A radius that gives the energy away

A large amount of energy $E$ released at a point in a gas of density $\rho_0$ drives a shock outwards. While the shock is strong enough for the ambient pressure to be negligible, the problem contains three dimensional quantities — $E$, $\rho_0$ and $t$ — and exactly one length can be made from them:

$$R = C\left(\frac{Et^2}{\rho_0}\right)^{1/5}.$$

There is no second group, so the shape of the answer is fixed and the only thing left to determine is $C$, a number of order one.

That is the strongest result dimensional analysis produces anywhere in this subject, and it is also its clearest limit: the exponent is free and the constant is not.

What a thin shell gets #

To find $C$ something has to be solved. The cheapest model that is a model rather than a fit: put all the swept-up mass in a shell at the shock, moving at the post-shock velocity $2\dot R/(\gamma+1)$, and the interior at the post-shock pressure $2\rho_0\dot R^2/(\gamma+1)$.

Adding the kinetic and internal energies gives

$$E = \frac{16\pi\gamma}{3(\gamma+1)^2(\gamma-1)}\,\rho_0R^3\dot R^2,$$

which is an ordinary differential equation, and its solution is the similarity law with $C = (25/4A)^{1/5}$.

Recovering the exponent rather than assuming it #

The virtue of having an equation rather than a formula is that the equation can be integrated forward from a small initial radius and asked what it does.

It does this: a log–log slope of $0.39983$ against $0.4$, and a prefactor of $0.90721$ against a closed form of $0.90702$. The two ten-thousandths are the integration’s memory of where it started, which the similarity solution has no equivalent of because it begins at zero radius and infinite speed.

Nothing told the equation about $t^{2/5}$. It came out, which is what makes the similarity law a consequence rather than a fit — and which is the same discipline this collection applies to every assertion it makes about a solved field.

Reading the dimension count #

The counting deserves to be done rather than asserted, because the whole result rests on it and because the argument is short enough to check by eye.

The quantities are $E$ with dimensions $ML^2T^{-2}$, $\rho_0$ with $ML^{-3}$, $t$ with $T$, and $R$ with $L$. That is four quantities in three independent dimensions — mass, length and time — so Buckingham’s theorem gives $4 - 3 = 1$ dimensionless group.

Finding it is elimination. Mass appears only in $E$ and $\rho_0$, so it can only be removed by the combination $E/\rho_0$, with dimensions $L^5T^{-2}$. Then $L^5T^{-2}$ against $R^5t^{-2}$ is the group, and there is nothing else to form.

What makes it powerful is what is absent from the list. The chemistry of the explosive does not appear because the only thing it does is deliver $E$. The initial radius does not appear because it has been assumed negligible — a hypothesis, and a checkable one. The ambient pressure does not appear because the strong-shock limit removed it, which is another hypothesis. And $\gamma$ does not appear because it is already dimensionless, so it can sit inside $C$ without disturbing the counting.

That last is the important one and it is why $C$ depends on $\gamma$. Dimensional analysis never says a dimensionless parameter is absent; it says it cannot be constructed from the dimensional ones, and any dimensionless parameter already in the problem is free to appear in the answer.

The energy that is not kinetic #

The thin-shell model divides the energy into a kinetic part in the shell and an internal part in the interior, and the ratio between them is worth having because it explains where the model goes wrong.

The two terms are $1/(\gamma+1)^2$ and $1/[(\gamma+1)(\gamma-1)]$, so their ratio is $(\gamma-1)/(\gamma+1)$ — the same $\varepsilon$ that sets the shock’s density ratio. At $\gamma = 1.4$ that is one to six: most of the energy of a blast is heat in the interior, not motion in the shell.

That is the physical reason the thin-shell model is low at small $\gamma$. As $\gamma\to1$ the fraction of energy in the interior grows without bound, and a model that treats the interior as uniform at the post-shock pressure is putting it in the wrong place — the real interior has a strong pressure gradient and most of its energy near the centre.

It is also why the exact solution matters. Sedov’s profiles show the density collapsing towards the shock and the temperature rising steeply towards the centre, so the blast is a thin dense shell around a hot near-vacuum. The thin-shell model has the first half right and the second half quite wrong, and the constant is the measurement of the difference.

What the law does not contain #

Two absences worth naming explicitly, because both are counter-intuitive.

The speed of sound. It appears nowhere, because in the strong-shock limit the ambient pressure is negligible and the ambient sound speed is built from it. The shock’s own Mach number is therefore enormous and falling, and there is no acoustic scale in the problem at all until there is.

