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Barenblatt on dimensional analysis

An overview of dimensional analysis, with some simple examples.

Overview

In dimensional analysis, the relation between some quantity of interest, \(a\), and the parameters on which it depends, \(a_1, \ldots, a_n\):

\[ a = f(a_1, \ldots, a_n) \]

is transformed into a relation between a dimensionless quantity of interest, \(\Pi\), and several dimensionless parameters, \(\Pi_1, \ldots, \Pi_m\).

\[ \Pi = f(\Pi_1, \ldots, \Pi_m) \]

The Buckingham Π-theorem states the following:

If there are \(n\) dimensional governing parameters in the original relation, of which a subset \(k\) have independent dimensions, then the number of dimensionless parameters in the new relation, \(m\), is given by \(n - k\).

The concept of independent dimensions is explained below.

The advantage of this is that when \(k\) is greater than zero, the new dimensionless relation consists of fewer parameters than the original dimensional one.

Dimensions

Measurement of any physical quantity consists of the direct comparison of that quantity, such as the mass of water in a glass, the length of a ruler, or the half-life of radium, with some appropriate standard, such as the gram, the metre or the year. The standard is called the unit of measurement.

In general there is more than one unit for any given physical quantity: lengths, for example, can be measured with reference to the Angström, the inch, the metre, the light year, the parsec, and many other pre-defined standards. The covariance principle of physics - that physical laws should be the same for all observers - implies that physical laws should be the same for observers using different magnitudes of basic units, and therefore the form of a physical relationship should not vary in the face of changes from one set of basic units to another. Only the magnitudes of the quantities involved change.

In changing from one set of units to another therefore, physical relations do not change, but each quantity in the relation changes its magnitude by some factor. The function that determines the factor by which the numerical value of a given quantity changes upon passing from one set of units to another is called the dimension function, or dimension, of that quantity. It is customary to denote the dimension of a quantity \(x\) by \([x]\). For example, if the unit of length, \(l\), is changed by a factor \(L\), then the dimension of length, \([l]\), is given by \(L\):

\[ [l] = L \]

We say that length has dimension \(L\). Analogously, mass has dimension \(M\), time has dimension \(T\), velocity has dimension \(LT^{-1}\), acceleration has dimension \(LT^{-2}\), etc. It should be emphasised that \(L\), \(M\) and \(T\) are nothing more than abstract, positive numbers, and the dimensions of length, mass, time, velocity and acceleration are functions of these numbers:

\[ \begin{align} [m] &= M \\ [t] &= T \\ [v] &= LT^{-1} \\ [a] &= LT^{-2} \end{align} \]

Quantities whose numerical values are identical in all systems of units are called dimensionless.

Independent dimensions

The quantities \(a_1, \ldots, a_k\) are said to have independent dimensions if none of them has a dimension function that can be represented in terms of a product of powers of the dimensions of the remaining quantities. Therefore, we can re-write our original relation describing the dependence of the quantity of interest, \(a\), on its governing parameters as:

\[ a = f(a_1, \ldots, a_k, b_1, \ldots, b_m) \]

where the governing parameters, \(a_1, \ldots, a_k, b_1, \ldots, b_m\), are divided up in such a way that parameters \(a_1, \ldots, a_k\) have independent dimensions, while the dimensions of parameters \(b_1, \ldots, b_m\) can be expressed as powers of the dimensions of the parameters \(a_1, \ldots, a_k\). Such a division may always be made; in some special cases \(k = 0\) or \(m = 0\), but in general this is not so.

