3 The Buckingham \(\Pi\)-theorem and the dimension matrix

Learning outcomes

By the end of this week you should be able to:

• state and apply the Buckingham \(\Pi\)-theorem to determine the number of dimensionless groups;

• construct and use the dimension matrix to find the dimensionless complexes systematically;

• interpret the physical meaning of dimensionless groups such as the Reynolds number and the drag coefficient.

3.1 Motivating example: the undamped pendulum

In the previous chapter you met an example of a damped pendulum, we begin this chapter by exploring a simpler problem, the undamped pendulum, in more detail. Consider a simple pendulum: a weight of mass \(m\) suspended from a rigid, massless rod of length \(l\), free to swing in a vertical plane under the influence of gravity. We wish to determine the period of oscillation \(\tau\).

Let \(\theta\) denote the angle between the rod and the downward vertical. Newton’s second law applied to the tangential component of the bob’s motion gives the equation of motion \[m l \ddot{\theta} = -m g \sin\theta,\] where \(g\) is the gravitational acceleration and \(\ddot{\theta} = \dd^2\theta/\dd t^2\). The mass \(m\) cancels immediately, yielding \[\begin{equation} \ddot{\theta} = -\frac{g}{l}\sin\theta. \label{eq:pendulum} \end{equation}\]

This already tells us something important: the motion is independent of the bob’s mass. The governing parameters that remain are \(l\), \(g\) and the initial swing angle, \(\alpha\), (the amplitude). For small oscillations, \(\sin\theta \approx \theta\), and [eq:pendulum] reduces to the equation of simple harmonic motion, \[\ddot{\theta} = -\frac{g}{l}\,\theta,\] whose solution is \(\theta(t) = \alpha \cos(\omega t)\) with angular frequency \(\omega = \sqrt{g/l}\). The period is therefore \[\tau = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{l}{g}}.\]

We can arrive at essentially the same result — up to the dimensionless numerical prefactor \(2\pi\) — without solving the differential equation at all, using only dimensional considerations. Since \(m\) has cancelled, the period \(\tau\) can depend only on \(l\) and \(g\) (ignoring the angle \(\alpha\) for the moment). The dimensions of these quantities are \[[\tau] = \mathrm{T}, \qquad [l] = \mathrm{L}, \qquad [g] = \mathrm{L\,T}^{-2}.\]

The only combination of \(l\) and \(g\) that has the dimension of time is \(\sqrt{l/g}\): \[\left[\sqrt{\frac{l}{g}}\right] = \left(\frac{\mathrm{L}}{\mathrm{L\,T}^{-2}}\right)^{\!1/2} = \mathrm{T}.\] Hence \(\tau\) must be proportional to \(\sqrt{l/g}\), and we can write \[\tau = C \sqrt{\frac{l}{g}},\] where \(C\) is a dimensionless constant that dimensional analysis alone cannot determine. Comparison with the exact small-angle solution shows that \(C = 2\pi\).

3.2 The \(\Pi\)-theorem (Buckingham’s theorem)

In this section we introduce a theorem that formalises the techniques illustrated above and develop a systematic method to identify the dimensionless complexes of the governing parameters. The theorem states that if a problem contains \(n\) variables that have \(k\) independent dimensions then there are \(m = n - k\) dimensionless complexes that control the behaviour of the system.

We shall introduce this theorem by describing the procedure of dimensional analysis aided by it, which consists of the following steps. In fact, we have already applied this procedure when using rescaling in the pendulum problem. Here we formulate it in the general case, identifying the corresponding steps in the analysis of the pendulum.

1. Identify the governing parameters of the system or phenomenon of interest. This is a step that requires intuition, common sense, background knowledge and creativity.

   The governing parameters may include:

   1. Directly measurable quantities and those to be determined (e.g. the period, length and bob mass of a pendulum).

   2. Quantities we believe to be (or have stronger reasons to consider) important. In the case of the pendulum, these could be the friction parameter or the swing angle \(\alpha\). Very often, the list of important parameters emerges gradually, via trial and error.

