Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁: Visual walkthrough

Step 0 — What are we even looking at?

WHAT. Draw a box (a control volume) that straddles the shock. Gas flows in from the left in state 1 and leaves on the right in state 2. The shock itself is the thin vertical line in the middle.

WHY. A control volume is just a fixed region of space we agree to watch. Whatever crosses its left face must be accounted for on its right face. That bookkeeping is the whole game — no forces, no magic, just "what goes in must come out."

PICTURE. In the figure below, blue is the fast upstream gas, orange is the slow squeezed downstream gas, and the red line is the shock.

Figure — Normal shock properties — M₂, P₂ - P₁, T₂ - T₁, ρ₂ - ρ₁, P₀₂ - P₀₁

Every quantity carries a subscript telling you which side it lives on:

Step 1 — Mass cannot pile up (continuity)

WHAT. Count the kilograms of gas crossing each face every second. In through the left face equals out through the right face: $\underbrace{\rho_1 u_1}_{\substack{\text{kg/s per m}^2\\ \text{entering}}}=\underbrace{\rho_2 u_2}_{\substack{\text{kg/s per m}^2\\ \text{leaving}}}\tag{1}$

Term by term: $ρ$ is how much gas is in each cubic metre, $u$ is how many metres of that gas sweep past per second. Multiply them and you get kilograms sweeping past per second, per square metre of face. That number must be identical on both sides — otherwise mass would build up inside the box forever.

WHY. Nothing is created or destroyed inside a paper-thin shock. This is the tightest constraint we have and it links $ρ$ to $u$ : if the gas slows down ( $u_{2} < u_{1}$ ), it must get denser ( $ρ_{2} > ρ_{1}$ ) to keep the product fixed.

PICTURE. The same "mass current" arrow, thick on both sides, but on the right the gas is slower (short arrow) and denser (dots crowded together).

Step 2 — Momentum changes only by a net push

WHAT. The net pressure force on the box equals the change in momentum flowing out: $\underbrace{P_1+\rho_1 u_1^2}_{\text{"push" on left}}=\underbrace{P_2+\rho_2 u_2^2}_{\text{"push" on right}}\tag{2}$

Term by term: $P$ is the static pressure pushing on the face. $ρ u^{2}$ is the momentum flux — the momentum ( $ρ u$ , mass current) carried across the face at speed $u$ . Their sum is called the impulse function, and it is the same on both sides.

WHY. Newton's second law for a flow: (net force) = (momentum out) − (momentum in). The only forces on our frictionless, walls-do-no-work box are the pressures on the two faces. So the whole balance reduces to equation (2). This is what forces $P$ to rise as $u$ falls.

PICTURE. Two competing effects drawn as arrows: pressure pushing left-to-right, momentum being carried across. The sum stays level like a see-saw in balance.

Step 3 — Energy is carried, not lost (adiabatic)

WHAT. No heat enters, no shaft does work, so the total energy per kilogram is conserved: $\underbrace{c_p T_1}_{\text{stored heat}}+\underbrace{\tfrac12 u_1^2}_{\text{motion}}=\underbrace{c_p T_2}_{\text{stored heat}}+\underbrace{\tfrac12 u_2^2}_{\text{motion}}\tag{3}$

Term by term: $c_{p} T$ is the enthalpy — the thermal energy each kilogram carries. $\frac{1}{2} u^{2}$ is the kinetic energy each kilogram carries. As the flow slows ( $u$ drops), that lost motion energy has nowhere to go but into heat, so $T$ rises. Energy just changes clothes.

WHY. The shock is too thin to leak heat and does no work on a piston. So the sum of thermal + kinetic energy is frozen. This is exactly the statement that stagnation temperature $T_{0}$ is conserved (see Stagnation Properties T0 and P0).

PICTURE. A two-bar "energy budget": on the left, tall motion bar + short heat bar. On the right the bars have swapped heights, but the total column height is identical.

Step 4 — Rewrite everything in Mach number

WHAT. We have three laws but they mix $P, ρ, u, T$ . The clean move is to translate into Mach number using two facts about an ideal gas: $a^{2} = γ R T, u = M a \Rightarrow ρ u^{2} = ρ γ R T M^{2} = γ P M^{2} .$

That last identity used $P = ρR T$ to swap $ρR T$ for $P$ . Now feed $ρ u^{2} = γ P M^{2}$ into the momentum law (2): $P_1\underbrace{(1+\gamma M_1^2)}_{\text{depends on }M_1}=P_2\underbrace{(1+\gamma M_2^2)}_{\text{depends on }M_2}\;\Rightarrow\;\boxed{\frac{P_2}{P_1}=\frac{1+\gamma M_1^2}{1+\gamma M_2^2}}\tag{4}$

WHY this tool and not another? Mach number is the one dimensionless dial that measures "how supersonic" the flow is. Because a shock's whole personality is set by how far past sound the flow is, expressing everything in $M$ makes the answer depend on $M_{1}$ alone — every other variable ( $P, T, ρ$ ) becomes a consequence. This is why we chose Mach number over, say, absolute velocity.

