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The Buckingham π theorem and the atomic bomb

Diego Carranza tells you to stop worrying and dimensionally analyse the bomb

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Atomic bomb

On 16 July 1945, the first nuclear test, ‘Trinity’, was carried out and with it the nuclear age began. The explosion was huge, but the actual calculation of the amount of energy released was rather difficult due to the large number of physical and chemical processes involved in the detonating reaction; even the rough estimates were far from accurate. It was not until the publication of the photographs of the explosion that scientists became aware of its magnitude. With just these photographs and some clever mathematical arguments, British physicist GI Taylor,  Soviet physicist LI Sedov and Hungarian–American mathematician John von Nuemann estimated independently an energy of about 17 kilotons of TNT.  Taylor published this result in 1950, with the US Army not at all thrilled that this sensitive piece of information was now in the public domain. Although the estimates of Taylor, Sedov and von Neumann required the use of some complex mathematics, dimensional analysis and the Buckingham π theorem allow us to come to the same conclusion with a minimum amount of knowledge of physics.

In order to understand the importance of the Buckingham π theorem, we have to realise that since the pioneering work of Isaac Newton, physics and maths have been intimately connected.  From that point onwards, understanding nature was not only a matter of knowing that a projectile would fall to earth some metres away, but also of being able to predict the exact point it would reach and the time it would spend in the air. Mathematics has proved to be a very powerful ally when describing the laws ruling the universe, but the difficulty involved can escalate very quickly as one moves away from idealised situations, which are sometimes not realistic at all.  If we want to tackle some serious or cutting edge problems it is very likely that we will have to draw upon computers. However, if one simply wants to gain an insight into what is happening physically speaking, some basic linear algebra can give us valuable information.  This aim lies at the heart of the Buckingham π theorem.

Units and dimensions

When we speak about the mass or speed of an object, it is not enough to give a number but we must also specify the units in which the quantity is measured. For example, we can say that the mass is 5kg and speed is 30 miles per hour; of course, equivalent units such as pounds or km/h are equally valid. As you can observe, velocity consists of two units: one expressing length (km) and the other time (h). These are known as dimensions. In modern physics, all quantities can be expressed in terms of seven fundamental and independent dimensions.  For our purposes it will be sufficient to consider only three: time ($T\hspace{0.5mm}$), length ($L\hspace{0.5mm}$) and mass ($M\hspace{0.5mm}$). For instance, density ($\rho$) has dimension mass over cubic length and energy ($E\hspace{0.5mm}$) has dimension mass times squared length divided by squared time.  From now on, we will denote dimensions using square brackets, $[\;\;]$, so the previous examples can be represented as:

\begin{eqnarray*}
[\rho] = \frac{M}{L^{3}} = ML^{-3},  \quad [E\hspace{0.5mm}] = \frac{ML^2}{T\hspace{0.5mm}^2} = ML^2T\hspace{0.5mm}^{-2}.
\end{eqnarray*}

It is important to note that although different systems of units can be employed, the physical dimensions remain the same.

It is not difficult to realise that there are certain combinations of quantities such that when their dimensions are considered they cancel each other out. Taking a time $t$, a length $\ell$ and an acceleration $a$—which has dimensions of length divided by squared time—the combination $at\hspace{0.5mm}^2/\ell$ is dimensionless.

Edgar Buckingham (1867--1940), after whom the Buckingham π theorem is named.

Edgar Buckingham (1867–1940), after whom the Buckingham π theorem is named.

These kind of quantities will be of great importance, since the Buckingham π theorem is expressed in terms of them.

The simple pendulum

Now that we have a clearer notion of what physical dimensions are, we are ready to understand the Buckingham π theorem.  There are several ways in which it can be stated, some of them actually quite sophisticated, but for this article we will use a simple version.  To help illustrate it, we will consider the classical example of the simple pendulum—a mass attached to an inextensible string moving from side to side, as shown overleaf.

Let’s assume we want to study a physical system (the pendulum) involving $n$ variables denoted as $a_1, a\hspace{0.2mm}_2, \dots, a_n$.  In principle, all of these variables are related via some function in such a way that any one of them, say $a_1$, can be obtained from the others.

Mathematically, this can be written as

\begin{eqnarray*}
a_1 = f\hspace{0.5mm}(a\hspace{0.2mm}_2, a\hspace{0.2mm}_3, \dots, a_n).
\end{eqnarray*}

The function $f$ can be quite simple or really complicated depending on how many variables are included in the problem: indeed, the trickiest step is to identify the relevant ones.  In the case of the pendulum, we want to understand how the mass $m$ moves as time $t$ passes, so it is natural to include both. Likewise, the length of the string $\ell$ must play an important role, as well as $\theta$, the angle it makes with the vertical. And we must also remember that the pendulum’s movement is possible thanks to the action of gravity, which is taken into account through the acceleration $g$ produced by the Earth. So we have five variables to consider.

Model of a simple pendulum.

Model of a simple pendulum.

