The Forces on a Fluid Parcel: Pressure, Viscosity and Gravity

Post 10 derived the material derivative \(D\mathbf{u}/Dt\), the \(a\) in \(F = ma\), and Post 11 derived the constraint \(\nabla \cdot \mathbf{u} = 0\) from conservation of mass. This post supplies the \(F\): the forces that act on a single fluid parcel.

It starts with what a force is and what its unit means, then names the forces that reach a parcel: pressure, viscous friction and gravity. Each one is derived separately and turned into its share of the parcel's acceleration. The post ends with the three shares added together.

A force changes motion; it does not maintain it. A cart on ice with zero friction that nobody pushes keeps its velocity. Everyday objects only seem to stop on their own because friction pushes back on them, unnoticed. A push changes the velocity: it makes an object faster, slower, or turns it. The unit of force, the newton (N), is defined by exactly this: 1 N is the push that makes 1 kg faster by 1 m/s every second. Written as a formula, this is Newton's second law:

Where \(F\) denotes the force (N), \(m\) denotes the mass being pushed (kg), and \(a\) denotes its acceleration (m/s²), how fast its velocity changes.

The unit, piece by piece. Velocity is measured in metres per second, m/s. Acceleration is how much the velocity changes per second, so its unit is (m/s)/s = m/s², read "metres per second, per second". The s² is shorthand for dividing by seconds twice, not a "squared second" with a meaning of its own. Force is mass times acceleration, so its unit is kg·m/s², and this unit gets its own name:

Where \(=:\) denotes "the left side is given the name on the right side", the mirror image of the \(:=\) from Post 10. Read kg·m/s² as "kilograms, having their velocity changed by so many m/s, every second". Dividing by kilograms gives \(\frac{\text{N}}{\text{kg}} = \frac{\text{kg}\cdot\text{m/s}^2}{\text{kg}} = \frac{\text{m}}{\text{s}^2}\): back an acceleration, by the definition of the newton. This identity is what turns every force in this post into an acceleration.

A force has a strength and a direction, so we draw it as an arrow. In one dimension, the direction is just a sign: + means right, − means left. Two people pull on the cart, one with +5 N and the other with −3 N. Forces add: +5 + (−3) = +2 N, so the cart is pulled with 2 N to the right. If both pull with 5 N in opposite directions, the sum is 0 and the cart's velocity does not change. It can still be moving; it just keeps its velocity, like the cart on ice.

In three dimensions, a force has one number per direction: \(x\), \(y\) and \(z\). For example, \(\mathbf{F}_1 = (5,\; 0,\; 2)\) N pulls with 5 N along \(x\), 0 N along \(y\) and 2 N along \(z\). A second force is \(\mathbf{F}_2 = (-3,\; 1,\; 0)\) N. Two forces are added direction by direction:

\(\mathbf{F}_1 + \mathbf{F}_2 = (5 + (-3),\; 0 + 1,\; 2 + 0) = (2,\; 1,\; 2)\) N.

The sum of all forces on an object is the net force, and only the net force enters Newton's law:

The length of an arrow is the force's total strength, its magnitude. This gives: \(|\mathbf{F}| = \sqrt{F_x^2 + F_y^2 + F_z^2}\). The net force above has a magnitude of \(\sqrt{2^2 + 1^2 + 2^2} = \sqrt{9} = 3\) N.

Two big forces can almost cancel: 1000 N to the right and 999 N to the left leave only 1 N. Pressure forces cancel like this all the time, as the next sections show.

A fluid parcel is surrounded by fluid on all sides. Forces reach it in two ways.

• Through its faces. The neighbouring fluid touches every face of the parcel. Pressure pushes onto each face, at right angles to it. Viscous friction drags along each face, parallel to it. Forces like these, which need contact with a neighbour, are surface forces.
• Through its bulk. Gravity pulls on every kilogram inside the parcel, without any contact.