PICTURE. A conversion chart: the messy $P, ρ, u, T$ box on the left, an arrow labelled " $u = M a, a^{2} = γ R T$ ", and a tidy $M$ -only box on the right.

Do the same substitution on energy (3). Since $T_{0} = T (1 + \frac{γ - 1}{2} M^{2})$ is conserved: $\frac{T_2}{T_1}=\frac{1+\frac{\gamma-1}{2}M_1^2}{1+\frac{\gamma-1}{2}M_2^2}\tag{5}$ Here $1 + \frac{γ - 1}{2} M^{2}$ is the factor that turns a static temperature $T$ into its stagnation value $T_{0}$ — it is the "kinetic energy tax" written in Mach language.

And mass (1), using $ρ u = \frac{P M γ}{R T}$ : $\frac{P_1 M_1}{\sqrt{T_1}}=\frac{P_2 M_2}{\sqrt{T_2}}\tag{6}$

Step 5 — Fold the three into one

WHAT. We now have three Mach-flavoured equations (4), (5), (6). Substitute the pressure ratio (4) and the temperature ratio (5) into the mass equation (6), then square both sides to remove the square roots. All the $P$ 's and $T$ 's cancel and we are left with a single equation relating only $M_{1}$ and $M_{2}$ .

WHY. Three equations, but we want one clean relationship. Eliminating $P$ and $T$ is the standard algebra move: keep combining until only the two quantities you care about survive.

PICTURE. Three streams (mass, momentum, energy) flowing into a funnel and dripping out as one equation.

That single equation is a quadratic in $M_{2}^{2}$ . A quadratic has two roots:

Step 6 — Check the physics: supersonic in, subsonic out

WHAT. Plug numbers into (7) and see what comes out. For air ( $γ = 1.4$ ):

$M_{1}$	$M_{2}$ from (7)
$1.0$	$1.000$
$2.0$	$0.577$
$5.0$	$0.415$
$\to \infty$	$\to 0.378$

WHY. A formula is only trustworthy if it behaves. Every entry with $M_{1} > 1$ gives $M_{2} < 1$ : supersonic always becomes subsonic. This is not put in by hand — it falls out of (7), and it is enforced by the second law (a shock the other way would decrease entropy, which is forbidden).

PICTURE. The curve $M_{2} (M_{1})$ crossing the line $M_{2} = M_{1}$ exactly once at $M_{1} = 1$ , then bending below it and flattening toward $0.378$ .

Step 7 — The edge cases (never leave the reader stranded)

WHAT & WHY & PICTURE, all three limits on one figure:

The strong-shock cap on Mach has a famous partner: density also saturates at $\frac{γ + 1}{γ - 1} = 6$ (see the Rankine–Hugoniot Relations). Both ceilings come from the same source — mass conservation plus a finite velocity drop.

The one-picture summary

This final figure compresses the whole journey: three conservation laws (mass, momentum, energy) enter on the left; the ideal-gas + Mach translation squeezes them in the middle; a single quadratic drops out on the right; its physical root is equation (7); and its behaviour is the supersonic-in / subsonic-out curve. From that one relation, every property ratio in the parent note follows by back-substitution.

Recall Feynman: the whole walkthrough in plain words

Picture gas rushing into a box faster than sound. Three unbreakable rules apply. Rule one: kilograms in equal kilograms out — so if the gas slows down inside, it must get denser. Rule two: the only thing that can change the gas's momentum is the pressure pushing on the two ends — so as it slows, pressure must climb. Rule three: no heat sneaks in or out — so the motion energy the gas loses turns straight into heat, and it warms up. To make these rules easy to combine, we rewrite all of them using one number, the Mach number, which just says how much faster than sound the gas is going. When we blend the three rewritten rules and cancel the leftovers, we're left with a single tidy equation: the downstream Mach number depends only on the upstream Mach number. It has two answers — "nothing happened" (boring) and "a real shock" (the one we want). The real one always turns fast-than-sound flow into slower-than-sound flow, never the reverse, because reversing it would tidy up entropy, and nature never tidies entropy for free.

Recall Self-test

Why do we rewrite the three laws in terms of Mach number? ::: Because a shock's entire behaviour is set by how far past sound the flow is; using $M$ makes the answer depend on $M_{1}$ alone and makes $P, T, ρ$ mere consequences. The quadratic for $M_{2}^{2}$ has two roots — what is the fake one and why? ::: $M_{2} = M_{1}$ , the "no shock" solution; it satisfies the algebra but represents the flow passing through unchanged. As $M_{1} \to \infty$ , what does $M_{2}$ approach for air? ::: $M_{2} \to \frac{γ - 1}{2 γ} \approx 0.378$ — downstream Mach floors out, never reaching zero. Why can't a normal shock occur in subsonic flow? ::: It would require $M_{1} < 1 \to M_{2} > 1$ , meaning $Δ s < 0$ , which the second law forbids.