Transformation to dimensionless parameters

We have established that the dimensions of the governing parameters \(b_1, \ldots, b_m\) can be expressed as products of powers of the dimensions of the governing parameters \(a_1, \ldots, a_k\):

\[ \begin{align} [b_1] &= [a_1]^{p_1} \ldots [a_k]^{r_1} \\ &\vdots \\ [b_i] &= [a_1]^{p_i} \ldots [a_k]^{r_i} \\ &\vdots \\ [b_m] &= [a_1]^{p_m} \ldots [a_k]^{r_m} \end{align} \]

That is, each of these governing parameters, \(b_i\), has dimensions equal to the product \([a_1]^{p_i}\ldots [a_k]^{r_i}\). This means that the governing parameter \(b_i\) itself can differ from the product \(a_1^{p_i}\ldots a_k^{r_i}\) only by a numerical factor, and not in dimension. Let us call this dimensionless numerical factor \(\Pi_i\). What we are saying is that:

\[ \begin{align} b_1 &= \Pi_1(a_1^{p_1} \ldots a_k^{r_1}) \\ &\vdots \\ b_i &= \Pi_i(a_1^{p_i} \ldots a_k^{r_i}) \\ &\vdots \\ b_m &= \Pi_m(a_1^{p_m} \ldots a_k^{r_m}) \end{align} \]

We can now use these equations to remove the governing parameters \(b_1, \ldots, b_m\) from the original relation, replacing them instead with \(\Pi_1, \ldots, \Pi_m\). Hence:

\[ a = g(a_1, \ldots, a_k, \Pi_1, \ldots, \Pi_m) \]

We have re-written the relation for \(a\) as a function of the dimensional governing parameters \(a_1, \ldots, a_k\), and the dimensionless governing parameters \(\Pi_1, \ldots, \Pi_m\), rather than of the dimensional governing parameters \(a_1, \ldots, a_k, b_1, \ldots, b_m\).

This new relation shows it is possible to express the dimension of \(a\) as a product of the powers of the dimensions of the parameters \(a_1, \ldots, a_k\), since the only other parameters on which \(a\) depends are dimensionless. Hence we can also write:

\[ [a] = [a_1]^p \ldots [a_k]^r \]

and therefore:

\[ a = \Pi a_1^p \ldots a_k^r \]

Consequently the relation for \(a\) becomes:

\[ \Pi a_1^p \ldots a_k^r = g(a_1, \ldots, a_k, \Pi_1, \ldots, \Pi_m) \]

or:

\[ \Pi = h(a_1, \ldots, a_k, \Pi_1, \ldots, \Pi_m) \]

The final stage in the process is to remove the remaining dimensional governing parameters \(a_1, \ldots, a_k\) from the relation in order that a new relation consisting entirely of the dimensionless quantities \(\Pi_1, \ldots, \Pi_m\) and \(\Pi\). By doing so, we will have removed \(k\) out of the original \(n\) parameters from the equation, and thus simplified it by some degree. Our ability to do this rests on the following important principle.

Because the parameters \(a_1, \cdots, a_k\) have independent dimensions, no parameter in the set has dimensions that can be expressed in terms of a product of the powers of the dimensions of any combination of the other parameters. This means that it is always possible to pass to a new system of units chosen such that only one parameter, say \(a_1\), changes its numerical value, while all the other parameters remain unchanged. The values of the dimensionless parameters \(\Pi_1, \ldots, \Pi_m\), and the dimensionless value of the function, \(\Pi\), remain unchanged under such a transformation. However, it follows from this fact that the function \(h\) must in fact be independent of the argument \(a_1\). If this were not the case, it would not be possible to change \(a_1\) and only \(a_1\) without also changing the value of \(\Pi\). In exactly the same way, it can be shown the the function \(h\) is also independent of the arguments \(a_2, \ldots, a_k\). So in fact:

\[ \Pi = \Phi(\Pi_1, \ldots, \Pi_m) \]

Thus, we have reached our final goal. An original relation for a dimensional quantity of interest \(a\) depending on \(n\) dimensional governing parameters has been replaced with a relation for a dimensionless parameter of interest \(\Pi\) depending on only \(m\) dimensionless governing parameters, where \(m = n - k\), and \(k\) is the number of governing parameters in the original relation that had independent dimensions.

