Dimensional Analysis

Exercises

1. Exercise 1 Worked answer

   [de Mestre] Consider a projectile, launched with initial velocity $v_0$, at an angle $\theta$. To study its motion we may guess that these are the relevant quantities.

   quantity	dimensional formula
   horizontal position $x$	$L^1{M}^0{T}^0$
   vertical position $y$	$L^1{M}^0{T}^0$
   initial speed $v_0$	$L^1{M}^0{T}^{-1}$
   angle of launch $\theta$	$L^0{M}^0{T}^0$
   acceleration due to gravity $g$	$L^1{M}^0{T}^{-2}$
   time $t$	$L^0{M}^0{T}^1$

   1. Show that $\{gt/v_0,gx/v_0^2,gy/v_0^2,\theta \}$ is a complete set of dimensionless products. (Hint. One way to go is to find the appropriate free variables in the linear system that arises but there is a shortcut that uses the properties of a basis.)

   2. These two equations of motion for projectiles are familiar: $x=v_0\cos (\theta )t$ and $y=v_0\sin (\theta )t-(g/2)t^2$. Manipulate each to rewrite it as a relationship among the dimensionless products of the prior item.

   Back to Exercise 1

   Answer.

   1. Assuming that this

      $$(L^1{M}^0{T}^0{)}^{p_1}(L^1{M}^0{T}^0{)}^{p_2}(L^1{M}^0{T}^{-1}{)}^{p_3}(L^0{M}^0{T}^0{)}^{p_4}(L^1{M}^0{T}^{-2}{)}^{p_5}(L^0{M}^0{T}^1{)}^{p_6}$$

      equals $L^0{M}^0{T}^0$ gives rise to this linear system

      $$\begin{array}{rcrcrcrcrcr}{p}_1& +& p_2& +& p_3& +& p_5& & & =& 0\\ & & & & -p_3& -& 2p_5& +& p_6& =& 0\end{array}$$

      (there is no restriction on $p_4$). The natural parametrization uses the free variables to give $p_3=-2p_5+p_6$ and $p_1=-p_2+p_5-p_6$. The resulting description of the solution set

      $$\{\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\\ p_4\\ p_5\\ p_6\end{array}\right)=p_2\left(\begin{array}{c}-1\\ 1\\ 0\\ 0\\ 0\\ 0\end{array}\right)+p_4\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right)+p_5\left(\begin{array}{c}1\\ 0\\ -2\\ 0\\ 1\\ 0\end{array}\right)+p_6\left(\begin{array}{c}-1\\ 0\\ 1\\ 0\\ 0\\ 1\end{array}\right)\mid p_2,p_4,p_5,p_6\in \mathrm{ℝ}\}$$

      gives $\{y/x,\theta ,xt/{v_0}^2,v_{0}t/x\}$ as a complete set of dimensionless products (recall that “complete” in this context does not mean that there are no other dimensionless products; it simply means that the set is a basis). This is, however, not the set of dimensionless products that the question asks for.

      There are two ways to proceed. The first is to fiddle with the choice of parameters, hoping to hit on the right set. For that, we can do the prior paragraph in reverse. Converting the given dimensionless products $gt/v_0$, $gx/v_0^2$, $gy/v_0^2$, and $\theta$ into vectors gives this description (note the ?’s where the parameters will go).

      $$\{\left(\begin{array}{c}{p}_1\\ p_2\\ p_3\\ p_4\\ p_5\\ p_6\end{array}\right)=\underset{―}{\mathit{\text{?}}}\left(\begin{array}{c}0\\ 0\\ -1\\ 0\\ 1\\ 1\end{array}\right)+\underset{―}{\mathit{\text{?}}}\left(\begin{array}{c}1\\ 0\\ -2\\ 0\\ 1\\ 0\end{array}\right)+\underset{―}{\mathit{\text{?}}}\left(\begin{array}{c}0\\ 1\\ -2\\ 0\\ 1\\ 0\end{array}\right)+p_4\left(\begin{array}{c}0\\ 0\\ 0\\ 1\\ 0\\ 0\end{array}\right)\mid p_2,p_4,p_5,p_6\in \mathrm{ℝ}\}$$

      The $p_4$ is already in place. Examining the rows shows that we can also put in place $p_6$, $p_1$, and $p_2$.