   3. Relevant fundamental constants. For example, include \(g\) or \(G\) if gravity is important, \(\hbar\) to allow for quantum effects, \(c\) the speed of light for relativistic effects, etc.

   Pendulum: the governing parameters are \(\tau, l, g, m\).

2. Identify parameters with independent dimensions. These are parameters whose dimensions cannot be expressed in terms of the dimensions of the remaining ones. To put it differently, parameters with independent dimensions cannot be combined into a dimensionless complex.

   For example, \(l\), \(\tau\) and \(m\) are independent regarding their dimensions, but \(l\), \(\tau\) and \(g\) are not: \[\tau \sqrt{\frac{g}{l}}, \qquad \frac{\tau^{2} g}{l}, \qquad \left(\frac{\tau^{2} g}{l}\right)^{\!m}\] are dimensionless combinations for any value of \(m\).

   Pendulum: \(n = 4\) governing parameters, \(\tau, l, m, g\), and \(k = 3\) of them have independent dimensions.

   The number of parameters with independent dimensions cannot be larger than the number of basic units, but can be smaller if not all the basic units appear in the governing parameters. Most often, the basic units required in the course will be: M (mass), L (length), T (time interval), \(\Theta\) (temperature). Thus, the number of parameters with independent dimensions will not normally exceed four, but can be smaller.

   In the case of the pendulum, \(\Theta\) does not appear, and this number is \(k = 3\). If mass is excluded (in anticipation that the period of oscillations is independent of the bob’s mass), \(k = 2\): L and T are the only basic units involved.

   Let us denote parameters with independent dimensions as \(a_1, a_2, \ldots, a_k\) (\(k\) of them), and the remaining parameters as \(b_1, b_2, \ldots, b_m\) (\(m\) of them).

3. Write the desired relation between the governing parameters in a generic form, \[f(a_1, a_2, \ldots, a_k, b_1, b_2, \ldots, b_m) = 0.\]

   We recall that the dimensions of all \(b_i\) can be expressed in terms of the dimensions of \(a_j\): \[\begin{aligned} [b_1] &= [a_1]^{p_1} \cdots [a_k]^{r_1}, \\ &\;\;\vdots \\ [b_i] &= [a_1]^{p_i} \cdots [a_k]^{r_i}, \\ &\;\;\vdots \\ [b_m] &= [a_1]^{p_m} \cdots [a_k]^{r_m}. \end{aligned}\]

   Pendulum: \(f(\tau, m, l, g) = 0\), with \[a_1 = \tau, \quad a_2 = m, \quad a_3 = l, \quad b_1 = g, \qquad [b_1] = [a_3] / [a_1]^{2}.\] Among \(n = 4\) governing parameters, \(k = 3\) have independent dimensions, hence there is \(m = n - k = 1\) dimensionless complex.

4. Demand that the desired relation is independent of the units of measurement.

   In other words, the relation must remain unchanged when the parameters are rescaled. Since \[f(a_1, a_2, \ldots, a_k, b_1, b_2, \ldots, b_m) = 0\] must be true independently of the units chosen, we can divide all the arguments by \(a_1\) — this is equivalent to changing the basic units so as to make the magnitude of \(a_1\) unity: \[a_1 = 1\,[a_1].\]

   For example, we can select \(\ell = 9.8~\mathrm{m}\) as the unit length. Then the gravitational acceleration will have unit magnitude: \[g = 1\,\frac{\ell}{\mathrm{s}^{2}}, \qquad \ell = 9.8~\mathrm{m}.\] Likewise, if we select \(M_{\odot} = 2 \times 10^{33}~\mathrm{g}\) as the unit mass, the mass of the sun becomes equal to unity: \[M_{\mathrm{sun}} = 1\,M_{\odot}.\]

