Linear Algebra/Topic: Dimensional Analysis

From Wikibooks, the open-content textbooks collection

There are no reviewed revisions of this page, so it may not have been checked for quality.

``You can't add apples and oranges, the old saying goes. It reflects our experience that in applications the quantities have units and keeping track of those units is worthwhile. Everyone has done calculations such as this one that use the units as a check. $60\,\frac{\text{sec}}{\text{min}} \cdot 60\,\frac{\text{min}}{\text{hr}} \cdot 24\,\frac{\text{hr}}{\text{day}} \cdot 365\,\frac{\text{day}}{\text{year}} = 31\,536\,000\,\frac{\text{sec}}{\text{year}}$ However, the idea of including the units can be taken beyond bookkeeping. It can be used to draw conclusions about what relationships are possible among the physical quantities.

To start, consider the physics equation: $\text{distance}=16\cdot(\text{time})^2$ . If the distance is in feet and the time is in seconds then this is a true statement about falling bodies. However it is not correct in other unit systems; for instance, it is not correct in the meter-second system. We can fix that by making the 16 a dimensional constant. $\text{dist}=16\,\frac{\text{ft}}{\text{sec}^2}\cdot (\text{time})^2$ For instance, the above equation holds in the yard-second system. $\text{distance in yards}=16\,\frac{(1/3)\,\text{yd}}{\text{sec}^2} \cdot (\text{time in sec})^2 =\frac{16}{3}\,\frac{\text{yd}}{\text{sec}^2} \cdot (\text{time in sec})^2$ So our first point is that by ``including the units we mean that we are restricting our attention to equations that use dimensional constants.

By using dimensional constants, we can be vague about units and say only that all quantities are measured in combinations of some units of length L, mass M, and time T. We shall refer to these three as dimensions^[Index] (these are the only three dimensions that we shall need in this Topic). For instance, velocity could be measured in \text{feet}/\text{second} or \text{fathoms}/\text{hour}, but in all events it involves some unit of length divided by some unit of time so the dimensional formula of velocity is L/T. Similarly, the dimensional formula of density is M/L^3. We shall prefer using negative exponents over the fraction bars and we shall include the dimensions with a zero exponent, that is, we shall write the dimensional formula of velocity as L^1M^0T^{-1} and that of density as L^{-3}M^1T^0.

In this context, ``You can't add apples to oranges becomes the advice to check that all of an equation's terms have the same dimensional formula. An example is this version of the falling body equation:~d-gt^2=0. The dimensional formula of the d term is L^1M^0T^0. For the other term, the dimensional formula of g is L^1M^0T^{-2} (g is the dimensional constant given above as 16\,\text{ft}/\text{sec}^2) and the dimensional formula of t is L^0M^0T^1, so that of the entire gt^2 term is L^1M^0T^{-2}(L^0M^0T^1)^2=L^1M^0T^0. Thus the two terms have the same dimensional formula. An equation with this property is dimensionally homogeneous.

Quantities with dimensional formula L^0M^0T^0 are dimensionless. For example, we measure an angle by taking the ratio of the subtended arc to the radius

Image:Linear Algebra Book2 ch2.22

which is the ratio of a length to a length L^1M^0T^0/L^1M^0T^0 and thus angles have the dimensional formula L^0M^0T^0.

The classic example of using the units for more than bookkeeping, using them to draw conclusions, considers the formula for the period of a pendulum. $p = --some expression involving the length of the string, etc.--$ The period is in units of time L^0M^0T^1. So the quantities on the other side of the equation must have dimensional formulas that combine in such a way that their L's and M's cancel and only a single T remains. The table on page~\pageref{table:Dimen} has the quantities that an experienced investigator would consider possibly relevant. The only dimensional formulas involving L are for the length of the string and the acceleration due to gravity. For the L's of these two to cancel, when they appear in the equation they must be in ratio, e.g., as (\ell/g)^2, or as \cos(\ell/g), or as (\ell/g)^{-1}. Therefore the period is a function of \ell/g.

This is a remarkable result:~with a pencil and paper analysis, before we ever took out the pendulum and made measurements, we have determined something about the relationship among the quantities.

To do dimensional analysis systematically, we need to know two things (arguments for these are in ^[1], Chapter~II and~IV). The first is that each equation relating physical quantities that we shall see involves a sum of terms, where each term has the form $m_1^{p_1}m_2^{p_2}\cdots m_k^{p_k}$ for numbers m_1, \ldots, m_k that measure the quantities.

