Oxygen makes a boom's crack, nitrogen its rise time

A boom is aged in the thin air it starts in follows a sonic boom’s signature down refracted rays and finds that the nonlinear ageing which turns an aeroplane’s pressure field into an N-wave is done mostly in the stratosphere, near the aeroplane, where a pressure disturbance distorts thin air fastest. It ends on the other thing a listener hears. An N-wave has two shocks, and how loud it sounds depends less on its peak pressure than on how fast that pressure arrives — the rise time. A shock that takes a tenth of a millisecond to rise is a crack; one that takes several milliseconds is a thud. The earlier essay expected the rise time to be decided at the bottom of the atmosphere, in the dense, wet air near the ground.

This essay computes the structure of such a shock, and finds that the air near the ground decides it through two of its molecules, which do quite different jobs.

A shock that is too weak to be thin #

The discontinuity that has a thickness solves the inside of a shock with viscosity and heat conduction and finds a thickness of a few mean free paths at a Mach number of two — a fraction of a micrometre in air at sea level. A sonic boom’s shock is extraordinarily weak by comparison. A pressure jump of 50 pascals on an atmosphere of a hundred thousand is a shock Mach number of 1.0002, and the thickness of a viscous shock grows as one over its strength. Taylor’s solution for a weak shock in a viscous gas is a logistic curve whose rise, from 10 to 90 per cent of the jump, takes

$$t_{\text{rise}} = \frac{4\ln 9\;\rho\,\delta}{\beta\,\Delta p},$$

with $\delta$ the diffusivity of sound and $\beta = (\gamma + 1)/2$ the coefficient of nonlinearity. For air at 20 °C the standard absorption model gives $\delta = 3.77\times10^{-5}$ square metres per second, and a 50-pascal shock rises in 3.3 microseconds. That is what the classical theory predicts for a boom, and measured booms rise hundreds of times more slowly.

The reason is that the weak shock is thick enough to feel a slower process. Oxygen and nitrogen molecules store energy in vibration as well as in motion and rotation, and when a pressure wave compresses them the vibration catches up only after a delay — a relaxation time, as in the gas that has not finished being shocked behind a strong shock. For as long as the vibration lags, the gas is effectively stiffer and sound travels slightly faster in it than the speed a small signal otherwise travels at. A process like this absorbs sound most strongly at frequencies near the inverse of its relaxation time, and it disperses sound: high frequencies travel faster than low ones by an amount $\Delta c$ that is small but, for a weak shock, decisive.

Two relaxing gases, and the equation that carries them #

The augmented Burgers equation carries a plane wave in retarded time $\tau$ over a distance $\sigma$ with nonlinear steepening, thermoviscous diffusion, and one term for each relaxing gas:

$$\frac{\partial p}{\partial\sigma} = \frac{\beta}{2\rho c^3}\frac{\partial p^2}{\partial\tau} + \frac{\delta}{2c^3}\frac{\partial^2 p}{\partial\tau^2} + \sum_\nu p_\nu, \qquad \Big(1 + \theta_\nu\frac{\partial}{\partial\tau}\Big)p_\nu = \frac{\Delta c_\nu}{c^2}\,\theta_\nu\frac{\partial^2 p}{\partial\tau^2}.$$

Every constant in it is taken from ISO 9613-1, the international standard for the absorption of sound in air. It gives the relaxation frequencies of oxygen and nitrogen from the temperature and the water content, and the height of each process’s absorption peak, from which its dispersion $\Delta c_\nu$ follows. At 20 °C and 50 per cent humidity oxygen relaxes at 35 kilohertz, a relaxation time $\theta_O$ of 4.5 microseconds, with $\Delta c_O = 0.115$ metres per second; nitrogen relaxes at 332 hertz, $\theta_N = 0.48$ milliseconds, with $\Delta c_N = 0.022$ metres per second.

