Lesson 12: Dimension Analysis

Objectives

Describe the difference between quantities, dimensions, and units.
Distinguish between fundamental and derived dimensions.
Apply rules of dimension arithmetic to find the dimension of an operation, or describe why the operation is invalid.
Given a model’s equation and the dimensions of certain quantities, use dimension algebra to solve for the dimension of another quantity.

1. Quantities, Dimensions, and Units

Objective. Describe the difference between quantities, dimensions, and units.

Dimension reference.

Common quantities.

Quantity	Dimension
3 minutes	$T$
14.2 kg	$M$
$0.78	$B$
44 mm	$L$
1.2 ft $^{3}$	$L^{3}$
$98.6^{\circ}$ F	$\Theta$

Dimension symbols.

Dimension	Symbol	Example units
Length	$L$	foot, mile, meter, kilometer
Mass	$M$	gram, pound, ton, stone
Time	$T$	second, hour, day, century
Temperature	$\Theta$	kelvin, degree F, degree C
Money	$B$	dollar, euro, yen

Population is a dimension with the symbol $P$ (people, dogs, mice). Other dimensions are amount of substance $N$ (mole), electric current $I$ (ampere), luminous intensity $J$ (candela), and angle $A$ (degree, radian).

Definition 1 (Quantity).

A quantity is the numerical value of a measurement together with its units, if it has any.

For example, 80 kilograms, 3 minutes, and $0.78 are quantities. The word “heavy” is not a quantity, because it has no numerical value, and the definition requires the numerical value of a measurement.

Definition 2 (Dimension).

A dimension is an inherent way of measuring a quantity.

The foot is not a dimension, because it is one particular scale for measuring length. The inherent way of measuring, which the definition requires, is length itself, $L$ .

Definition 3 (Unit).

A unit is a way to assign a numerical value to the measurement of a quantity.

Length is not a unit, because it assigns no numerical value to a measurement. A length gets its number only once a unit such as the foot or the meter is chosen.

Bracket notation, $[\text{quantity}] = \text{dimension symbol}$ , expresses the dimension of a quantity. For example, $[3\ \text{feet}] = L$ and $[2\ \text{days}] = T$ .

A dimensionless quantity is a pure number that does not have a dimension. The symbol “1” denotes a dimensionless quantity and is written as $[\text{number}] = 1$ . For example, $[6] = 1$ .

Example 1

A 2.5-hour flight burns 1,400 gallons of fuel. State the quantity, the dimension, and the units of (a) the fuel burned and (b) the flight time.

ShowHide solution

By the definition of a quantity, the fuel burned is the quantity 1,400 gallons, and its unit is the gallon. Because a gallon measures volume, a length cubed like the 1.2 ft $^{3}$ in the table, the dimension is $[1{,}400\ \text{gallons}] = L^{3}$ .
The flight time is the quantity 2.5 hours, and its unit is the hour. Because an hour measures time, the dimension is $[2.5\ \text{hours}] = T$ .

Example 2

Classify each item as a quantity, a dimension, or a unit:

the kilogram
26.2 miles
mass
$4.50

ShowHide solution

The kilogram is a unit, because it assigns a numerical value to a measurement of mass.
The distance 26.2 miles is a quantity, because it is a numerical value, 26.2, with its unit, miles. Its dimension is $[26.2\ \text{miles}] = L$ .
Mass is a dimension, because it is an inherent way of measuring a quantity, with the symbol $M$ .
The amount $4.50 is a quantity, because it is a numerical value, 4.50, with its unit, dollars. Its dimension is $[\$4.50] = B$ .

Next: 2. Fundamental and Derived Dimensions Lesson map

2. Fundamental and Derived Dimensions

Objective. Distinguish between fundamental and derived dimensions.

Definition 4 (Fundamental dimension).

A fundamental dimension is a dimension that cannot be expressed in terms of other dimensions.

Area is not a fundamental dimension, because it can be expressed in terms of length as $L^{2}$ , and the definition requires a dimension that cannot be expressed in terms of others.

Definition 5 (Derived dimension).

A derived dimension is a dimension other than a fundamental dimension that can be expressed as a product of integer powers of fundamental dimensions.

The expression $L^{1/2}$ is not a derived dimension, because its exponent $\frac{1}{2}$ is not an integer, and the definition requires integer powers of fundamental dimensions.

