1.10 Dimensional Analysis

Definition

Dimensional analysis is a method used to examine the dimensions of physical quantities.

It can be used to check whether the terms in a physical equation are dimensionally consistent.

Dimensional analysis cannot prove that an equation is completely correct. However, it can show that an equation is incorrect if its terms have inconsistent dimensions.

Dimensions

The dimension of a physical quantity describes its physical nature.

In mechanics, most physical quantities can be expressed using three fundamental dimensions.

Physical QuantityDimension
MassM
LengthL
TimeT
Table 1.10.1 — Fundamental Dimensions. Fundamental dimensions used in mechanics.

Dimensions are written inside square brackets. For example,

[mass]=M[\text{mass}] = M

[length]=L[\text{length}] = L

[time]=T[\text{time}] = T

The symbols M, L, and T represent dimensions, not SI units.

For example, meter and kilometer are different units, but both have the same dimension:

LL

Dimensions of Derived Quantities

The dimensions of a derived quantity can be obtained from the equation used to define it.

For example, area is calculated by multiplying one length by another.

Area=Length×LengthArea = Length \times Length

Therefore,

[Area]=L×L=L2[Area] = L \times L = L^2

Similarly,

Volume=Length×Length×LengthVolume = Length \times Length \times Length

Therefore,

[Volume]=L3[Volume] = L^3

Some common derived quantities are shown below.

Physical QuantitySI UnitDimension
Aream²L²
Volumem³L³
Velocitym s⁻¹LT⁻¹
Accelerationm s⁻²LT⁻²
ForceNMLT⁻²
Momentumkg m s⁻¹MLT⁻¹
EnergyJML²T⁻²
PowerWML²T⁻³
Table 1.10.2 — Dimensions of Derived Quantities. SI units and dimensions of common derived physical quantities.

Dimensional Consistency

A valid physical equation must be dimensionally consistent.

This means that both sides of the equation must have exactly the same dimensions.

Only quantities having the same dimensions can be added or subtracted.

For example,

3m+2m=5m3\,m + 2\,m = 5\,m

is meaningful because both quantities have the dimension of length.

However,

3m+2s3\,m + 2\,s

has no physical meaning because length and time have different dimensions.

Example 1 – Verifying an Equation

For an object moving with constant acceleration, the final velocity is given by

v=u+atv = u + at

where

v=final velocityv = \text{final velocity}

u=initial velocityu = \text{initial velocity}

a=accelerationa = \text{acceleration}

t=timet = \text{time}

We can use dimensional analysis to check whether this equation is dimensionally consistent.

The dimensions are

[v]=LT−1[v] = LT^{-1}

[u]=LT−1[u] = LT^{-1}

[a]=LT−2[a] = LT^{-2}

[t]=T[t] = T

Therefore,

[at]=[a][t]=(LT−2)(T)=LT−1[at] = [a][t] = (LT^{-2})(T) = LT^{-1}

Thus,

[u]=[at]=[v][u] = [at] = [v]

Both sides have the same dimensions.

Therefore, the equation is dimensionally consistent.

Example 2 – Detecting an Incorrect Equation

Suppose the following equation is proposed for displacement.

s=ut+as=ut+a

where

s=displacements= \text{displacement}

u=initial velocityu = \text{initial velocity}

a=accelerationa = \text{acceleration}

t=timet =\text{time}

The dimensions are

[s]=L[s] = L

[ut]=(LT−1)(T)=L[ut] = (LT^{-1})(T) = L

[a]=LT−2[a] = LT^{-2}

Since

[ut]≠[a][ut] \neq [a]

the two terms on the right-hand side cannot be added. Therefore, the equation is dimensionally inconsistent.

Example 3 – Finding the Dimensions of Force

Newton’s Second Law states

F=maF=ma

where

F=resultant forceF = \text{resultant force}

m=massm = \text{mass}

a=accelerationa = \text{acceleration}

The dimensions are

[m]=M[m] = M

[a]=LT−2[a] = LT^{-2}

Therefore,

[F]=[m][a]=(M)(LT−2)=MLT−2[F] = [m][a] = (M)(LT^{-2}) = MLT^{-2}

Hence,

[F]=MLT−2[F] = MLT^{-2}

Example 4 – Finding the Dimensions of Kinetic Energy

The kinetic energy of a moving object is given by

Ek=12mv2E_k = \frac{1}{2}mv^2

where

Ek=kinetic energyE_k = \text{kinetic energy}

m=massm = \text{mass}

v=velocityv = \text{velocity}

The constant

12\frac{1}{2}

has no dimensions. Therefore,

[Ek]=12[m][v]2[E_k] = \frac{1}{2}[m][v]^2

Since

[m]=M[m] = M

and

[v]=LT−1[v] = LT^{-1}

then

[Ek]=ML2T−2[E_k] = ML^2T^{-2}

Uses of Dimensional Analysis

Dimensional analysis can be used to

  • check whether an equation is dimensionally consistent,
  • identify dimensionally incorrect equations,
  • determine the dimensions of derived quantities,
  • verify whether quantities can be added or subtracted,
  • help detect mistakes in calculations.

Limitations of Dimensional Analysis

Dimensional consistency is a necessary condition for a physical equation, but it is not sufficient to prove that the equation is correct.

For example,

s=ut+12at2s = ut + \frac{1}{2}at^2

and

s=ut+3at2s = ut + 3at^2

are both dimensionally consistent because every term has the dimension of length.

However, dimensional analysis cannot determine whether the numerical coefficient should be

12,3,\frac{1}{2},\,3,

or another dimensionless number.

Similarly, different physical quantities may have the same dimensions.

For example,

  • Work
  • Energy
  • Torque

all have the dimensions

ML2T−2ML^2T^{-2}

but they represent different physical concepts.

Figure 1.10.1 — Dimensional Analysis. A systematic method for checking the dimensional consistency of a physical equation.

Key Points

  • Dimensions describe the physical nature of a quantity.
  • The fundamental dimensions used in mechanics are M, L, and T.
  • Derived quantities are expressed as combinations of these fundamental dimensions.
  • Both sides of a valid physical equation must have identical dimensions.
  • Only quantities with the same dimensions can be added or subtracted.
  • Dimensional analysis can identify incorrect equations.
  • A dimensionally consistent equation is not necessarily a correct physical equation.

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