Each force is turned into an acceleration by dividing it by the parcel's mass. The forces add (Equation (3)), and every one of them is divided by the same mass, so the accelerations add too. The parcel's acceleration is therefore the sum of three shares, one per force:

Where \(\mathbf{a}\) denotes the parcel's total acceleration (m/s²), \(\mathbf{a}_{\text{pressure}}\) denotes the acceleration caused by pressure, \(\mathbf{a}_{\text{fric}}\) the acceleration caused by viscous friction, and \(\mathbf{a}_{\text{grav}}\) the acceleration caused by gravity (all in m/s², all arrows). Each share is its force divided by the parcel's mass. The rest of the post derives the three shares one at a time:

1. Pressure. Start from the push of a fluid on a single face. Compare the pushes on the left and right faces of a small cube: only their difference moves it. Turn that difference into a slope of the pressure, and divide by the cube's mass.
2. Friction. Start from the friction between two solids, then apply the same idea to thin layers of fluid sliding past each other. Compare the friction on the two sides of a parcel: again, only the difference counts. Divide by the mass.
3. Gravity. Gravity pulls on every kilogram inside the parcel, so its force grows with the mass. Dividing by the mass leaves the same acceleration for every fluid.

The molecules of a fluid constantly bump into every surface they touch. Each bump is a tiny shove, but billions of them every second add up to one steady push. A bigger face is hit by more molecules and feels more force, but the force per square metre stays the same. That ratio describes the fluid itself rather than the face, and it is the pressure:

Where \(p\) denotes the pressure (N/m²), \(F\) denotes the force the fluid exerts on a face (N), \(A\) denotes the area of that face (m²), and \(\iff\) denotes "is equivalent to". The unit N/m² is called the pascal (Pa). For example, 1 Pa on a face of 0.2 m² gives 1 × 0.2 = 0.2 N; the m² cancels, since N/m² · m² = N.

Pressure always pushes straight onto a face, at right angles to it. Pressure itself has no direction: it is one number per place, a field \(p(\mathbf{x}, t)\), like the density field \(\rho(\mathbf{x}, t)\) of Post 11. The direction comes from the face: the same pressure pushes a parcel's left face toward \(+x\) and its right face toward \(-x\).

Air at sea level has a pressure of about 101 300 Pa. On a fingernail of 1 cm² = 0.0001 m², that is 101 300 × 0.0001 ≈ 10 N, the weight of a 1 kg bag of sugar.

Newtons and pascals are related the same way as mass and density:

Where \(V\) denotes a volume (m³). A newton measures the total force on a chosen face or object; a pascal measures the force per m², the intensity of the fluid's push. Pressure and density belong to the fluid; force and mass belong to the chunk we choose to look at. A force in N still has a direction; a pressure in Pa does not.

How large atmospheric pressure is. Let 101 300 Pa push on a 1 m² face of a 1 kg object, with vacuum on the other side. By Equation (5), the force is 101 300 × 1 = 101 300 N, and by Equation (1), the acceleration is 101 300 / 1 = 101 300 m/s². After 1 s, the object would move at 101 300 m/s, about 300 times the speed of sound in air (101 300 / 340 ≈ 300). A fingernail is not crushed because it is pushed from the other side almost equally hard, and the two forces cancel as in Equation (3).

Take the cube of fluid with 1 cm sides from the interactive widget in the previous subsection, so each face has an area of 0.01 × 0.01 = 0.0001 m². The fluid on its left is at 101 300 Pa and pushes the left face with 101 300 × 0.0001 = +10.13 N. The fluid on its right is at 101 200 Pa and pushes the right face with 101 200 × 0.0001 = 10.12 N, toward \(-x\), so −10.12 N. The net force is +0.01 N, about the weight of one gram, and it points toward the lower pressure:

Where \(F_{\text{net},x}\) denotes the net force along \(x\) (N), \(p_{\text{left}}\) denotes the pressure on the left face (Pa), \(p_{\text{right}}\) denotes the pressure on the right face (Pa), and \(A\) denotes the area of each face (m²). The minus sign in front of \(p_{\text{right}}\) comes from the fact that it pushes toward \(-x\).

The difference \(p_{\text{left}} - p_{\text{right}}\) becomes a slope by the move from Post 11: for a thin parcel, the difference of a quantity across it is approximately its slope times the width, and this becomes exact as the width shrinks to zero. With the left face at \(x\) and the right face at \(x + \Delta x\), that is \(p(x+\Delta x) - p(x) \approx \frac{\partial p}{\partial x}\,\Delta x\). Equation (6) takes "left minus right", the reverse order, which flips the sign: \(p_{\text{left}} - p_{\text{right}} \approx -\frac{\partial p}{\partial x}\,\Delta x\). Multiplying by the face area and using \(\Delta x \cdot A = \Delta V\) gives

Where \(\partial p/\partial x\) denotes the slope of the pressure along \(x\) at one frozen instant (Pa/m), \(\Delta V\) denotes the volume of the parcel (m³), and the minus sign means that the push points toward falling pressure. The \(\Delta\) marks a small chunk that will later shrink, like \(\Delta x\) and \(\Delta t\).