Symbol test. A dimension is fundamental when its symbol is a single letter with no exponent, such as $L$ , $M$ , or $T$ . A dimension is derived when its symbol is a product of two or more fundamental dimensions, or one fundamental dimension with an integer exponent other than 0 and 1, such as $L^{2}$ , $LT^{-1}$ , or $MLT^{-2}$ .

Negative exponents. The dimension $T^{-1}$ means $1/T$ . In $LT^{-1}$ the exponent applies to the $T$ alone, which makes $LT^{-1}$ length divided by time.

Derived dimensions.

Dimension	Intuition	Symbol	Example units
Area	length $\times$ width	$L^{2}$	square foot
Volume	length $\times$ width $\times$ height	$L^{3}$	cubic meter, liter, gallon
Velocity	distance per time	$LT^{-1}$	kilometers per hour
Acceleration	velocity per time	$LT^{-2}$	feet per second squared
Density	mass per volume	$ML^{-3}$	ounces per cubic inch
Frequency	per time	$T^{-1}$	per minute
Force	mass $\times$ acceleration	$MLT^{-2}$	newton
Pressure	force per area	$ML^{-1}T^{-2}$	pascal
Energy	force $\times$ distance	$ML^{2}T^{-2}$	joule
Wages	money per time	$BT^{-1}$	dollars per hour
Population density	population per area	$PL^{-2}$	people per square mile
Population growth	population per time	$PT^{-1}$	people per year

Example 3

Classify each dimension as fundamental or derived:

T, \qquad L^{3}, \qquad B, \qquad LT^{-2}, \qquad \Theta, \qquad BT^{-1}.

ShowHide solution

By the definition of a fundamental dimension, $T$ , $B$ , and $\Theta$ are fundamental, because none can be expressed in terms of other dimensions: each symbol is a single letter with no exponent. The dimension $L^{3}$ is derived, because it raises the fundamental dimension $L$ to the power 3. Both $LT^{-2}$ and $BT^{-1}$ are derived, because each is a product of two fundamental dimensions.

Example 4

Give the dimension of each quantity, and classify the dimension as fundamental or derived:

12 seconds
3 square meters
9.8 meters per second squared
40 people

ShowHide solution

The dimension is $[12\ \text{seconds}] = T$ , which is fundamental, because time cannot be expressed in terms of other dimensions.
The dimension is $[3\ \text{square meters}] = L \cdot L = L^{2}$ , which is derived, because it is the fundamental dimension $L$ raised to the power 2.
The dimension is $[9.8\ \text{meters per second squared}] = L/T^{2} = LT^{-2}$ , which is derived, because it is a product of integer powers of the two fundamental dimensions $L$ and $T$ .
The dimension is $[40\ \text{people}] = P$ , which is fundamental, because population cannot be expressed in terms of other dimensions.

Rule 6 (Exponent rules).

For every base $a$ with $a > 0$ and exponents $r$ and $s$ ,

a^{r}a^{s} = a^{r+s}, \qquad \frac{a^{r}}{a^{s}} = a^{r-s}, \qquad (a^{r})^{s} = a^{rs}, \qquad a^{-r} = \frac{1}{a^{r}}, \qquad a^{0} = 1.

Dimension symbols follow the same rules. A fundamental dimension absent from one side of an equation has the exponent 0 on that side.

Example 5

Simplify each product of dimensions:

\text{(a)}\ L^{2}T^{-3} \cdot LT, \qquad\qquad \text{(b)}\ M^{2}L^{-3} \cdot M^{-2}L^{3}.

ShowHide solution

(a) Add the exponents of each fundamental dimension, by the exponent rule $a^{r}a^{s} = a^{r+s}$ :

L^{2}T^{-3} \cdot LT = L^{2+1}\,T^{-3+1} = L^{3}T^{-2}.

(b) Add the exponents of each fundamental dimension in the same way:

M^{2}L^{-3} \cdot M^{-2}L^{3} = M^{2-2}\,L^{-3+3} = M^{0}L^{0} = 1.

The product is dimensionless.

Example 6

Find $[\text{pressure}]$ in fundamental dimensions.

ShowHide solution

Pressure is force per area. Find $[\text{force}]$ first, then divide it by $[\text{area}]$ . Force is mass times acceleration, and the table gives $[\text{acceleration}] = LT^{-2}$ :

[\text{force}] = [\text{mass}]\cdot[\text{acceleration}] = M \cdot LT^{-2} = MLT^{-2}.

