What a signal travels at

Air gets out of the way of an aeroplane. It does not do this out of politeness, and it does not do it because it can see what is coming: it does it because the aeroplane has already pushed on the air in front of it, and that push has travelled ahead as a pressure disturbance.

The whole of compressible flow follows from asking how fast that push travels, and from noticing that the answer is finite.

The disturbance is the sound #

There is no separate mechanism to learn. The pressure disturbance that warns the air is a sound wave, and the speed at which the warning arrives is the speed of sound, because those are two descriptions of one thing.

That identification is the useful part. It converts a question about aerodynamics — how much notice does the fluid get? — into a question about a gas, which has a clean answer.

Consider a plane wave running down a long tube of still gas, and step into the frame moving with the wave. In that frame the gas approaches the wave at speed $a$ and leaves at $a - \mathrm{d}u$, having had its pressure raised by $\mathrm{d}p$ and its density by $\mathrm{d}\rho$. Mass and momentum both have to balance across the front, and the pressure is a function of density, so

$$a^2 = \frac{\mathrm{d}p}{\mathrm{d}\rho}$$

which says that the speed of a small disturbance is set by how stiff the gas is against being squeezed. A stiffer gas carries the news faster, and a heavier one carries it more slowly.

Which derivative, and Newton’s mistake #

The derivative has to be taken at constant something, and choosing the wrong something is a famous error rather than a hypothetical one.

Newton took it at constant temperature, which gives $a = \sqrt{p/\rho}$ and a sea-level answer near 290 metres per second. The measured value is about 340. The discrepancy was known, embarrassing, and unexplained for over a century, and various fudges were proposed to close it.

Laplace supplied the resolution: the compressions and rarefactions in a sound wave happen too fast for heat to move between them. The gas is not held at constant temperature; it is compressed adiabatically, and it heats up where it is squeezed. The correct derivative is taken along an isentrope, giving $\mathrm{d}p/\mathrm{d}\rho = \gamma p/\rho$ and

$$a = \sqrt{\gamma \frac{p}{\rho}} = \sqrt{\gamma R T}$$

The ratio of specific heats $\gamma$ is 1.4 for air, and $\sqrt{1.4}$ is 1.183 — which is very nearly exactly the factor by which Newton’s answer was short.

This is worth dwelling on because it is the first appearance of the theme that runs through the whole of this field. The equations do not decide which process a gas is undergoing. The same conservation laws admit an isothermal wave and an adiabatic one, and picking the right hypothesis is a physical judgement rather than an algebraic step. It comes back at the shock, where mass, momentum and energy admit two jumps and only the second law rules one of them out.

The absent variable #

$a = \sqrt{\gamma R T}$ has no pressure in it, and no density either. Both were present in the previous line and both cancelled.

That absence is more surprising than it looks, and it is the claim this essay’s second figure exists to test.

The reason is a cancellation rather than a coincidence. Raising the pressure at fixed temperature raises the density in exactly the same proportion — that is what the equation of state says — so $p/\rho$ does not move. Sound is carried by a competition between stiffness and inertia, and compressing a gas increases both by the same factor.

So the common statement that sound travels more slowly in thin air is false as stated, and true by accident. It travels more slowly high up because it is colder high up. An aircraft climbing into the stratosphere at fixed true airspeed does creep closer to the speed of sound — from 340.3 m/s at sea level to 295.1 m/s at the tropopause — and the reason is a temperature falling by 71.5 K, not a pressure falling to a quarter.

What this makes possible, and what it makes impossible #

A finite signal speed has an immediate consequence that nothing in incompressible flow prepares anybody for.

In an incompressible fluid, the pressure field satisfies Laplace’s equation, which is elliptic: every point influences every other point instantly. Moving a body anywhere in the fluid changes the pressure everywhere at once. That is a strong and slightly alarming statement, and it is exactly the statement that $a \to \infty$ makes. The incompressible assumption is not “density is constant”; it is “the news arrives instantly”, and the two are the same assumption viewed from different sides.

Give the news a finite speed and the character of the problem changes. The fluid ahead of a body now learns about it at a definite moment, at a definite distance, and if the body moves faster than the news, some of the fluid never learns at all. What that does to the flow is the next rung and it is where the subject stops being a correction to what came before.

The number, at the numbers that matter #

The Mach number is the ratio of the flow speed to this signal speed, and it is the one number that decides whether density has to be treated as a variable.