The force in Equation (7) depends on \(\Delta V\), which is our arbitrary choice of parcel size, and \(F = ma\) asks for the acceleration. A bigger parcel feels more force, but it also has proportionally more mass. Dividing by the mass \(m = \rho\,\Delta V\) removes the size:

Where \(\rho\) denotes the density (kg/m³) and \(a_{x,\text{pressure}}\) denotes pressure's share of the parcel's acceleration along \(x\) (m/s²). Check for water, \(\rho = 1000\) kg/m³: the cube has a mass of \(1000 \cdot 10^{-6} = 0.001\) kg, so 0.01 N / 0.001 kg = 10 m/s², and the formula gives \(-\frac{1}{1000} \cdot (-10\,000) = 10\) m/s².

Read \(-\frac{1}{\rho}\frac{\partial p}{\partial x}\) as (push per m³) ÷ (kilograms per m³) = push per kilogram, which is an acceleration by Equation (2): how hard each m³ is pushed, shared among the kilograms in it.

The force on an infinitely small parcel is zero. As \(\Delta V \to 0\), Equation (7) goes to zero. Force is a total, like mass, and a point has no mass either. What survives are two size-independent versions: the force per volume, \(-\partial p/\partial x\) in N/m³, called a force density, and the force per mass, \(-\frac{1}{\rho}\,\partial p/\partial x\) in m/s². An actual force always needs a chosen chunk: force = (per m³) · \(\Delta V\) = (per kg) · \(\rho\,\Delta V\).

Each pair of opposite faces runs its own tug of war: the faces at right angles to \(y\) give \(-\frac{1}{\rho}\,\partial p/\partial y\), and those at right angles to \(z\) give \(-\frac{1}{\rho}\,\partial p/\partial z\). Stacking the three slopes gives the gradient from Post 10:

Where \(\nabla p\) denotes the pressure gradient (Pa/m) and \(\mathbf{a}_{\text{pressure}}\) denotes pressure's share of the acceleration as an arrow (m/s²). Each entry of \(\nabla p\) is positive along a direction in which the pressure rises, so \(\nabla p\) points uphill in pressure, and the minus sign makes parcels go downhill. For example, in water with \(\partial p/\partial x = -10\,000\) Pa/m, \(\partial p/\partial y = +5\,000\) Pa/m and \(\partial p/\partial z = 0\), pressure's share is \(\mathbf{a}_{\text{pressure}} = -\frac{1}{1000}\,(-10\,000,\; 5\,000,\; 0) = (10,\; -5,\; 0)\) m/s².

A book shoved across a table slows down and stops. By Newton's law, something must push it backward: in this case the table. Both surfaces have microscopic bumps that catch on each other, and billions of catches add up to one steady backward push. This friction has three properties:

1. It exists only while the surfaces slide against each other.
2. It acts along the surface. Pressure acts onto it.
3. It acts on both sides, equally hard and in opposite directions: the table holds the book back, and the book drags the table forward. Friction slows the faster side and drags the slower one along, so it tries to equalize their speeds, and it disappears once they are equal.

In short, friction fights speed differences between things that touch.

Two observations show that fluids have friction too. Stirred water keeps swirling for a while, but stirred honey stops almost at once. And on a river, leaves near the bank drift slowly while leaves in the middle race along, with the speed rising gradually in between.

Picture the fluid as thin layers, like the cards in a deck. Neighbouring layers that move at different speeds rub against each other, with the same three properties as the friction between solids. Roughly, this comes from the molecules: neighbouring molecules cling to each other a little, and when layers slide past each other, these connections stretch and tug, pulling the faster layer back and the slower one forward. Honey's molecules cling more strongly than water's. Viscosity is how strongly a fluid resists its layers sliding past each other.

