squeeze (the local flow of momentum in the direction perpendicular to the plates) is equal to the force per unit area pushing outward at the edge of the plates. The flow of momentum is the same in all directions.

You can measure the pressure by putting a homogenous solid at the point in the liquid, and noting how much it compresses. You can also do it by noting the very slight change in density of the fluid with pressure.

The pressure of fluids is not an exact description of the fluid when there is viscosity, but it is close to perfect, and the viscosity and the pressure are independent ideas which can be treated separately.

For the purpose of your class, you can think of the pressure as the amount a tiny box-spring will compress if you place a scale model in the fluid, moving along with the water. This pressure is the same in all directions in the fluid, despite the answer you got earlier.

A gedanken experiment to illustrate the scalarness of pressure:

• take a rubber glove over a glass container and make it hermetic so that water will not penetrate.

• take the glass container under water. Measure the inward curvature of the rubber surface and take it as a rough measurement of the force that water is exerting over the surface opening of the container

Now question yourself: how does this curvature varies in function of different spatial orientations of the mouth of the glass container that is hermetically covered with a rubber surface?

it is a property of all liquids and gases that they don't care about the orientation of your container, it will exert the same pressure (eventually you might end up learning that this is because liquids and gases are phases with rotational symmetry).

pressure in liquids and gases is a scalar because of this rotational invariance. If you have taken a class of linear algebra you might be familiar with matrices and how they are understood as operators over vectors. You can think of pressure in general as such linear operators that can be represented as $3 \times 3$ matrices that are a property of the material. In the case of liquids and gases, this operator multiplies vectors in the same way as a scalar would, so they can be think as the identity matrix times a scalar, called scalar pressure.

Since you are in a fluid dynamics class, which is about liquids, the above applies

When we say that the force on a little unit area is independent of orientation, we're making an approximation; that the fluid is static. More generally, the force does depend on the direction in which the small area is placed.

This orientation dependent force is called the stress tensor. The stress tensor is not a scalar. The nonscalar part of the stress is sometimes called a "shear" which is also measured in terms of force per unit area (i.e. see "shear strength"). If this seems strange, perhaps it's worthwhile to review the wikipedia entry for pressure:

"A closely related quantity is the stress tensor $\hat{\sigma}$, which relates the vector force $\mathbf{F}$ to the vector area $\mathbf{A}$ via $$\mathbf{F}=\hat{\sigma}\mathbf{A}\,$$ This tensor may be divided up into a scalar part (pressure) and a traceless tensor part shear. The shear tensor gives the force in directions parallel to the surface, usually due to viscous or frictional forces."

The pressure is the average of the stress tensor over all orientations, and in a fluid which isn't moving, this is the only stress that is allowed. The reason is rotational invariance, the only rotational invariant tensor is proportional to the diagonal tensor.

So pressure is the scalar part of the stress tensor, and it is the only part of the stress tensor that exists in fluids.

• -1: This answer is grossly misleading, or else wrong. The leading "shear" in the case of fluids only appears as viscous drag, which is clearly negligible, and vanishes for small velocity gradient limit (as it must by Galilean invariance). – Ron Maimon Dec 14 '11 at 5:30
• Carl:The assumption made is not that the fluid is static, but incompressible. – Antillar Maximus Dec 14 '11 at 5:36
• @Ron M; If you can buy a quart of fluid at the local supermarket it's going to have (non negligible) viscous drag, LOL. Even air has viscosity. @SAntillar; hard to see where you came up with "incompressible". Air is manifestly compressible and exhibits pressure rather well. – Carl Brannen Dec 14 '11 at 20:35
• @Carl: The reason I was unhappy with the answer is that it claims that the pressure is not really a scalar, and introduces complications, when the simple answer is not wrong--- the pressure is the scalar trace part of the stress energy tensor, and it is a scalar. So it seems that you are saying "it's a scalar, except its more complicated than that", when it's a scalar and its not more complicated than that. I don't think the answer is wrong, but it is misleading. Student's are most confused by things that they are told are more or less true, with some nebulous confusing caveats thrown in. – Ron Maimon Dec 15 '11 at 4:09
• @Ron; I agree that "pressure" as it's usually defined is a scalar. But the OP defined pressure in terms of the force on a small unit area (which has an orientation) and so is not generally a scalar. You have 13K reputation points, so modify my answer already if you can improve it. – Carl Brannen Dec 17 '11 at 4:25

Because it doesn't behave like a vector -- for example, the standard vector norm of the pressure doesn't mean anything, but the sum of pressure components does. Just because something has a bunch of components doesn't mean it's a vector. Pressure refers to the diagonal components of the stress tensor (actually their sum, the trace) -- this arises naturally in terms of momentum fluxes.

As an analogy, consider the fact that the divergence of a vector field is not a vector. Analogously, divergence is just the diagonal components of the tensor derivative. One level further, the dot product is not a vector, for the same reason.