Divide $[\text{force}]$ by $[\text{area}] = L^{2}$ , also from the table, and subtract the exponents of $L$ :

[\text{pressure}] = \frac{[\text{force}]}{[\text{area}]} = \frac{MLT^{-2}}{L^{2}} = M\,L^{1-2}\,T^{-2} = ML^{-1}T^{-2}.

Next: 3. Dimension Arithmetic Lesson map

3. Dimension Arithmetic

Objective. Apply rules of dimension arithmetic to find the dimension of an operation, or describe why the operation is invalid.

Rule 7 (Dimension arithmetic).

Adding, subtracting, and comparing. When adding, subtracting, or comparing (for example, with $=$ , $\geq$ , or $<$ ) two quantities, they must have the same dimension, although their units may be different. The resulting dimension is equal to that of the quantities being added, subtracted, or compared.
Multiplying and dividing. When multiplying and dividing quantities, they can have any dimensions and any units. The resulting dimension is equal to the product or quotient of the dimensions of the quantities being multiplied or divided.
Exponents. In an exponential expression $a^{r}$ , the exponent $r$ must be dimensionless.

Complete answer. An answer is complete when each fundamental dimension appears at most once. The answer $ML^{-1}T^{-2}$ is complete. The answer $ML^{2}L^{-3}T^{-2}$ has the same dimension as $ML^{-1}T^{-2}$ but is not complete, because $L$ appears twice.

Equivalent notations. A quotient and a negative exponent are both correct notations. The answers $1/L$ and $L^{-1}$ are both complete, as are $M/(LT^{2})$ and $ML^{-1}T^{-2}$ .

Sum of like quantities. Because a sum takes the one dimension its terms share, $[5\ \text{ft} + 3\ \text{ft}] = L$ . In the same way, $[60\ \text{min} - 15\ \text{min}] = T$ , because subtracting one time from another leaves a time.

Functions of a quantity. Because the functions $e^{x}$ , $2^{x}$ , $\sin$ , $\cos$ , and $\ln$ take a dimensionless input and give a dimensionless output, the values $e^{2}$ , $\sin(\pi/6)$ , and $\ln 5$ are pure numbers with dimension 1.

The input to $\sin$ or $\cos$ is a pure number in radians, such as the input $2\pi(x - h)/P$ of a Lesson 8 sine model. Because that input is dimensionless, the difference $x - h$ shares its dimension with the period $P$ .

The dimension $A$ belongs to an angle stated with its unit, such as 45 degrees.

Because the value of pnorm(x, m, s) depends only on the pure number $(x - m)/s$ (Lesson 9 sheet, section 2), the mean $m$ and the spread $s$ share the dimension of $x$ .

Example 7

Identify each computation or comparison as valid or invalid:

$3.2\ \text{meters} + 4.5\ \text{inches}$
$2.1\ \text{hours} - 3.675\ \text{meters}$
$18\ \text{feet/hour} < 22\ \text{miles/second}$
$5^{\circ}\ \text{Fahrenheit} = \$2.18$

ShowHide solution

Quantities added, subtracted, or compared must share one dimension.

The sum is valid with dimension $L$ , because both quantities are lengths, although their units differ:
$[3.2\ \text{meters}] = [4.5\ \text{inches}] = L.$
The difference is invalid, because $[2.1\ \text{hours}] = T$ and $[3.675\ \text{meters}] = L$ differ.
The comparison is valid with dimension $LT^{-1}$ , because both quantities are velocities:
$[18\ \text{feet/hour}] = [22\ \text{miles/second}] = LT^{-1}.$
The comparison is invalid, because $[5^{\circ}\ \text{Fahrenheit}] = \Theta$ and $[\$2.18] = B$ differ.

Example 8

Find the dimension of each quantity:

the distance a car covers at 55 miles per hour for 0.5 hours
$(3)(2\ \text{m})(6\ \text{cm})$
$8\ \text{kg} / 3\ \text{km}^{3}$
$45\ \text{km}^{2} / ((9\ \text{km})(200\ \text{m}))$

ShowHide solution

The dimension of a product or quotient is the product or quotient of the dimensions.

