Pascal's law — pressure transmits equally: Visual walkthrough

We need only two prerequisites, and we will re-explain both from scratch:

Pressure — force per unit area — what "pressure" even means.
Hydrostatic pressure — p = p0 + ρgh — why pressure grows with depth.

Step 1 — What "pressure" means (the thing we are about to spread around)

WHAT. Before we can transmit pressure, we must know what one unit of it is. Take a flat patch of area $A$ . A fluid touching that patch pushes on it with some total force $F$ , aimed straight into the patch (perpendicular). We define:

WHY per area? Because the same total push feels very different on a pin-tip versus a dinner plate. Dividing by $A$ strips out "how big the patch is" and leaves the pure "how hard is the fluid pressing" — a property of the point, not the patch.

PICTURE. In the figure, the same fluid pushes on a small patch and a big patch. The arrows (force) always aim perpendicular into the surface — this "always perpendicular" fact is the seed of the whole derivation.

Figure — Pascal's law — pressure transmits equally

Step 2 — Why fluid force must be perpendicular

WHAT. Claim: a fluid at rest can only push straight into a surface, never sideways along it.

WHY. Suppose the fluid pushed at a slant. That slanted push has a component along the surface (a tangential, sliding push). A fluid has no rigidity — the tiniest sliding push makes it flow. But we said the fluid is at rest (static). So the sliding component must be zero. What's left points straight in.

PICTURE. The red slanted arrow is split into a green perpendicular part and a yellow tangential part. The yellow part would make the fluid slide — forbidden in a fluid at rest — so only green survives.

Recall Why does this matter so much?

Static fluid force perpendicular to any surface ::: Because any tangential (sideways) push would make the fluid flow, contradicting "at rest".

Step 3 — Build the tiny wedge (our test element)

WHAT. To find how pressure at one point behaves in different directions, we isolate a microscopic wedge of fluid: a triangular prism. It has a vertical face, a horizontal face, and a slanted face. We'll shrink it toward a single point.

Name the three pressures on the three faces:

$p_{x}$ on the vertical face (its push points horizontally, in the $x$ direction),
$p_{y}$ on the horizontal face (its push points vertically, in the $y$ direction),
$p_{n}$ on the slanted face (its push points perpendicular to the slant).

The slant makes an angle $θ$ (theta = "the tilt of the slanted face"). Call the face areas $d A_{x}$ , $d A_{y}$ , $d A_{n}$ (the small " $d$ " just means tiny).

WHY a wedge? Because a wedge has three faces pointing in three different directions. If pressure secretly changed with direction, this shape would reveal it as a leftover unbalanced force. A cube (all faces axis-aligned) couldn't test a slanted direction; the wedge can.

PICTURE. The wedge with its three labelled faces, the angle $θ$ at the base, and one perpendicular pressure-arrow pushing into each face.

Step 4 — The geometry: how the slant face relates to the others

WHAT. We need the areas linked. Let the slant face have area $d A_{n}$ . Basic right-triangle trigonometry (here $sin θ$ = opposite-over-hypotenuse, $cos θ$ = adjacent-over-hypotenuse, measured on the wedge's triangular cross-section) gives:

$d A_{x} = d A_{n} sin θ, d A_{y} = d A_{n} cos θ$

Term by term: $d A_{x}$ (the vertical face) is the slant face "shadowed" onto the vertical — it shrinks by the factor $sin θ$ . $d A_{y}$ (the horizontal face) is the slant "shadowed" onto the horizontal — factor $cos θ$ .

WHY do we need this? The three faces have different sizes, so their pressure forces ( $p \times A$ ) can't be compared until we know the size ratios. These two relations are those ratios.

PICTURE. The triangular cross-section with the hypotenuse $d A_{n}$ , and its two shadows $d A_{x}$ (vertical) and $d A_{y}$ (horizontal), each labelled with its $sin θ$ / $cos θ$ factor.

Step 5 — Balance the horizontal forces → $p_{x} = p_{n}$

WHAT. The wedge is at rest, so net force in each direction is zero. Look at the horizontal ( $x$ ) direction. Two faces push horizontally:

the vertical face pushes with force $p_{x} d A_{x}$ (fully horizontal),
the slanted face's push has a horizontal component $p_{n} d A_{n} sin θ$ .