For the second, observe that an easy way to construct a dimensionally homogeneous expression is by taking a product of dimensionless quantities or by adding such dimensionless terms. Buckingham's Theorem states that any complete relationship among quantities with dimensional formulas can be algebraically manipulated into a form where there is some function f such that $f(\Pi_1,\ldots,\Pi_n)=0$ for a complete set {\Pi_1,\ldots,\Pi_n} of dimensionless products. (The first example below describes what makes a set of dimensionless products `complete'.) We usually want to express one of the quantities, m_1 for instance, in terms of the others, and for that we will assume that the above equality can be rewritten $m_1=m_2^{-p_2}\cdots m_k^{-p_k}\cdot \hat{f}(\Pi_2,\ldots,\Pi_n)$ where \Pi_1=m_1m_2^{p_2}\cdots m_k^{p_k} is dimensionless and the products \Pi_2, \ldots, \Pi_n don't involve m_1 (as with f, here \hat{f} is just some function, this time of n-1 arguments). Thus, to do dimensional analysis we should find which dimensionless products are possible.

For example, consider again the formula for a pendulum's period.

$\begin{matrix} [[Image:Linear Algebra Book2 ch2.23]] \end{matrix}$

Failed to parse (unknown function\multicolumn): \begin{matrix} \multicolumn{1}{r}{''quantity''} &\multicolumn{1}{l}{ <math>l} ''dimensional'' \\ ''formula'' \end{matrix} } \\ \hline period p &L^0M^0T^1 \\ length of string \ell &L^1M^0T^0 \\ mass of bob m &L^0M^1T^0 \\ acceleration due to gravity g &L^1M^0T^{-2} \\ arc of swing \theta &L^0M^0T^0 \end{matrix}</math>

By the first fact cited above, we expect the formula to have (possibly sums of terms of) the form p^{p_1}\ell^{p_2}m^{p_3}g^{p_4}\theta^{p_5}. To use the second fact, to find which combinations of the powers~p_1, \ldots, p_5 yield dimensionless products, consider this equation. $(L^0M^0T^1)^{p_1}(L^1M^0T^0)^{p_2}(L^0M^1T^0)^{p_3} (L^1M^0T^{-2})^{p_4}(L^0M^0T^0)^{p_5} =L^0M^0T^0$ It gives three conditions on the powers. $<math>\begin{bmatrix} &\ &p_2 &\ & &+ &p_4 & & = &0 \\ & & & &p_3 & & & & = &0 \\ p_1& & & & &- &2p_4 & & = &0 \end{bmatrix}$ </math> Note that p_3 is 0 and so the mass of the bob does not affect the period. Gaussian reduction and parametrization of that system gives this ${\begin{bmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \end{bmatrix}= \begin{bmatrix} 1 \\ -1/2 \\ 0 \\ 1/2 \\ 0 \end{bmatrix}p_1+ \begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}p_5 o p_1,p_5\in\Re}$ (we've taken p_1 as one of the parameters in order to express the period in terms of the other quantities).

Here is the linear algebra. The set of dimensionless products contains all terms p^{p_1}\ell^{p_2}m^{p_3}a^{p_4}\theta^{p_5} subject to the conditions above. This set forms a vector space under the `+' operation of multiplying two such products and the `\cdot' operation of raising such a product to the power of the scalar (see \nearbyexercise{exer:DimLessProdsVecSp}). The term `complete set of dimensionless products' in Buckingham's Theorem means a basis for this vector space.

We can get a basis by first taking p_1=1, p_5=0 and then p_1=0, p_5=1. The associated dimensionless products are \Pi_1=p\ell^{-1/2}g^{1/2} and \Pi_2=\theta. Because the set {\Pi_1,\Pi_2} is complete, Buckingham's Theorem says that $p = \ell^{1/2}g^{-1/2}\cdot\hat{f}(\theta) = \sqrt{\ell/g}\cdot\hat{f}(\theta)$ where \hat{f} is a function that we cannot determine from this analysis (a first year physics text will show by other means that for small angles it is approximately the constant function \hat{f}(\theta)=2\pi).

Thus, analysis of the relationships that are possible between the quantities with the given dimensional formulas has produced a fair amount of information:~a pendulum's period does not depend on the mass of the bob, and it rises with the square root of the length of the string.

