The fireball is hollow

A radius that gives the energy away derives the law of a strong blast, $R = C(Et^2/\rho_0)^{1/5}$, from four quantities and one dimensionless group, and then estimates the one number the dimensions leave free with a thin-shell model: all the swept-up air in a shell at the shock, all of it moving at the speed the shock gives it, the interior at the pressure just behind the shock. The model gives 0.907 at $\gamma = 1.4$ against Sedov’s exact 1.033.

That essay quoted Sedov’s number and did not compute it. This one computes it, and the solution it comes from, because the solution is a stranger object than the constant suggests. It also has a consequence for the essay before, which the section on the thin shell below sets out.

Three equations in one variable #

Dimensional analysis is why the problem can be solved at all. The flow depends on the distance $r$ and the time $t$ only through the ratio $\xi = r/R(t)$, so every quantity has the same shape at every instant, stretched to fit the growing radius. Writing the velocity as $\dot R\,\varphi(\xi)$, the density as $\rho_0\psi(\xi)$ and the pressure as $\rho_0\dot R^2\pi(\xi)$ turns the equations of gas dynamics — mass, momentum and the entropy each parcel carries — into three ordinary differential equations in $\xi$. At the shock, $\xi = 1$, the strong-shock jump conditions fix all three:

$$\varphi = \frac{2}{\gamma+1}, \qquad \psi = \frac{\gamma+1}{\gamma-1}, \qquad \pi = \frac{2}{\gamma+1}.$$

The equations are integrated from there inwards, in the logarithms of the density and the pressure, because the density falls so steeply towards the centre — as the fifteenth power of $\xi$ at $\gamma = 1.2$ — that a step in the density itself overshoots zero long before the answer has converged. The energy inside the shock is then an integral over the profiles, and requiring it to equal $E$ fixes the constant: $\xi_0 = (\alpha^2\sigma I)^{-1/(\nu+2)}$, with $\nu$ the number of dimensions, $\alpha = 2/(\nu + 2)$ and $\sigma$ the area of a unit sphere in $\nu$ dimensions.

The reduction is worth a sentence of its own because it is where the dimensional argument pays out twice. Counting what matters gives one group and so fixes the exponent; the same count, applied to the whole flow rather than to the radius, says every profile is a function of $\xi$ alone, which is what turns partial differential equations in two variables into ordinary ones in one. The constant is the only thing the count leaves open, and it takes the whole solution to close it.

The shell and the core #

Just behind the shock the air is compressed sixfold and moving outward at five-sixths of the shock speed. Going inward, the density collapses. At nine-tenths of the shock radius it is a fifth of its post-shock value; at eight-tenths, 6.5 per cent; at half the radius, 0.17 per cent — a hundredth of the density of the undisturbed air outside. The gas velocity falls roughly in proportion to the radius, as a uniform expansion’s would. And the pressure does something the density does not: inside about six-tenths of the radius it stops changing, and sits at 0.366 of the post-shock value all the way to the centre.

A nearly uniform pressure over a nearly empty volume means a temperature that rises as the density falls. At half the radius the gas is two hundred times hotter than just behind the shock; at a tenth of the radius, tens of millions of times. At the centre the similarity solution’s temperature is infinite and its density zero, which is the solution’s way of saying that the first parcels hit by the explosion were shocked hardest, received the most entropy, and have been expanding ever since with it.

The interior of a strong blast is not compressed air. It is a hot, rarefied core at nearly uniform pressure, wrapped in a thin, dense shell. The fireball is hollow.

Where the mass and the energy are #

Counting from the centre outwards makes the hollowness quantitative. Nine-tenths of the swept-up air lies outside 0.844 of the shock radius, and half of it in the outer 4.2 per cent. The mass that ought to fill a sphere has been gathered into a skin.

The energy is distributed very differently, and the reason is the uniform pressure. Seventy-eight per cent of the blast’s energy is heat, and heat per unit volume is the pressure over $\gamma - 1$, so where the pressure is uniform the heat is spread by volume. Ten per cent of the energy lies inside 0.55 of the radius, where there is almost no mass at all, and half of it outside 0.898. The kinetic energy — 21.9 per cent of the total — is carried by the shell, where the mass and the speed both are.

That total is checked independently of the energy integral that fixed the constant. The mass inside the shock must equal the mass of the air that was there before it arrived, and integrating the density profile gives it to two parts in a million in all three geometries. The constant itself converges in the integration grid to one part in a million.