The distance is the speed times the time. Multiply $[55\ \text{miles per hour}] = LT^{-1}$ by $[0.5\ \text{hours}] = T$ :
$[\text{distance}] = LT^{-1} \cdot T = L.$
Multiply the dimensions of the three factors. The pure number 3 has dimension 1, and each length has dimension $L$ :
$[(3)(2\ \text{m})(6\ \text{cm})] = [3]\cdot[2\ \text{m}]\cdot[6\ \text{cm}] = 1 \cdot L \cdot L = L^{2}.$
Divide the dimension of the mass by the dimension of the volume:
$[8\ \text{kg} / 3\ \text{km}^{3}] = [8\ \text{kg}] / [3\ \text{km}^{3}] = M / L^{3} = ML^{-3}.$
The quotient has the dimension of a density.
Divide the dimension of the area by the dimension of the product of two lengths:
$[45\ \text{km}^{2}] / ([9\ \text{km}]\cdot[200\ \text{m}]) = L^{2}/(L \cdot L) = L^{2}/L^{2} = 1.$

Example 9

Identify each computation as valid or invalid, and give the dimension of each valid one:

$5^{8\ \text{ft}}$
$e^{7\ \text{ft} / 6\ \text{in}}$
$(4\ \text{km/hr})^{2}$
$e^{2} \cdot 4\ \text{m}^{2}$

ShowHide solution

An exponent must be dimensionless.

The power is invalid, because its exponent has dimension $[8\ \text{ft}] = L$ .
The power is valid and dimensionless, because its exponent is dimensionless:
$[7\ \text{ft} / 6\ \text{in}] = L/L = 1.$
The power is valid with dimension $L^{2}T^{-2}$ , because the exponent 2 is a pure number and squaring multiplies the dimension by itself:
$[(4\ \text{km/hr})^{2}] = LT^{-1} \cdot LT^{-1} = L^{2}T^{-2}.$
The product is valid with dimension $L^{2}$ , because the factor $e^{2}$ is a pure number:
$[e^{2} \cdot 4\ \text{m}^{2}] = [e^{2}]\cdot[4\ \text{m}^{2}] = 1 \cdot L^{2} = L^{2}.$

Invalid piece in a larger computation. A computation with one invalid piece is invalid as a whole.

Example 10

Identify each computation or comparison as valid or invalid, and give the dimension of each valid one:

$10\ \text{m}^{2} - (2\ \text{m})(40\ \text{cm})$
$6\ \text{hours} + 2\ \text{miles} + 5\ \text{minutes}$
$\dfrac{e^{0.5}}{(2\ \text{ft})(9\ \text{in})}$
$\$15/\text{hour} > \$90$
$9\ \text{m}^{2} + (4\ \text{m})(5^{3\ \text{cm}})$

ShowHide solution

The difference is valid with dimension $L^{2}$ , because both terms have dimension $L^{2}$ :
$[(2\ \text{m})(40\ \text{cm})] = L \cdot L = L^{2}.$
The sum is invalid, because its terms differ in dimension:
$[6\ \text{hours}] = [5\ \text{minutes}] = T, \qquad [2\ \text{miles}] = L.$
The quotient is valid with dimension $L^{-2}$ , because the exponent 0.5 is a pure number. It follows that $[e^{0.5}] = 1$ and
$\frac{[e^{0.5}]}{[2\ \text{ft}]\cdot[9\ \text{in}]} = \frac{1}{L \cdot L} = L^{-2}.$
The comparison is invalid, because $[\$15/\text{hour}] = BT^{-1}$ and $[\$90] = B$ differ.
The sum is invalid, because the exponent in $5^{3\ \text{cm}}$ has dimension $[3\ \text{cm}] = L$ . The term $9\ \text{m}^{2}$ is valid on its own, and the term $(4\ \text{m})(5^{3\ \text{cm}})$ is invalid.

Next: 4. Solving for an Unknown Dimension Lesson map

4. Solving for an Unknown Dimension

Objective. Given a model’s equation and the dimensions of certain quantities, use dimension algebra to solve for the dimension of another quantity.

A valid model equation obeys every rule of dimension arithmetic. Each rule gives an equation between dimensions, written in bracket notation, that the unknown dimension must satisfy.

Dimension algebra.

Put brackets around both sides: $[\text{left side}] = [\text{right side}]$ .
Give each term of a sum or difference the dimension of the side it sits on, with one bracket equation per term, as in $[y] = [mx] = [b]$ .
Split the dimension of each product or quotient into the product or quotient of its factors’ dimensions, as in $[d/t] = [d]/[t]$ and $[r^{2}] = [r]^{2}$ . Give a pure-number coefficient the dimension 1, as in $[4q] = [q]$ .
Set the dimension of every exponent to 1, as in $[kt] = 1$ for $2^{kt}$ .
Substitute the known dimensions. Divide to isolate the unknown, and simplify with the exponent rules.
Substitute the answer back and check that every term has the same dimension.