For the next example we try to determine the period of revolution of two bodies in space orbiting each other under mutual gravitational attraction. An experienced investigator could expect that these are the relevant quantities.

Failed to parse (lexing error): \begin{matrix} c@{}} [[Image:Linear Algebra Book2 ch2.24]] \end{matrix}

Failed to parse (unknown function\multicolumn): \begin{matrix} \multicolumn{1}{r}{''quantity''} &\multicolumn{1}{l}{ <math>l@{}} ''dimensional'' \\ ''formula'' \end{matrix} } \\ \hline period p &L^0M^0T^1 \\ mean separation r &L^1M^0T^0 \\ first mass m_1 &L^0M^1T^0 \\ second mass m_2 &L^0M^1T^0 \\ grav.\ constant G &L^3M^{-1}T^{-2} \end{matrix}</math>

To get the complete set of dimensionless products we consider the equation $(L^0M^0T^1)^{p_1}(L^1M^0T^0)^{p_2}(L^0M^1T^0)^{p_3}(L^0M^1T^0)^{p_4} (L^3M^{-1}T^{-2})^{p_5}=L^0M^0T^0$ which results in a system $<math>\begin{bmatrix} & &p_2 & & & & &+ &3p_5 = &0 \\ & & & &p_3 &+ &p_4 &- &p_5 = &0 \\ p_1 & & & & & & &- &2p_5 = &0 \end{bmatrix}$ </math> with this solution. ${\begin{bmatrix} 1 \\ -3/2 \\ 1/2 \\ 0 \\ 1/2 \end{bmatrix}p_1 +\begin{bmatrix} 0 \\ 0 \\ -1 \\ 1 \\ 0 \end{bmatrix}p_4 o p_1,p_4\in\Re}$

As earlier, the linear algebra here is that the set of dimensionless products of these quantities forms a vector space, and we want to produce a basis for that space, a `complete' set of dimensionless products. One such set, gotten from setting p_1=1 and p_4=0, and also setting p_1=0 and p_4=1 is {\Pi_1=pr^{-3/2}m_1^{1/2}G^{1/2},\,\Pi_2=m_1^{-1}m_2}. With that, Buckingham's Theorem says that any complete relationship among these quantities is stateable this form. $p = r^{3/2}m_1^{-1/2}G^{-1/2}\cdot\hat{f}(m_1^{-1}m_2) = \frac{r^{3/2}}{\sqrt{Gm_1}}\cdot\hat{f}(m_2/m_1)$

Remark. An important application of the prior formula is when m_1 is the mass of the sun and m_2 is the mass of a planet. Because m_1 is very much greater than m_2, the argument to \hat{f} is approximately 0, and we can wonder whether this part of the formula remains approximately constant as m_2 varies. One way to see that it does is this. The sun is so much larger than the planet that the mutual rotation is approximately about the sun's center. If we vary the planet's mass m_2 by a factor of x (e.g., Venus's mass is x=0.815 times Earth's mass), then the force of attraction is multiplied by x, and x times the force acting on x times the mass gives, since F=ma, the same acceleration, about the same center (approximately). Hence, the orbit will be the same and so its period will be the same, and thus the right side of the above equation also remains unchanged (approximately). Therefore, \hat{f}(m_2/m_1) is approximately constant as m_2 varies. This is Kepler's Third Law:~the square of the period of a planet is proportional to the cube of the mean radius of its orbit about the sun.

The final example was one of the first explicit applications of dimensional analysis. Lord Raleigh considered the speed of a wave in deep water and suggested these as the relevant quantities.

Failed to parse (unknown function\multicolumn): \begin{matrix} \multicolumn{1}{r}{''quantity''} &\multicolumn{1}{l}{ <math>l} ''dimensional'' \\ ''formula'' \end{matrix} } \\ \hline velocity of the wave v &L^1M^0T^{-1} \\ density of the water d &L^{-3}M^1T^0 \\ acceleration due to gravity g &L^1M^0T^{-2} \\ wavelength \lambda &L^1M^0T^0 \end{matrix}</math>

The equation $(L^1M^0T^{-1})^{p_1}(L^{-3}M^1T^0)^{p_2} (L^1M^0T^{-2})^{p_3}(L^1M^0T^0)^{p_4}=L^0M^0T^0$ gives this system $<math>\begin{bmatrix} p_1 &- &3p_2 &+ &p_3 &+ &p_4 = &0 \\ & &p_2 & & & & = &0 \\ -p_1 & & &- &2p_3 & & = &0 \end{bmatrix}$ </math> with this solution space ${\begin{bmatrix} 1 \\ 0 \\ -1/2 \\ -1/2 \end{bmatrix}p_1 o p_1\in \Re}$ (as in the pendulum example, one of the quantities d turns out not to be involved in the relationship). There is one dimensionless product, \Pi_1=vg^{-1/2}\lambda^{-1/2}, and so v is \sqrt{\lambda g} times a constant (\hat{f} is constant since it is a function of no arguments).