A steady shock is a profile that travels without changing, and the equation can be integrated once for it. Outside a thermoviscous layer a few microseconds thick, the nonlinear term must be balanced exactly by the relaxing gases, and what is left is a single ordinary differential equation for the pressure across the shock:

$$\frac{dp}{dt} = \frac{\sum_\nu r_\nu/\theta_\nu}{\sum_\nu \Delta c_\nu/c^2 - \frac{\beta}{2\rho c^3}(\Delta p - 2p)},$$

with $r_\nu$ each gas’s accumulated lag. If the denominator is positive from the foot of the shock, the profile is smooth all the way up: the shock is fully dispersed, with no discontinuity in it at all. That happens when

$$\Delta p < \Delta p^* = \frac{2\rho c\sum_\nu \Delta c_\nu}{\beta},$$

which for air at 20 °C is 94 pascals. Most sonic booms at the ground are weaker than that.

The first figure draws the result for a 50-pascal shock. At 90 per cent humidity it rises in 0.35 milliseconds; at 30 per cent in 1.03; at 10 per cent in 2.95 — nearly nine hundred times the viscous shock’s rise, which at this scale is a vertical line. Each profile has the same shape: a steep front that carries most of the jump, and then a long, slow approach to the top.

Oxygen removes the jump, nitrogen sets the time #

The shape says two processes are at work, and taking them one at a time says which does what.

With oxygen alone, the second figure’s green curve, the shock is fully dispersed — oxygen’s dispersion alone can smooth any shock below 79 pascals — and it rises in 31 microseconds, ten times the viscous shock and twenty times faster than the real one. With nitrogen alone the story reverses. Nitrogen’s dispersion is five times smaller and can smooth only shocks below 15 pascals, so a 50-pascal shock keeps a discontinuity of 35 pascals at its front, a viscous subshock microseconds thick, followed by a tail that takes most of a millisecond. Together, the discontinuity is gone, because the two dispersions add and their sum exceeds what the shock needs. But the tail stays.

So the two gases divide the work. The first seven-tenths of the jump arrive in 34 microseconds, carried by oxygen’s fast relaxation; from 10 to 50 per cent takes 24 microseconds. From 50 to 90 per cent takes 600 microseconds, and that part is nitrogen’s. The rise time that is measured, the 10-to-90 per cent interval by which booms are compared, is set almost entirely by the slow gas. Oxygen decides whether there is a discontinuity; nitrogen decides how long the rise lasts.

What water does #

Both relaxation rates depend on water. A vibrating oxygen or nitrogen molecule gives up its vibrational energy in a collision, and a water molecule, with its own low-lying vibrational states and a strong interaction, takes it far more readily than another oxygen or nitrogen molecule does. The third figure shows the consequence: oxygen’s relaxation frequency rises from 1.3 kilohertz at 5 per cent humidity to 80 kilohertz in saturated air, and nitrogen’s from 41 hertz to 650. The water content that matters is the molar fraction, so warm humid air, which holds more water at the same relative humidity, relaxes faster still.

The fourth figure is the rise time across the whole range of humidity. For a 50-pascal shock it falls from 2.95 milliseconds at 10 per cent humidity to 1.03 at 30, 0.52 at 60 and 0.35 at 90 — roughly in step with the nitrogen relaxation time, which is what the division of labour predicts. Halving the shock’s strength doubles its rise time at every humidity.

The dispersions are a different matter. In the standard model the heights of the absorption peaks, and so the $\Delta c_\nu$, depend on temperature and not on humidity, and the threshold below which a shock is fully dispersed, 94 pascals at 20 °C, is the same on a dry day and a wet one. Water decides how thick the shock is, but not whether it has a discontinuity.

What the ear gets #

A rise time is one number for a shape that has two parts, and the ear does not listen to either part alone. Its sensitivity peaks between one and four kilohertz, and what it hears of the front of a boom is how much of the jump survives at those frequencies. The spectrum of the front says it directly: the Fourier transform of the pressure’s rate of rise, divided by that of a perfect jump, which is one at low frequencies and falls where the front stops looking like a jump.

The sixth figure computes it. At 50 per cent humidity the front of a 50-pascal shock is 2.8 decibels below a perfect jump at one kilohertz and 4.0 at four. The nitrogen tail is responsible for about three of those decibels at every frequency above a few hundred hertz — the slowly arriving fifth of the jump simply does not count at those frequencies — and it costs no more at higher frequencies than lower ones. Where the front begins to fall steeply is oxygen’s business, and at 50 per cent humidity that is above ten kilohertz, where the ear is already losing interest. At 10 per cent humidity oxygen relaxes ten times more slowly, in 42 microseconds, and the front is 7.3 decibels down at one kilohertz and 35 down at four: the crack has gone.

