When gamma stops being a number

Every compressible essay on this site carries a $\gamma$, and it has been 1.4 in all of them. That number is a count: three ways a nitrogen molecule can translate and two ways it can rotate, so $\gamma = 1 + 2/f$ with $f = 5$.

Above about a thousand kelvin a third kind of motion opens. A diatomic molecule can vibrate, the energy in that mode is quantised, and it switches on gradually over a range of temperature set by the characteristic vibrational temperature — 3,390 K for nitrogen, 2,270 K for oxygen. The heat capacity follows the Einstein function

$$\frac{c_v}{R} = \frac52 + \frac{(\theta_v/T)^2 e^{\theta_v/T}}{(e^{\theta_v/T}-1)^2},$$

and $\gamma = 1 + R/c_v$ falls away from 1.4 accordingly.

Which equation the heat capacity is in #

Before computing anything, it is worth noticing where the change can possibly enter. The jump conditions are four relations and only one of them contains a heat capacity:

$$\rho_1u_1 = \rho_2u_2,\qquad p_1 + \rho_1u_1^2 = p_2 + \rho_2u_2^2,$$

$$h_1 + \tfrac12u_1^2 = h_2 + \tfrac12u_2^2,\qquad p = \rho RT.$$

Mass and momentum contain no thermodynamics whatever — they are statements about fluxes of mass and momentum, and a gas’s internal degrees of freedom are invisible to both. The equation of state contains $R$, which is a property of the molecular weight and not of the internal modes. Only the energy equation knows that the molecule can vibrate, through $h(T)$.

That single observation predicts the whole result, and the rest of the essay is measuring it.

Solving the jump without a constant γ #

The closed form for a normal shock assumes $h = c_pT$ with $c_p$ constant, so with $h(T)$ a general function there is no closed form. What there is, is a fixed point.

Guess the density ratio $r$. Then $u_2 = u_1/r$ follows from mass; $p_2$ follows from momentum; $h_2$ follows from energy; $T_2$ follows from inverting $h(T)$; and the thermal equation of state gives $r$ back. Iterate, with damping, because the map is stiff at high Mach number.

The enthalpy is $\int c_p\,\mathrm dT$ by Simpson’s rule and its inverse is a bisection, since $h$ is strictly increasing. Nothing in the loop assumes a constant $\gamma$, and the loop is the whole difference between this and the closed form.

The asymmetry, measured #

Across Mach 2 to 15, with the free stream at 220 K:

The pressure moves by at most four and a half per cent and the temperature by eighteen.

The mechanism is the one the equation count predicts. Energy going into the vibrational mode is energy not going into random translation, so the gas is cooler than a perfect-gas calculation says. Being cooler at the same pressure, it is denser. Being denser, it is slower — and the momentum balance, which fixed the pressure, is satisfied by the same pressure with a different pair of $(\rho, u)$.

What γ is counting #

It is worth spending a section on the number itself, because it is used throughout this collection as a property of air and it is not one.

Equipartition gives each quadratic degree of freedom $\tfrac12 kT$ of energy. A monatomic gas has three — the three components of translation — so $c_v = \tfrac32 R$ and $\gamma = 5/3$. A diatomic molecule adds two rotational degrees, giving $c_v = \tfrac52 R$ and $\gamma = 7/5 = 1.4$. Vibration would add two more — one kinetic, one potential — taking $c_v$ to $\tfrac72 R$ and $\gamma$ to $9/7 = 1.286$.

The reason $\gamma$ is 1.4 at room temperature rather than 1.286 is quantum mechanical: the vibrational energy levels are spaced by $k\theta_v$, and at 300 K almost no molecules have enough energy to reach the first excited state, so the mode is frozen out and contributes nothing. As the temperature rises a larger fraction can reach it, and the Einstein function is exactly the fraction.

So the curve in the first figure is a thermometer reading of a quantum effect, and its asymptote at high temperature is $9/7$ — which the computed value at 6,000 K, 1.288, is very close to. Air’s γ is not a material constant; it is a statement about which modes are accessible at the temperature in question, and the only reason it can be treated as a constant in ordinary aerodynamics is that ordinary aerodynamics never gets hot.

Reading the two columns #

The pressure and the temperature behave differently and the reason is worth restating in physical terms rather than by counting equations.

Momentum says the flux $p + \rho u^2$ is the same on both sides. That is a statement about how hard the gas pushes and how fast momentum is being carried, and neither depends on where the gas is keeping its internal energy. Whatever the molecule does internally, it arrives at the shock with a certain momentum flux and leaves with the same one.

Energy says the stagnation enthalpy is the same on both sides. That is a statement about where the energy went, because enthalpy includes the internal energy, and a molecule with more places to store energy reaches a lower temperature for the same enthalpy.