Two things about the ratio are worth stating carefully, because they are routinely conflated.

The speed of sound is a local property. It is set by the local temperature, and in an accelerating flow the temperature is falling, so the speed of sound falls with it. A stream tube that accelerates from 200 to 340 metres per second does not reach Mach 1 at 340 metres per second, because by then the gas has cooled and its own signal speed is lower. The Mach number climbs faster than the speed does, which is one reason the transition sneaks up.

The Mach number is a ratio of speeds and not a speed. It is the compressible member of the same family as the Reynolds number, and it behaves the same way: two aircraft at the same Mach number in different air are doing different speeds over the ground, and the same aircraft holding the same true airspeed through a climb is not holding its Mach number. Everything in this field that depends on compressibility depends on the ratio.

What the assumption is actually worth #

The threshold at Mach 0.3 is a convention with a number behind it, and the number can be computed rather than quoted.

Bringing a flow to rest isentropically raises its density by a factor that depends on the Mach number alone. At Mach 0.3 that factor is 1.046 — a 4.6 per cent change, which is comfortably below the uncertainty in most of the rest of an aerodynamic estimate. At Mach 0.6 it is 19.5 per cent, which is not. At Mach 1 it is 57.7 per cent, at which point pretending that density is constant is no longer an approximation but a different problem.

The important thing about the curve is that it is smooth. There is no threshold in it. The threshold is a decision about how much error is tolerable, taken once, by people who then wrote it in the textbooks, and it deserves to be treated as such — a flow at Mach 0.32 is not compressible in any sense a flow at Mach 0.28 is not.

The molecules are the messengers, and it shows in the number #

Here is the connection that makes the formula stop looking like an accident.

A gas has no way to transmit a pressure change except by its molecules colliding. Nothing else is there. So the speed at which a disturbance propagates cannot possibly exceed the speed at which the molecules themselves are moving, and one would expect it to be the same sort of size.

It is. The root-mean-square molecular speed in a perfect gas is $\sqrt{3RT}$, which at 288.15 K comes to 498.0 metres per second. The speed of sound at the same temperature is 340.3. The ratio is

$$\frac{a}{\sqrt{3RT}} = \sqrt{\frac{\gamma}{3}} = 0.683$$

exactly, with no fitting and nothing measured. Sound travels at roughly two thirds of the speed of the molecules carrying it, and the shortfall is entirely the factor $\gamma/3$ — a number assembled from how many ways a diatomic molecule can store energy.

That is why the temperature is the only variable. Temperature is molecular speed, in the sense that it is defined by it, and the speed of sound is a fixed fraction of that. Compressing the gas puts more molecules in the box without making any of them faster, so the messages travel no more quickly; heating it makes every molecule faster, and they do.

The subsonic panel is worth a second look, because it contains a result usually taught as a separate topic. The wavelengths ahead of a moving source are compressed and those behind are stretched, in the ratio $1 \pm M$, and that is the Doppler shift falling out of a figure drawn for another purpose. It is the same set of circles.

What the solver computes, and how it is checked #

The relations used above are the isentropic relations for a calorically perfect gas, and they are closed form. That is a different situation from most of this site, where a field is stepped on a grid and the check is that the answer solves the equations. Here the arithmetic cannot come out wrong, so the checks have to be aimed somewhere else.

They are aimed at consistency between routes. assertIsentropicConsistent takes the ratios at a given Mach number and verifies three things that were not used to produce them: that $p/\rho^\gamma$ is the same at the static and stagnation states, which is what isentropic means; that $h + u^2/2$ comes out the same at both, which is the energy equation; and that the local speed of sound recovers the Mach number it started from. All three agree to within $10^{-10}$ at every Mach number this site draws.

The check earns its place because there is a plausible way to get this wrong. The density ratio and the pressure ratio differ only in an exponent — $1/(\gamma-1)$ against $\gamma/(\gamma-1)$ — and writing the pressure exponent in the density slot produces a table that is smooth, monotone, entirely ordinary-looking, and violates the energy equation at every row. The gate feeds exactly that error to the assertion and requires it to be refused.

The state ratios, and why they are the working form #

In practice nobody carries $a$ around. What gets used is the set of ratios between a flowing state and the state that flow would reach if it were brought to rest without loss, because those are the quantities an instrument can read and a duct can impose. They are the compressible replacement for the statement that speed and pressure trade against each other, and they replace it rather than correcting it, because a third variable has joined the exchange.