And the size of the release. A blast from a point and a blast from a sphere a metre across converge to the same trajectory within a few radii, because the similarity solution is an attractor. That is why the law can be applied to a device whose actual geometry is complicated, and it is why it cannot be applied to the first few radii, where the geometry has not yet been forgotten.

What the constant is worth #

At $\gamma = 1.4$ the thin-shell model gives $C = 0.9070$ against Sedov’s exact $1.033$ — twelve per cent low. That is a respectable showing for a two-line energy balance.

The sweep across $\gamma$ is more interesting than the single number.

The model crosses the exact answer rather than approximating it. So agreeing at any one $\gamma$ would be a coincidence and not a validation, and quoting the $\gamma = 1.4$ error alone would suggest a systematic twelve per cent that could be corrected for. There is no such correction: the error changes sign.

That is a general point about one-parameter models checked at one value of their parameter, and it is worth the sweep every time.

The calculation the result is famous for #

With the exact constant, a radius and a time read off a photograph give the energy:

$$E = \frac{\rho_0R^5}{C^5t^2}.$$

At $R = 140$ metres, $t = 25$ milliseconds, $\rho_0 = 1.2$ kg m⁻³ and $C = 1.033$, that is $8.8\times10^{13}$ joules, or 21 kilotons.

The consistency in that figure is the useful part. The similarity law has one free constant, so each $(R, t)$ pair fixes it, and if the law holds every pair gives the same energy. A disagreement between two frames is the law announcing that it has stopped applying — which is how the late frames, where the ambient pressure stops being negligible, make themselves known.

That is the whole of the calculation which made the yield of the first atomic test a matter of public arithmetic from a published photograph with a scale on it, at a time when the number was classified.

The measurement that made it famous #

The Trinity calculation is worth a section on its own, because it is one of the very few places where a dimensional argument settled a question of consequence and because the story is usually told wrong.

The photographs were published in 1947 with a scale and a timing mark. The yield was classified. Taylor had derived the similarity law during the war for the British atomic project, and he applied it to the published frames: plot $\tfrac52\log R$ against $\log t$, read the intercept, and the energy follows.

The often-repeated version of the story has Taylor embarrassing the American authorities by publishing a classified number. What actually happened is more interesting for this collection’s purposes: the value he obtained was close enough to the real one to demonstrate that the physics of the problem contains no free parameters worth hiding. Once the exponent is known and the constant is a number of order one, the yield is a measurement of a radius and a time.

The same arithmetic is now standard practice in astrophysics, where supernova remnants are dated and their energies estimated from exactly this law. There the ambient density is the uncertain quantity rather than the energy, and the fifth-power dependence makes the inference correspondingly delicate: a factor of two in $\rho_0$ is a factor of two in $E$, and a factor of two in $R$ is a factor of thirty-two.

Why the fifth power makes the inference sharp #

That last observation deserves its own paragraph, because it cuts both ways.

$E\propto R^5t^{-2}$, so an error in the radius is amplified fivefold in the energy and an error in the time is doubled. A photograph gives the radius to a per cent or two and the time to better than that, so the inferred energy is good to perhaps ten per cent — which is remarkable for a measurement made with a ruler.

But it also means the law is a poor way to measure a radius from a known energy, and an excellent way to detect that the law has stopped holding. Two frames giving energies that differ by twenty per cent correspond to radii differing by four, which is far outside the reading error — so the consistency check in the figure above is a sensitive one, and its passing is evidence rather than a formality.

That asymmetry between a sharp inference in one direction and a blunt one in the other is a general feature of power laws with large exponents, and it is worth recognising: the exponent that makes a law useful for measurement is the same exponent that makes it fragile as a prediction.

What is not in the flow #

Everything except the energy and the ambient density.

This is the only essay of the group whose answer is remarkable for how little it needs. Not the device, not the chemistry, not the initial radius, not the shape of the release. None of them can appear, because none of them can be made dimensionless with the three quantities the problem contains.

And they genuinely do not appear: the self-similar solution forgets its initial conditions within a few shock radii, which is what makes the estimate work at all. That forgetting is the same phenomenon the sonic boom’s ageing exhibits, arrived at from a completely different beginning, and it is why both essays end with a wave that has two parameters and no memory.

What is missing is the constant. Dimensional analysis says the answer has one number in it that counting cannot supply, and supplying it needs the full self-similar solution — a set of ordinary differential equations for the profiles inside the blast, integrated from the shock inwards. That is a real calculation, and this collection quotes its result rather than performing it.

