MASSACHUSETTS INSTITUTE OF TECHNOLOGY
Department of Physics
Physics 8.01
Supplementary Notes 1: Dimensional Analysis
1.1 International System of System of Units
There are only four fundamental quantities (measurements) necessary to
specify all physical phenomena: length, time, mass and charge. All other quantities are
expressible in terms of these, constructed as a matter of convenience.
The basic system of units used throughout science and technology today is the
internationally accepted Système International (SI) (Table 1). It consists of four base
quantities and their corresponding base units: length (meter), mass (kilogram), time
(second), and electric current (ampere). The unit for electric charge, the coulomb, is
defined in terms of the ampere, and hence is referred to as a derived unit. In addition,
three other quantities, temperature, amount of substance, and luminous intensity are part
of the SI base quantities with corresponding units shown in Table 1.
Mechanics is based on just the first three of these quantities, the MKS or meterkilogram-second system. An alternative metric system to this, still widely used, is the socalled CGS system (centimeter-gram-second). For distance and time measurements,
British Imperial units (especially in the USA) based on the foot (ft), the yard (yd), the
mile (mi), etc., as units of length, and also the minute, hour, day and year as units of time.
Table 1 Système International (SI) System of Units
Base Quantity
Length
Mass
Time
Electric Current
Temperature
Amount of Substance
Luminous Intensity
Base Unit
meter (m)
kilogram (kg)
second (s)
ampere (A)
Kelvin (K)
mole (mol)
candela (cd)
We shall refer to the dimension of the base quantity by the quantity itself, for example
dim length [length] = L, dim mass [mass] M, dim time [time] T.
(1)
1.2 Dimensions of Commonly Encountered Quantities
Many physical quantities are derived from the base quantities by a set of algebraic
relations defining the physical relation between these quantities. The dimension of the
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derived quantity can always be written as a product of the powers of the dimensions of
the base quantities.
Example 1 Derived Dimensions of Mechanical Quantities
(i) The dimensions of velocity are given by the relationship
[velocity] = [length]/[time] = L T-1 .
(2)
(ii) Force is also a derived quantity and using the definition of force F = ma and
acceleration a = dv / dt , force has dimensions
[force] =
[mass][velocity]
.
[time]
(3)
We could express force in terms of mass, length, and time by the relationship
[force] =
[mass][length]
= M L T-2 .
2
[time]
(iii) The derived dimension of kinetic energy follows from the definition that K =
thus
[kineticenergy] = [mass][velocity]2 ,
(4)
1
2
mv 2 ,
(5)
which in terms of mass, length, and time is
[kineticenergy] =
[mass][length]2
= M L2 T-2
2
[time]
(6)
(iv) The derived dimension of work is
[work] = [force][length] ,
(7)
which in terms of our fundamental dimensions is
[work] =
[mass][length]2
= M L2 T-2
2
[time]
(8)
So work and kinetic energy have the same dimensions.
2
(v) Power is defined to be the rate of change in time of work so the dimensions are
[power] =
[work] [force][length] [mass][length]2
=
=
= M L2 T-3
[time]
[time]
[time]3
(9)
In Table 2 we list the derived dimensions of some common mechanical quantities
in terms of mass, length, and time.
Table 2 Dimensions of Some Common Mechanical Quantities
M mass , L length , T time
Quantity
Angle
Steradian
Area
Volume
Frequency
Velocity
Acceleration
Angular Velocity
Angular Acceleration
Density
Dimension
dimensionless1
dimensionless
L2
L3
T-1
L T-1
L T-2
T-1
T-2
M L-3
MKS unit
Dimensionless = radian
Dimensionless = radian2
m2
m3
s 1 = hertz = Hz
m s 1
m s 2
rad s 1
rad s 2
kg m 3
Momentum
M L T-1
kg m s 1
Angular Momentum
M L2 T-1
kg m 2 s 1
Force
M L T-2
kg m s2 = newton = N
Work, Energy
M L2 T-2
kg m 2 s 2 = joule = J
Torque
M L2 T-2
kg m 2 s 2
Power
M L2 T-3
kg m 2 s 3 = watt = W
Pressure
M L-1 T-2
kg m 1 s 2 = pascal= Pa
1
Even though angle and steradian are dimensionless quantities, it is often helpful to carry around a “unit”
associated with them, like the radian, to understand their role in an expression or to determine if a result
makes sense.