One advantage of this as follows. To determine the functional dependence of \(a\) on each of the governing parameters in the original relation, it is necessary to calculate or measure the function \(f\) for, say, 10 values of each governing parameter. (The choice of 10 here is arbitrary; a smaller number might suffice for smooth functions, while even 100 might not be sufficient for rapidly-varying functions.) Thus, it is necessary to carry out \(10^n = 10^{k+m}\) function evaluations. After applying dimensional analysis, the problem is reduced to to one of determining only \(10^m\) evaluations of the new function \(\Phi\), i.e. a factor of \(10^k\) fewer. The result is that the amount of work required to determine the desired function is reduced by as many orders of magnitude as there are governing parameters with independent dimensions.

In summary then, the procedure is as follows:

  1. Ascertain the parameters on which the quantity to be determined depends
  2. Select the parameters which have independent dimensions
  3. Transform the relation into one between dimensionless quantities

Examples

Drag force on a high-velocity sphere in a fluid

Ascertain the parameters on which the quantity to be determined depends

At high velocity, it intuitively seems that the internal friction in the gas - the viscosity - can be neglected, since the resistance to the motion of the sphere is due mainly to the inertia of the gas as it is pushed apart by the sphere. Therefore, the drag force, \(F\), depends on the fluid density, \(\rho\), the fluid pressure, \(p\), the speed of the body, \(u\), and the diameter of the sphere, \(d\).

\[ F = f(\rho, d, p, u) \]

Select the parameters that have independent dimensions

The dimensions of the governing parameters are:

\[ \begin{align} [F] &= LMT^{-2} \\ [\rho] &= L^{-3}M \\ [d] &= L \\ [p] &= L^{-1}MT^{-2} \\ [u] &= LT^{-1} \end{align} \]

We find \(\rho\), \(d\) and \(u\) have independent dimensions, so \(k = 3\) and \(m = 1\).

Transform the relation into one between dimensionless quantities

The dimension of \(p\) can be expressed as a product of powers of the dimensions of \(\rho\), \(d\) and \(u\), these being the quantities with independent dimensions:

\[ [p] = [\rho]^{x_1} [d]^{y_1} [u]^{z_1} \]

with \(x_1 = 1\), \(y_1 = 0\) and \(z_1 = 2\). Hence \(p\) differs from the product \(\rho u^2\) by a numerical factor only, call this \(\Pi_1\).

\[ p = \Pi_1\rho u^2 \]

We substitute this in the expression for the drag force:

\[ F = g(\rho, d, u, \Pi_1) \]

The same applies to \(F\), whose dimensions can depend only on powers of the dimensions of  \(\rho\), \(d\) and \(u\), because \(\Pi_1\) is dimensionless. Hence:

\[ [F] = [\rho]^x[d]^y[u]^z \]

where \(x = 1\), \(y = 2\), \(z = 2\). Hence \(F\) is only different from the product \(\rho u^2 d^2\) by a numerical factor, \(\Pi\), and:

\[ F = \Pi \rho d^2 u^2 \]

Now we have:

\[ \Pi \rho d^2 u^2 = g(\rho, d, u, \Pi_1) \]

or:

\[ \Pi = h(\rho, d, u, \Pi_1) \]

Since \(\rho\), \(d\) and \(u\) have independent dimensions, we can switch to a system of units in which the numerical value of \(\rho\) changes, but not those of \(u\) or \(d\); since upon doing this the values of the dimensionless quantities \(\Pi\) and \(\Pi_1\) do not change, the function \(h\) must in fact be independent of \(\rho\). The same can be said for \(u\) and \(d\). Hence:

\[\Pi = \Phi(\Pi_1) \]

Substituting this back into the formula for the drag force gives:

\[ F = \rho d^2 u^2 \Pi(\Pi_1) \]

or:

\[ F = \rho d^2 u^2 \Pi\left(\frac{p}{\rho u^2}\right) \]

So, we have gone from a general problem where the quantity of interest (\(F\)) has an unknown dependence on four governing parameters, \(\rho\), \(d\), \(p\) and \(u\), to one where it has an unknown dependence on only one, \(p/\rho u^2\). If we were investigating experimentally, we would now only to measure \(F\) at (say) ten values of \(p/\rho u^2\), rather than at (say) ten values of \(\rho\) for each of 9say) ten values of \(d\), for each of ... etc.