Setting them equal (they oppose each other): $p_{x} d A_{x} = p_{n} d A_{n} sin θ$

Now substitute $d A_{x} = d A_{n} sin θ$ from Step 4: $p_{x} (d A_{n} sin θ) = p_{n} d A_{n} sin θ$

The $d A_{n} sin θ$ cancels on both sides (it's nonzero for any real wedge): $p_{x} = p_{n}$

WHY does only the horizontal component of the slant force count? Because we are balancing the $x$ -direction alone. The slant push points perpendicular to the slant; only its shadow onto the $x$ -axis, which is the $sin θ$ fraction, can fight the vertical face's horizontal push.

WHY did gravity not appear? Gravity pulls on the wedge's volume $\sim (length)^{3}$ , but the pressure forces act on areas $\sim (length)^{2}$ . As we shrink the wedge to a point, volume vanishes faster than area, so the weight term becomes negligible next to the pressure forces. That is exactly why the result is a statement about a point.

PICTURE. The horizontal force arrows: green $p_{x} d A_{x}$ pushing right, the red slant force with its dashed green horizontal shadow pushing left, drawn equal.

Step 6 — Balance the vertical forces → $p_{y} = p_{n}$ , and the punchline

WHAT. Repeat in the vertical ( $y$ ) direction: $p_{y} d A_{y} = p_{n} d A_{n} cos θ$ Substitute $d A_{y} = d A_{n} cos θ$ ; the $d A_{n} cos θ$ cancels: $p_{y} = p_{n}$

Combining Steps 5 and 6:

WHY this is the whole foundation. We chose $θ$ arbitrarily. Since $p_{n}$ (any slanted direction) always equals $p_{x}$ and $p_{y}$ , pressure at that point is one single number, direction-blind. So when you push a fluid, there's no direction it can "hide" the extra pressure — it must show up equally everywhere pointing every way.

PICTURE. The point at the wedge's shrinking tip, with equal-length arrows radiating in all directions — pressure the same whichever way you look.

Step 7 — From "equal in all directions" to "transmits undiminished"

WHAT. Isotropy is about directions at one point. Pascal's full law is about spreading a change to every point. We bridge them with the hydrostatic relation. For a fluid column of depth $h$ below a surface held at pressure $p_{0}$ : $p = p_{0} + ρ g h$ Term by term: $p_{0}$ = pressure at the top surface. $ρ$ (rho) = the fluid's density (mass per volume). $g$ = gravity's strength. $h$ = depth below the surface. So $ρ g h$ = the extra squeeze from the weight of fluid stacked above depth $h$ .

Now increase the surface pressure by an amount $Δ p$ (delta-p = "the change in pressure we apply"). The new pressure at the same depth $h$ is: $p_{new} = (p_{0} + Δ p) + ρ g h = p_{old} (p_{0} + ρ g h) + Δ p$

Subtracting: $p_{new} - p_{old} = Δ p at every depth h .$

WHY does it come out equal everywhere? Look at what changed and what didn't. The term $ρ g h$ depends on the fluid's own weight and the depth — we changed neither, so it's frozen. The only thing that moved is $p_{0} \to p_{0} + Δ p$ , and it moved by the same $Δ p$ regardless of $h$ . So every point rises by exactly $Δ p$ . That is Pascal's law.

PICTURE. Two identical water columns side by side; the right one has $+ Δ p$ pressed on top. At three depths the pressure bars both grow by the same yellow chunk $Δ p$ , while the blue $ρ g h$ part is unchanged.

