Lesson Name: Dimensional Analysis
Test Drive Lesson

Instructional Block

Dimensional Analysis

Learning Objectives

By the end of this section, you will be able to:

Find the dimensions of a mathematical expression involving physical quantities.
Determine whether an equation involving physical quantities is dimensionally consistent.

The dimension of any physical quantity expresses its dependence on the base quantities as a product of symbols (or powers of symbols) representing the base quantities. Table 1 lists the base quantities and the symbols used for their dimension. For example, a measurement of length is said to have dimension L or L¹, a measurement of mass has dimension M or M¹, and a measurement of time has dimension T or T¹. Like units, dimensions obey the rules of algebra. Thus, area is the product of two lengths and so has dimension L², or length squared. Similarly, volume is the product of three lengths and has dimension L³, or length cubed. Speed has dimension length over time, L/T or LT^–1. Volumetric mass density has dimension M/L³ or ML^–3, or mass over length cubed. In general, the dimension of any physical quantity can be written as $L^aM^bT^cI^dΘ^eN^fJ^g$ for some powers $a,b,c,d,e,f,$ and $g$ . We can write the dimensions of a length in this form with $a = 1$ and the remaining six powers all set equal to zero: $L^1=L^1M^0T^0I^0Θ^0N^0J^0$ Any quantity with a dimension that can be written so that all seven powers are zero (that is, its dimension is $L^0M^0T^0I^0Θ^0N^0J^0$ ) is called dimensionless (or sometimes “of dimension 1,” because anything raised to the zero power is one). Physicists often call dimensionless quantities pure numbers.

Table 1 Base Quantities and Their Dimensions

Physicists often use square brackets around the symbol for a physical quantity to represent the dimensions of that quantity. For example, if $r$ is the radius of a cylinder and $h$ is its height, then we write $[r]=L$ and $[h]=L$ to indicate the dimensions of the radius and height are both those of length, or L. Similarly, if we use the symbol $A$ for the surface area of a cylinder and $V$ for its volume, then $[A] = L^2$ and $[V] = L^3$ . If we use the symbol $m$ for the mass of the cylinder and $ρ$ for the density of the material from which the cylinder is made, then $[m]=M$ and $[\rho]=ML^{−3}$

The importance of the concept of dimension arises from the fact that any mathematical equation relating physical quantities must be dimensionally consistent, which means the equation must obey the following rules:

Every term in an expression must have the same dimensions; it does not make sense to add or subtract quantities of differing dimension (think of the old saying: “You can’t add apples and oranges”). In particular, the expressions on each side of the equality in an equation must have the same dimensions.
The arguments of any of the standard mathematical functions such as trigonometric functions (such as sine and cosine), logarithms, or exponential functions that appear in the equation must be dimensionless. These functions require pure numbers as inputs and give pure numbers as outputs.

If either of these rules is violated, an equation is not dimensionally consistent and cannot possibly be a correct statement of physical law. This simple fact can be used to check for typos or algebra mistakes, to help remember the various laws of physics, and even to suggest the form that new laws of physics might take. This last use of dimensions is beyond the scope of this text, but is something you may learn later in your academic career.

EXAMPLE 1: Using Dimensions to Remember an Equation

Suppose we need the formula for the area of a circle for some computation. Like many people who learned geometry too long ago to recall with any certainty, two expressions may pop into our mind when we think of circles: $\pi r^2$ and $2\pi r$ $.$ One expression is the circumference of a circle of radius r and the other is its area. But which is which?

Strategy

One natural strategy is to look it up, but this could take time to find information from a reputable source. Besides, even if we think the source is reputable, we shouldn’t trust everything we read. It is nice to have a way to double-check just by thinking about it. Also, we might be in a situation in which we cannot look things up (such as during a test). Thus, the strategy is to find the dimensions of both expressions by making use of the fact that dimensions follow the rules of algebra. If either expression does not have the same dimensions as area, then it cannot possibly be the correct equation for the area of a circle.

Solution

We know the dimension of area is $L^2$ . Now, the dimension of the expression $\pi r^2$ is

$[\pi r^2] = [\pi] \cdot [r]^2 = 1 \cdot L^2 = L^2$

since the constant $π$ is a pure number and the radius $r$ is a length. Therefore, $\pi r^2$ has the dimension of area. Similarly, the dimension of the expression $2 \pi r$ is

$[2 \pi r]=[2]·[\pi]·[r]=1·1·L=L,$

since the constants $2$ and $π$ are both dimensionless and the radius $r$ is a length. We see that $2\pi r$ has the dimension of length, which means it cannot possibly be an area.