Extra

With a bit of additional knowledge we can take this further. We happen to know, for a fluid, that the ratio \(p/\rho\) is constant, being proportional to \(c^2\) (\(c\) being the speed of sound in the fluid). Hence we can write:

\[ F = \rho d^2 u^2 \Pi_0\left(\frac{u}{c}\right) \]

where \(u/c\) is the Mach number. We also know that:

\[ F = \frac{1}{2} AC_d\rho u^2 \]

where \(A\) is the cross-sectional area of the object (not the surface area), and \(C_d\) is the drag coefficient. Thus the drag coefficient is:

\[ C_d = \frac{8}{\pi} \Pi_0\left(\frac{u}{c}\right) \]

That is, the drag coefficient is a function of only the Mach number; this functional form can be measured experimentally.

Drag force on a slow-moving thin plate in a fluid

Ascertain the parameters on which the quantity to be determined depends

In this case the fluid compressibility can be ignored for low speeds, but viscosity is essential. The drag force (per unit width), \(F\), consequently depends on the relative speed of the plate, \(u\), its length, \(l\), and the fluid's density, \(\rho\), and (kinematic) viscosity, \(\nu\):

\[ F = f(\rho, u, l, \nu) \]

Select the parameters that have independent dimensions

The dimensions of the quantity of interest and of the governing parameters are:

\[ \begin{align} [F] &= LMT^{-2} \\ [\rho] &= L^{-3}M \\ [\nu] &= L^2T^{-1} \\ [l] &= L \\ [u] &= LT^{-1} \end{align} \]

The dimensions of viscosity can be expressed in terms of the dimensions of the remaining governing parameters, which have independent dimensions. Thus \(k = 3\) and \(m = 1\).

Transform the relation into one between dimensionless quantities

The dimension of viscosity can be formed from the dimensions of the product of length and speed, so the viscosity must only differ from that product by a numerical factor. For reasons that will be clear later, we will call it \(1/\Pi_1\).

\[ \nu = \left(\frac{1}{\Pi_1}\right)lu \]

such that:

\[ \Pi_1 = \frac{lu}{\nu} \]

As before:

\[ F = g(\rho, l, u, \Pi_1) \]

The dimensions of force can be formed from the dimensions of the remaining dimensional governing parameters in such a way that force can only differ from the product \(\rho l u^2\) by a numerical factor, \(\Pi\):

\[ F = \Pi \rho l u^2 \]

Consequently:

\[ \Pi \rho l u^2 = g(\rho, l, u, \Pi_1) \]
\[ \Pi = h(\rho, l, u, \Pi_1) \]

Because the dimensional arguments of the function \(h\) have independent dimensions, and we can change to alternative system of units in which one changes its value without changing the others, and obviously without changing the dimensionless values, the function \(h\) must in fact be independent of them:

\[ \Pi = \Phi(\Pi_1) \]

Therefore:

\[ F = \rho l u^2 \Pi(\Pi_1) \]

or:

\[ F = \rho l u^2 \Pi \left(\frac{lu}{\nu}\right) \]

As with the high-velocity sphere, the unknown function of four dimensional governing parameters has been reduced to an unknown function of a single dimensionless parameter.

The dimensionless parameter in this case is the Reynolds number, \(\mathrm{Re}\):

\[ \mathrm{Re} = \frac{lu}{\nu} \]

Period of a pendulum

Now consider an example of the special case that arises when all the dimensional parameters in the governing equation have independent dimension, \(k = n\), \(m = 0\). In this situation, \(\Pi\) becomes an unknown constant rather than an unknown function of \(m\) dimensionless parameters.