The river shows how this shapes a flow. The water touching the bank clings to it and stands still, because the bank does not move. The standing layer slows the next layer through friction, that layer slows the next one, and so on, so the middle is fastest. The effect works both ways: the fast inner layers also pull the outer ones forward. The flow is steady when whatever drives the river downstream balances the friction; the gravity section shows what that drive is.

A layer touches two neighbours, one on each side. If it is slower than their average, it is pulled forward; if it is faster, it is held back. If its speed is exactly their average, as in a straight-line speed profile, it feels friction on both faces, but the two cancel and there is no net force.

View the river from above. The flow goes along \(x\), \(y\) is the distance from the bank, and the speed \(u_x(y)\) is 0 at the bank. At one boundary between two layers, the friction depends on three things:

1. The area of the boundary. A bigger boundary feels more friction, so, as with pressure, friction is measured per m².
2. The velocity slope, not the speed difference. The speed difference between neighbouring "layers" depends on how finely we slice the river. If the speed rises by 1 m/s over 10 cm, layers 1 cm thick differ by 0.1 m/s, and layers 0.5 cm thick by 0.05 m/s, although the water is the same. The slope \(\partial u_x/\partial y\), here 1 / 0.1 = 10 (m/s)/m, does not depend on how we slice.
3. The fluid's viscosity.

Experiments give Newton's law of viscosity (I did not learn how to derive this, so we just take it as given here):

Where \(\tau\) denotes the shear stress, the friction per m² along the boundary, \(\mu\) denotes the dynamic viscosity of the fluid (Pa·s), and \(\partial u_x/\partial y\) denotes the velocity slope across the flow. The law holds exactly for Newtonian fluids, which include water, air, oil and honey; for ketchup or toothpaste, the friction is not proportional to the slope. The Navier–Stokes equations assume a Newtonian fluid. Equation (10) is also the simplest case of the law: a flow along \(x\) whose speed changes only along \(y\).

\(\tau\) plays the same role as \(p\) before: a push per m² on a face. \(F = \tau A\) along the face mirrors \(F = pA\) onto the face. So \(\tau = 10\) Pa is an intensity: 1 cm² of boundary feels 10 × 0.0001 = 0.001 N.

Friction comes in pairs. Like the book and the table, the two layers at a boundary feel the same friction in opposite directions: the slower layer is pulled forward, the faster one is held back. As a formula, the layer on the smaller-\(y\) side gets \(+\tau A\) and the layer on the larger-\(y\) side gets \(-\tau A\). Between the bank and the middle, the speed grows with \(y\), so \(\tau > 0\) and the slower layer is pulled forward. Past the middle, the speed falls with \(y\), so \(\tau < 0\), and the same rule again holds the faster layer back.

Units. The velocity slope has the unit (m/s)/m = 1/s; 1/s is simply what is left after the metres cancel, and the slope is also called the shear rate. \(\mu\) is the friction per m² at a slope of 1, that is, at 1 m/s per metre, so its unit is Pa/(1/s) = Pa·s; this is pure bookkeeping. Water has \(\mu \approx 0.001\) Pa·s, honey about 10 Pa·s (strongly dependent on temperature).

Take a thin parcel in the half of the river between the bank and the middle. At its bank-side boundary, the slower neighbour holds it back, toward \(-x\). At its middle-side boundary, the faster neighbour pulls it forward, toward \(+x\). It is a tug of war, and the strength of each pull is set by the velocity slope at that boundary. With layers 1 cm apart and speeds in m/s:

In the second row, the bank-side boundary has a slope of (0.5 − 0.4) / 0.01 = 10/s, and the middle-side boundary a slope of (0.7 − 0.5) / 0.01 = 20/s. Those are two different values of \(\tau\), just as \(p_{\text{left}} \neq p_{\text{right}}\) for pressure. The net friction therefore depends on how much the velocity slope changes across the parcel.

Put the bank-side boundary at \(y\) and the middle-side boundary at \(y + \Delta y\), both with area \(A\). Each boundary hands out a pair of forces, one to each of the two layers it separates:

Our parcel collects one half from each of its two boundaries:

Where \(\tau(y)\) and \(\tau(y+\Delta y)\) denote the shear stress at the two boundaries (Pa), and \(F_{\text{fric},x}\) denotes the net friction force on the parcel along \(x\) (N).