We rule out $2\pi r$ because it is not dimensionally consistent with being an area. We see that $πr2$ is dimensionally consistent with being an area, so if we have to choose between these two expressions, $\pi r^2$ is the one to choose.

Significance

This may seem like kind of a silly example, but the ideas are very general. As long as we know the dimensions of the individual physical quantities that appear in an equation, we can check to see whether the equation is dimensionally consistent. On the other hand, knowing that true equations are dimensionally consistent, we can match expressions from our imperfect memories to the quantities for which they might be expressions. Doing this will not help us remember dimensionless factors that appear in the equations (for example, if you had accidentally conflated the two expressions from the example into $2 \pi r^2$ $,$ then dimensional analysis is no help), but it does help us remember the correct basic form of equations.

EXAMPLE 2: Checking Equations for Dimensional Consistency

Consider the physical quantities $s$ , $v$ , $a$ , and $t$ with dimensions $[s]=L$ $,$ $[v] = LT^{-1}$ , $[a]=LT^{−2}$ $,$ and $[t]=T$ $.$ Determine whether each of the following equations is dimensionally consistent: (a) $s=vt+0.5at^2$ $;$ (b) $s=vt^2+0.5at$ $;$ and (c) $v=\sin(at^2/s)$ $.$

Strategy

By the definition of dimensional consistency, we need to check that each term in a given equation has the same dimensions as the other terms in that equation and that the arguments of any standard mathematical functions are dimensionless.

Solution

a. There are no trigonometric, logarithmic, or exponential functions to worry about in this equation, so we need only look at the dimensions of each term appearing in the equation. There are three terms, one in the left expression and two in the expression on the right, so we look at each in turn:

$[s] = L\\ [vt] = [v] \cdot [t] = LT^{-1} \cdot T = LT^0 = L\\ [0.5at^2] = [a] \cdot [t]^2 = LT^{-2} \cdot T^2 = LT^0 = L$

All three terms have the same dimension, so this equation is dimensionally consistent.

b. Again, there are no trigonometric, exponential, or logarithmic functions, so we only need to look at the dimensions of each of the three terms appearing in the equation:

$[s] = L\\ [vt^2] = [v] \cdot [t]^2 = LT^{-1} \cdot T^2 = LT^0 = LT\\ [at] = [a] \cdot [t] = LT^{-2} \cdot T = LT^{-1}$

None of the three terms has the same dimension as any other, so this is about as far from being dimensionally consistent as you can get. The technical term for an equation like this is nonsense.

c. This equation has a trigonometric function in it, so first we should check that the argument of the sine function is dimensionless:

$\bigg[ {at^2 \over s} \bigg] = {[a] \cdot [t]^2 \over [s]} = { LT^{-2} \cdot T^2 \over L} = {L \over L} = 1$

The argument is dimensionless. So far, so good. Now we need to check the dimensions of each of the two terms (that is, the left expression and the right expression) in the equation:

$[v] = LT^{-1}\\ \bigg[ \sin \bigg({ at^2 \over s } \bigg) \bigg] = 1$ .

The two terms have different dimensions—meaning, the equation is not dimensionally consistent. This equation is another example of “nonsense.”

Significance

If we are trusting people, these types of dimensional checks might seem unnecessary. But, rest assured, any textbook on a quantitative subject such as physics (including this one) almost certainly contains some equations with typos. Checking equations routinely by dimensional analysis save us the embarrassment of using an incorrect equation. Also, checking the dimensions of an equation we obtain through algebraic manipulation is a great way to make sure we did not make a mistake (or to spot a mistake, if we made one).

One further point that needs to be mentioned is the effect of the operations of calculus on dimensions. We have seen that dimensions obey the rules of algebra, just like units, but what happens when we take the derivative of one physical quantity with respect to another or integrate a physical quantity over another? The derivative of a function is just the slope of the line tangent to its graph and slopes are ratios, so for physical quantities v and t, we have that the dimension of the derivative of v with respect to t is just the ratio of the dimension of v over that of t:

$\bigg[ {dv \over dt} \bigg] = {[v] \over [t]}$

Similarly, since integrals are just sums of products, the dimension of the integral of v with respect to t is simply the dimension of v times the dimension of t:

By the same reasoning, analogous rules hold for the units of physical quantities derived from other quantities by integration or differentiation.

No hints found for this question

Check All That Apply

A student is trying to remember some formulas from geometry. In what follows, assume $A$ is area, $V$ is volume, and all other variables are lengths. Determine which formulas are dimensionally consistent. Of the options below, choose the 3 options that are correct.