Sedov’s constant, and a table that was wrong #

The integrated constant for a spherical blast at $\gamma = 1.4$ is 1.0328, which is Sedov’s 1.033. For a monatomic gas, $\gamma = 5/3$, it is 1.1517, the value astrophysicists use for supernova remnants; for a cylindrical blast at $\gamma = 1.4$, 1.0040. All three agree with the values quoted wherever the solution is used. The constant rises with $\gamma$, and the reason is physical: a gas with a larger $\gamma$ holds less internal energy per unit of pressure, so a given energy buys more pressure and drives the shock further. At $\gamma = 1.2$ the constant is 0.897; at 1.3, 0.974.

The essay before carried a table of these constants, and three of its four entries were wrong: it gave 1.163, 1.086 and 0.944 at $\gamma$ = 1.2, 1.3 and 5/3, falling where the true constant rises. They had been typed in rather than computed, and only the entry at 1.4 was right. From them it drew a conclusion that the integration reverses. It said the thin-shell model’s error ran from −32 per cent at $\gamma = 1.2$ to +7 per cent at 5/3, crossing zero, so that no correction factor could exist and agreeing at any one $\gamma$ would be a coincidence. With the constants computed, the model is between 12.1 and 12.3 per cent low at every $\gamma$. The error is a steady bias that one factor corrects. The table, the conclusion and the check that enforced it have been corrected, and the general point the essay drew — that one value of a parameter cannot tell a bias from a crossing — survives with its sign reversed: only the sweep can.

Why the thin shell is steadily wrong #

The exact solution says what the model gets wrong, and why by the same fraction at every $\gamma$. The model fills the interior with gas at the post-shock pressure. The exact interior sits at about a third of it, and averaged over the volume the pressure is 0.476 of the model’s at $\gamma = 1.4$ — and 0.485, 0.480 and 0.470 at 1.2, 1.3 and 5/3. Since most of the energy is heat, the model counts about twice the heat the blast has for a given radius and speed, and so, for a given energy, puts the shock at a smaller radius. The average-pressure ratio hardly moves with $\gamma$, and that steadiness is the steadiness of the bias.

The kinetic energy is the lesser error. The model moves all the mass at the post-shock speed; the exact solution’s mass is almost all in the shell and moving nearly that fast, so its kinetic energy is between 74 and 87 per cent of the model’s — the gap wider at larger $\gamma$, but in a share of the energy that is only 14 to 28 per cent. The two errors together make the constant’s twelve per cent.

A line, a plane or a point #

The same equations describe blasts in fewer dimensions. A planar one is driven by energy deposited on a plane — a sudden sheet of heating — and grows as $t^{2/3}$; a cylindrical one by energy along a line — a lightning channel, an exploding wire — and grows as $t^{1/2}$. Their constants at $\gamma = 1.4$ are 0.976 and 1.004. The hollowing is weaker the fewer the dimensions: half the mass lies outside 0.876 of the radius for a planar wave, 0.937 for a cylindrical one and 0.958 for a sphere, because a sphere has the most volume far from the centre to fill from the least mass near it. The kinetic share rises with the dimension too, from 18 per cent to 21 and 22.

The cylindrical case has an application that makes it more than a curiosity, which the last section takes up.

A shell four per cent thick behind a shock a micrometre thick #

The shell is thin, but it is not the shock. The jump itself — the sixfold compression that the jump conditions allow and the second law insists on — happens across a few molecular mean free paths, a fraction of a micrometre in air at sea level. The shell of gathered air behind it is a flow structure, set by how fast the gas just behind the shock is left behind by the shock itself, and its thickness scales with the radius: half the mass within the outer 4.2 per cent at every instant. For the fireball Taylor measured, 140 metres across at 25 milliseconds, that is a shell six metres deep behind a shock a millionth of a metre thick. Neither length appears in the other’s physics, which is why the strong-shock solution can treat the shock as a surface and still resolve the shell.

The shell’s thinness is also what makes the thin-shell model as good as it is. Its assumption that all the mass is at the shock is wrong by four per cent in radius for half the mass, and costs little; its assumption that the interior is at the post-shock pressure is wrong by a factor of two, and costs everything the model loses.

What a wrong constant does to a reading #

The point of the constant is to read an energy off a radius and a time, $E = \rho_0R^5/(C^5t^2)$, and the fifth power makes the reading sensitive to it. Taylor used $\gamma = 1.4$, and air behind a strong shock is not quite a gas of constant $\gamma$: dissociation lowers its effective value towards 1.3. Using the integrated constant at 1.3, 0.974, in place of 1.033 raises the inferred energy by a factor of $(1.033/0.974)^5 = 1.34$ — a third more energy for the same photograph. At an effective $\gamma$ of 1.2 the factor is two.