The dimensions of our variables are as follows:
\begin{eqnarray*}
[m\hspace{1pt}] = M, \ [\ell\hspace{1pt}] = L, \ [t\hspace{0.5mm}] = T, \ [\theta\hspace{0.5mm}] = 1 \ \ \text{and} \ \  [g\hspace{1pt}] = LT^{\hspace{3pt}-2}.
\end{eqnarray*}

Firstly, we note that since $\theta$ is a pure number, its dimensions are denoted simply as 1. Secondly, we see that although the dimensions associated with each variable are different, those corresponding to $g$ can be expressed in terms of the others. On the other hand, the dimensions of $m, \ell$ and $t$ are fundamental, so there is no way that any of them can be expressed in terms of the others (you can think of this as being like a basis of a vector space). We say that they are dimensionally independent.

So, returning to the general case, assume that we list the $n$ variables in such a way that $a_1, a\hspace{0.2mm}_2, \dots, a_j$ are dimensionally independent, while the remaining ones, $a_{j+1},\dots, a_n$, can be written as combinations of the others.  Having done this, the π quantities are constructed as follows. Take $a_{j+i}$ and divide it by a product of powers of $a_1, \dots, a_j$:
\begin{eqnarray*}
\pi_{i} = \frac{a_{j+i}}{a_1^{q_i}a_2^{r_i}\dots a_j^{z_i}}, \ \ \textrm{for} \ \ i=1,\dots, n-j.
\end{eqnarray*}

The powers $q_i, r_i, \dots, z_i$ must then be chosen such that $\pi_i$ has no dimensions. For the case of the pendulum, we have five variables, of which three are independent: hence $n=5$ and $j=3$.  So there are two π quantities:
\begin{eqnarray*}
\pi_1 = \frac{\theta}{m^{q_1}\ell^{r_1}t\hspace{0.2mm}^{s_1}} \quad \rightarrow \quad [\pi_1] = \frac{1}{M^{q\hspace{0.2mm}_1}L^{r_1}T\hspace{0.5mm}^{s_1}}, \\
\pi_2 = \frac{g}{m^{q_2}\ell^{r_2}t\hspace{0.2mm}^{s_2}} \quad \rightarrow \quad [\pi_2] = \frac{L^{1-r_2}}{M^{q\hspace{0.2mm}_2}T\hspace{0.5mm}^{s_2+2}}.
\end{eqnarray*}

We want both $\pi_1$ and $\pi_2$ to be dimensionless, so for $\pi_1$ it is evident that $q_1 = r_1=s_1= 0$ do the job: this means that $\theta$ is by itself a $\pi$ quantity.  On the other hand, the powers appearing in $\pi_2$ must satisfy the following system of linear equations:

\begin{equation*}
\begin{cases}
s\hspace{0.3mm}_2 + 2 =0 \\
r_2 – 1 =0 \\
q_2 = 0.
\end{cases}
\end{equation*}

Its solution is easy to obtain: $q_2=0, \ r_2 = 1$ and $s\hspace{0.3mm}_2=-2$. Thus the dimensionless quantities governing the motion of the pendulum are:

\begin{eqnarray*}
\quad \pi_1 = \theta \ \ \text{and} \ \ \pi_2 = \frac{gt\hspace{0.5mm}^2}{\!\ell}.
\end{eqnarray*}

The Buckingham π theorem then establishes that our physical system can be completely described in terms of this set of dimensionless quantities. In general, this means that there exists a function $F$ such that

\begin{eqnarray*}
F\hspace{0.5mm}(\pi_1, \pi_2,\dots, \pi_{n-j}) = 0.
\end{eqnarray*}

The Buckingham π theorem proves to be a priceless—and simple—tool

In our case we have,
\begin{eqnarray*}
F\hspace{0.5mm}\left(\theta, \frac{gt{\hspace{0.5mm}^2}}{\!\ell}\right) = 0.
\end{eqnarray*}

The first interesting thing to observe is that the mass no longer appears. This seems counterintuitive, but it is explained by the fact that under the action of gravity all bodies fall with the same acceleration regardless of their mass. The other point concerns the function $F$, expressing the relationship between $\pi_1$ and $\pi_2$. The Buckingham π theorem only establishes that such a function exists, but it does not provide any further information about the form of $F$. This must be obtained in some other way: through an experiment, for example. The theorem’s aim is different: the advantage it gives is that it tells us which parameters are important ($gt\hspace{0.5mm}^2/\ell$) and which ones can be ignored (the mass).  Actually, since the motion of our pendulum is oscillatory, we can guess a trigonometric function like sine or cosine must be involved. For those who have previously studied the pendulum, the form of $F$ might not have come as a great surprise since the exact solution for its oscillation about a small angle is

\begin{eqnarray*}
\theta = \theta_{\text{max}} \sin\bigg(\sqrt{\frac{gt\hspace{0.5mm}^2}{\!\ell}}\bigg),
\end{eqnarray*}

where $\theta_{\text{max}}$ is the angle the string makes with the vertical just before the mass is released.  Taking into account that only a small amount of physical intuition was necessary, that is not bad for a first attempt.