As for pressure, the difference across the thin parcel becomes a slope, \(\tau(y+\Delta y) - \tau(y) \approx \frac{\partial \tau}{\partial y}\,\Delta y\), and \(\Delta y \cdot A = \Delta V\):

Where \(\partial \tau/\partial y\) denotes the slope of the shear stress along \(y\) (Pa/m = N/m³). When the parcel shrinks, the force goes to zero, as it did for pressure, but the force density \(\partial \tau/\partial y\) survives.

The Navier–Stokes equations are written in terms of the velocity, so we insert Equation (10) (\(\tau = \mu \frac{\partial u_x}{\partial y}\)) for \(\tau\). If \(\mu\) is constant, meaning the same fluid at the same temperature everywhere, it can be pulled out of the slope:

Where \(\partial^2 u_x/\partial y^2\) denotes the slope of the velocity slope (1/(m·s)).

The second derivative is exactly the neighbour-average idea that I've mentioned from the start of this section. With three layers \(\Delta y\) apart, it is Post 2's three-point stencil, the same one that drives the heat equation:

Where \(u_{\text{here}}\) denotes the parcel's speed, \(u_{\text{above}}\) and \(u_{\text{below}}\) denote the speeds of its neighbours at \(y + \Delta y\) and \(y - \Delta y\) (m/s), and \(\Delta y\) denotes the spacing between the layers (m).

So friction pulls every parcel's speed toward the average speed of its neighbours, and \(\mu\) says how strongly. That is why the honey's swirl dies out, and why the fast middle of the river is always held back.

The friction force still has to be divided by the mass, for three reasons:

1. The left side of the Navier–Stokes equations is the parcel's acceleration, the material derivative of Post 10. A force alone does not say how fast the velocity changes; that needs \(a = F/m\).
2. The force depends on the arbitrary parcel size, but the mass scales the same way, so the size cancels and the result is a statement about a single point.
3. All terms must be accelerations to be added, and the pressure term, Equation (8), already is one.

With \(m = \rho\,\Delta V\), we get:

Where \(a_{x,\text{fric}}\) denotes friction's share of the parcel's acceleration along \(x\) (m/s²). Read the right side piece by piece:

• \(\partial^2 u_x/\partial y^2\) is how far the parcel lags behind its neighbours' average speed (Equation (14)). Positive means it is slower than them.
• \(\mu\) is how sticky the fluid is. It turns that lag into a friction push on every m³ of fluid (N/m³).
• \(1/\rho\) shares that push among the \(\rho\) kilograms in each m³. Push per kilogram is an acceleration.

The ratio gets its own name, the kinematic viscosity \(\nu := \mu/\rho\) (m²/s). \(\mu\) says how strong the friction is; \(\nu\) says how quickly it changes the fluid's speed, because it already accounts for the mass that has to be moved. For water, \(\nu \approx 0.001/1000 = 10^{-6}\) m²/s.

In a real river, \(u_x\) also changes with the depth \(z\): the riverbed is a no-slip wall too, so the water is slow at the bottom and fast at the surface. And \(u_x\) changes along \(x\) itself, for example where the river narrows (Post 11). A cube-shaped parcel has six neighbours. Each opposite pair plays the game of Equation (14), and the forces add:

Where \(\nabla^2\) denotes the Laplacian, the sum of the three second slopes. Each second slope compares the parcel with its two neighbours along one direction, as in Equation (14). Their sum compares it with all six neighbours at once, and it is positive when the parcel is slower than they are on average.

The same reasoning applies to \(u_y\) and \(u_z\), so friction's share of the acceleration is an arrow:

Where \(\nabla^2 \mathbf{u}\) denotes the Laplacian applied to each velocity component separately, and \(\mathbf{a}_{\text{fric}}\) denotes friction's share of the acceleration as an arrow (m/s²).

Pressure and friction act on the parcel's faces. Gravity acts on the whole parcel: Earth pulls on every kilogram inside it, even at its very centre, without touching it. A force like this is a body force. Since every kilogram is pulled, two buckets of water weigh twice as much as one, and the force grows with the mass:

Where \(F_{\text{grav}}\) denotes the gravitational force on the parcel (N), \(m = \rho\,\Delta V\) denotes the parcel's mass (kg), and \(g \approx 9.81\) N/kg denotes Earth's pull on each kilogram. A 1 cm cube of water has a mass of 0.001 kg, so it is pulled down with \(0.001 \cdot 9.81 \approx 0.0098\) N.