      The second way to proceed, following the hint, is to note that the given set is of size four in a four-dimensional vector space and so we need only show that it is linearly independent. That is easily done by inspection, by considering the sixth, first, second, and fourth components of the vectors.

   2. The first equation can be rewritten

      $$\frac{gx}{{v_0}^2}=\frac{gt}{v_0}\cos \theta$$

      so that Buckingham’s function is $f_1({\mathrm{\Pi }}_1,{\mathrm{\Pi }}_2,{\mathrm{\Pi }}_3,{\mathrm{\Pi }}_4)={\mathrm{\Pi }}_2-{\mathrm{\Pi }}_1\cos ({\mathrm{\Pi }}_4)$. The second equation can be rewritten

      $$\frac{gy}{{v_0}^2}=\frac{gt}{v_0}\sin \theta -\frac{1}{2}{\left(\frac{gt}{v_0}\right)}^2$$

      and Buckingham’s function here is $f_2({\mathrm{\Pi }}_1,{\mathrm{\Pi }}_2,{\mathrm{\Pi }}_3,{\mathrm{\Pi }}_4)={\mathrm{\Pi }}_3-{\mathrm{\Pi }}_1\sin ({\mathrm{\Pi }}_4)+(1/2){{\mathrm{\Pi }}_1}^2$.

2. Exercise 2 Worked answer

   [Einstein] conjectured that the infrared characteristic frequencies of a solid might be determined by the same forces between atoms as determine the solid’s ordinary elastic behavior. The relevant quantities are these.

   quantity	dimensional formula
   characteristic frequency $\nu$	$L^0{M}^0{T}^{-1}$
   compressibility $k$	$L^1{M}^{-1}{T}^2$
   number of atoms per cubic cm $N$	$L^{-3}{M}^0{T}^0$
   mass of an atom $m$	$L^0{M}^1{T}^0$

   Show that there is one dimensionless product. Conclude that, in any complete relationship among quantities with these dimensional formulas, $k$ is a constant times ${\nu }^{-2}{N}^{-1/3}{m}^{-1}$. This conclusion played an important role in the early study of quantum phenomena.

   Back to Exercise 2

   Answer. Consider

   $$(L^0{M}^0{T}^{-1}{)}^{p_1}(L^1{M}^{-1}{T}^2{)}^{p_2}(L^{-3}{M}^0{T}^0{)}^{p_3}(L^0{M}^1{T}^0{)}^{p_4}=(L^0{M}^0{T}^0)$$

   which gives these relations among the powers.

   $$\begin{array}{rcrcrcrcr}& & p_2& -& 3p_3& & & =& 0\\ & & -p_2& & & +& p_4& =& 0\\ -p_1& +& 2p_2& & & & & =& 0\end{array}\underset{\begin{array}{}\end{array}}{\overset{{\rho }_1\leftrightarrow {\rho }_3\phantom{(}}{⟶}}\underset{\begin{array}{}\end{array}}{\overset{{\rho }_2+{\rho }_3\phantom{(}}{⟶}}\begin{array}{rcrcrcrcr}-p_1& +& 2p_2& & & & & =& 0\\ & & -p_2& & & +& p_4& =& 0\\ & & & & -3p_3& +& p_4& =& 0\end{array}$$

   This is the solution space (because we wish to express $k$ as a function of the other quantities, we take $p_2$ as the parameter).

   $$\{\left(\begin{array}{c}2\\ 1\\ 1/3\\ 1\end{array}\right)p_2\mid p_2\in \mathrm{ℝ}\}$$

   Thus, ${\mathrm{\Pi }}_1={\nu }^{2}k{N}^{1/3}m$ is the dimensionless combination, and we have that $k$ equals ${\nu }^{-2}{N}^{-1/3}{m}^{-1}$ times a constant (the function $\hat{f}$ is constant since it has no arguments).

