The shock layer is Hill's vortex, and it heats the nose a tenth harder than Newton says

A shock that lies on the body found the bow shock’s standoff from a mass balance. Take the shock concentric with a sphere, the gas behind it compressed by a ratio $\varepsilon = \rho_\infty/\rho_2$ and uniform, and require the mass that crosses the shock within an angle of the axis to leave through the annulus at that angle: $\varepsilon(1 + \Delta/R)^2 = 2\Delta/R$, a standoff of 0.101 body radii for a perfect gas compressed six-fold. The balance knows how much gas is in the layer and nothing about how it moves there, and the most consequential number about a blunt nose at hypersonic speed is a property of the motion: how fast the gas accelerates away from the stagnation point along the surface. That velocity gradient sets the thickness of the boundary layer at the nose, and through it the heat flux, which decides how thick a re-entry vehicle’s heat shield has to be.

This essay keeps the shock concentric and the density constant and solves for the motion. The surprise is that the problem has an exact solution, and that it belongs to a family met somewhere quite different: the one rotational solution, Hill’s spherical vortex, a ball of spinning fluid travelling through fluid at rest.

A rotational flow with no viscosity in it #

The flow behind a curved shock is rotational. Each streamline crosses the shock at a different angle, so each gains a different amount of entropy while keeping the same total enthalpy, and Crocco’s theorem turns that gradient of entropy across the streamlines into vorticity. The layer is therefore not a potential flow, and the usual exact solutions of ideal-flow theory are not available for it.

There is a family that is. In a steady axisymmetric flow of constant density, Euler’s equations hold whenever the vorticity divided by the distance from the axis is a function of the stream function alone. The simplest choice is a constant, and with the Stokes stream function written as $\psi = \sin^2\chi\,f(r)$ — $\chi$ the angle from the stagnation line, $r$ the distance from the sphere’s centre — the equation for $f$ is linear and its general solution is

$$f(r) = a\,r^4 + b\,r^2 + \frac{c}{r}.$$

The $r^4$ term carries the uniform vorticity; $r^2$ is a uniform stream; $c/r$ is a doublet. Hill’s vortex is the member with $c = 0$ fitted to a stream outside it. The shock layer is another member, fitted to a shock.

Three conditions at the shock, one at the body #

With the shock’s radius and the free stream’s speed as units, the strong shock gives three conditions just behind it. Mass: the gas crossing the shock is compressed by $1/\varepsilon$ and its normal velocity cut by $\varepsilon$, so $f(1) = \varepsilon/2$. Tangential velocity: a shock changes only the velocity normal to it, so the tangential velocity behind is the free stream’s, $U\sin\chi$, which makes $f'(1) = 1$. Tangential momentum: the pressure just behind the shock is Rankine–Hugoniot’s, $p_2 = \rho_\infty U^2(1 - \varepsilon)\cos^2\chi$ in the strong-shock limit, and the tangential momentum equation along the shock must produce its gradient; written out, that requires

$$f''(1) = \frac{1 - 2\varepsilon + 2\varepsilon^2}{\varepsilon}.$$

Three linear equations fix $a$, $b$ and $c$, and the body is the streamline $\psi = 0$, the root of $f$ inside the shock. For a perfect gas with $\gamma = 1.4$ in the hypersonic limit, $\varepsilon = 1/6$, and something tidy happens: $c$ vanishes exactly, $f = a r^4 + b r^2$ with $a = 5/12$ and $b = -1/3$, and the body’s radius is $\sqrt{-b/a} = \sqrt{0.8} = 0.8944$ of the shock’s. The shock layer of a sphere in a perfect gas is a piece of Hill’s vortex — no doublet, the same $r^4$ and $r^2$ — cut off by the shock outside and the body inside.

The solution checks itself in a way the construction does not arrange. The pressure at the stagnation point follows directly from the stagnation streamline: behind the normal part of the shock, then brought to rest. It can also be reached by a different path — from the shock at thirty-four degrees off the axis, where Rankine–Hugoniot gives the pressure, inward across the layer with the full radial momentum equation, then along the body, a streamline, by Bernoulli. The two agree to three parts in a hundred billion. A function that satisfied the shock’s conditions without satisfying the momentum equations inside would not do that.

How far the shock stands off #

The exact layer puts the shock 0.118 body radii ahead of the nose at $\varepsilon = 1/6$, against 0.101 from the uniform balance and 0.143 measured — Billig’s correlation of sphere data in the hypersonic limit. Allowing the gas to move as the equations require recovers about two-fifths of the gap between the balance and the measurement. The rest is in what both models hold fixed: the real shock is not concentric with the sphere but flatter near the axis, and the gas expands as it accelerates round the shoulder, so its density falls and the layer must thicken to carry the same mass. Hayes and Probstein’s formula, $\varepsilon/(1 + \sqrt{8\varepsilon/3})$ in shock radii, is an approximation to the same exact layer and agrees with it to five per cent at $\varepsilon = 1/6$ and ten at 0.3.