   So, we adjust the basic units in such a manner that the numerical magnitude of \(a_1\) is equal to unity: this operation is called “rescaling”; in this case, rescaling \(l\): \[f(1, \ldots) = 0.\] Change the units again, now to make the second argument, the former \(a_2\), have unit magnitude. Since the first argument, 1, is dimensionless, it will not be affected, but the remaining ones, shown with dots, may change: \[f(1, 1, \ldots) = 0.\]

We can repeat this procedure exactly \(k\) times, and the first \(k\) arguments of \(f\) will be replaced by unity, whereas the remaining \(m\) arguments will become dimensionless. Denoting the latter as \(\Pi_i\), \(1 \leq i \leq m\), we obtain the desired relation in the form \[f(1, 1, \ldots, 1, \Pi_1, \Pi_2, \ldots, \Pi_m) = 0,\] or \[\Phi(\Pi_1, \Pi_2, \ldots, \Pi_m) = 0,\] where \(\Phi\) is some function — it does not matter for us what the form of this function is: we only need to know that some relation between the \(\Pi_i\)’s exists.

Pendulum: \[f(1, m, l, g\tau^{2}) = 0, \quad f(1, 1, l, g\tau^{2}) = 0, \quad f\!\left(1, 1, 1, \frac{g\tau^{2}}{l}\right) = 0; \qquad m = n - k = 1, \quad \Pi_1 = \frac{g\tau^{2}}{l}.\]

Of course, all the \(k\) unit transformations can be performed all at once by changing all the relevant basic units in a coordinated manner. How to do this? We shall introduce a convenient method to find such a transformation in the next section.

The arguments presented are in fact the very idea of the proof of the \(\Pi\)-theorem, or Buckingham’s theorem,¹ which can be formulated as follows:

Since the dimensions of both \(a_i\) and \(b_j\) are of the form of a power-law monomial (the product of the powers of the basic units), the dimensionless parameters \(\Pi\) only contain powers of \(a_i\).

The reduction in the number of arguments from \(k + m\) to \(m\), even if \(k \leq 4\), is a very significant advantage. It should be noted that a function of one variable can be tabulated in just one line (1D), a table of a function of two variables would take a whole page (2D), for three variables a book would be required to print such a table (3D), and four variables require a whole library (4D).²

3.3 Applying the \(\Pi\)-theorem: the dimension matrix

The \(\Pi\)-theorem suggests the most efficient way to perform dimensional analysis: we have to identify dimensionless complexes of the governing parameters whose number is equal to the number of the parameters minus the number of those with independent dimensions, or the number of the basic units involved. In a sense, the number of dimensionless parameters is the number of “degrees of freedom”. To introduce a regular method to accomplish this, consider the pendulum again.

The relation between \(\tau\) and the four governing parameters, \(l\), \(m\), \(g\), can be written as \[f(\tau, m, l, g) = 0,\] with their dimensions being \[[\tau] = \mathrm{T}, \qquad [l] = \mathrm{L}, \qquad [m] = \mathrm{M}, \qquad [g] = \mathrm{L\,T}^{-2}.\] The number of parameters with independent dimensions is three, e.g. \(\tau\), \(l\), \(m\). Since the total number of the governing parameters is four, there must be a single dimensionless complex \(\Pi\), and the relation between the parameters can be reduced to \[f(1,1,1,\Pi) \equiv \Phi(\Pi) = 0, \qquad \text{or} \qquad \Pi = \text{const}, \quad [\Pi] = 1.\] Since the dimensions of all the parameters are power-law monomials of the basic units, \(\Pi\) must be a power-law monomial of the governing parameters themselves: \[\begin{equation} \Pi = \tau^{k_1}\, l^{k_2}\, m^{k_3}\, g^{k_4}, \label{eq:Pi-monomial} \end{equation}\] with the exponents \(k_1, k_2, k_3, k_4\) to be determined from the requirement that \(\Pi\) is dimensionless, \[[\Pi] = 1 = \mathrm{M}^{0}\,\mathrm{L}^{0}\,\mathrm{T}^{0}.\]