3. Exercise 3 Worked answer

   [Giordano, Wells, Wilde] The torque produced by an engine has dimensional formula $L^2{M}^1{T}^{-2}$. We may first guess that it depends on the engine’s rotation rate (with dimensional formula $L^0{M}^0{T}^{-1}$), and the volume of air displaced (with dimensional formula $L^3{M}^0{T}^0$).

   1. Try to find a complete set of dimensionless products. What goes wrong?

   2. Adjust the guess by adding the density of the air (with dimensional formula $L^{-3}{M}^1{T}^0$). Now find a complete set of dimensionless products.

   Back to Exercise 3

   Answer.

   1. Setting

      $$(L^2{M}^1{T}^{-2}{)}^{p_1}(L^0{M}^0{T}^{-1}{)}^{p_2}(L^3{M}^0{T}^0{)}^{p_3}=(L^0{M}^0{T}^0)$$

      gives this

      $$\begin{array}{rcrcrcr}2p_1& & & +& 3p_3& =& 0\\ p_1& & & & & =& 0\\ -2p_1& -& p_2& & & =& 0\end{array}$$

      which implies that $p_1=p_2=p_3=0$. That is, among quantities with these dimensional formulas, the only dimensionless product is the trivial one.

   2. Setting

      $$(L^2{M}^1{T}^{-2}{)}^{p_1}(L^0{M}^0{T}^{-1}{)}^{p_2}(L^3{M}^0{T}^0{)}^{p_3}(L^{-3}{M}^1{T}^0{)}^{p_4}=(L^0{M}^0{T}^0)$$

      gives this.

      $$\begin{array}{c}\begin{array}{rcrcrcrcr}2p_1& & & +& 3p_3& -& 3p_4& =& 0\\ p_1& & & & & +& p_4& =& 0\\ -2p_1& -& p_2& & & & & =& 0\end{array}\hfill \\ \hfill \underset{\begin{array}{l}{\rho }_1+{\rho }_3\end{array}}{\overset{(-1/2){\rho }_1+{\rho }_2\phantom{(}}{⟶}}\underset{\begin{array}{}\end{array}}{\overset{{\rho }_2\leftrightarrow {\rho }_3\phantom{(}}{⟶}}\begin{array}{rcrcrcrcr}2p_1& & & +& 3p_3& -& 3p_4& =& 0\\ & & -p_2& +& 3p_3& -& 3p_4& =& 0\\ & & & & (-3/2)p_3& +& (5/2)p_4& =& 0\end{array}\end{array}$$

      Taking $p_1$ as parameter to express the torque gives this description of the solution set.

      $$\{\left(\begin{array}{c}1\\ -2\\ -5/3\\ -1\end{array}\right)p_1\mid p_1\in \mathrm{ℝ}\}$$

      Denoting the torque by $\tau$, the rotation rate by $r$, the volume of air by $V$, and the density of air by $d$ we have that ${\mathrm{\Pi }}_1=\tau r^{-2}{V}^{-5/3}{d}^{-1}$, and so the torque is $r^2{V}^{5/3}d$ times a constant.

4. Exercise 4 Worked answer

   [Tilley] Dominoes falling make a wave. We may conjecture that the wave speed $v$ depends on the spacing $d$ between the dominoes, the height $h$ of each domino, and the acceleration due to gravity $g$.

   1. Find the dimensional formula for each of the four quantities.

   2. Show that $\{{\mathrm{\Pi }}_1=h/d,{\mathrm{\Pi }}_2=dg/v^2\}$ is a complete set of dimensionless products.

   3. Show that if $h/d$ is fixed then the propagation speed is proportional to the square root of $d$.

   Back to Exercise 4

   Answer.