As the three examples above show, dimensional analysis can bring us far toward expressing the relationship among the quantities. For further reading, the classic reference is ^[2]---this brief book is delightful. Another source is ^[3]. A description of dimensional analysis's place in modeling is in ^[4]. }}

[edit] Exercises

}}

[edit] Exercise 1

Problem

Consider a projectile, launched with initial velocity v_0, at an angle \theta. An investigation of this motion might start with the guess that these are the relevant quantities. ^[5]

Failed to parse (unknown function\multicolumn): \begin{matrix} \multicolumn{1}{r}{''quantity''} &\multicolumn{1}{l}{ <math>l} ''dimensional'' \\ ''formula'' \end{matrix} } \\ \hline horizontal position x &L^1M^0T^0 \\ vertical position y &L^1M^0T^0 \\ initial speed v_0 &L^1M^0T^{-1} \\ angle of launch \theta &L^0M^0T^0 \\ acceleration due to gravity g &L^1M^0T^{-2} \\ time t &L^0M^0T^1 \end{matrix}</math>

Show that {gt/v_0,gx/v_0^2,gy/v_0^2,\theta}

is a complete set of dimensionless products. (Hint. This can be done by finding the appropriate free variables in the linear system that arises, but there is a shortcut that uses the properties of a basis.)

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.

Answer

This relationship

$(L^1M^0T^0)^{p_1}(L^1M^0T^0)^{p_2}(L^1M^0T^{-1})^{p_3} (L^0M^0T^0)^{p_4}(L^1M^0T^{-2})^{p_5}(L^0M^0T^1)^{p_6}=L^0M^0T^0$ gives rise to this linear system Failed to parse (PNG conversion failed; check for correct installation of latex, dvips, gs, and convert): <math>\begin{bmatrix} p_1 &+ &p_2 &+ &p_3 &\ &\ &+ &p_5 & & = &0 \\ & & & & & & & & & &0 = &0 \\ & & & &-p_3 & & &- &2p_5 &+ &p_6 = &0 \end{bmatrix}

</math> (note that 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 ${\begin{bmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \\ p_6 \end{bmatrix} =p_2\begin{bmatrix} -1 \\ 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix} +p_4\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} +p_5\begin{bmatrix} 1 \\ 0 \\ -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} +p_6\begin{bmatrix} -1 \\ 0 \\ 1 \\ 0 \\ 0 \\ 1 \end{bmatrix} o p_2, p_4, p_5, p_6\in\Re }$ gives {y/x,\theta,xt/{v_0}^2,v_0t/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). Failed to parse (lexing error): {\begin{bmatrix} p_1 \\ p_2 \\ p_3 \\ p_4 \\ p_5 \\ p_6 \end{bmatrix} =\underline{\ \text{''?''}\ } \begin{bmatrix} 0 \\ 0 \\ -1 \\ 0 \\ 1 \\ 1 \end{bmatrix} +\underline{\ \text{''?''}\ } \begin{bmatrix} 1 \\ 0 \\ -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} +\underline{\ \text{''?''}\ } \begin{bmatrix} 0 \\ 1 \\ -2 \\ 0 \\ 1 \\ 0 \end{bmatrix} +p_4\begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \end{bmatrix} o p_2, p_4, p_5, p_6\in\Re }

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.

[edit] Exercise 2

Problem: 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(\Pi_1,\Pi_2,\Pi_3,\Pi_4)=\Pi_2-\Pi_1\cos(\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(\Pi_1,\Pi_2,\Pi_3,\Pi_4)

=\Pi_3-\Pi_1\sin(\Pi_4)+(1/2){\Pi_1}^2.