To derive equations for \(k_i\) resulting from \([\Pi] = 1\), construct what is known as the dimension matrix: \[\begin{array}{c|cccc} & \tau & l & m & g \\ & (k_1) & (k_2) & (k_3) & (k_4) \\ \hline \mathrm{M} & 0 & 0 & 1 & 0 \\ \mathrm{L} & 0 & 1 & 0 & 1 \\ \mathrm{T} & 1 & 0 & 0 & -2 \end{array}\]

The columns of the dimension matrix correspond to the governing parameters, and the rows correspond to each of the basic units involved. The elements of the matrix are the exponents of the basic units in the dimension of the corresponding parameter. Column by column, the elements express the following relations: \[[\tau] = \mathrm{M}^{0}\,\mathrm{L}^{0}\,\mathrm{T}^{1}, \qquad [l] = \mathrm{M}^{0}\,\mathrm{L}^{1}\,\mathrm{T}^{0}, \qquad [m] = \mathrm{M}^{1}\,\mathrm{L}^{0}\,\mathrm{T}^{0}, \qquad [g] = \mathrm{M}^{0}\,\mathrm{L}^{1}\,\mathrm{T}^{-2}.\]

Now, consider the rows. If the exponents of M, L and T should each be equal to zero in \(\Pi\) of the form [eq:Pi-monomial], we have the following equations for \(k_i\): \[\begin{aligned} \mathrm{M}: &\quad 0 \cdot k_1 + 0 \cdot k_2 + 1 \cdot k_3 + 0 \cdot k_4 = 0, \\ \mathrm{L}: &\quad 0 \cdot k_1 + 1 \cdot k_2 + 0 \cdot k_3 + 1 \cdot k_4 = 0, \\ \mathrm{T}: &\quad 1 \cdot k_1 + 0 \cdot k_2 + 0 \cdot k_3 - 2 \cdot k_4 = 0, \end{aligned}\] three simultaneous equations for four unknowns — the number of the equations is equal to the number of the basic units (\(=\) the number of parameters with independent dimensions \(k\)), whereas the number of the unknowns is equal to the number of the governing parameters, \(n\). Thus, one of the exponents can be chosen as convenient. Since we seek a relation of the form \(\tau = F(\ldots)\), it is reasonable to select \[k_1 = 1.\] The first equation yields \(k_3 = 0\), i.e. the bob mass does not enter the desired relation. The remaining two equations then reduce to \[k_2 + k_4 = 0, \qquad k_1 - 2k_4 = 0,\] which yields \(k_4 = \tfrac{1}{2}k_1 = \tfrac{1}{2}\), \(k_2 = -k_4 = -\tfrac{1}{2}\), and we arrive at \[\Pi = \tau\, l^{-1/2}\, g^{1/2},\] or \[\boxed{\tau = \Pi \sqrt{\frac{l}{g}},}\] where \(\Pi\) is a dimensionless constant.

Remarks.

• Since none of the quantities having independent dimensions can be dimensionless (Section 1.5), none of the \(a_i\) can be dimensionless.

• The \(\Pi\)-theorem is just another formulation of the fact that any equation can be written in a dimensionless form using the units characteristic of the phenomenon it describes (rather than chosen arbitrarily), and the theorem specifies how many dimensionless parameters will occur in the dimensionless equation.

• The number of parameters with independent dimensions is always equal to the rank of the dimension matrix, i.e. the number of rows or columns in the highest-order (largest) non-zero minor of the matrix. The rank of the dimension matrix may happen to be less than \(k\). This would mean that we made an error when identifying parameters with independent dimensions and in fact there are fewer of them. This error can easily be corrected should it occur.

