Dimensions

Introduction to Dimensions

Welcome to Dimensions in Mechanics! Have you ever solved a long, complicated physics problem, looked at your final answer, and wondered: "Is this formula even remotely possible?"

Dimensional analysis is your ultimate built-in safety net. It allows you to check whether physical equations are valid, figure out relationships between physical quantities from scratch, and find the units of unfamiliar constants. In Mechanics 2, we break every physical measurement down to its fundamental building blocks.

Don't worry if this seems a bit abstract at first! Once you learn the basic rules, working with dimensions feels just like solving straightforward algebraic puzzles.

1. Fundamental Dimensions

In classical mechanics, almost every quantity can be expressed in terms of three fundamental base dimensions:

• Mass: represented by the symbol \( [M] \)
• Length: represented by the symbol \( [L] \)
• Time: represented by the symbol \( [T] \)

We use square brackets, such as \( [X] \), to mean "the dimensions of quantity \(X\)".

Important distinction: Dimensions are not the same as units. Units tell us how much of something we have (like metres, inches, or light-years), but dimensions tell us the fundamental physical nature of the quantity (all of those units have the dimension of Length, \( [L] \)).

Memory Aid: Remember the acronym MLT (Mass, Length, Time) — these three letters are all you need for the mechanics syllabus!

Key Takeaway

Every mechanical quantity is built from combinations of \( [M] \), \( [L] \), and \( [T] \).

2. Derived Dimensions

We find the dimensions of other physical quantities by looking at their defining formulas.

Step-by-Step Derivations of Common Quantities

1. Area and Volume:
• \(\text{Area} = \text{length} \times \text{width} \implies [\text{Area}] = [L] \times [L] = [L^2]\)
• \(\text{Volume} = \text{length} \times \text{width} \times \text{height} \implies [\text{Volume}] = [L^3]\)

2. Density:
• \(\text{Density} = \frac{\text{Mass}}{\text{Volume}} \implies [\text{Density}] = \frac{[M]}{[L^3]} = [M L^{-3}]\)

3. Velocity and Acceleration:
• \(\text{Velocity } (v) = \frac{\text{Distance}}{\text{Time}} \implies [v] = \frac{[L]}{[T]} = [L T^{-1}]\)
• \(\text{Acceleration } (a) = \frac{\text{Change in Velocity}}{\text{Time}} \implies [a] = \frac{[L T^{-1}]}{[T]} = [L T^{-2}]\)

4. Force:
• Using Newton's Second Law, \( F = ma \):
• \([F] = [m][a] = [M] \times [L T^{-2}] = [M L T^{-2}]\)

5. Momentum and Impulse:
• \(\text{Momentum } (p) = mv \implies [p] = [M] \times [L T^{-1}] = [M L T^{-1}]\)
• \(\text{Impulse } (I) = F \Delta t \implies [I] = [M L T^{-2}] \times [T] = [M L T^{-1}]\)
Notice that Momentum and Impulse have identical dimensions!

6. Work, Energy, and Power:
• \(\text{Work } (W) = \text{Force} \times \text{Distance} \implies [W] = [M L T^{-2}] \times [L] = [M L^2 T^{-2}]\)
• Kinetic Energy \(\left(\frac{1}{2}mv^2\right)\) and Potential Energy \((mgh)\) also have dimensions of \( [M L^2 T^{-2}] \).
• \(\text{Power } (P) = \frac{\text{Work}}{\text{Time}} \implies [P] = \frac{[M L^2 T^{-2}]}{[T]} = [M L^2 T^{-3}]\)

7. Pressure and Stress:
• \(\text{Pressure } (P) = \frac{\text{Force}}{\text{Area}} \implies [P] = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}]\)

Dimensionless Quantities

Some quantities have no dimensions at all. They are pure numbers, and we say their dimension is \( 1 \) (or \( [M^0 L^0 T^0] \)).

Common dimensionless quantities include:
• Pure numbers (such as \( \frac{1}{2} \), \( 2 \), \( \pi \))
• Angles in radians (defined as \(\frac{\text{arc length}}{\text{radius}} = \frac{[L]}{[L]} = 1\))
• Coefficient of friction \(\mu\) (defined as \(\frac{\text{Friction Force}}{\text{Normal Reaction}} = \frac{[M L T^{-2}]}{[M L T^{-2}]} = 1\))

Quick Review: Essential Dimensions Table

• Velocity: \( [L T^{-1}] \)
• Acceleration: \( [L T^{-2}] \)
• Force: \( [M L T^{-2}] \)
• Work / Energy: \( [M L^2 T^{-2}] \)
• Power: \( [M L^2 T^{-3}] \)
• Momentum / Impulse: \( [M L T^{-1}] \)
• Density: \( [M L^{-3}] \)
• Pressure: \( [M L^{-1} T^{-2}] \)

3. Dimensional Homogeneity (Consistency)

In everyday life, you cannot add \( 3 \text{ kilograms} \) to \( 5 \text{ seconds} \) — the idea makes no physical sense! Similarly, in any physically valid equation, every single term being added, subtracted, or equated must have the exact same dimensions.

This fundamental rule is known as the Principle of Dimensional Homogeneity.

Worked Example: Checking an Equation

Show that the kinematic equation \( v^2 = u^2 + 2as \) is dimensionally consistent.