Blast wave

We return now to perhaps the most famous example of how the Buckingham π theorem can be used to understand a complex physical system: the energy released by a nuclear explosion like the Trinity test.  Of course, a similar process can also be applied to estimate the energy released by any huge explosion, like a star ejecting its outer layers into space during a supernova event. A phenomenon of this type passes through different stages, each one of them characterised by particular physical processes, but for the sake of simplicity, we can think of it simply as a hemispherical fireball expanding into the atmosphere.

A photo of the Trinity test, taken 0.025s after the explosion.

A photo of the Trinity test, taken 0.025s after the explosion.

First, we have to identify the relevant variables in our problem. As the explosion is a dynamical process, time $t$ must play a role, as must the radius of the fireball $R$. One might also think that the mass $m$ of the bomb should be included, but it is in fact equivalent to the energy $E$ (recall the famous Einstein equation $E = mc\hspace{0.2mm}^2$), so it is unnecessary to include both. An additional variable, which is perhaps not as evident at first sight, is the density $\rho$ of the surrounding environment. The reason is simple: the shock wave will expand more easily in a rarefied atmosphere than in a dense one. Of course, there can be lots of additional variables, but as long as they do not have a great relevance to the problem, we can neglect them.

Once we are confident in our choice of variables, we can start to construct the $\pi$ quantities. We have four quantities with dimensions:
\begin{eqnarray*}
[R\hspace{0.5mm}] = L, \ [t\hspace{0.5mm}] = T, \ [\rho] = ML^{-3} \ \  \text{and} \ \ [E\hspace{0.5mm}] = ML^2T\hspace{0.5mm}^{-2}.
\end{eqnarray*}
The dimensionally independent variables this time correspond to $R, t$ and $\rho$. (Of course, just as for the basis of a vector space, the choice of independent variables is not unique and we could have chosen $R, t$ and $E$ to play this role: this would lead to the same final result.) Since we have four variables and three of them are dimensionally independent, we will have only one $\pi$ quantity:
\begin{eqnarray*}
\pi_1 = \frac{E}{R\hspace{0.2mm}^qt\hspace{0.3mm}^r\rho^s} \quad \rightarrow \quad [{\pi_1}] = \frac{L^{3s-q+2}}{T\hspace{0.3mm}^{r+2}M\hspace{0.3mm}^{s-1}}.
\end{eqnarray*}
To apply the Buckingham π theorem this must be dimensionless and so we have the following system of equations:
\begin{equation*}
\begin{cases}
3s -q + 2 =0 \\
r + 2 = 0 \\
s – 1 = 0.
\end{cases}
\end{equation*}
This system is only satisfied when $q = 5, \ r=-2$ and $s=1$, so the theorem guarantees that the energy released by the bomb can be expressed as
\begin{eqnarray*}
F\hspace{0.5mm}\bigg(\frac{E\hspace{0.5mm}t\hspace{0.5mm}^2}{R\hspace{0.5mm}^5\rho}\bigg) = 0,
\end{eqnarray*}

where $F$ is some function.  Again, we do not know what $F$ is, but on this occasion we do not need any further information about it: the quantity $E\hspace{0.5mm}t\hspace{0.5mm}^2/R\hspace{0.5mm}^5\rho$ must satisfy the above equation at all times.  The only way in which this is possible is if it is constant.  Therefore, we set it equal to $C$ and solve for the energy to obtain
\begin{eqnarray*}
E = C\hspace{0.5mm}\frac{\rho R\hspace{0.5mm}^5}{t\hspace{0.5mm}^2}.
\end{eqnarray*}
In reality, the constant $C$ must be obtained by carrying out an experiment or using a physical model.  Surprisingly, its typical values have been found to range between $0.9$ and $1.1$, so setting $C = 1$ is a reasonable assumption.

So what is our estimate for the emitted energy of the Trinity test?  From the photo on the previous page, we can observe that 0.025s after the explosion the radius of the fireball was approximately $R = 130$ m.  We also know that the density of air at 20°C is $\rho = 1.20$kg/$\text{m}^3$.  Inserting these values into our formula, we compute an energy of $7.13 \times 10^{13}$J, which is similar to 17.1 kilotons of TNT, the same as Taylor’s result.

To sum up, we have found a formula to estimate the energy released in a nuclear explosion without a single bit of nuclear physics, but only some guesswork, elementary maths and the invaluable help of the Buckingham π theorem.

The range of applicability of the Buckingham π theorem is not restricted to these few examples, but using more powerful versions it proves to be a priceless—and simple—tool in the study of thermodynamic and gravitational systems.  Although it is usually underestimated, it makes explicit some of the deepest connections between physics and mathematics, as well as highlights some fundamental properties inherent to natural phenomena.

Diego is a PhD student at Queen Mary University of London, working on the mathematical aspects of the general theory of relativity using conformal methods.
Twitter  @diego_carranza_    + More articles by Diego Antonio

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