# How to explain the isotropic nature of pressure?

Shown in figure is a pipe in which a fluid is static. The pressure at A is equal to pressure at B since it is at the same level and pressure is isotropic.

The pressure at A could be defined as the downward thrust acting on a unit area of cross section that extends into the screen (i.e the plane of the mentioned cross section is perpendicular to the plane of the drawn figure).

How can the pressure at B be defined in such a manner ? [It doesn't make sense to say that the same definition may be taken for B when a vertical column similar to A is not to be found directly above B.][What I am talking about is how would the orientation of thrust and the cross sectional area be ?]

In other words, how can the isotropic nature of pressure be explained considering the definition of pressure as thrust (Not force) /unit area ?

• Why downward ? Is all fluid continuously flowing downwards because of downward force ? – Poutnik Sep 21 '19 at 4:54
• @Poutnik That makes sense. Please explain how is the case then ? – Zam Sep 21 '19 at 5:11
• Please resize and crop the gigantic figure. – user4552 Sep 22 '19 at 15:04

## 1 Answer

### Short answer

Wikipedia: Pressure  (symbol $$p$$ or $$P$$) is the force applied perpendicular to the surface of an object per unit area

So direction of the force caused by pressure is determined by surface orientation.

If pressures at A and B were different, horizontal pressure force would be pushing fluid from higher pressure to lower one to equalise the pressure.

### Longer answer

Pressure is not a force and it has no direction.

Yes, it can be said it is a thrust downwards on a given area. Or upwards. Or leftwards. Or right/front/rear/any direction-wards.

Pressure can be defined by 2 ways:

• A force over an oriented area $$\mathrm{d}\vec{F_\mathrm{p}}=p \cdot \mathrm{d}\vec{A}$$

• A work over a volume $$\mathrm{d}W=p \cdot \mathrm{d}V$$

$$\mathrm{1\ Pa = 1\ Nm^{-2} = 1\ Jm^{-3}}$$

Acceleration $$\vec a$$ of a fluid element is

$$\vec a=\vec g - \frac {\vec {\nabla p} }{\rho}$$

where $$\vec g$$ is gravity acceleration and the second term is pressure gradient divided by fluid density.

In the static fluid case, horizontal forces caused by pressure are mutually cancelled and vertical forces caused by pressure gradient are cancelled by the layer force of weight.

If pressures at A and B ( on the same horizontal level) were different, then fluid would flow horizontally until they were the same, as the net force acting on thought fluid segment would be nonzero and horizontal.

Hurricanes are attempts to equalise horizontal pressure gradients, turned to circular motion by the Coriolis force. If the Earth had not rotate, they would not have existed, nor the pressure highs and lows, and gradients would have been cancelled quickly.

The same vertically, if $$\mathrm{d} p <> - \rho \cdot g \cdot \mathrm{d}h$$

If you push fluid away in any direction you do work against the pressure

• by raising its position potential energy by displacement or

• raising its compression potential energy, or

• raising its kinetic energy.

In real cases, it is usually combination of above.

• So how would you explain pressure at points A and B ? – Zam Sep 22 '19 at 3:07
• @Zam See the update. If net force on fluid segment is nonzero, fluid moves. – Poutnik Sep 22 '19 at 4:51