Drop a small steel ball bearing into a tall jar of thick oil in the school laboratory. After the first few millimetres it sinks at a slow, steady pace. The oil drags on the ball. The exact flow of oil round a sphere is hard mathematics, first solved by George Stokes in 1851. Yet a student can find the form of the drag in five minutes.
A formula from dimensions alone
Suppose you know which quantities a result depends on, but not the formula. The principle of homogeneity then limits the formula severely. Write the result as a pure number times a product of powers of those quantities. The dimensions of the two sides must be the same, and that fixes the powers. This method is called dimensional analysis.
Try it on a simple pendulum. Its period \(T\) might depend on the mass \(m\) of the bob, the length \(l\) of the string and the acceleration due to gravity \(g\). Write \(T = k\,m^al^bg^c\), with \(k\) a pure number. In the steps below the letter T in upright type is the dimension of time; the italic \(T\) is the period.
\[ T = k\,m^{a}\,l^{b}\,g^{c} \]
Why. the period may depend on the mass of the bob, the length of the string and \(g\); \(k\) is a pure number.
\[ \text{T} = \text{M}^{a}\,\text{L}^{b}\,(\text{LT}^{-2})^{c} \]
Why. write the dimensions of both sides; a pure number has none, and \([g] = \text{LT}^{-2}\).
\[ \text{M}^{0}\,\text{L}^{0}\,\text{T}^{1} = \text{M}^{a}\,\text{L}^{b + c}\,\text{T}^{-2c} \]
Why. collect the powers of M, L and T on the right.
\[ a = 0, \quad b + c = 0, \quad -2c = 1 \]
Why. equate the powers of M, of L and of T on the two sides.
\[ a = 0, \quad c = -\tfrac{1}{2}, \quad b = \tfrac{1}{2} \]
Why. the third equation gives \(c\), and then the second gives \(b\).
\[ T = k\,\sqrt{l/g} \]
Read it. put the powers back. The mass has disappeared, and dimensions cannot give the pure number \(k\).
The result says three things before any experiment. The period does not depend on the mass of the bob. It grows as the square root of the length, so four times the length gives twice the period. And a weaker gravity gives a longer period. Only the number \(k\) is left open. The figure below lets you time the pendulum and find that number.
Every setting of the figure gives the same ratio, \(T/\sqrt{l/g} = 6.28\). That is the number \(2\pi\), fixed by timing real pendulums or by the full theory of simple harmonic motion. So \(T = 2\pi\sqrt{l/g}\), the formula of the chapter Dynamics of SHM.
For a simple pendulum, dimensional analysis gives \(a = 0\) as the power of the mass of the bob. What does this mean?
The period \(T\) of a block on a spring depends only on the mass \(m\) and the spring constant \(k_s\), whose dimensions are \(\text{MT}^{-2}\). What form must the period have?
Back to the ball in the oil
Now return to the phenomenon. The drag \(F\) on a slow sphere may depend on the viscosity \(\eta\) of the oil, the radius \(r\) and the speed \(v\). The viscosity has dimensions \(\text{ML}^{-1}\text{T}^{-1}\), as the chapter Viscosity shows. The example below follows the steps of the pendulum.
Assume the drag on a small sphere moving slowly through a fluid is \(F = k\,\eta^ar^bv^c\). Find the powers. Then use Stokes' value \(k = 6\pi\) to find the drag on a sphere of radius \(1\,\text{mm}\) moving at \(0.1\,\text{m/s}\) through a liquid of viscosity \(0.1\,\text{Pa s}\).
Dimensions: \(\text{MLT}^{-2} = (\text{ML}^{-1}\text{T}^{-1})^a\,\text{L}^b\,(\text{LT}^{-1})^c = \text{M}^a\,\text{L}^{-a + b + c}\,\text{T}^{-a - c}\).
Powers of M: \(a = 1\). Powers of T: \(-a - c = -2\), so \(c = 1\). Powers of L: \(-a + b + c = 1\), so \(b = 1\).
So \(F = k\,\eta rv\). The full calculation of the flow gives \(k = 6\pi\), which dimensions could never give.
Number: \(F = 6\pi\eta rv = 6 \times 3.14 \times 0.1 \times 10^{-3} \times 0.1 = 1.9 \times 10^{-4}\,\text{N}\), the value of the physics chapter.
The phenomenon is answered. The drag grows in proportion to the radius and to the speed, and the viscosity multiplies both. Stokes needed the full theory of the flow only for the number \(6\pi\).