Ascertain the parameters on which the quantity to be determined depends

Consider a simple pendulum, the period of which we shall denote by \(\theta\). By considering the physics of the pendulum we can see that the parameters the period can possibly depend on are the length of the pendulum, \(l\), the mass of the bob, \(m\), and the acceleration due to gravity \(g_0\). Consequently:

\[ \theta = f(g_0, l, m) \]

Select the parameters that have independent dimensions

There are three governing parameters, \(n = 3\). The dimensions of the quantities are as follows:

\[ \begin{align} [\theta] &= T \\ [g_0] &= LT^{-2} \\ [l] &= L \\ [m] &= M \end{align} \]

All three of the governing parameters are independent, so \(k = 3\) and \(m = 0\). Hence the first stage of the transformation process, to eliminate the parameters without independent dimensions, \(b_1, \ldots, b_m\), from the right-hand-side, replacing them with dimensionless parameters \(\Pi_1, \ldots, \Pi_m\), has effectively been done, and:

\[ \theta = g(g_0, l, m) \]

where in this instance the functional form of \(g\) is the same as that of \(f\).  

Transform the relation into one between dimensionless quantities

It remains to replace the quantity of interest, \(\theta\), with its dimensionless equivalent, \(\Pi\). The dimensions of \(\theta\) can be expressed as a product of the powers of the dimensions of the governing parameters:

\[ [\theta] = [g_0]^p [l]^q [m]^r \]

where we can determine that \(p = -1/2\), \(q = 1/2\) and \(r = 0\). So \(\theta\) must only differ from the product \(l^{1/2}/g_0^{1/2}\) by a numerical factor, \(\Pi\), and not in dimension; hence:

\[ \theta = \Pi \sqrt{\frac{l}{g_0}} \]

Some introductions to dimensional analysis, understandably, end here, asserting that \(\Pi\) is a dimensionless constant, and identifying it with the familiar constant \(k = 2\pi\). Strictly speaking, I think, at this point we only know that \(\Pi\) is dimensionless, not that it is constant. So we continue the analysis as above, next writing:

\[ \Pi \sqrt{\frac{l}{g_0}} = g(g_0, l, m) \]

and

\[ \Pi = h(g_0, l, m) \]

Given that we can move to a system of units in which only \(g_0\) changes its value, but not \(l\) or \(m\), (because these parameters have independent dimensions), and that in doing so the dimensionless quantity \(\Pi\) must not change, we can say that the function \(h\) is independent of \(g_0\). In exactly the same way we can see that it is also independent of \(l\) and \(m\), and hence:

\[ \Pi = \mathrm{const} \]

The dimensionless parameter of interest, \(\Pi\), is therefore a constant. Hence we retain:

\[ \theta = \Pi \sqrt{\frac{l}{g_0}} \]

As mentioned already, this is the correct formula for the period of a simple pendulum, with \(\Pi = 2\pi\). This has allowed us to derive the equation for the period of a pendulum without recourse to calculus.

Radius of a nuclear explosion

This is another example where \(k = n\) and \(m = 0\).

In a nuclear explosion, a large amount of energy, \(E\), is released, and a strong shock wave develops in the atmosphere. The size of the region of energy release can be considered small compared with the radius of the shock wave, and the time taken to release the energy small compared with time over which the shock wave proceeds, and finally the pressure behind the shock is so large that the ambient atmospheric pressure can be considered small. These quantities are therefore neglected, leaving the energy \(E\), the initial air density, \(\rho\), and the time \(t\) as the only governing parameters:

\[ r = f(\rho, E, t) \]

Select the parameters that have independent dimensions

The dimensions of the quantity of interest and governing parameters are:

\[ \begin{align} [\rho] &= ML^{-3} \\ [E] &= ML^2T^{-2} \\ [r] &= L \\ [t] &= [T] \end{align} \]

So \(k = n = 3\) and \(m = 0\).