3.4 Worked example: nuclear explosion

A nuclear explosion can be considered with an appropriate and reasonable idealisation as an instantaneous release of a certain energy, \(E\), which causes an expanding shock wave to propagate outwards. As the amount of energy released is very large, pressure inside the expanding, very hot region is by far larger than that in the surrounding medium, say, air. This leads to a rapid expansion of the hot gas region, the fireball. Since the expansion is so fast, the gas energy lost during the expansion at this early stage is negligible, so \(E = \text{const}\) is a good approximation at early stages of the expansion, where \(E\) is the total energy of the fireball. Obviously, this energy is very close to the energy released in the explosion: the original air energy within the heated region is negligible in comparison with the explosion energy. However, the ambient gas mass density \(\rho\) cannot be neglected because the mass of the gas produced in the explosion (the explosion debris) is negligible and the hot air is largely the ambient air heated by the radiation produced in the explosion and swept up by the outer boundary of the expanding hot region.

Apart from \(E\) and \(\rho\), the governing parameters of the problem are \(t\), time elapsed after the energy release, and \(R\), the size of the hot region.

At an early stage after the energy release the hot region expands so rapidly that, when the lower part of it encounters the Earth surface (if the explosion charge is located on top of a tower or is dropped from an airplane and explodes in air), the upper part is not affected and remains spherical for a relatively long time. Thus, it is not unreasonable to assume that the expanding hot region is spherically symmetric, hence it is called a fireball.

Applying the dimension matrix.The governing parameters and their dimensions are: \[[E] = \mathrm{M\,L}^{2}\,\mathrm{T}^{-2}, \qquad [\rho] = \mathrm{M\,L}^{-3}, \qquad [t] = \mathrm{T}, \qquad [R] = \mathrm{L},\] and our goal is to find how \(R\) depends on the other parameters. The dimension matrix has the form \[\begin{array}{c|cccc} & E & \rho & t & R \\ & (k_1) & (k_2) & (k_3) & (k_4) \\ \hline \mathrm{M} & 1 & 1 & 0 & 0 \\ \mathrm{L} & 2 & -3 & 0 & 1 \\ \mathrm{T} & -2 & 0 & 1 & 0 \end{array}\] and equations for \(k_i\) follow as \[\begin{aligned} \mathrm{M}: &\quad k_1 + k_2 = 0, \\ \mathrm{L}: &\quad 2k_1 - 3k_2 + k_4 = 0, \\ \mathrm{T}: &\quad -2k_1 + k_3 = 0, \end{aligned}\] and we put \(k_4 = 1\). This leads to \[\begin{aligned} \mathrm{M}: &\quad k_2 = -k_1, \\ \mathrm{L}: &\quad 2k_1 - 3k_2 + 1 = 0, \\ \mathrm{T}: &\quad k_3 = 2k_1, \end{aligned}\] which solves to yield \[k_1 = -\tfrac{1}{5}, \qquad k_2 = \tfrac{1}{5}, \qquad k_3 = -\tfrac{2}{5}.\] We have thus derived the following dimensionless complex of the governing parameters: \[\Pi = E^{-1/5}\,\rho^{1/5}\,t^{-2/5}\,R,\] which leads to the desired result: \[\begin{equation} \boxed{R = \Pi\left(\frac{Et^{2}}{\rho}\right)^{\!1/5}.} \label{eq:sedov-taylor} \end{equation}\]

This is the famous expansion law of an explosion fireball known as the Sedov–Taylor solution. It was first derived in 1941 by Geoffrey Ingram Taylor (1886–1975) in the UK, John von Neumann (1903–1957) in the USA and Leonid Ivanovich Sedov (1907–1999) in the USSR (published in 1946, but derived earlier).

In 1950, Sir Geoffrey Taylor published a paper where he analysed photographs of the first atomic explosion in New Mexico declassified in 1947. Sir Geoffrey found that the expansion of the fireball was described by [eq:sedov-taylor] to an impressive degree of accuracy, and obtained, from fitting it to the observed expansion, the explosion energy \(E\). This paper caused a scandal in the USA military and intelligence establishment since, unlike the photographs of the explosion, the charge energy remained highly classified.

This solution also describes explosions of another kind and scale — explosions of young, hot, massive, rapidly evolving stars, leading to the appearance of extremely bright supernova stars. The supernova of 1054 AD at its peak brightness outshone the Moon and was recorded in historical chronicles (the supernova exploded 6,500 light years away). Its remnant is now known as the Crab nebula.