Example 11

Suppose

y = mx + b,

where $[y] = L$ and $[m] = MLT^{-2}$ . Find $[x]$ and $[b]$ .

ShowHide solution

Put brackets around both sides of the equation:

[y] = [mx + b].

Giving each term of the sum the dimension of $y$ finds $[b]$ directly:

[y] = [mx] = [b] = L.

Split the product $mx$ into the dimensions of its factors and substitute $[m]$ :

[m]\cdot[x] = L, \qquad MLT^{-2}\cdot[x] = L.

Divide both sides by $MLT^{-2}$ to isolate $[x]$ and subtract the exponents:

[x] = \frac{L}{MLT^{-2}} = L^{1-1}\,M^{0-1}\,T^{0-(-2)} = M^{-1}T^{2}.

As a check, substituting back gives $MLT^{-2}\cdot M^{-1}T^{2} = L$ , the given $[y]$ .

Example 12

Suppose

p = 2q - rs,

where $[p] = M$ and $[r] = ML^{-3}$ . Find $[q]$ and $[s]$ .

ShowHide solution

Give each term of the difference the dimension of $p$ , as in Example 11:

[p] = [2q] = [rs] = M.

Because the coefficient 2 is a pure number, the term $2q$ gives $[q]$ :

[2q] = [2]\cdot[q] = [q] = M.

Split the product $rs$ , substitute $[r]$ , and divide to isolate $[s]$ :

[r]\cdot[s] = ML^{-3}\cdot[s] = M, \qquad [s] = \frac{M}{ML^{-3}} = M^{1-1}\,L^{0-(-3)} = L^{3}.

Example 13

Suppose

v = u + \frac{d}{t},

where $[v] = LT^{-1}$ and $[d] = L$ . Find $[u]$ and $[t]$ .

ShowHide solution

Giving each term of the sum the dimension of $v$ finds $[u]$ directly:

[v] = [u] = [d/t] = LT^{-1}.

Split the quotient $d/t$ , substitute $[d]$ , and divide to isolate $[t]$ :

\frac{[d]}{[t]} = \frac{L}{[t]} = LT^{-1}, \qquad [t] = \frac{L}{LT^{-1}} = L^{1-1}\,T^{0-(-1)} = T.

Example 14

Newton’s law of gravitation gives the force between two masses $m_1$ and $m_2$ a distance $r$ apart as

F = \frac{G\,m_1 m_2}{r^{2}},

where

[F] = MLT^{-2}, \qquad [m_1] = [m_2] = M, \qquad [r] = L.

Find $[G]$ .

ShowHide solution

Put brackets around both sides, split the right side into the dimensions of its factors, and substitute the known dimensions:

[F] = \frac{[G]\cdot[m_1]\cdot[m_2]}{[r]^{2}}, \qquad MLT^{-2} = \frac{[G]\cdot M^{2}}{L^{2}}.

Multiply both sides by $L^{2}$ and divide by $M^{2}$ to isolate $[G]$ :

[G] = \frac{MLT^{-2}\cdot L^{2}}{M^{2}} = M^{1-2}\,L^{1+2}\,T^{-2} = M^{-1}L^{3}T^{-2}.

As a check, substituting back gives $M^{-1}L^{3}T^{-2}\cdot M^{2} / L^{2} = MLT^{-2}$ , the given $[F]$ .

Example 15

Radiocarbon dating models the mass of carbon-14 left in a dead organism as

a = c \cdot 2^{-kt},

where $[a] = M$ and $[t] = T$ . Find $[c]$ and $[k]$ .

ShowHide solution

Set the dimension of the exponent $-kt$ to 1, and split it into its factors $-1$ , $k$ , and $t$ . The pure number $-1$ has $[-1] = 1$ :

[-kt] = [-1]\cdot[k]\cdot[t] = [k]\cdot T = 1, \qquad [k] = \frac{1}{T} = T^{-1}.

Because a pure number raised to a dimensionless power is a pure number, $[2^{-kt}] = 1$ . Split the product on the right side to find $[c]$ :

[a] = [c]\cdot[2^{-kt}] = [c]\cdot 1 = [c], \qquad [c] = M.

Lesson map