$V=\pi r^2h$

$A=2\pi r^2 + 2\pi rh$

$V=\pi d^3/6$

$V=\pi d^2$

$V=0.5bh$

Answer	Rubric	% Of Choosen
$V=\pi r^2h$	Correct	29 %
$A=2\pi r^2 + 2\pi rh$	Correct	29 %
$V=\pi d^3/6$	Correct	30 %
$V=\pi d^2$	In-Correct	8 %
$V=0.5bh$	In-Correct	5 %

Check All That Apply

Consider the physical quantities s, v, a, and t with dimensions $[s]=L, [v]=LT^{-1}, [a]=LT^{-2}, \text{ and }[t]=T$ . Determine whether each of the following equations is dimensionally consistent.

Choose 3 correct options.

$v^2=2as$

$a=\frac{v}{t}$

$s=vt^2+0.5at^2$

$v=\frac{s}{t}$

Answer	Rubric	% Of Choosen
$v^2=2as$	Correct	33 %
$a=\frac{v}{t}$	Correct	33 %
$s=vt^2+0.5at^2$	In-Correct	0 %
$v=\frac{s}{t}$	Correct	33 %

Multiple Choice

Consider the physical quantities m, s, v, a, and t with dimensions $[m] = M,\; [s] = L,\; [v] = LT^{–1}, \;[a] = LT^{–2}, \text{ and } [t] = T$ .

Assuming each of the following equations is dimensionally consistent, find the dimension of the quantity on the left-hand side of the equation:

$F = ma$

$[F]=ML^{-2}$

$[F]=LT^{-2}$

$[F]=MLT^{-2}$

$[F]=MLT^{-1}$

Answer	Rubric	% Of Choosen
$[F]=ML^{-2}$	In-Correct	0 %
$[F]=LT^{-2}$	In-Correct	0 %
$[F]=MLT^{-2}$	Correct	0 %
$[F]=MLT^{-1}$	In-Correct	0 %

No hints found for this question

Multiple Choice

$K = 0.5mv^2$

$[K]=ML^2T^{-2}$

$[K]=MLT^{-2}$

$[K]=L^2T^{-2}$

$[K]=ML^2T$

Answer	Rubric	% Of Choosen
$[K]=ML^2T^{-2}$	Correct	0 %
$[K]=MLT^{-2}$	In-Correct	0 %
$[K]=L^2T^{-2}$	In-Correct	0 %
$[K]=ML^2T$	In-Correct	0 %

No hints found for this question

Multiple Choice

$p = mv$

$[P]=ML^2T^{-1}$

$[P]=M^2LT^{-1}$

$[P]=MLT^{-1}$

$[P]=MLT$

Answer	Rubric	% Of Choosen
$[P]=ML^2T^{-1}$	In-Correct	0 %
$[P]=M^2LT^{-1}$	In-Correct	0 %
$[P]=MLT^{-1}$	Correct	0 %
$[P]=MLT$	In-Correct	0 %

No hints found for this question

Multiple Choice

$W = mas$

$[W]=ML^2T^{-2}$

$[W]=ML^2T$

$[W]=ML^{-2}T^{-1}$

$[W]=L^2M^{-2}$

Answer	Rubric	% Of Choosen
$[W]=ML^2T^{-2}$	Correct	0 %
$[W]=ML^2T$	In-Correct	0 %
$[W]=ML^{-2}T^{-1}$	In-Correct	0 %
$[W]=L^2M^{-2}$	In-Correct	0 %

No hints found for this question

Multiple Choice

$L = mvr$

$[L]=ML^2T^{-1}$

$[L]=M^{-2}L^2T^{-1}$

$[L]=MLT^{-2}$

$[L]=M^2LT^{-1}$

Answer	Rubric	% Of Choosen
$[L]=ML^2T^{-1}$	Correct	0 %
$[L]=M^{-2}L^2T^{-1}$	In-Correct	0 %
$[L]=MLT^{-2}$	In-Correct	0 %
$[L]=M^2LT^{-1}$	In-Correct	0 %

No hints found for this question

Multiple Choice

Suppose quantity $s$ is a length and quantity $t$ is a time. Suppose the quantities $v$ and $a$ are defined by $v=ds/dt$ and $a=dv/dt$ . What is the dimension of $v$ ?

$[v]=L^2T^{2}$

$[v]=LT$

$[v]=LT^{-1}$

$[v]=L^{-2}T^{-1}$

Answer	Rubric	% Of Choosen
$[v]=L^2T^{2}$	In-Correct	0 %
$[v]=LT$	In-Correct	0 %
$[v]=LT^{-1}$	Correct	0 %
$[v]=L^{-2}T^{-1}$	In-Correct	0 %

No hints found for this question

Multiple Choice

What is the dimension of the quantity $a$ ?