Dividing by the mass, as for pressure and friction:

Where \(a_{\text{grav}}\) denotes gravity's share of the parcel's acceleration (m/s²); N/kg is the same as m/s² by Equation (2). The mass cancels: twice the mass means twice the pull, but also twice the inertia. So every parcel gets the same 9.81 m/s², whether it is water or air.

Gravity only pulls down. With \(z\) pointing up:

This arrow is simply added to the shares of pressure and friction, and it only changes the \(z\)-component. That is the whole term.

If gravity pulls every parcel down at 9.81 m/s², why does a parcel in the middle of a pool not fall to the ground (I know this question might seem odd, but it makes sense if you think about it)? Because the water below pushes back. In a pool at rest, nothing moves, so every parcel's acceleration is zero, and friction is zero too, because there are no speed differences. The pressure share must therefore cancel gravity exactly:

The \(z\)-component of \(\mathbf{g}\) is -9.81, so the \(z\)-component of Equation (21) reads \(0 = -\frac{1}{\rho}\,\frac{\partial p}{\partial z} - g\), or

Where \(\partial p/\partial z\) denotes how fast the pressure changes going up (Pa/m). The minus sign means that the pressure grows going down: each layer of water has to hold up the weight of all the water above it. In water, that is \(1000 \cdot 9.81 \approx 9\,800\) Pa more for every metre of depth. This counter-pressure of a fluid at rest is the hydrostatic pressure.

When the water moves, the pressure is the hydrostatic part plus an extra caused by the motion: \(p = p_{\text{rest}} + p_{\text{stir}}\), where \(p_{\text{rest}}\) denotes the hydrostatic pressure and \(p_{\text{stir}}\) the extra from the motion (both in Pa). The hydrostatic part still cancels gravity, exactly as in Equation (21), so the shares of pressure and gravity together are

Where the first step uses that the slope of a sum is the sum of the slopes. Gravity and the hydrostatic pressure cancel each other, and only \(p_{\text{stir}}\) moves anything. Many papers therefore drop \(\mathbf{g}\) and simply write \(p\) for \(p_{\text{stir}}\).

This only works if the hydrostatic pressure holds up every parcel's weight exactly. That is the case when the density is the same everywhere and the fluid has no free surface that can move or tilt, for example in a closed, completely filled tank, in a pipe, or in a simulated box of fluid. Otherwise gravity does real work. A river's surface slopes downhill, so at the same depth there is slightly more water above a point upstream than downstream, and that pressure difference pushes the river along against friction. Warm air is lighter than the cool air around it, so the counter-pressure of the cool air is stronger than the warm air's weight, and the warm air rises.

Without gravity, nothing keeps a closed box of fluid moving: stir it once, and friction evens everything out (Equation (17)). Simulations therefore replace \(\mathbf{g}\) with a general push on every kilogram, written \(\mathbf{f}\) (m/s²). Like gravity, it is a body force, and it takes gravity's place in the sum. Keeping gravity is one choice, \(\mathbf{f} = \mathbf{g}\), whereas no drive at all is \(\mathbf{f} = \mathbf{0}\).

Each force has been turned into its share of the parcel's acceleration, and by Equation (3) the shares add:

Where \(\mathbf{a}\) denotes the parcel's total acceleration (m/s²), and the three terms are the shares of pressure (Equation (9)), friction (Equation (17)) and gravity (Equation (20)), each in m/s². In general, a body force \(\mathbf{f}\) takes the place of \(\mathbf{g}\).

• Pressure: fluid is pushed downhill in pressure, and light fluid much faster than heavy fluid.
• Friction: every parcel's velocity is pulled toward the average of its neighbours.
• Gravity: every parcel is pulled down at the same 9.81 m/s², whatever the fluid. At rest, the hydrostatic pressure balances it exactly.

The next post sets this sum equal to the material derivative \(D\mathbf{u}/Dt\) from Post 10 and adds \(\nabla \cdot \mathbf{u} = 0\) from Post 11, which gives the full Navier–Stokes equations.