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1.3 Dimensional Analysis
There are many phenomena in nature that can be explained by simple relationships
between the observed phenomena. When trying to find a dimensional correct formula for
a quantity from a set of given quantities, an answer that is dimensionally correct will
scale properly and is generally off by a constant of order unity.
Consider a simple pendulum consisting of a massive bob suspended from a fixed
point by a string. Let T denote the time (period of the pendulum) that it takes the bob to
complete one cycle of oscillation. How does the period of the simple pendulum depend
on the quantities that define the pendulum and the quantities that determine the motion?
What possible quantities are involved? The length of the pendulum l , the mass of
the pendulum bob m , the gravitational acceleration g , and the initial angular amplitude
of the bob 0 are all possible quantities that may enter into the formula for the period of
the swing. Have we included every possible quantity? We can never be sure but let’s first
work with this set and if we need more than we will have to think harder!
Our problem is then to find a function f such that
(
T = f l, m, g, 0
)
(10)
We first make a list of the dimensions of our quantities as shown in Table 3. Choose the
set: mass, length, and time, to use as the base dimensions.
Table 3 Dimensions of quantities that may describe the period of pendulum
Name of Quantity
Time of swing
Length of pendulum
Mass of pendulum
Gravitational acceleration
Angular amplitude of swing
Symbol
t
l
m
g
0
Dimensional Formula
T
L
M
L T-2
No dimension
We begin by writing the period as a product of these given quantities that have
dimensions, (thus the initial angular amplitude of swing cannot enter into our expression),
with each given quantity raised to a rational power,
T = bl X mY g Z
(11)
where b is a dimensionless constant. Our first observation is that since the period has
only dimensions of time, the mass of the bob cannot enter into our relationship since
neither length of the gravitational acceleration can remove the dimension of the
pendulum mass. Therefore the power Y = 0 , hence
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T = bl X g Z
(12)
A dimensional analysis of Eq. (12) yields
[T ] = [L] X [L T 2 ]Z = [L] X 2Z [T ]2Z
(13)
Let’s focus on the length. In order to eliminate length, we see that the condition
0= X +Z
(14)
must be satisfied in Eq.(13). In a similar fashion, when we consider time, Eq.(13)
requires that
1 = 2Z .
(15)
It is a rather straightforward exercise to solve these two equations to find that
X = Z = 1 / 2
(16)
Thus a dimensionally correct formula for the period of the pendulum is
T =b l/g.
(17)
Since the angular amplitude 0 is dimensionless, it may or may not appear. We
can account for this by introducing some function y( 0 ) into our relationship, which is
beyond the limits of this type of analysis. Then the period for a complete swing is
T = y( 0 ) l / g
(18)
We shall discover later on by solving the problem exactly that y( 0 ) is independent of
the angular amplitude 0 for very small amplitudes and is equal to y( 0 ) = 2 ,
T = 2 l / g
(19)
Example 2 Work and Kinetic Energy
A constant force of magnitude F acts over a distance d on an object of mass m . Find the
final speed v of the body.
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Solution: Here the list of quantities that might be involved is given for us. We write the
speed as
v = bF X d Y mZ
(20)
where X , Y , and Z are rational numbers. A dimensional analysis yields
[L T 1 ] = [ M L T 2 ] X [L]Y [ M ]Z
(21)
Therefore we have the following set of algebraic relations
1= X +Y
(22)
1 = 2X
(23)
0= X +Z
(24)
Solving this set of equations we have that
X = 1/ 2, Z = 1/ 2, X = 1/ 2, Y = 1/ 2 .
(25)
v = b Fd / m .
(26)
So the final speed is
Later on that using the work–kinetic energy theorem we will discover that the constant
b = 2 , and therefore
v = 2Fd / m .
(27)
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