Shock Waves

The question

Whisper, and the pressure in the air changes by a few billionths of an atmosphere. A lightning bolt or a supersonic jet changes it by a lot more — and you hear a crack, not a rumble. What happens when gas is pushed faster than it can get out of the way?

Sound that catches itself up

Sound is a pressure disturbance that travels at the speed of sound, $a = \sqrt{\gamma R T}$ — about 340 m/s in room air. For a whisper, that is the whole story: every part of the wave travels at the same speed, and the wave keeps its shape.

A loud wave is different. Where the gas is compressed, it is a little hotter, so sound travels a little faster there; and the gas itself is moving forward, carrying the wave with it. So the crest of a strong pressure pulse moves faster than its foot. It catches up, and the front of the wave gets steeper and steeper.

WatchWhy a loud sound turns into a shock

Strength of the pulseu = 0.25 a₀ · peak Δp ≈ 41%

Crest speed1.30 a₀

Foot speed1.00 a₀

Shock forms aftert = 0.27stronger pulses steepen sooner

Challenges0/2

Watch a pulse break into a shock.
Make a pulse strong enough to break before t = 0.15.

A pressure pulse travelling to the right, plotted as the gas velocity u it carries, in a frame moving at the speed of sound. Every part of the pulse moves at the local sound speed plus the local gas speed — and both are larger where the gas is compressed. So the crest gains on the foot and the front steepens. At the breaking time the profile would fold over (dashed); nature forbids that, and a shock forms instead, placed so the two lobes it cuts off have equal areas.

The steepening stops only when the front becomes so thin — a few molecular mean free paths, less than a micrometre — that viscosity and heat conduction can balance it. That nearly discontinuous jump is a . Shocks are not exotic: any sufficiently strong compression must end up as one.

What a shock does

To an observer riding on a shock, gas streams in from ahead at a supersonic speed — the shock's $M_1$ — and leaves behind it, slower and subsonic. Whatever happens inside the thin front, mass, momentum and energy must be conserved across it. Those three balances, the , fix every jump once the Mach number is known:

$\frac{\cshock{p_2}}{\cshock{p_1}} = 1 + \frac{2\gamma}{\gamma+1}\left(M_1^2 - 1\right), \qquad \frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2 + 2}.$

ExploreWhat a shock does to gas

Shock Mach numberM = 3.00

Ratio of specific heats γγ = 1.40 · air

Pressure× 10.3

Temperature× 2.68300 K → 804 K

Density× 3.86

Mach behind (wave frame)0.48always subsonic

Challenges0/2

Find the shock Mach number that heats 300 K air to 1600 K — a detonation’s von Neumann temperature.
Get the density ratio within 5% of its limit.

The jump conditions across a normal shock, from conservation of mass, momentum and energy alone. Pressure and temperature rise without limit as the shock gets stronger; density cannot exceed (γ+1)/(γ−1) — six times for air. Logarithmic vertical scale.

Notice the asymmetry. The pressure and temperature rise without limit as the shock strengthens, but the density saturates at $(\gamma+1)/(\gamma-1)$ : no shock can compress air more than six-fold. A strong shock spends its energy heating the gas rather than squeezing it. That is precisely what a detonation needs.

Estimate first

Air at 300 K is hit by a Mach 5 shock. What temperature does it reach, in kelvin?

1,00010,000100,000

Drag along the line to place your guess — each tick is ten times the last.

Shocks on demand: the shock tube

To study shocks and detonations in the laboratory, researchers use a deceptively simple device: a long tube divided by a thin diaphragm, with high-pressure "driver" gas on one side and low-pressure "driven" gas on the other. Burst the diaphragm, and the high-pressure gas acts like a piston that has suddenly started moving. A shock runs into the driven gas, an expansion fan runs back into the driver, and a contact surface — where the two gases meet — follows the shock at the gas speed.

ExperimentBursting a diaphragm: the shock tube

Diaphragm pressure ratio p₄/p₁10 : 1

press Burst after changing

Driver temperature T₄/T₁1.0 : 1

Shock Mach number1.61theory

Pressure behind the shock2.8 × p₁

Temperature behind the shock418 Kfrom 300 K

Challenges0/3

Burst the diaphragm and find the shock, the contact surface and the expansion fan on the x–t diagram.
Make a shock faster than Mach 3.
Heat the gas behind the shock above 1000 K — hot enough to ignite hydrogen.

A closed tube with high-pressure gas on the left and low-pressure gas on the right, computed live by solving the Euler equations. When the diaphragm bursts, a shock wave runs right, an expansion fan runs left, and a contact surface between the two gases follows the shock. The lower panel is an x–t diagram: time runs downward, and the brightness marks steep gradients — like a schlieren photograph taken at every instant.

The simulation above is not an animation: it solves the compressible Euler equations — conservation of mass, momentum and energy — on 360 cells, live, using the same kind of finite-volume method that research codes use. Later in the course, the same solver, with chemistry added, will run a detonation and then a whole rotating detonation engine.

Going deeper
Going deeper: how the solver captures a shock

The solver divides the tube into cells and tracks the mass, momentum and energy in each. At every cell face it estimates the flux of each quantity by solving an approximate Riemann problem — exactly the shock-tube problem, in miniature, between the states on either side of the face. This one uses the HLL flux of Harten, Lax and van Leer, with a "MUSCL" reconstruction that makes it second-order accurate in smooth regions and a minmod limiter that keeps it from oscillating at shocks. Because it conserves mass, momentum and energy exactly, it automatically gets the right shock speed and jumps, even though the shock is smeared over two or three cells instead of a micrometre.

Check your intuition

A shock tube is run with a much higher pressure ratio across the diaphragm. Which of these does NOT keep increasing without limit?

Recap

What to carry forward

Strong compressions steepen because their crests travel faster; they end as shock waves, thin jumps in pressure, density and temperature.
The Rankine–Hugoniot relations fix the jumps from the Mach number alone. Pressure and temperature rise without limit; density saturates at $(\gamma+1)/(\gamma-1)$ .
A Mach 5 shock heats air from 300 K to about 1700 K — hot enough to ignite fuel. In a detonation, the shock is the igniter.
A shock tube makes shocks on demand: shock, contact surface and expansion fan.

Next: put a fuel behind the shock. How does a detonation choose how fast to run?