Step 8 — Edge & degenerate cases (never leave a gap)

WHAT / WHY, case by case:

$θ \to 0$ (slant becomes horizontal). Then $sin θ \to 0$ , and $d A_{x} \to 0$ : the vertical face vanishes. The horizontal balance $p_{x} = p_{n}$ becomes $0 = 0$ — vacuously true, no information lost; the wedge degenerates into a flat slab and the $y$ -balance still gives $p_{y} = p_{n}$ . ✅
$θ \to 9 0^{\circ}$ (slant becomes vertical). Now $cos θ \to 0$ , $d A_{y} \to 0$ : the horizontal face vanishes. By symmetry the $x$ -balance survives and $p_{x} = p_{n}$ still holds. ✅ Together the two limits show the result is not an accident of one tilt.
Compressible fluid (gas). If the fluid can be squashed, part of your applied $Δ p$ goes into compressing the fluid rather than pushing on the far wall — transmission is no longer perfectly undiminished or instant. Pascal's law needs an incompressible fluid. See Incompressibility & continuity equation.
Not enclosed (open surface, escape path). If the fluid can flow out, pushing just moves fluid instead of raising pressure everywhere. The law needs an enclosed (confined) fluid.
Zero applied change ( $Δ p = 0$ ). Then $p_{new} = p_{old}$ everywhere — the law trivially holds and predicts no change, exactly as expected.

PICTURE. Three mini-panels: the flat-wedge limit, a gas partly absorbing the push (spring-like), and an open container leaking the push away.

The one-picture summary

Everything above in a single flow: static fluid → forces are perpendicular → tiny wedge → geometry links the face areas → force balance in $x$ and $y$ cancels $θ$ → pressure isotropic ( $p_{x} = p_{y} = p_{n}$ ) → feed into $p = p_{0} + ρ g h$ → the applied $Δ p$ rides on top unchanged → transmits undiminished to every point. And the payoff: same $Δ p$ over a bigger area $\Rightarrow$ bigger force — the hydraulic press $F_{2} = F_{1} A_{2} / A_{1}$ (see Hydraulic systems — brakes, lifts, jacks).

Recall Feynman: the whole walkthrough in plain words

Imagine a tiny wedge of water, so small it's basically a dot. Water at rest can't push sideways (it'd just slide), so it only pushes straight in on each face. Write down "pushes left = pushes right" and "pushes up = pushes down" for this wedge. When you plug in the shapes of the faces, the tilt angle cancels itself out and you're left with a shocking, simple fact: the water pushes exactly as hard in every direction at that dot. No direction is special. Now stack up a tall column of water — the deeper you go, the more weight sits on top, adding $ρ g h$ . If you press a piston on the surface with some extra squeeze $Δ p$ , that squeeze doesn't fade going down, because the weight part hasn't changed — only the top pressure did, and it moved by $Δ p$ everywhere. So the same extra pressure shows up at every point. Finally: press a skinny piston, and that same pressure lands on a fat piston with a huge area — pressure times big area = huge force. You lifted a car with your finger. The catch? The skinny piston has to travel far while the fat one barely moves — you swapped distance for force, and energy stays honest.

Recall Quick self-test

Which term in $p = p_{0} + ρ g h$ carries the applied change $Δ p$ ? ::: $p_{0}$ (the surface term); $ρ g h$ is unchanged. Why does $θ$ disappear from the wedge result? ::: The same factor $d A_{n} sin θ$ (or $cos θ$ ) sits on both sides and cancels. What breaks Pascal's law? ::: A compressible fluid (absorbs the change) or an un-enclosed fluid (lets it escape).

Connections

Parent: Pascal's law — the summary this page derives.
Pressure — force per unit area — the definition $p = F / A$ from Step 1.
Hydrostatic pressure — p = p0 + ρgh — supplies the $ρ g h$ term held constant in Step 7.
Incompressibility & continuity equation — why the fluid must not compress.
Hydraulic systems — brakes, lifts, jacks — the payoff $F_{2} = F_{1} A_{2} / A_{1}$ .
Conservation of energy in machines — why distance-for-force keeps energy honest.
Archimedes' principle — buoyancy — also rests on isotropic fluid pressure.

Visual walkthrough — Pascal's law — pressure transmits equally

Step 1 — What "pressure" means (the thing we are about to spread around)

Step 2 — Why fluid force must be perpendicular

Step 3 — Build the tiny wedge (our test element)

Step 4 — The geometry: how the slant face relates to the others

Step 5 — Balance the horizontal forces → $p_{x} = p_{n}$

Step 6 — Balance the vertical forces → $p_{y} = p_{n}$ , and the punchline

Step 7 — From "equal in all directions" to "transmits undiminished"

Step 8 — Edge & degenerate cases (never leave a gap)

The one-picture summary

Connections

Concept Map