Supernova explosions drive the turbulence in the interstellar gas whose scale is close to the size of the fireball at a time when its expansion speed \(\dd R/\dd t\) becomes comparable to the speed of sound in the ambient gas — at the early stages of supersonic expansion no perturbation can propagate ahead of the shock, except for light.

3.5 Worked example: terminal velocity of a raindrop

A raindrop of mass \(m\) falls through air under gravity until it reaches a constant speed — the terminal velocity \(v\). At terminal velocity the gravitational force is exactly balanced by air resistance, and the drop no longer accelerates. We wish to determine how \(v\) depends on the governing parameters using the dimension matrix.

The relevant governing parameters are:

• \(v\), the terminal velocity;

• \(m\), the mass of the drop;

• \(g\), the gravitational acceleration;

• \(\rho\), the density of air;

• \(A\), the cross-sectional area of the drop.

Step 1.There are \(n = 5\) governing parameters. Their dimensions are: \[[v] = \mathrm{L\,T}^{-1}, \qquad [m] = \mathrm{M}, \qquad [g] = \mathrm{L\,T}^{-2}, \qquad [\rho] = \mathrm{M\,L}^{-3}, \qquad [A] = \mathrm{L}^{2}.\] There are \(k = 3\) base dimensions (M, L, T), so the \(\Pi\)-theorem predicts \(m = n - k = 5 - 3 = 2\) dimensionless groups.

However, we can simplify. At terminal velocity the weight \(mg\) is balanced by drag, so the combination \(mg\) always appears together as a single force. Physically, only the weight \(W = mg\) matters, not \(m\) and \(g\) separately. Replacing them with \(W\) reduces the parameter list to four: \(v\), \(W\), \(\rho\) and \(A\).

Step 2 (revised).Now \(n = 4\) governing parameters with dimensions: \[[v] = \mathrm{L\,T}^{-1}, \qquad [W] = \mathrm{M\,L\,T}^{-2}, \qquad [\rho] = \mathrm{M\,L}^{-3}, \qquad [A] = \mathrm{L}^{2}.\] Still \(k = 3\) base dimensions, so \(m = n - k = 4 - 3 = 1\) — a single dimensionless group. The dimension matrix is: \[\begin{array}{c|cccc} & v & W & \rho & A \\ & (k_1) & (k_2) & (k_3) & (k_4) \\ \hline \mathrm{M} & 0 & 1 & 1 & 0 \\ \mathrm{L} & 1 & 1 & -3 & 2 \\ \mathrm{T} & -1 & -2 & 0 & 0 \end{array}\]

Step 3.We seek \(\Pi = v^{k_1}\,W^{k_2}\,\rho^{k_3}\,A^{k_4}\) with \([\Pi] = 1\). The equations for \(k_i\) are: \[\begin{aligned} \mathrm{M}: &\quad k_2 + k_3 = 0, \\ \mathrm{L}: &\quad k_1 + k_2 - 3k_3 + 2k_4 = 0, \\ \mathrm{T}: &\quad -k_1 - 2k_2 = 0. \end{aligned}\] Since we want \(v\) as a function of the other parameters, set \(k_1 = 1\). Then: \[\begin{aligned} \mathrm{T}: &\quad k_2 = -\tfrac{1}{2}k_1 = -\tfrac{1}{2}, \\ \mathrm{M}: &\quad k_3 = -k_2 = \tfrac{1}{2}, \\ \mathrm{L}: &\quad 1 - \tfrac{1}{2} - \tfrac{3}{2} + 2k_4 = 0 \quad\Rightarrow\quad k_4 = \tfrac{1}{2}. \end{aligned}\]

Step 4.The single dimensionless group is: \[\Pi = v\,W^{-1/2}\,\rho^{1/2}\,A^{1/2} = v\sqrt{\frac{\rho A}{W}}.\] Since \(m = 1\), the \(\Pi\)-theorem gives \(\Pi = \text{const}\), i.e. \[\boxed{v = C\sqrt{\frac{W}{\rho A}} = C\sqrt{\frac{mg}{\rho A}},}\] where \(C\) is a dimensionless constant that must be determined experimentally or by solving the full fluid mechanics problem.