The wrong table would have moved the reading the other way. Its constant of 1.086 at $\gamma = 1.3$ lowers the inferred energy by a fifth, and its 1.163 at 1.2 halves it. A reader correcting Taylor’s number for real-gas effects with that table would have been sent in the wrong direction by roughly the size of the correction itself. That is the practical cost of a constant typed in rather than computed, and it is invisible in the one case, $\gamma = 1.4$, that every account checks.

The same sensitivity is why the geometry matters to a reading. A blast near the ground is not spherical: it reflects, forms a Mach stem along the surface, and above the stem behaves nearly as a hemisphere carrying twice the energy per unit solid angle. Reading it as a sphere misstates the energy by a factor of two before any constant enters, which is why the late frames of a ground-level film are the ones to distrust — and why, far away, what reaches an observer is not this self-similar shape at all but the two-shock signature every distant explosion and every supersonic aircraft decays into.

What the hollow core means for a real explosion #

The first photographs of a nuclear fireball, from which G. I. Taylor read its energy, show a bright sphere with a sharp edge, and the similarity solution says what the brightness is: a thin shell of shocked air at its outer edge, hot enough to glow, around a core hotter still and far thinner. The core’s extreme temperature is also the solution’s first failure. In a real explosion the centre is filled with the vaporised device, whose mass the similarity solution neglects, and radiation carries heat outward from the hottest gas far faster than the gas moves. Both flatten the core’s temperature to something finite.

The hollowness is also why the blast wave outruns everything behind it and then weakens as it does. Nearly all the momentum is in the shell, and once the shock has slowed to the point where the air’s own pressure matters, the shell detaches from the hot core and propagates as an ordinary pressure wave, while the core — nearly empty and at low pressure by then — is left behind, rises by its buoyancy, and becomes the familiar mushroom. The strong-shock solution describes the stage before that separation, and what a shock costs in entropy is exactly what the core has been given and can never return.

What the picture cannot show #

A strong shock. The ambient pressure is neglected, which is exact only while the shock is much stronger than the air’s own pressure. Late in a real blast the solution fails from the outside in.

A perfect gas of constant $\gamma$. Air at the temperatures of a fireball’s core dissociates and ionises, and its effective $\gamma$ falls towards 1.2 or below; the constants above show how much that moves the radius.

A point release and no radiation. The mass of the source is neglected and so is radiative transfer, which in a very energetic blast carries more heat than the flow does in the first milliseconds.

The convention the numbers depend on #

The similarity constant $\xi_0$ is defined by $R = \xi_0(Et^2/\rho_0)^{1/(\nu+2)}$, with $E$ the energy, and for the planar and cylindrical waves $E$ is per unit area and per unit length. Profiles are given as fractions of their values just behind the shock, and positions as fractions of the shock radius. “Kinetic share” is the kinetic energy as a fraction of the total inside the shock.

Who found it, and when #

Leonid Sedov found the exact solution in closed form in 1946, and John von Neumann independently in 1941 in a report that stayed classified; G. I. Taylor had integrated the same equations numerically in 1941 and published in 1950, with the Trinity test’s photographs and the energy he read off them. The constants quoted above have been recomputed many times since, and the closed form is now a standard test problem for the codes that compute blasts. The self-similarity that makes it possible is the first kind, fixed by dimensions alone; an exponent dimensions cannot give is the second kind, where an implosion rather than an explosion has an exponent the equations must find.

Still open: a nose is a cylindrical explosion #

A blunt body flying at hypersonic speed deposits energy in the air along its path at a rate equal to its drag. Seen from the ground, each slice of air the body passes through is struck by an impulse and left to expand — a cylindrical blast, with the drag per unit length as its energy and the distance behind the nose divided by the flight speed as its time. The analogy predicts that the bow shock’s radius grows as the square root of the distance behind the nose, with a coefficient set by the drag coefficient to the quarter power and by the cylindrical constant 1.004 above, and that the flow behind a shock that lies on the body carries a hollow, hot core along the wake. The calculation that follows puts those numbers against a computed bow shock and asks how many nose diameters downstream the analogy takes to become right, and whether the heated core it predicts is the entropy layer that the nose’s curved shock is known to leave on the body.