Units and Measurements • Topic 2 of 3

Dimensional Analysis

Every derived quantity in mechanics can be expressed using just three base quantities — mass, length and time — written with the symbols $[M]$, $[L]$ and $[T]$. (For the full SI set we add $[A]$ for current, $[K]$ for temperature, $[\\text{mol}]$ and $[\\text{cd}]$.) The dimensions of a physical quantity are the powers to which these base quantities are raised to represent it. The expression that shows those powers is the dimensional formula.

For example, speed is length per time, so its dimensional formula is $[M^0L^1T^{-1}]$, usually shortened to $[LT^{-1}]$. Acceleration is speed per time, giving $[LT^{-2}]$. Force is mass times acceleration, so:

$$[\\text{Force}] = [M] \\times [LT^{-2}] = [M^1L^1T^{-2}]$$

From force you can quickly build others: work or energy is force times distance, $[M^1L^2T^{-2}]$; power is energy per time, $[M^1L^2T^{-3}]$; pressure is force per area, $[M^1L^{-1}T^{-2}]$. A few quantities such as strain, refractive index and angle are pure ratios and are dimensionless, written $[M^0L^0T^0]$.

The single most useful rule is the principle of homogeneity: every term that is added, subtracted or equated in a physically correct equation must have the same dimensions. You cannot add a length to a time any more than you can add metres to seconds. This principle drives three powerful applications:

  • Checking equations: if the dimensions on the two sides disagree, the equation is certainly wrong. (Agreement does not guarantee correctness, because dimensionless constants are invisible to the method.)
  • Converting units between systems using the relation $n_1[M_1^a L_1^b T_1^c] = n_2[M_2^a L_2^b T_2^c]$.
  • Deriving relations by assuming a quantity depends on others as a product of powers and matching dimensions on both sides.

The method has real limitations. It cannot find dimensionless constants (the $\\tfrac{1}{2}$ in $\\tfrac{1}{2}mv^2$ is undetectable), cannot handle equations with trigonometric, exponential or logarithmic functions (their arguments must themselves be dimensionless), and fails when a quantity depends on more than three other quantities since only three equations are available in mechanics.

Deeper Insight — homogeneity is a free error-detector: Treat dimensional analysis as a habit, not a chapter. Before trusting any derived formula in an exam, glance at its dimensions on both sides. If you wrongly write $v = u + \\tfrac{1}{2}at^2$, the term $at^2$ has dimensions $[LT^{-2}][T^2] = [L]$, which cannot be added to a velocity $[LT^{-1}]$ — the mistake reveals itself instantly. This single reflex catches a large fraction of careless algebra errors and costs you nothing.

Dimensional formulae of common mechanical quantitiesDimensional FormulaeQuantityDimensional FormulaVelocity[M0 L1 T-1]Acceleration[M0 L1 T-2]Force[M1 L1 T-2]Work / Energy[M1 L2 T-2]Power[M1 L2 T-3]Pressure[M1 L-1 T-2]Principle of homogeneity: all added terms share dimensionsPrinciple of Homogeneitys = ut + (1/2) a t^2[L] = [LT-1][T] = [L][LT-2][T2] = [L] ✓
Solved Examples
1
Worked Example
Find the dimensional formula of force, and hence of work.
Solution
  1. Force $=$ mass $\\times$ acceleration $= [M] \\times [LT^{-2}]$.
  2. So $[\\text{Force}] = [M^1L^1T^{-2}]$.
  3. Work $=$ force $\\times$ displacement $= [M^1L^1T^{-2}] \\times [L]$.
  4. $= [M^1L^2T^{-2}]$.

Answer: $[\\text{Force}] = [M^1L^1T^{-2}]$, $[\\text{Work}] = [M^1L^2T^{-2}]$.

2
Worked Example
Check whether the equation $v^2 = u^2 + 2as$ is dimensionally correct.
Solution
  1. $[v^2] = [LT^{-1}]^2 = [L^2T^{-2}]$.
  2. $[u^2] = [L^2T^{-2}]$.
  3. $[2as] = [LT^{-2}][L] = [L^2T^{-2}]$ (the $2$ is dimensionless).
  4. All three terms have dimensions $[L^2T^{-2}]$.

Answer: Dimensionally consistent — the equation passes the homogeneity test.