$[a]=L^{-1}T^{-2}$

$[a]=LT^{-2}$

$[a]=L^{-1}T$

$[a]=LT^{2}$

Answer	Rubric	% Of Choosen
$[a]=L^{-1}T^{-2}$	In-Correct	0 %
$[a]=LT^{-2}$	Correct	0 %
$[a]=L^{-1}T$	In-Correct	0 %
$[a]=LT^{2}$	In-Correct	0 %

No hints found for this question

Multiple Choice

What are the dimensions of $\int v\; dt$ ?

$\big[\int v\;dt \big]=L$

$\big[\int v\;dt \big]=LT^{-2}$

$\big[\int v\;dt \big]=L^{-1}$

$\big[\int v\;dt \big]=LT$

Answer	Rubric	% Of Choosen
$\big[\int v\;dt \big]=L$	Correct	0 %
$\big[\int v\;dt \big]=LT^{-2}$	In-Correct	0 %
$\big[\int v\;dt \big]=L^{-1}$	In-Correct	0 %
$\big[\int v\;dt \big]=LT$	In-Correct	0 %

No hints found for this question

Multiple Choice

What are the dimensions of $\int a\; dt$ ?

$\big[ \int a\: dt\big]=L^{-2}T^{-1}$

$\big[ \int a\: dt\big]=LT^{-2}$

$\big[ \int a\: dt\big]=L^2T$

$\big[ \int a\: dt\big]=LT^{-1}$

Answer	Rubric	% Of Choosen
$\big[ \int a\: dt\big]=L^{-2}T^{-1}$	In-Correct	0 %
$\big[ \int a\: dt\big]=LT^{-2}$	In-Correct	0 %
$\big[ \int a\: dt\big]=L^2T$	In-Correct	0 %
$\big[ \int a\: dt\big]=LT^{-1}$	Correct	0 %

No hints found for this question

Multiple Choice

What are the dimensions of $\frac{da}{dt}$ ?

$\big[\frac{da}{dt}\big]=LT$

$\big[\frac{da}{dt}\big]=L^{-3}T$

$\big[\frac{da}{dt}\big]=L^{-1}T^{-3}$

$\big[\frac{da}{dt}\big]=LT^{-3}$

Answer	Rubric	% Of Choosen
$\big[\frac{da}{dt}\big]=LT$	In-Correct	0 %
$\big[\frac{da}{dt}\big]=L^{-3}T$	In-Correct	0 %
$\big[\frac{da}{dt}\big]=L^{-1}T^{-3}$	In-Correct	0 %
$\big[\frac{da}{dt}\big]=LT^{-3}$	Correct	0 %

No hints found for this question

Multiple Choice

Suppose [V] = L³, [ $\rho$ ]=ML-3, and [t]=T. What is the dimension of $\int{\rho dV}$ ?

$ML^{-3}$

$ML^3$

$L$

$M$

Answer	Rubric	% Of Choosen
$ML^{-3}$	In-Correct	0 %
$ML^3$	In-Correct	0 %
$L$	In-Correct	0 %
$M$	Correct	0 %

No hints found for this question

Multiple Choice

What is the dimension of dV/dt?

$\frac{L^3}{T}$

$\frac{L^{-3}}{T^2}$

$LT^2$

$L^3T$

Answer	Rubric	% Of Choosen
$\frac{L^3}{T}$	Correct	0 %
$\frac{L^{-3}}{T^2}$	In-Correct	0 %
$LT^2$	In-Correct	0 %
$L^3T$	In-Correct	0 %

No hints found for this question

Numeric	Multiple Choice	Check All That Apply	Fill-In	Free Response	Ungraded	Instructional Block	Numeric + Units	Total
0	0	0	0	0	0	0	0

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Lesson Name: Dimensional Analysis
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Dimensional Analysis

Learning Objectives

Table 1 Base Quantities and Their Dimensions

EXAMPLE 1: Using Dimensions to Remember an Equation

EXAMPLE 2: Checking Equations for Dimensional Consistency

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Less effort, better results

The all-in-one learning platform built for physics educators and their students.

Lesson Name: Dimensional Analysis Test Drive Lesson

Table 1 Base Quantities and Their Dimensions

EXAMPLE 1: Using Dimensions to Remember an Equation​

Simplify how you teach physics.

Start with Key2PHYSICS, our all-in-one Learning Platform for Physics Educators. It has the tools you need to teach physics—lessons, assessments, and learning reports—all in one place.

Lesson Name: Dimensional Analysis
Test Drive Lesson

EXAMPLE 1: Using Dimensions to Remember an Equation