So the pressure is set by a bookkeeping that does not care about the molecule and the temperature by one that does. It is the same division the shock’s own structure has between the jump conditions and the transport, and the same one Mach independence has between the quantities that saturate and the quantities that do not.

The one place the pressure does move #

The four per cent is not zero, and it is worth saying where it comes from, because a reader who has followed the argument above should be asking why it is not exactly zero.

Mass and momentum relate four quantities — $\rho_1u_1$, $\rho_2u_2$, $p_1$, $p_2$ — and close only when the density ratio is known. The density ratio comes from the equation of state applied to a temperature that the energy equation supplied. So the caloric behaviour reaches the pressure through the density, at second hand, and the four per cent is the size of that indirect route.

It is small because the pressure ratio at high Mach number is dominated by $\rho_1u_1^2$, which is a free-stream quantity, and only weakly affected by what happens behind. That is the same reason the pressure ratio grows without bound while the pressure coefficient saturates: both are statements that the momentum flux arriving is what sets the pressure.

What that means for a calculation #

The practical division is unusually clean.

Forces are nearly safe. A perfect-gas calculation of the pressure distribution on a hypersonic body is within a few per cent, and the pressure coefficient’s hypersonic limit barely moves. Lift, drag and moment estimates survive.

Heating is not. The convective heat transfer is driven by the temperature difference between the gas and the wall, so a fifteen per cent error in the gas temperature is a fifteen per cent error in the driving potential, before anything about the transport properties is considered. And a fifteen per cent error in a quantity that thermal protection is sized on is a design error.

That split — right about the force, wrong about the heat — is the sentence to carry away, and it follows from a count of which equations contain $c_p$.

The shock layer, which follows the density #

The density ratio’s twenty-seven per cent is the third column and it is the one with a geometric consequence.

The shock standoff distance is a function of the density ratio and of nothing else: $\varepsilon(1+\Delta/R)^2 = 2\Delta/R$ with $\varepsilon = \rho_\infty/\rho_2$. A perfect gas asymptotes to $\rho_2/\rho_1 = 6$; with vibration the ratio passes 7 by Mach 12 and keeps rising.

So a real shock layer at these speeds is substantially thinner than the perfect-gas calculation says — which is the same factor of two that essay’s γ sweep brackets, arrived at from the other end.

The model naming its own limit #

There is a number in the sweep that has to be reported, because it undermines the sweep.

Oxygen begins to dissociate at about 2,500 K. The computed temperature behind the shock crosses that at Mach 8, so every case above Mach 8 in the tables here is outside the range of the model that produced it.

Dissociation absorbs far more energy than vibration and drives the effective $\gamma$ lower still, so the true temperature at Mach 15 is lower than either curve and the true density ratio is higher than either. The model is being used here for the difference it makes rather than for the number it returns, and the figure says where it stops.

That is the honest way to use a model outside its range, and it is worth stating as a general practice: report the direction and the order of magnitude of a correction, name the next effect, and say which way it goes.

Why the ladder is a ladder #

Air at increasing temperature switches on new physics in a definite order, and the order is worth having.

Below 800 K: translation and rotation only. $\gamma = 1.4$, and everything in this collection’s compressible field is applicable.

800 to 2,500 K: vibration fills. $\gamma$ falls to about 1.3. This is the regime the calculation here covers.

2,500 to 4,000 K: oxygen dissociates. Large energy absorption, effective $\gamma$ towards 1.2.

4,000 to 9,000 K: nitrogen dissociates. Lower still.

Above 9,000 K: ionisation, and the gas becomes a plasma with electrical properties as well.

Each step absorbs energy at nearly constant temperature, which is why the effective $\gamma$ keeps falling and why the shock layer keeps getting thinner. A re-entry vehicle passes through all five on the way down.

Where else this shows up in the collection #

Three results elsewhere on this site change when γ does, and it is worth knowing which.

The detonation speed is bracketed rather than predicted for exactly this reason: the unburnt gas has one γ and the products have another, and the model has room for one. That essay’s forty per cent bracket and this one’s four per cent are two measurements of the same difficulty, and they differ in size because a detonation changes the gas’s composition as well as its temperature.

The shock standoff halves between γ = 1.4 and 1.2, which is the largest sensitivity to γ anywhere in this collection. Combined with the density-ratio result here, the two essays close on the same conclusion from opposite directions.

And the speed of sound is $\sqrt{\gamma RT}$, so a hot gas has a sound speed that is not simply proportional to the square root of its temperature. At 3,000 K the correction is about four per cent, which is small — but it is the quantity every Mach number in a hypersonic flow field is computed against, so it propagates.