The volume \(Q\) of liquid flowing each second through a narrow tube depends on three things. They are the pressure difference per unit length \(P/l\), the viscosity \(\eta\) and the radius \(r\). Dimensional analysis gives \(Q \propto (P/l)\,r^c/\eta\). What is \(c\)?
Carrying a unit from one system to another
The older CGS system measures length in centimetres, mass in grams and time in seconds. Its unit of force is the dyne and its unit of energy the erg. Dimensions convert such units in one line. A quantity is the same in both systems, so \(n_1u_1 = n_2u_2\). The unit of a quantity with dimensions \(\text{M}^a\text{L}^b\text{T}^c\) is \(M^aL^bT^c\) in the base units of its system.
For the newton, \(a = 1\), \(b = 1\) and \(c = -2\). From SI to CGS, the mass ratio is \(1\,\text{kg}/1\,\text{g} = 10^3\), the length ratio \(1\,\text{m}/1\,\text{cm} = 10^2\), and the time ratio 1. So \(1\,\text{N} = 10^3 \times 10^2 = 10^5\,\text{dyne}\). For the joule the length ratio is squared: \(1\,\text{J} = 10^3 \times (10^2)^2 = 10^7\,\text{erg}\). For viscosity, \(\text{ML}^{-1}\text{T}^{-1}\), one poise is \(10^{-3} \times 10^{2} = 0.1\,\text{Pa s}\).
The gravitational constant is \(G = 6.67 \times 10^{-11}\) in SI units, with dimensions \(\text{M}^{-1}\text{L}^3\text{T}^{-2}\). Its CGS value is \(a \times 10^{-8}\,\text{dyne cm}^2/\text{g}^2\). Enter \(a\).
Castor oil has a viscosity of about \(10\,\text{poise}\). What is this in pascal seconds?
What the method cannot do
Dimensional analysis is fast, but it has firm limits. It never gives the pure number, such as the \(2\pi\) of the pendulum or the \(6\pi\) of Stokes. It works only for a product of powers, so it cannot build a sum such as \(s = ut + \tfrac12at^2\). It cannot find a sine, an exponential or a logarithm. In mechanics it has only three equations, one each for M, L and T, so at most three unknown powers can be found.
It also cannot tell apart two quantities with the same dimensions. Work and torque are both a force times a length, \(\text{ML}^2\text{T}^{-2}\). Yet work is an energy and torque is a turning effect, as Fig 5.1 shows. The same is true of pressure and energy per unit volume.
Which of these relations could not be derived by dimensional analysis alone, even apart from a pure number?
The method trusts you to list every quantity that matters. Omit one, and the powers cannot be matched, or they are wrong. It also cannot see a dependence on a pure number. The angle of swing of a pendulum is dimensionless, so dimensions cannot say whether the period depends on it. In fact it does, a little, for swings much larger than about 15°.
In 1945 the first atomic bomb was exploded in the desert of New Mexico. Its energy was a military secret. In 1947 photographs of the growing fireball were published, each with its time and a scale of length. The British physicist G. I. Taylor argued that the radius of the fireball could depend only on the energy, the time and the density of the air. Dimensions then fix the form, \(R \propto (Et^2/\rho)^{1/5}\). From the photographs he found an energy of about 17 kilotons of TNT. The secret value was about 20 kilotons.
Dimensional analysis gives the form of Stokes' law for a sphere in a fluid. It also checks the \(r^4\) of Poiseuille's law for flow through a narrow tube. The chapter Dynamics of SHM finds the period of a pendulum and of a spring. The waves chapter finds that a wave on a string travels at a speed proportional to \(\sqrt{T/\mu}\). The CGS values of \(G\), the dyne and the poise all follow from the conversion rule.
- Expecting dimensions to give the \(2\pi\) of the pendulum.Pure numbers have no dimensions, so the method is blind to them. Experiment or full theory supplies them.
- Leaving out a quantity that matters, such as \(g\) for the pendulum.List every quantity the result could depend on before matching powers.
- Converting the joule to ergs with the length ratio only once.Energy has \(\text{L}^2\), so the ratio 100 is squared: \(1\,\text{J} = 10^3 \times 10^4 = 10^7\,\text{erg}\).
- Using dimensional analysis for a sum such as \(v^2 = u^2 + 2as\).The method builds one product of powers. A sum of terms must come from the physics itself.