3
Worked Example
The time period $T$ of a simple pendulum may depend on its length $l$, mass $m$ and $g$. Derive the form of $T$ using dimensions.
Solution
  1. Assume $T = k\\, l^a m^b g^c$, where $k$ is a dimensionless constant.
  2. Dimensions: $[T] = [L]^a [M]^b [LT^{-2}]^c = [M^b L^{a+c} T^{-2c}]$.
  3. Match powers: $M$: $b = 0$; $T$: $-2c = 1 \\Rightarrow c = -\\tfrac{1}{2}$; $L$: $a + c = 0 \\Rightarrow a = \\tfrac{1}{2}$.
  4. So $T = k\\, l^{1/2} g^{-1/2} = k\\sqrt{\\dfrac{l}{g}}$.

Answer: $T = k\\sqrt{\\dfrac{l}{g}}$; $T$ is independent of mass. (Experiment gives $k = 2\\pi$.)

4
Worked Example
Find the dimensional formula of the gravitational constant $G$ from $F = \\dfrac{G m_1 m_2}{r^2}$.
Solution
  1. Rearrange: $G = \\dfrac{F r^2}{m_1 m_2}$.
  2. $[F] = [M^1L^1T^{-2}]$, $[r^2] = [L^2]$, $[m_1 m_2] = [M^2]$.
  3. $[G] = \\dfrac{[M^1L^1T^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$.

Answer: $[G] = [M^{-1}L^3T^{-2}]$.

5
Worked Example
Convert a force of $1\\,\\text{newton}$ into dynes using dimensional analysis ($[F] = [M^1L^1T^{-2}]$).
Solution
  1. $n_2 = n_1 \\left(\\dfrac{M_1}{M_2}\\right)^1 \\left(\\dfrac{L_1}{L_2}\\right)^1 \\left(\\dfrac{T_1}{T_2}\\right)^{-2}$ with SI $\\to$ CGS.
  2. $\\dfrac{M_1}{M_2} = \\dfrac{\\text{kg}}{\\text{g}} = 10^3$; $\\dfrac{L_1}{L_2} = \\dfrac{\\text{m}}{\\text{cm}} = 10^2$; time unit unchanged.
  3. $n_2 = 1 \\times 10^3 \\times 10^2 \\times 1 = 10^5$.

Answer: $1\\,\\text{N} = 10^{5}\\,\\text{dyne}$.

6
Worked Example
Is the equation $y = a\\sin(\\omega t)$ dimensionally meaningful, and what must the argument $\\omega t$ satisfy?
Solution
  1. The argument of a sine must be dimensionless, so $[\\omega t] = [M^0L^0T^0]$.
  2. Since $[t] = [T]$, we need $[\\omega] = [T^{-1}]$ (angular frequency).
  3. Then $\\sin(\\omega t)$ is a pure number, so $[y] = [a]$, both being lengths.

Answer: Yes; it requires $[\\omega] = [T^{-1}]$ so that $\\omega t$ is dimensionless.

Key Points

  • Dimensions are the powers of base quantities $[M]$, $[L]$, $[T]$ that build a derived quantity; the dimensional formula shows them.
  • Key formulae: force $[M^1L^1T^{-2}]$, energy $[M^1L^2T^{-2}]$, power $[M^1L^2T^{-3}]$, pressure $[M^1L^{-1}T^{-2}]$.
  • Principle of homogeneity: every added/equated term must have identical dimensions.
  • Uses: checking equations, converting units between systems, and deriving relations by matching powers.
  • Limits: cannot find dimensionless constants, cannot handle trig/exp/log terms, and fails with more than three dependencies.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The dimensional formula of force is:
Explanation: Force $=$ mass $\\times$ acceleration $= [M][LT^{-2}] = [M^1L^1T^{-2}]$.
Q2.Which quantity is dimensionless?
Explanation: Strain is a ratio of two lengths, so $[M^0L^0T^0]$.
Q3.The principle of homogeneity states that, in a correct equation:
Explanation: Terms added, subtracted or equated must share the same dimensions.
Q4.A limitation of dimensional analysis is that it cannot:
Explanation: Pure numbers like the $\\tfrac{1}{2}$ in $\\tfrac{1}{2}mv^2$ are invisible to the method.
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