Why the ambient pressure is negligible, and when it stops being #

The hypothesis running through everything above is the strong-shock limit: $p_2\gg p_1$, so the ambient pressure drops out of the jump conditions and cannot appear in the dimensional count.

That is excellent early and fails late. As the shock slows towards the ambient sound speed the pressure ratio approaches one, the ambient pressure becomes a fourth dimensional quantity, and a second dimensionless group appears — at which point the answer is a function rather than a power law, and dimensional analysis stops determining anything.

The transition happens when the shock Mach number falls to a few, which for a kiloton-scale blast in air is at a few hundred metres. After that the wave decays into an ordinary acoustic disturbance, and its far-field form is an N-wave — two shocks with a linear fall between them — arrived at by exactly the mechanism that essay describes.

So a blast wave’s life has three stages: a fireball nobody models with gas dynamics, a self-similar phase where this essay applies, and an acoustic phase where the previous one does.

The exponent is a fingerprint of what is conserved #

The counting above assumed the energy survives, and it is worth asking what happens when it does not — because the answer is another power law, obtained by the identical argument, with a different exponent.

Let the shocked gas get hot enough to radiate efficiently. Its internal energy then leaves the system as light, and $E$ is no longer a constant of the problem. What is still conserved is the shell’s radial momentum, since radiation carries essentially none away, so the quantity to put in the dimension list is an impulse rather than an energy.

Run the same elimination. Momentum over density has dimensions $L^4T^{-1}$, against $R^4t^{-1}$, so again there is exactly one group and again the shape of the answer is fixed:

$$R \propto \left(\frac{I t}{\rho_0}\right)^{1/4}.$$

Two fifths becomes one quarter, and the change came from swapping which conserved quantity was available rather than from any new physics of shocks. That is the useful generalisation: the exponent of a self-similar expansion is not a fact about blast waves, it is a fact about what the problem is allowed to keep.

Which turns the exponent into a diagnostic. A supernova remnant expanding as $t^{2/5}$ is adiabatic and its energy can be inferred by the arithmetic above; one expanding as $t^{1/4}$ has begun radiating away its interior and that arithmetic no longer applies to it. Astronomers use the transition between the two as the marker of the remnant’s age, and the stage before it — before enough ambient mass has been swept up to matter — is a third law again, with the ejecta simply coasting at constant speed.

A blast’s whole history is therefore a sequence of power laws, and each change of slope is a change in the list of quantities that the counting is allowed to use.

The three regimes of a dimensional argument #

It is worth generalising, because the pattern recurs and the failure modes are the same each time.

More quantities than dimensions plus one, and there are several groups: the answer is an unknown function of them and counting has narrowed the problem without solving it. That is the usual case.

Exactly one group, and the answer is fixed up to a constant. That is here, and it is rare and valuable.

And a group that turns out to be irrelevant, which is the trap. If the answer happens not to depend on one of the groups, the counting cannot know, and a whole dimensionless parameter can be carried through a correlation for years while the physics does not use it. The reverse trap — dropping a quantity that does matter — is what the strong-shock hypothesis is, and it is why the hypothesis has to be stated with the result.

Where else the same shape appears #

Self-similar solutions of this kind are one of the few genuinely general tools in nonlinear fluid mechanics, and this collection has several.

The Blasius layer is the same idea applied to a boundary layer: a problem with no length in it, so the profile at every station is the same profile stretched.

The Kolmogorov scales are a dimensional argument of exactly the first kind — two quantities, one length — and the constant there is measured rather than computed for the same reason.

And the Taylor–Aris dispersion constant is the opposite case: dimensional analysis gives the form and the constant is computed exactly, because the cell problem is linear.

The blast wave sits between them. The form is free, the constant is computable in principle, and the computation is a nonlinear boundary-value problem rather than a linear one — which is why Taylor and Sedov’s independent solutions in the 1940s were an achievement rather than an exercise.

The model limit #

Three.

The strong-shock hypothesis, discussed above, which bounds the applicable range at both ends.

Spherical symmetry, which a real explosion in air near the ground does not have — the reflection off the ground produces a Mach stem, which is the configuration two essays ago, and the wave becomes hemispherical with a very different constant.

And the thin shell. The model’s error is not a bias to be corrected: it crosses zero within the range of $\gamma$ that matters, so the honest statement is a bound of a third rather than a correction factor. For any application where the constant matters, Sedov’s value is the one to use, and the model here is worth having because it says where the two-fifths comes from.