Chapter 1 Introduction to dimensional analysis

To motivate the discussion, consider the following way to approach some questions.

1. 1.

   Let us start with a simple one – vibrations of a stretched string.

   Question 1.1.

   A guitar string has a mass of 10 g and length 0.25 m. When a tension of 100 N is applied to stretch it, what is the fundamental frequency of its natural vibrations?

   Answer 1.1.

   The units of the answer we seek, the frequency of vibrations ($f$), is Hz, i.e. 1/s. Noting that 1 N = 1 kg m/s², there is only one way to combine the units of tension, mass and length to get the units of 1/s, viz.

   $f=C\sqrt{{\displaystyle \frac{100\text{ kg m/s}\text{<sup class="ltx\_sup">2</sup>}}{0.25\text{ m}\times 0.01\text{ kg}}}}=200C{\displaystyle \frac{1}{\text{s}}},$

   (1.1)

   where $C$ is a constant that does not have any units. This constant $C$ cannot be predicted in this method, and so this approach appears to be futile. But a moment’s thought reveals that the method produces another result. It produces the manner in which the independent parameters (the tension, the mass and length of the string) must appear in this calculation for frequency. So, if the tension were $\tau$, the string mass and length were $m$ and $l$, respectively, then the frequency of its natural vibration would be related to it through the following relation:

   $f=C\sqrt{{\displaystyle \frac{\tau }{ml}}}.$

   (1.2)

   In this case the exact answer is

   $f={\displaystyle \frac{1}{2}}\sqrt{{\displaystyle \frac{\tau }{ml}}},$

   (1.3)

   so $C$ is the constant $1/2$.

   (Did you notice that we presumed that the frequency depends on and only on the tension, the string mass and length?)

2. 2.

   Consider astronomy for the next example.

   Question 1.2.

   An extrasolar planet is observed to orbit its host star in ${10}^6$ s. The star has a mass of ${10}^{30}$ kg. Estimate the distance of the planet from the star. The gravitational force of attraction is quantified by the gravitational constant $G=6.67\times {10}^{-11}$ m³/(kg s²). Assume a circular orbit.

   Answer 1.2.

   We know that the relation between the period of orbit and the radius of orbit does not involve the mass of the planet itself, but depends on the mass of the host star. We wish to determine the radius, $R$, which has SI units of m. There is only one combination of the period, $T$ = 10⁶ s, the stellar mass, $M$ = ${10}^{30}$ kg, and $G$,

   $R=C{\left(6.67\times {10}^{-11}{\displaystyle \frac{{\text{m}}^3}{\text{kg s}\text{<sup class="ltx\_sup">2</sup>}}}\times {10}^{30}\text{ kg}\times {({10}^6\text{ s})}^2\right)}^{1/3}=4.05C\times {10}^{10}\text{ m},$

   (1.4)

   where $C$ is a constant without units. Again, without knowing $C$, we cannot evaluate $R$, but we seem to have discovered how this radius depends on the stellar mass, the gravitational constant and the orbital period. Indeed,

   $R=C\times {(GM{T}^2)}^{1/3}.$

   (1.5)

   Note that we arrived at this expression without any knowledge of orbital mechanics. Using orbital mechanics, one can determine $C=1/{(4{\pi }^2)}^{1/3}$.

3. 3.

   Let us now consider an example from astronomy for which we may not know how to compute the answer.

   Question 1.3.

   A black hole has such a strong gravitational pull that even light cannot escape it’s vicinity. The edge of this region is called the event horizon. Estimate the radius of the event horizon for a super-massive black hole of mass $M={10}^{39}$ kg.

   Answer 1.3.

   The gravitational force is characterized by the mass of the black hole and the gravitational constant $G=6.67\times {10}^{-11}$ m³/(kg s²). Light is characterized by its speed, $c=3\times {10}^8$ m/s. The event horizon radius, $R$, has units of $m$, which can be constructed only by the combination

   $R=C{\displaystyle \frac{6.67\times {10}^{-11}\text{ m}\text{<sup class="ltx\_sup">3</sup>}\text{/(kg s}\text{<sup class="ltx\_sup">2</sup>}\text{) }\times {10}^{39}\text{ kg}}{{(3\times {10}^8\text{ m/s})}^2}}=7.41\times {10}^{11}C\text{ m}.$

   (1.6)

   Again, $C$ is a constant, which is indeterminate without the application of the theory of general relativity. Yet, for any black hole, we can deduce that the event horizon radius is given by

   $R=C{\displaystyle \frac{GM}{c^2}}.$

   (1.7)

   This equation implies that a black hole of twice the mass has twice as large an event horizon. And we deduced that without using general relativity.