So the two molecules divide the listening too. The measured rise time is nitrogen’s, and a front that a rise-time meter scores at 0.62 milliseconds can still contain most of its jump in 34 microseconds and sound sharp. How sharp is oxygen’s, and oxygen’s rate is the one that water changes most — by sixty times between 5 per cent humidity and saturation, where nitrogen’s changes by sixteen. A dry day softens a boom mostly through oxygen.

Why oxygen disperses five times more than nitrogen #

The two dispersions differ by a factor of 5.3, and the reason is written into the standard’s own formula for them. Each contains a Boltzmann factor: $e^{-2239.1/T}$ for oxygen and $e^{-3352/T}$ for nitrogen, and the numbers in the exponents are the temperatures, in kelvin, that correspond to one quantum of each molecule’s vibration. A relaxing gas can disperse sound only to the extent that its molecules are actually exchanging energy with the vibration, and at room temperature a small fraction of them are excited — far more of oxygen’s, whose quantum is smaller. The ratio of the two Boltzmann factors at 20 °C is 45; the standard’s coefficients take back a factor of about eight for the different strengths of the two absorption processes; what is left is the factor of 5.3 between $\Delta c_O$ and $\Delta c_N$.

The same factors make the dispersion depend on temperature, sharply. At 0 °C the two dispersions fall to 0.073 and 0.010 metres per second, and the strength below which a shock is continuous falls from 94 pascals to 60; at −20 °C it falls to 35. A boom that would arrive fully dispersed on a warm day keeps a discontinuity at its front on a winter one, and cold air also holds far less water: 50 per cent humidity is 1.15 per cent water by molecule count at 20 °C and 0.06 per cent at −20 °C. Both effects push the same way. The rise times of booms measured in winter and in summer are not directly comparable, and the reason is in two exponents.

At frequencies far below its relaxation frequency, a relaxing gas behaves as though it had an extra viscosity that acts only on compression — the viscosity nobody uses, which for air at audible frequencies is mostly this. A weak shock’s front, whose frequencies run from below nitrogen’s relaxation to above oxygen’s, straddles both regimes, which is why neither a bulk viscosity nor a frequency-independent absorption can describe it.

Strength, and where the discontinuity comes back #

The fifth figure follows the strength. Below the limit the rise time is inversely proportional to the jump — 3.3 milliseconds at 10 pascals, 1.6 at 20, 0.62 at 50, 0.24 at 94 — the same law the viscous shock obeys, displaced upwards by the ratio of the relaxing and thermoviscous time scales. Above it, a discontinuity reappears at the front: 5.9 pascals of a 100-pascal shock, 56 of a 150-pascal one, 106 of a 200-pascal one. Once the subshock carries the 10 per cent point, the rise time measures the subshock plus a short dispersed tail, and it collapses towards the viscous value — 31 microseconds at 150 pascals, 16 at 200.

That transition is where the everyday picture of a boom comes from. The booms measured close under a supersonic aircraft, or from a fast low pass, are strong enough to keep a viscous subshock and crack; the booms that arrive at the ground from cruise altitude are usually weaker, fully dispersed, and their fronts are the slow, relaxing structure drawn here. It is also why a boom shaped to be quiet can gain more than its reduced peak pressure suggests: a weaker front is a longer-rising one, and the rise time falls as one over the strength twice over — once through the peak, once through the shape.

What this does to the edge of the carpet #

The earlier essay found that the rays reaching the edge of the boom carpet arrive with 0.56 of the under-track overpressure. In the fully dispersed regime the rise time goes as one over the strength, so at the same humidity the edge’s rise time is 1.8 times the under-track one: a 50-pascal boom rising in 0.62 milliseconds under the track becomes a 28-pascal boom rising in 1.14 milliseconds at the edge. That is the steady-state part of the answer to the earlier essay’s question, and it comes from the strength alone. The rumble at the edge is partly this: a weaker boom whose front has spread, before any account is taken of the longer path the edge rays spend in the wet air near the ground.