What does not change is anything derived from mass and momentum alone: the momentum theorem, Betz’s limit, the Borda–Carnot loss, and every control-volume result in the applied field. Those are safe at any temperature, which is a large part of why they are the results this collection trusts most.

Equilibrium, and when there is not any #

One large assumption underlies everything above: that the internal modes are in equilibrium with the translational temperature — that is, that the gas has had time to redistribute its energy.

It has not, always. Vibrational relaxation takes a definite number of collisions, and behind a strong shock the gas is travelling fast enough that it may cross the whole shock layer before vibration has filled. Then the gas is vibrationally frozen, the effective $\gamma$ is nearer 1.4 than 1.3, and the answer is the perfect-gas one after all.

Which case applies is a ratio of a relaxation time to a flow time — a Damköhler number, which is exactly the shape every regime question on this site takes. At high altitude, where collisions are rare, flows are frozen; at low altitude they are in equilibrium; and in between they are neither, and the composition has to be carried as a set of extra transported variables.

The heat that is not thermal, and the surface that decides #

The essay’s conclusion is that the pressure survives and the heating does not, and once dissociation is in the picture the heating fails in a further way that has nothing to do with temperature at all.

A dissociated gas is storing an enormous amount of energy chemically. Breaking an oxygen molecule costs about 5 eV and a nitrogen molecule about 9.8, and that energy is carried along in the atoms themselves rather than in their motion. A thermometer in that gas does not see it. It is invisible to every quantity in the tables above.

What decides whether it reaches the vehicle is not a fluid-mechanical question. It is whether the surface will let the atoms recombine on it. A metal will: atoms adsorb, find partners, and release the full dissociation energy at the wall as heat. A well-chosen glass largely will not: the atoms bounce off still dissociated, are swept downstream, and carry their chemical energy away with them.

The difference is not a correction. Between a fully catalytic wall and a non-catalytic one the stagnation-point heating at some re-entry conditions differs by roughly a factor of two, which is a larger effect than everything else in this essay put together — and it depends on the material of the tile rather than on anything about the flow.

So the thermal protection of a re-entry vehicle is partly a chemistry problem. The Shuttle’s tiles carried a glass coating chosen with low catalytic efficiency among its requirements, and the point was demonstrated in flight rather than argued: patches of a deliberately catalytic coating were applied to selected tiles on an early mission, and they ran measurably hotter than their neighbours in the same flow. A catalytic contaminant on an otherwise inert surface heats a streak behind itself, for the same reason.

It also couples straight back to the frozen-or-equilibrium question above, and inverts the intuition. A frozen boundary layer over a non-catalytic wall is the benign case: the chemical energy never recombines and leaves with the gas. A layer in equilibrium recombines within the gas near the surface whatever the wall is made of, so the energy is released there and conducted in regardless — and the wall’s catalycity stops mattering.

Which is a strange place for the argument to arrive. The whole essay has been about counting a molecule’s internal modes and finding that the count reaches the temperature and not the pressure. The last mode counted turns out to reach neither: it is a store of energy that the flow variables cannot see, and whose delivery is decided at the surface by chemistry the fluid equations do not contain.

What is not in the flow #

The molecule’s internal degrees of freedom.

A velocity field, a density field and a pressure field are a complete description of a perfect gas’s motion. They are not a complete description of a real one, because the same $(p, \rho, u)$ can correspond to different amounts of energy locked in vibration and dissociation, and the amount decides what happens next.

That is a genuinely different kind of missing information from the essays gathered around it. Elsewhere the answer needs a boundary, an observer, a history — all of them external. Here it needs something inside the fluid, at a scale the continuum description has abstracted away, and the abstraction is exactly the one the continuum hypothesis makes and normally gets away with.

The model limit #

Three.

Equilibrium vibration only. No dissociation, no ionisation, no relaxation. The first of those is already active over most of the sweep.

Two species with fixed proportions. Air is treated as 79 per cent nitrogen and 21 per cent oxygen with no argon, no trace gases and no change of composition — which is exactly what dissociation would violate.

And a thermally perfect gas throughout, $p = \rho RT$, which is excellent at these pressures and would not be at the densities behind a shock in a shock tube’s driven section. The imperfection being modelled is caloric — the heat capacity — and not thermal, and the two are different and often confused. A thermally imperfect gas has $p \neq \rho RT$, which happens at high density and low temperature and is what a van der Waals equation is for; a calorically imperfect gas obeys the ideal equation of state and has a heat capacity that depends on temperature, which is what happens here. Air behind a shock is calorically imperfect and thermally nearly perfect, and getting those the wrong way round leads to correcting the wrong equation.