The standoff is nearly proportional to $\varepsilon$, and the dependence is the one a ratio of specific heats that stops being a number makes consequential: a gas that dissociates behind the shock compresses further, $\varepsilon$ falls towards a twelfth or less, and the shock moves in. At $\varepsilon = 0.05$ the exact layer is 0.039 body radii thick.

How fast the gas leaves the stagnation point #

The velocity along the body near the stagnation point grows in proportion to the distance from it, $u = (du/ds)\,s$, and the constant of proportionality is the stagnation-point velocity gradient. In the exact layer it is $f'(R_b)/R_b^2$ in units of $U/R_s$. Newton’s estimate takes the pressure on the body to fall as $\cos^2\chi$ from its stagnation value and the gas at the stagnation density to accelerate under it by Bernoulli, which gives

$$\frac{du}{ds} = \frac{1}{R_b}\sqrt{\frac{2(p_s - p_\infty)}{\rho_s}} = \frac{U}{R_b}\sqrt{2\varepsilon\left(1 - \tfrac{\varepsilon}{2}\right)}.$$

At $\varepsilon = 1/6$ the exact layer’s gradient is 0.667 $U/R_b$ and Newton’s 0.553: the gas leaves the stagnation point 21 per cent faster than the Newtonian estimate says. As $\varepsilon$ falls the ratio does not go to one; it tends to $\sqrt{4/3} = 1.155$, and the exact gradient to $\sqrt{8\varepsilon/3}$ against Newton’s $\sqrt{2\varepsilon}$. The Newtonian estimate is wrong by a fixed factor even in the limit where Newton’s sine-squared law is supposed to become exact.

The reason is that Newton’s law is a statement about pressure on the body, and the gradient depends on how that pressure falls with angle and on the density of the gas it accelerates. The exact layer’s pressure falls faster than $\cos^2\chi$ near the axis — to 0.83 of its stagnation value at twenty degrees, where $\cos^2\chi$ is 0.883 — and that faster fall is the steeper gradient seen as a pressure. Newton’s law gets the pressure on the body right in the sense that matters for drag, the integral over the whole face; it does not get the curvature of the pressure at the axis right, and the heating depends on the curvature.

The centrifugal pressure Newton left out #

There is a classical name for what Newton’s law misses. Newton imagined each particle of the stream striking the surface, losing its normal momentum and sliding off along it; the pressure is the normal momentum destroyed. Busemann pointed out in 1933 that a particle sliding along a curved surface has to be held on it, and the force that holds it is a pressure difference across the layer: on a convex body the layer’s centrifugal tendency lowers the pressure on the surface below Newton’s value, more the further the gas has travelled round. The Newton–Busemann pressure for a sphere falls faster than $\cos^2\chi$ for exactly that reason, and in the limit of an infinitely thin layer it can even go negative round the shoulder, which is the limit’s way of saying the layer would leave the body.

The exact layer contains the centrifugal term without having been told about it. Its momentum equation across the layer balances the pressure gradient against the curvature of the streamlines, and that is why its pressure on the body falls below $\cos^2\chi$ near the axis. But it also contains the vorticity, which Newton–Busemann theory does not: the slow gas next to the body has less centrifugal force to be held against, and the layer has finite thickness. So the exact layer’s correction to Newton is of the Busemann kind and smaller than Busemann’s thin-layer version, and it is the one that survives at a finite density ratio. Hypersonic theory is a succession of these corrections to Newton, and the only theory simple enough to optimise is the first term of all of them.

What it does to the heating #

A nose’s convective heat flux at the stagnation point is set by its boundary layer, whose thickness there goes as $\sqrt{\nu/(du/ds)}$: a steeper gradient thins the layer and steepens the temperature gradient across it. Fay and Riddell’s correlation for the stagnation point, the formula re-entry heating is designed with, makes the heat flux proportional to the square root of the velocity gradient, all else equal. A gradient 1.206 times Newton’s is therefore a heat flux $\sqrt{1.206} = 1.098$ times the Newtonian estimate — ten per cent — and in the thin-layer limit $(4/3)^{1/4} = 1.075$.

The heat flux is driven by the difference between the recovery enthalpy and the wall’s, and a wall left to itself heats itself most of the way to the stagnation temperature, which is what makes a re-entry nose a heat-transfer problem rather than a temperature one: the shield is cooled by ablating, and the flux it must absorb is the one the gradient sets. A cooler wall takes more of it, and holds its boundary layer on longer against a pressure rise, as a cold wall holds its layer on longer showed; neither changes the factor here, because both multiply the two estimates alike.

Ten per cent is not a refinement on a heat shield. Ablative shields are sized by the heat load integrated over the entry, and their thickness grows with it; a ten-per-cent underestimate at the stagnation point, where the load is largest, is a ten-per-cent thinner margin where the margin matters most. Designers know this, which is why the velocity gradient is taken from numerical solutions of the shock layer rather than from Newton once a design is past its first sketch. The exact layer says where the difference comes from: the vorticity a curved shock puts into the gas.