Is the air given time to do it? #

A steady profile is the one the equation settles on after the shock has travelled far enough to forget where it started, and it exists at all only because the equations allow a jump to be a travelling state of its own. How far is enough can be estimated for each gas separately: the distance over which a dispersion $\Delta c_\nu$ acting for a time $\theta_\nu$ reshapes the front, $c^2\theta_\nu/\Delta c_\nu$. For oxygen at 50 per cent humidity it is 4.6 metres; the oxygen structure is set up almost as soon as a boom enters any layer of air. For nitrogen it is 2.6 kilometres, and at 10 per cent humidity 11.8.

That is the most important caveat on the numbers here. A boom descending through the lowest two or three kilometres of the atmosphere, where the humidity and temperature change with height, does not have time to establish nitrogen’s steady structure at every level; the rise time at the ground is partly a memory of the drier, colder air above. The steady values are therefore bounds on what the relaxing air can do rather than predictions of a particular boom, and the full answer needs the augmented Burgers equation marched along each ray through a measured atmosphere.

What the calculation was checked against #

The seventh figure is the ledger. With one relaxing gas, the steady profile has a closed form, and its 10-to-90 per cent rise time is $2\ln 9\,\theta\,(\Delta c/c^2)/(\beta\Delta p/2\rho c^3)$; the numerical profile for nitrogen alone at three strengths below its own limit reproduces it to a part in $10^7$. The profile reaches the full jump. Quadrupling the number of integration steps moves no rise time by more than a part in $10^5$. And the rise time falls with humidity at every level tested while staying more than twenty times the viscous value — which is the essay’s claim, stated as a test.

What the picture cannot show #

Turbulence. Measured boom rise times are usually longer than the steady relaxation value for the same strength and humidity, and much more variable from one boom to the next. The accepted explanation is scattering by turbulence in the atmospheric boundary layer, which distorts the front differently along each ray; it is not in the model, which describes still air.

A steady shock in uniform air. The atmosphere changes with height, and nitrogen’s structure takes kilometres to establish.

The standard absorption model’s constants. ISO 9613-1 is an empirical fit to laboratory absorption measurements, accurate to about ten per cent over the ranges it was fitted to, and it is used here as the source of every relaxation constant.

A plane wave. The earlier essays’ ray-tube spreading and ageing are left out, so each profile is the local steady state of a shock of the stated strength, wherever it is.

The convention the numbers depend on #

The rise time is the interval between 10 and 90 per cent of the jump, the convention used in boom measurements. Relative humidity is converted to the molar fraction of water by the standard’s own formula; at 20 °C and 50 per cent the air is 1.15 per cent water by molecule count. The relaxation time is $\theta_\nu = 1/2\pi f_\nu$, with $f_\nu$ the relaxation frequency. The strengths are pressure jumps on a sea-level atmosphere, and the nonlinear coefficient is $\beta = 1.2$, which is the ratio of specific heats entering through $(\gamma + 1)/2$.

Who found it, and when #

Michael Lighthill worked out the structure of a weak shock in a relaxing gas in 1956, including the fully dispersed case in which there is no discontinuity at all, and Polyakova, Soluyan and Khokhlov gave the general theory in 1962. Allan Pierce and Jongmin Kang applied it to sonic booms around 1990 and showed that molecular relaxation, not viscosity, sets the rise times of booms at the ground, while leaving part of the measured spread to turbulence. The absorption constants are those of Henry Bass, Louis Sutherland and their colleagues, standardised as ISO 9613-1 in 1993. The augmented Burgers equation in the form used here, with several relaxation processes, is Robin Cleveland’s, from his 1995 thesis on the propagation of sonic booms.

Still open: the rise time along a ray #

The two remaining steps are the ones this essay identified and did not take. The first is to march the augmented Burgers equation along a boom’s ray through a measured atmosphere — humidity and temperature changing with height — from the altitude at which the N-wave forms to the ground, and to find how much of the steady rise time the nitrogen relaxation actually reaches in the last two or three kilometres. The second is the turbulent boundary layer, whose effect on the rise time is measured to be comparable to relaxation’s. A calculation that carries a front through a layer of random refractive-index fluctuations of stated strength and scale, and asks how far the rise time is lengthened and how widely it is scattered, would say how much of the spread in measured booms is the air’s molecules and how much its eddies.