[edit] Exercise 3

Problem: ^[6]

conjectured that the infrared characteristic frequencies of a solid may be determined by the same forces between atoms as determine the solid's ordanary elastic behavior. The relevant quantities are

Failed to parse (unknown function\multicolumn): \begin{matrix} \multicolumn{1}{r}{''quantity''} &\multicolumn{1}{l}{ <math>l} ''dimensional'' \\ ''formula'' \end{matrix} } \\ \hline characteristic frequency \nu &L^0M^0T^{-1} \\ compressibility k &L^1M^{-1}T^2 \\ number of atoms per cubic cm N &L^{-3}M^0T^0 \\ mass of an atom m &L^0M^1T^0 \end{matrix}</math>

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.

Answer

We consider $(L^0M^0T^{-1})^{p_1}(L^1M^{-1}T^2)^{p_2} (L^{-3}M^0T^0)^{p_3}(L^0M^1T^0)^{p_4}=(L^0M^0T^0)$ which gives these relations among the powers. $<math>\begin{bmatrix} & &p_2 &- &3p_3 & & = &0 \\ & &-p_2 & & &+ &p_4 = &0 \\ -p_1 &+ &2p_2 & & & & = &0 \end{bmatrix}$ \;\grstep{\rho_1\leftrightarrow\rho_3} \;\grstep{\rho_2+\rho_3}\; $\begin{bmatrix} -p_1 &+ &2p_2 & & & & = &0 \\ & &-p_2 & & &+ &p_4 = &0 \\ & & &\ &-3p_3 &+ &p_4 = &0 \end{bmatrix}$ </math> This is the solution space (because we wish to express k as a function of the other quantities, p_2 is taken as the parameter). ${\begin{bmatrix} 2 \\ 1 \\ 1/3 \\ 1 \end{bmatrix}p_2 o p_2\in\Re}$ Thus, \Pi_1=\nu^2kN^{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).

[edit] Exercise 4

Problem

The torque produced by an engine has dimensional formula L^2M^1T^{-2}. We may first guess that it depends on the engine's rotation rate (with dimensional formula L^0M^0T^{-1}), and the volume of air displaced (with dimensional formula L^3M^0T^0). ^[7]

Try to find a complete set of dimensionless products.

What goes wrong?

Adjust the guess by adding the density of the air

(with dimensional formula L^{-3}M^1T^0). Now find a complete set of dimensionless products.

Answer

Setting

$(L^2M^1T^{-2})^{p_1}(L^0M^0T^{-1})^{p_2}(L^{3}M^0T^0)^{p_3} =(L^0M^0T^0)$ gives this $<math>\begin{bmatrix} 2p_1 & & &+ &3p_3 = &0 \\ p_1 & & & & = &0 \\ -2p_1 &- &p_2 & & = &0 \end{bmatrix}$ </math> 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.

Setting

$(L^2M^1T^{-2})^{p_1}(L^0M^0T^{-1})^{p_2}(L^{3}M^0T^0)^{p_3} (L^{-3}M^1T^0)^{p_4}=(L^0M^0T^0)$ gives this. $<math>\begin{bmatrix} 2p_1 & & &+ &3p_3 &- &3p_4 = &0 \\ p_1 & & & & &+ &p_4 = &0 \\ -2p_1 &- &p_2 & & & & = &0 \end{bmatrix}$ \;\grstep[\rho_1+\rho_3]{(-1/2)\rho_1+\rho_2} \;\grstep{\rho_2\leftrightarrow\rho_3} $\begin{bmatrix} 2p_1 & & &+ &3p_3 &- &3p_4 = &0 \\ & &-p_2 &+ &3p_3 &- &3p_4 = &0 \\ &\ & & &(-3/2)p_3 &+ &(5/2)p_4 = &0 \end{bmatrix}$ </math> Taking p_1 as parameter to express the torque gives this description of the solution set. ${\begin{bmatrix} 1 \\ -2 \\ -5/3 \\ -1 \end{bmatrix}p_1 o p_1\in\Re}$ 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 \Pi_1=\tau r^{-2} V^{-5/3} d^{-1}, and so

the torque is r^2V^{5/3}d times a constant.

[edit] Exercise 5

Problem

Dominoes falling make a wave. We may conjecture that the wave speed v depends on the the spacing d between the dominoes, the height h of each domino, and the acceleration due to gravity g. ^[8]

Find the dimensional formula for each of the four quantities.
Show that {\Pi_1=h/d,\Pi_2=dg/v^2}

is a complete set of dimensionless products.

Show that if h/d is fixed then

the propagation speed is proportional to the square root of d.

Answer

These are the dimensional formulas.