Transform the relation into one between dimensionless quantities

The product \((Et^2/\rho)^{1/5}\) can be formed from which \(r\) differs only by a numerical factor, \(\Pi\):

\[ \Pi = \left(\frac{\rho}{Et^2}\right)^{1/5} r \]

Therefore:

\[ \Pi \left(\frac{Et^2}{\rho}\right)^{1/5} = f(\rho, E, t) \]

and:

\[ \Pi = h(\rho, E, t) \]

Since all the governing parameters in the function on the right have independent dimensions, \(\Pi\) must be constant. Hence:

\[ r = \Pi \left(\frac{Et^2}{\rho}\right)^{1/5} \]

Thus it is possible to infer the energy released in an explosion from a time series of measurements of the radius of the shock front, if the value of the constant can be found. In 1950 G. I. Taylor famously published a calculation of the energy released in the 1945 Trinity test, based on a series of unclassified high-speed photographs of the expansion of the fireball taken by J. Mack. The solution to the appropriate problem in gas dynamics showed that the constant has a value close to unity, and the energy value Taylor obtained was \(10^{21}\) ergs; this caused, in Taylor's words, "much embarrassment" in the US government, as the figure was classified.

Flow rate of a viscous fluid through a circular pipe (Poiseuille's Formula)

Consider a viscous fluid undergoing steady flow through a pipe of circular cross-section. The rate of volume flow, i.e. the volume \(V\) of fluid passing a section of the pipe in time \(t\), will depend on:

  • The coefficient of dynamic viscosity, \(\eta\),
  • The radius \(r\) of the pipe, and
  • The pressure gradient \(p/l\), where \(p\) is the difference in pressure between the two ends of the pipe, and \(l\) is its length.

Hence:

\[ \frac{V}{t} = f\left(\eta, r, \frac{p}{l}\right) \]

The dimensions are:

\[ \begin{align} [V/t] &= L^3 T^{-1} \\ [\eta] &= ML^{-1}T{-1} \\ [r] &= L \\ [p/l] &= LM^{-2}T^{-2} \end{align} \]

As in the previous two example, \(k = n = 3\) and \(m = 0\).

The dimensions on both sides must be the same:

\[ [V/l] = [\eta]^x [r]^y [p/l]^z \]
\[ L^3 T^{-1} = (ML^{-1}T^{-1})^x (L)^y (LM^{-2}T^{-2})^z \]

Equating the indices on both sides and solving gives:

\[ \begin{align} x &= -1 \\ y &= 4 \\ z &= 1 \end{align} \]

and: We can write:

\[ \frac{V}{l} = \Pi \frac{r^4}{\eta} \left(\frac{p}{l}\right) \]

where \(\Pi\) is a dimensionless parameter. To show that \(\Pi\) is a constant:

\[ \Pi \frac{r^4}{\eta} \left(\frac{p}{l}\right) = f\left(\eta, r, \frac{p}{l}\right) \]
\[ \Pi = g\left(\eta, r, \frac{p}{l}\right) \]

Since \(\eta\), \(r\) and \(p/l\) have independent dimensions:

\[ \Pi = \mathrm{const} \]

In fact the constant has value \(\pi/8\), so:

\[ \frac{V}{l} = \frac{\pi p r^4}{8\eta l} \]

This is known as Poiseuille's formula.

Stokes's Law

Consider a similar situation to the first example, of a sphere with radius \(r\) in a fluid flow with velocity \(u\). This time we assume the fluid is viscous, with coefficient of (dynamic) viscosity \(\eta\). To calculate the force due to viscous drag only, we can ignore the mass of the sphere and the density of the fluid.

Hence:

\[ F = f(\eta, r, u) \]

The dimensions are:

\[ \begin{align} [F] &= MLT^{-2} \\ [\eta] &= ML^{-1}T^{-1} \\ [r] &= L \\ [u] &= LT^{-1} \end{align} \]

Again, \(k = 3\), \(m = 0\).

\[ F = \Pi r \eta u \]

and as abov we can demonstrate that \(\Pi = \mathrm{const}\). As it happens the value is \(6\pi\), so:

\[ F = 6\pi \eta r u \]

which is Stokes's Law.

References

Barenblatt, G. I. (1996): Scaling, Self-Similarity, and Intermediate Asymptotics. Cambridge: Cambridge University Press. 386pp.