The three curves are the same relation raised to three powers, so they cannot cross and cannot change order. Temperature carries the exponent 1, density $1/(\gamma-1) = 2.5$, pressure $\gamma/(\gamma-1) = 3.5$ — which is why a compressible flow’s pressure changes are always the largest and its temperature changes the most modest, and why an aerodynamicist who ignores the temperature usually gets away with it and one who ignores the density does not.

The same competition in other media #

The perfect-gas form is a special case of the line before it — a signal speed is the square root of a stiffness over a density — and running that generally is worth doing, because it produces a result no intuition survives.

Water is about eight hundred times denser than air, so on the inertia alone sound should crawl through it. It travels at about 1,500 metres a second, four times faster, because water’s resistance to being compressed is greater by very much more than eight hundred. Steel is denser still and faster still. Stiffness wins, and it wins by more than density loses.

Now mix the two. A liquid carrying a few per cent of gas bubbles has essentially water’s density and essentially the gas’s compressibility, since the bubbles do all of the squeezing. Both factors move the wrong way at once, and the sound speed of a bubbly mixture falls to a few tens of metres a second — slower than in either constituent, by a wide margin, and low enough that a modest flow in such a mixture is transonic.

That single fact carries two of this collection’s essays. It is why an aerated spillway cushions a collapsing cavity, and it is why injecting air into a pipeline is a standard defence against water hammer: the pressure rise a stopped column produces is proportional to the signal speed, and the bubbles have removed most of it.

Where the model stops #

Three limits, and the first two are close enough to matter for real air.

$\gamma$ is not a constant. It is 1.4 for air only while the vibrational modes of the diatomic molecules stay unexcited, which fails above roughly 600 K. Everything on this site takes $\gamma = 1.4$, which is fine through the Mach numbers drawn here and wrong for re-entry, where the gas dissociates and then ionises and the very idea of a single ratio of specific heats stops applying.

A perfect gas is an idealisation. The relation $p = \rho R T$ is the first term of a series, and air near liquefaction is not described by it. That is remote from atmospheric flight and central to cryogenic wind tunnels.

A sound wave is small by assumption. The derivation above took the disturbance to be infinitesimal, so that $\mathrm{d}p/\mathrm{d}\rho$ could stand in for a finite ratio. A finite disturbance travels faster than $a$, because the compressed gas behind the front is hotter and its own signal speed is higher. That is not a defect of the model; it is the mechanism by which compression waves steepen into a shock, and it is why the subject has discontinuities in it at all.

What the picture cannot show #

The circles in the first figure are exact, and everything else about the drawing is a convenience.

A real pressure pulse is not a circle in a plane but a sphere in space, and its amplitude falls as it expands. Nothing about the amplitude is drawn here — the circles are all the same weight, so the figure says where the news has reached and is silent about how loud it is. That silence is appropriate for the argument, which is entirely about arrival times, and it would be misleading in any figure about attenuation.

The figure also draws pulses as discrete events, at intervals. Real disturbance is continuous, and the discrete circles are a sampling of it. Nothing in the argument depends on the sampling, which is easy to check by drawing more of them: the envelope in the supersonic case is the same line.

Who found it, and when #

Newton published the constant-temperature calculation in the Principia in 1687, and it was wrong by a factor of 1.183. He knew it disagreed with measurement and proposed corrections involving the finite size of air particles and the presence of water vapour, which were ingenious and not the answer.

Laplace supplied the adiabatic argument in 1816, some hundred and thirty years later. The gap is worth noticing: what was missing was not mathematics but thermodynamics, which did not yet exist in usable form. The ratio of specific heats had to be understood before the speed of sound could be computed, and the speed of sound had in fact been used to measure $\gamma$ for decades afterwards, which is the sort of loop a subject closes when it becomes coherent.

Where the ladder goes next #

The signal has a speed. The next question is what happens when the source is going faster than the signal it is making — which is not a matter of degree, because the fluid ahead is then in a state no subsonic flow has an analogue for. It has not been told anything at all.

That is the next rung, and it is where the Mach cone comes from and where the equations change type. The other direction from here is what stays constant along an accelerating flow once density is free to change, which is where Bernoulli’s equation gets replaced rather than corrected.