   1. These are the dimensional formulas.

      quantity	dimensional formula
      speed of the wave $v$	$L^1{M}^0{T}^{-1}$
      separation of the dominoes $d$	$L^1{M}^0{T}^0$
      height of the dominoes $h$	$L^1{M}^0{T}^0$
      acceleration due to gravity $g$	$L^1{M}^0{T}^{-2}$

   2. The relationship

      $$(L^1{M}^0{T}^{-1}{)}^{p_1}(L^1{M}^0{T}^0{)}^{p_2}(L^1{M}^0{T}^0{)}^{p_3}(L^1{M}^0{T}^{-2}{)}^{p_4}=(L^0{M}^0{T}^0)$$

      gives this linear system.

      $$\begin{array}{rcrcrcrcr}{p}_1& +& p_2& +& p_3& +& p_4& =& 0\\ & & & & & & 0& =& 0\\ -p_1& & & & & -& 2p_4& =& 0\end{array}\underset{\begin{array}{}\end{array}}{\overset{{\rho }_1+{\rho }_4\phantom{(}}{⟶}}\begin{array}{rcrcrcrcr}{p}_1& +& p_2& +& p_3& +& p_4& =& 0\\ & & p_2& +& p_3& -& p_4& =& 0\end{array}$$

      Taking $p_3$ and $p_4$ as parameters, we can describe the solution set in this way.

      $$\{\left(\begin{array}{c}0\\ -1\\ 1\\ 0\end{array}\right)p_3+\left(\begin{array}{c}-2\\ 1\\ 0\\ 1\end{array}\right)p_4\mid p_3,p_4\in \mathrm{ℝ}\}$$

      That gives $\{{\mathrm{\Pi }}_1=h/d,{\mathrm{\Pi }}_2=dg/v^2\}$ as a complete set.

   3. Buckingham’s Theorem says that $v^2=dg\cdot \hat{f}(h/d)$ and so, since $g$ is a constant, if $h/d$ is fixed then $v$ is proportional to $\sqrt{d}$.

5. Exercise 5 Worked answer

   Prove that the dimensionless products form a vector space under the $\overrightarrow{+}$ operation of multiplying two such products and the $\overrightarrow{\cdot }$ operation of raising such the product to the power of the scalar. (The vector arrows are a precaution against confusion.) That is, prove that, for any particular homogeneous system, this set of products of powers of $m_1$, …, $m_k$

   $$\{m_1^{p_1}\dots m_k^{p_k}\mid p_1\text{, }\dots \text{, }{p}_k\text{ satisfy the system}\}$$

   is a vector space under:

   $$m_1^{p_1}\dots m_k^{p_k}\overrightarrow{+}{m}_1^{q_1}\dots m_k^{q_k}=m_1^{p_1+q_1}\dots m_k^{p_k+q_k}$$

   and

   $$r\overrightarrow{\cdot }(m_1^{p_1}\dots m_k^{p_k})=m_1^{r{p}_1}\dots m_k^{r{p}_k}$$

   (assume that all variables represent real numbers).

   Back to Exercise 5

   Answer. Checking the conditions in the definition of a vector space is routine.

6. Exercise 6 Worked answer

   The advice about apples and oranges is not right. Consider the familiar equations for a circle $C=2\pi r$ and $A=\pi r^2$.

   1. Check that $C$ and $A$ have different dimensional formulas.

   2. Produce an equation that is not dimensionally homogeneous (i.e., it adds apples and oranges) but is nonetheless true of any circle.

   3. The prior item asks for an equation that is complete but not dimensionally homogeneous. Produce an equation that is dimensionally homogeneous but not complete.

   (Just because the old saying isn’t strictly right doesn’t keep it from being a useful strategy. Dimensional homogeneity is often used to check the plausibility of equations used in models. For an argument that any complete equation can easily be made dimensionally homogeneous, see [Bridgman], Chapter I, especially page 15.)

   Back to Exercise 6

   Answer.

   1. The dimensional formula of the circumference is $L$, that is, $L^1{M}^0{T}^0$. The dimensional formula of the area is $L^2$.

   2. One is $C+A=2\pi r+\pi r^2$.

   3. One example is this formula relating the the length of arc subtended by an angle to the radius and the angle measure in radians: $\ell -r\theta =0$. Both terms in that formula have dimensional formula $L^1$. The relationship holds for some unit systems (inches and radians, for instance) but not for all unit systems (inches and degrees, for instance).