Failed to parse (unknown function\multicolumn): \begin{matrix} \multicolumn{1}{r}{''quantity''} &\multicolumn{1}{l}{ <math>l} ''dimensional'' \\ ''formula'' \end{matrix} } \\ \hline speed of the wave v &L^1M^0T^{-1} \\ separation of the dominoes d &L^1M^0T^0 \\ height of the dominoes h &L^1M^0T^0 \\ acceleration due to gravity g &L^1M^0T^{-2} \end{matrix}</math>

The relationship

$(L^1M^0T^{-1})^{p_1}(L^1M^0T^0)^{p_2}(L^1M^0T^0)^{p_3} (L^1M^0T^{-2})^{p_4}=(L^0M^0T^0)$ gives this linear system. $<math>\begin{bmatrix} p_1 &+ &p_2 &+ &p_3 &+ &p_4 = &0 \\ & & & & & &0 = &0 \\ -p_1 & & & & &- &2p_4 = &0 \end{bmatrix}$ \;\grstep{\rho_1+\rho_4}\; $\begin{bmatrix} p_1 &+ &p_2 &+ &p_3 &+ &p_4 = &0 \\ & &p_2 &+ &p_3 &- &p_4 = &0 \end{bmatrix}$ </math> Taking p_3 and p_4 as parameters, the solution set is described in this way. ${\begin{bmatrix} 0 \\ -1 \\ 1 \\ 0 \end{bmatrix}p_3 +\begin{bmatrix} -2 \\ 1 \\ 0 \\ 1 \end{bmatrix}p_4 o p_3,p_4\in\Re}$ That gives {\Pi_1=h/d,\Pi_2=dg/v^2} as a complete set.

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\,}.

[edit] Exercise 6

Problem

Prove that the dimensionless products form a vector space under the \vec{+} operation of multiplying two such products and the \vec{\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, \ldots, m_k Failed to parse (lexing error): {m_1^{p_1}\dots m_k^{p_k} o \text{p_1, \ldots, p_k satisfy the system}}

is a vector space under: $m_1^{p_1}\dots m_k^{p_k} \vec{+} m_1^{q_1}\dots m_k^{q_k} = m_1^{p_1+q_1}\dots m_k^{p_k+q_k}$ and $r\vec{\cdot} (m_1^{p_1}\dots m_k^{p_k}) = m_1^{rp_1}\dots m_k^{rp_k}$ (assume that all variables represent real numbers).

Answer

Checking the conditions in the definition of a vector space is

routine.

[edit] Exercise 7

Problem

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.

Check that C and A have different dimensional formulas.
Produce an equation that is not

dimensionally homogeneous (i.e., it adds apples and oranges) but is nonetheless true of any circle.

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 as a check on the plausibility of equations used in models. For an argument that any complete equation can easily be made dimensionally homogeneous, see ^[9], Chapter~I, especially page~15.)

Answer

The dimensional formula of the circumference is L,

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

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

[edit] References

↑ Bridgman (Linear Algebra/Book2/References)
↑ Bridgman (Linear Algebra/Book2/References)
↑ Giordano83 (Linear Algebra/Book2/References)
↑ Giordano82 (Linear Algebra/Book2/References)
↑ deMestre (Linear Algebra/Book2/References)
↑ Einstein1911 (Linear Algebra/Book2/References)
↑ Giordano83 (Linear Algebra/Book2/References)
↑ Tilley (Linear Algebra/Book2/References)
↑ Bridgman (Linear Algebra/Book2/References)

[0] Bridgman (Linear Algebra/Book2/References)

[1] Bridgman (Linear Algebra/Book2/References)

[2] Giordano83 (Linear Algebra/Book2/References)

[3] Giordano82 (Linear Algebra/Book2/References)

[4] Mestre (Linear Algebra/Book2/References)

[5] Einstein1911 (Linear Algebra/Book2/References)

[6] Giordano83 (Linear Algebra/Book2/References)

[7] Tilley (Linear Algebra/Book2/References)

[8] Bridgman (Linear Algebra/Book2/References)

[1]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

[9]

Linear Algebra/Topic: Dimensional Analysis

From Wikibooks, the open-content textbooks collection

Contents

[edit] Exercises

[edit] Exercise 1

[edit] Exercise 2

[edit] Exercise 3

[edit] Exercise 4

[edit] Exercise 5

[edit] Exercise 6

[edit] Exercise 7

[edit] References

Views

Personal tools

Navigation

Community

Create a book

Search

Toolbox