4. 4.

   Let us now look at an example, from the realm of quantum mechanics, where we may not know another way of obtaining the answer.

   Question 1.4.

   The energy of an electron orbiting a proton in a hydrogen atom consists of its electrostatic part and its kinetic part. The electrostatic energy depends on the charge of the electron, $e=1.6\times {10}^{-19}$ C, the permittivity of free space, ${ϵ}_0=8.85\times {10}^{-12}$ C² s²/ (kg m³), the electronic mass $m=9.1\times {10}^{-31}$ kg, and the distance between the distance between the electron and the proton. Quantum mechanics says that the energy is quantized, i.e. it can take on only discrete values. It further introduces uncertainty about where the electron’s position. This uncertainty is quantified by the Planck constant, $h=6.62\times {10}^{-34}$ kg m²/s. Estimate the electrostatic energy of the electron in its ground state, i.e. lowest discrete energy level.

   Answer 1.4.

   The units of energy are kg m²/s². Let us presume that the electrostatic energy of the electron in its ground state depends on $e$, ${ϵ}_0$, $h$ and $m$. It is a little tricky to see but there is only one way to combine these parameters to make the units of energy, but here it is.

   $E=C\times {\displaystyle \frac{(9.1\times {10}^{-31}\text{ kg}){(1.6\times {10}^{-19}\text{ C})}^4}{{(6.62\times {10}^{-34}\text{ kg m}\text{<sup class="ltx\_sup">2</sup>}\text{/s})}^2{(8.85\times {10}^{-12}\text{ C}\text{<sup class="ltx\_sup">2</sup>}\text{ s}\text{<sup class="ltx\_sup">2</sup>}\text{/(kg m}\text{<sup class="ltx\_sup">3</sup>}\text{)})}^2}}=1.73\times {10}^{-17}C{\displaystyle \frac{\text{ kg m}\text{<sup class="ltx\_sup">2</sup>}}{\text{ s}\text{<sup class="ltx\_sup">2</sup>}}}.$

   (1.8)

   Here, again, $C$ is a constant without units, which cannot be determined without application of quantum mechanics. Regardless, the ground state electrostatic energy of the electron depends on the properties of the hydrogen atom as

   $E=C{\displaystyle \frac{m{e}^4}{h^2{ϵ}_0^2}}.$

   (1.9)

All we used in these examples is that the units match and without using any knowledge of the underlying physics. Indeed, general relativity and quantum mechanics are very advanced topics outside the scope of this module, but they all must respect units. The numerical constant $C$ cannot be determined simply by using the units of quantities, but do not let that disappoint you. For example, a single experimental measurement of the frequency of a vibrating string under tension can yield the $C$ for Question 1.1.

Here is another example.

Question 1.5 demonstrates that the number of parameters required to represent a phenomenon can be reduced by judicious re-organization of the parameters.

These examples showed a few application in an ad hoc manner of the principle that units must match, but the principle is so much more powerful. A systematic study of the consequences of this principle is called dimensional analysis. The primary application of dimensional analysis lies in reducing the number of parameters required to describe a phenomenon, as shown in Question 1.5. Fewer experiments need to be performed to span the parameter space that is reduced using dimensional analysis. When a scale model of a wing, a car, an aeroplane or a rocket is tested in a wind tunnel, it is dimensional analysis at work behind the scenes.

Here we develop this subject in the following order:

1. 1.

   Units,

   1. (a)

      SI Units,

   2. (b)

      Scales and systems of units,

   3. (c)

      Conversion between units,

2. 2.

   Dimensions,

   1. (a)

      Definition of base and derived dimensions,

   2. (b)

      Dimensional consistency,

   3. (c)

      Dimensionless parameters,

3. 3.

   Dimensionless forms of equations

   1. (a)

      Physical relations between dimensionless parameters,

   2. (b)

      Buckingham’s Pi theorem,

4. 4.

   Applications,

   1. (a)

      Dimensionless representation of experimental data, and

   2. (b)

      Scale models.

The principle of matching of units is a profound one, which can be exploited to our advantage if treated systematically. The aims of the course are as follows (from Dr John P Longley’s Lecture Notes)
:

• •

  To introduce and illustrate the use of dimensional analysis.

• •

  To develop an understanding of the principles of dimensional consistency.

• •

  To develop techniques required to form non-dimensional quantities and relationships.

• •

  To explain how dimensional analysis can be used to simplify problems by reducing the number of parameters.

• •

  To correlate experimental data and to assist in the design of models for testing using dimensional principles.