A nose, in numbers #

Take a hemispherical nose of half a metre radius entering at seven kilometres a second, where the gas behind the shock is dissociating and compressed perhaps tenfold, $\varepsilon \approx 0.1$. The exact layer’s gradient there is $0.517\,U/R_b$, about 7,200 per second; Newton’s is $0.436\,U/R_b$, about 6,100. The stagnation heat flux is in the ratio of their square roots, 1.088, so an estimate built on Newton’s gradient is nine per cent low before any of the hard parts of the problem — the chemistry, the catalytic wall, the radiation from the shock layer — have been touched.

The same gradient carries the reason noses are blunt at all. It is proportional to $U/R_b$, so the heat flux at the stagnation point, going as its square root, falls as $1/\sqrt{R_b}$: a nose twice as large is heated $1/\sqrt 2$ as hard per unit area. Allen and Eggers made that the founding argument of re-entry design in the 1950s, against the instinct that a sharp nose, with less drag, should be better. The constant-density layer changes the constant in that scaling and not the scaling; the blunt nose is right whichever constant is used, and the exact constant says how blunt. A heating figure is also only as good as its local model, and the spot a local theory cannot see — where another shock’s jet strikes the surface — can exceed the stagnation point’s heating several times over, which no correction to the stagnation gradient anticipates.

The vorticity, seen across the layer #

The tangential velocity is the free stream’s just behind the shock and falls towards the body: at $\varepsilon = 1/6$ the gas sliding along the body near the axis moves at two-thirds of the speed it had just behind the shock. That shear across the layer is the uniform vorticity of the solution, $-10a$ in its units, and its sign is the one Crocco’s theorem gives: the gas that crossed the shock near the axis, where the shock is normal and the entropy rise largest, lies next to the body and moves slowest. The gas beside a hypersonic body remembers its nose for the same reason far downstream — the high-entropy gas from the nose stays on the body — and the exact layer is where that entropy layer begins.

A potential-flow calculation inside the layer would have the tangential velocity growing towards the body, as it does round any convex surface in irrotational flow. The vorticity reverses that, and the reversal and the steeper gradient at the stagnation point are the same fact: the gas next to the body is slower, so it must accelerate harder along the surface to carry the mass out.

The convention, and what the model leaves out #

The density ratio $\varepsilon$ is the strong-shock value, $(\gamma - 1)/(\gamma + 1)$ for a perfect gas, so every number here is for the hypersonic limit, where the flow round a blunt body stops depending on the Mach number. Lengths are in body radii unless stated, gradients in $U/R_b$. The shock is a concentric sphere — the model’s largest simplification, since the real shock is flatter on the axis and leaves the body’s shape at the shoulder. The density is constant, which is right at the axis and wrong round the shoulder where the gas expands. The gas is inviscid, so the boundary layer whose heating the gradient drives is not in the model, and the heating ratio rests on Fay and Riddell’s square-root dependence holding with all its other factors equal between the two estimates. And the flow is in chemical equilibrium or frozen — a single $\varepsilon$ — where a real entry flow is neither.

Who solved it #

Lighthill published the constant-density solution for a sphere in 1957, in the first part of his study of the dynamics of a dissociating gas, as the simplest model of the shock layer in which the compression is so large that the density variation inside it can be neglected. Hayes and Probstein carried it, with its approximate standoff formula, into their Hypersonic Flow Theory of 1959, where the stagnation-point velocity gradient and its excess over the Newtonian value are discussed alongside the numerical blunt-body solutions that were then becoming possible. Hill’s spherical vortex is from 1894. Fay and Riddell’s heating correlation is from 1958. Billig’s standoff correlation, from shadowgraphs of spheres and cylinders, is from 1967.

How it was checked #

The mass flux and tangential velocity just behind the shock, read back from the solution at three angles, match the strong shock’s to rounding. The stagnation pressure reached by the path across the layer and along the body matches the stagnation streamline’s to $2.6 \times 10^{-11}$ at $\varepsilon = 1/6$ from thirty-four degrees off the axis, and to $7 \times 10^{-11}$ at $\varepsilon = 0.1$ from fifty-two degrees — a momentum balance across the layer that the construction does not impose. At $\varepsilon = 1/6$ the doublet’s coefficient is zero to rounding and the body’s radius $\sqrt{0.8}$ to ten figures. The ratio of the exact gradient to Newton’s rises steadily with $\varepsilon$ from 1.159 at $\varepsilon = 0.01$, within two per cent of $\sqrt{4/3}$. And the model refuses a density ratio of zero, a negative one, and one too large for a strong shock.

Still open: the layer that thins round the shoulder #

The exact layer’s two largest approximations are linked: the shock is concentric and the density is constant, and the real flow relaxes both together round the shoulder, where the gas expands and the shock falls back. The next calculation keeps the constant density near the axis and lets the shock’s radius of curvature differ from the body’s — a shock of radius $R_s$ ahead of a body of radius $R_b$ with $R_s/R_b$ found rather than set to one plus the standoff — by solving for the shock shape that makes the same Hill family satisfy the shock’s conditions to second order in the angle. That gives the standoff and the stagnation-point gradient a second geometric parameter, and asks how much of the remaining gap to the measured 0.143 the shock’s own curvature accounts for.