Step 1: Find the dimensions of the left-hand side (LHS):
\( [v^2] = ([L T^{-1}])^2 = [L^2 T^{-2}] \)

Step 2: Find the dimensions of each term on the right-hand side (RHS):
• Term 1: \( [u^2] = ([L T^{-1}])^2 = [L^2 T^{-2}] \)
• Term 2: The number \( 2 \) is dimensionless. Therefore:
\( [2as] = 1 \times [a] \times [s] = [L T^{-2}] \times [L] = [L^2 T^{-2}] \)

Step 3: Compare all terms:
Since LHS = \( [L^2 T^{-2}] \) and both terms on the RHS are \( [L^2 T^{-2}] \), all terms have identical dimensions. Therefore, the equation is dimensionally consistent.

Common Mistake to Avoid

A dimensionally consistent equation is not guaranteed to be completely correct in the real world. For instance, writing \( v^2 = u^2 + 999as \) is dimensionally consistent, but physically incorrect because the constant is wrong. Dimensional analysis checks the structure, not the numerical coefficients!

4. Deducing Formulae Using Dimensions

One of the most powerful applications in Mechanics 2 is finding how physical quantities depend on each other. If we know which variables affect a certain quantity, we can set up an equation with unknown index powers (\(\alpha, \beta, \gamma\)) and solve for them.

General Method:

1. Express the quantity \( Q \) as a product of powers: \( Q = k A^\alpha B^\beta C^\gamma \), where \( k \) is a dimensionless constant.
2. Replace each quantity with its dimensions.
3. Equate indices for \( [M] \), \( [L] \), and \( [T] \) on both sides.
4. Solve the simultaneous equations to find \( \alpha \), \( \beta \), and \( \gamma \).

Worked Example: The Simple Pendulum

The period of oscillation \( T \) of a simple pendulum is believed to depend on its length \( l \), the mass of the bob \( m \), and the acceleration due to gravity \( g \). Find a formula for \( T \).

Step 1: Set up the relationship
Assume \( T = k m^\alpha l^\beta g^\gamma \), where \( k \) is a dimensionless constant.

Step 2: Write in dimensional form
\( [T] = [M]^\alpha \times [L]^\beta \times [L T^{-2}]^\gamma \)
\( [M^0 L^0 T^1] = [M]^\alpha [L]^{\beta + \gamma} [T]^{-2\gamma} \)

Step 3: Equate indices
• For \( [M] \): \( 0 = \alpha \implies \alpha = 0 \)
• For \( [T] \): \( 1 = -2\gamma \implies \gamma = -\frac{1}{2} \)
• For \( [L] \): \( 0 = \beta + \gamma \implies \beta = -\gamma = -\left(-\frac{1}{2}\right) = \frac{1}{2} \)

Step 4: Substitute the powers back
\( T = k m^0 l^{1/2} g^{-1/2} \)
\( T = k \sqrt{\frac{l}{g}} \)

Did you know? Because \( \alpha = 0 \), the mass of the bob has no effect on the time period of a simple pendulum! Dimensions helped us discover this surprising physical fact without doing an experiment.

Key Takeaway

Equating powers of \( [M] \), \( [L] \), and \( [T] \) generates up to three linear equations, allowing you to determine the unknown indices of up to three variables.

5. Finding Dimensions of Unknown Constants

Some physical constants carry dimensions (they are not pure numbers). We can rearrange equations to determine their dimensions and units.

Worked Example: Newton's Universal Law of Gravitation

Newton's law of gravitation is given by \( F = \frac{G m_1 m_2}{r^2} \), where \( F \) is force, \( m_1, m_2 \) are masses, \( r \) is distance, and \( G \) is the gravitational constant. Find the dimensions of \( G \).

Step 1: Rearrange for \( G \)
\( G = \frac{F r^2}{m_1 m_2} \)

Step 2: Substitute known dimensions
\( [G] = \frac{[F] [r^2]}{[m_1] [m_2]} = \frac{[M L T^{-2}] [L^2]}{[M] [M]} \)

Step 3: Simplify the expression
\( [G] = \frac{[M L^3 T^{-2}]}{[M^2]} = [M^{-1} L^3 T^{-2}] \)

This tells us immediately that the SI base units of \( G \) are \( \text{kg}^{-1} \text{m}^3 \text{s}^{-2} \).

6. Limitations of Dimensional Analysis

While dimensional analysis is a fantastic tool, it does have specific limitations that you should remember for your exam:

1. Dimensionless constants cannot be found: Dimensional analysis cannot find the numerical value of constants like \( k = 2\pi \) or \( k = \frac{1}{2} \). These must be determined experimentally or through deeper mathematical derivation.
2. Cannot distinguish between different quantities with the same dimensions: For example, Work and Torque both have dimensions \( [M L^2 T^{-2}] \), but Work is a scalar while Torque is a vector.
3. Limited to three variables: Since we only have three independent base dimensions in mechanics (\( M, L, T \)), we can only form three simultaneous equations. If a quantity depends on four or more independent variables, dimensional analysis alone cannot determine all the powers.
4. Trigonometric, logarithmic, and exponential functions: The inputs (arguments) to functions like \( \sin(\theta) \), \( \cos(\theta) \), \( e^x \), or \( \ln(x) \) must always be dimensionless pure numbers.

7. Exam Tips & Common Pitfalls Summary

• Always use square brackets when writing dimensional statements (e.g. \( [v] = [L T^{-1}] \)).
• Watch your minus signs when moving dimensions from the denominator to the numerator (e.g. \(\frac{1}{T^2} = T^{-2}\)).
• Do not treat constants as zero: A dimensionless constant has dimension \( 1 \), not \( 0 \). Setting it to \( 0 \) would make the whole term multiply to zero!
• Check index algebra carefully: When multiplying powers with the same base, add exponents: \( [L] \times [L^{-2}] = [L^{-1}] \).