3.6 Worked example: drag force on a sphere

We shall now consider a slightly more complicated example which will give rise to two dimensionless complexes rather than one. The drag force \(F_D\) on a sphere depends on its diameter \(L\), velocity \(V\), the fluid density \(\rho\), and the dynamic viscosity \(\mu\). This is the foundation of fluid dynamic similarity.

Step 1.There are \(n = 5\) governing parameters: \(F_D\), \(L\), \(V\), \(\rho\) and \(\mu\). Their dimensions are: \[[F_D] = \mathrm{M\,L\,T}^{-2}, \qquad [L] = \mathrm{L}, \qquad [V] = \mathrm{L\,T}^{-1}, \qquad [\rho] = \mathrm{M\,L}^{-3}, \qquad [\mu] = \mathrm{M\,L}^{-1}\,\mathrm{T}^{-1}.\] There are \(k = 3\) base dimensions: M, L and T. Hence there are \(m = n - k = 5 - 3 = 2\) dimensionless complexes.

Step 2.The dimension matrix has the form \[\begin{array}{c|ccccc} & F_D & L & V & \rho & \mu \\ & (k_1) & (k_2) & (k_3) & (k_4) & (k_5) \\ \hline \mathrm{M} & 1 & 0 & 0 & 1 & 1 \\ \mathrm{L} & 1 & 1 & 1 & -3 & -1 \\ \mathrm{T} & -2 & 0 & -1 & 0 & -1 \end{array}\] and we seek \(\Pi = F_D^{k_1}\,L^{k_2}\,V^{k_3}\,\rho^{k_4}\,\mu^{k_5}\) with \([\Pi] = 1\). The equations for \(k_i\) follow as \[\begin{aligned} \mathrm{M}: &\quad k_1 + k_4 + k_5 = 0, \\ \mathrm{L}: &\quad k_1 + k_2 + k_3 - 3k_4 - k_5 = 0, \\ \mathrm{T}: &\quad -2k_1 - k_3 - k_5 = 0, \end{aligned}\] three equations for five unknowns. Since \(m = 2\), two of the exponents can be chosen freely.

Step 3 — first \(\Pi\)-group.Put \(k_1 = 1\) and \(k_5 = 0\). Then \[\begin{aligned} \mathrm{M}: &\quad k_4 = -k_1 = -1, \\ \mathrm{T}: &\quad k_3 = -2k_1 = -2, \\ \mathrm{L}: &\quad 1 + k_2 - 2 + 3 = 0 \quad\Rightarrow\quad k_2 = -2, \end{aligned}\] which gives the first dimensionless complex: \[\Pi_1 = \frac{F_D}{\rho\,V^{2}\,L^{2}}.\] This is recognised as the drag coefficient \(C_D\) (up to a conventional numerical factor).

Step 4 — second \(\Pi\)-group.Put \(k_1 = 0\) and \(k_5 = 1\). Then \[\begin{aligned} \mathrm{M}: &\quad k_4 = -k_5 = -1, \\ \mathrm{T}: &\quad k_3 = -k_5 = -1, \\ \mathrm{L}: &\quad k_2 - 1 + 3 - 1 = 0 \quad\Rightarrow\quad k_2 = -1, \end{aligned}\] which gives the second dimensionless complex: \[\Pi_2 = \frac{\mu}{\rho\,V\,L}.\] The reciprocal of \(\Pi_2\) is the Reynolds number, \(\mathrm{Re} = \rho V L / \mu\).

Result.The \(\Pi\)-theorem tells us that \[\boxed{\Phi(\Pi_1, \Pi_2) = 0, \qquad \text{or equivalently} \qquad C_D = C_D(\mathrm{Re}).}\]