Dimensional Analysis

Dimensions Of Physical Quantities

Dimensions of a physical quantity are the powers to which the fundamental quantities (like length, mass, time, etc.) must be raised to represent that quantity. For instance, velocity is distance divided by time. The dimension of velocity is therefore length divided by time.

The most common fundamental quantities and their symbols used in dimensional analysis are:

• Length (L)
• Mass (M)
• Time (T)
• Electric Current (A or I)
• Temperature ($ \Theta $ or K)
• Amount of Substance (N)
• Luminous Intensity (J)

When analyzing physical quantities, we usually focus on the mechanical dimensions (L, M, T). For example:

• Length: Has dimensions of length, represented as [L].
• Area: Length × Length = $ \text{L}^2 $.
• Volume: Length × Length × Length = $ \text{L}^3 $.
• Velocity: Distance / Time = $ \text{L T}^{-1} $.
• Acceleration: Velocity / Time = $ \text{L T}^{-2} $.
• Force: Mass × Acceleration = $ \text{M L T}^{-2} $.
• Energy (Work): Force × Distance = $ (\text{M L T}^{-2}) \times \text{L} = \text{M L}^2 \text{T}^{-2} $.
• Density: Mass / Volume = $ \text{M L}^{-3} $.
• Pressure: Force / Area = $ (\text{M L T}^{-2}) / \text{L}^2 = \text{M L}^{-1} \text{T}^{-2} $.

It is important to note that dimensions are different from units. While units are specific (e.g., metres, kilograms), dimensions are fundamental concepts (length, mass).

Dimensional Formulae And Dimensional Equations

A dimensional formula expresses the dimensions of a physical quantity in terms of the fundamental quantities. It is usually written in terms of L, M, and T, with exponents indicating the power to which each fundamental quantity is raised.

A dimensional equation is an equation that equates a physical quantity to its dimensional formula.

Examples of Dimensional Formulae:**

• Velocity: $ [\text{v}] = \text{L T}^{-1} $
• Acceleration: $ [\text{a}] = \text{L T}^{-2} $
• Force: $ [\text{F}] = \text{M L T}^{-2} $
• Work (Energy): $ [\text{W}] = \text{M L}^2 \text{T}^{-2} $
• Power: Work / Time = $ (\text{M L}^2 \text{T}^{-2}) / \text{T} = \text{M L}^2 \text{T}^{-3} $. So, $ [\text{P}] = \text{M L}^2 \text{T}^{-3} $.
• Pressure: $ [\text{P}] = \text{M L}^{-1} \text{T}^{-2} $
• Density: $ [\rho] = \text{M L}^{-3} $
• Momentum: Mass × Velocity = $ \text{M} \times (\text{L T}^{-1}) = \text{M L T}^{-1} $. So, $ [\text{p}] = \text{M L T}^{-1} $.

Important Notes on Dimensions:**

• Quantities that are dimensionless (like ratios, pure numbers, angles in radians) have dimensions of $ \text{M}^0 \text{L}^0 \text{T}^0 $.
• Constants like $ \pi $, 2, etc., are dimensionless.
• In addition and subtraction, only quantities with the same dimensions can be added or subtracted.
• In multiplication and division, dimensions are multiplied and divided like algebraic quantities.

Dimensional Analysis And Its Applications

Dimensional analysis is a powerful technique used to analyze the relationships between different physical quantities. It relies on the principle that a physically valid equation must have the same dimensions on both sides.

Checking The Dimensional Consistency Of Equations

The principle of homogeneity of dimensions states that for an equation to be dimensionally correct, the dimensions of all terms added or subtracted must be the same, and the dimensions of the left-hand side must be equal to the dimensions of the right-hand side.

Steps:**

1. Write down the dimensional formula for each physical quantity in the equation.
2. Check if the dimensions on both sides of the equation are identical.
3. If the dimensions match, the equation is dimensionally consistent. This does not guarantee the equation is physically correct, but if dimensions do not match, the equation is definitely incorrect.

Example: Checking $ v = u + at $

Where $ v $ is final velocity, $ u $ is initial velocity, $ a $ is acceleration, and $ t $ is time.

• Dimension of $ v $: $ [\text{v}] = \text{L T}^{-1} $
• Dimension of $ u $: $ [\text{u}] = \text{L T}^{-1} $
• Dimension of $ a $: $ [\text{a}] = \text{L T}^{-2} $
• Dimension of $ t $: $ [\text{t}] = \text{T} $

Now, consider the term $ at $:

$ [at] = [\text{a}] \times [\text{t}] = (\text{L T}^{-2}) \times \text{T} = \text{L T}^{-1} $

Since dimensions of $ u $ and $ at $ are both $ \text{L T}^{-1} $, they can be added. The dimension of the right side ($ u + at $) is $ \text{L T}^{-1} $.

The dimension of the left side ($ v $) is $ \text{L T}^{-1} $.

Since $ [\text{LHS}] = [\text{RHS}] = \text{L T}^{-1} $, the equation $ v = u + at $ is dimensionally consistent.

Deducing Relation Among The Physical Quantities

Dimensional analysis can be used to derive the possible form of an equation relating several physical quantities, provided we know the quantities involved and assume the relationship is a simple product of these quantities raised to some powers.

Steps:**

1. Identify the physical quantity whose formula is to be deduced and the quantities on which it depends.
2. Assume the formula is of the form $ Q = k \cdot x^a y^b z^c \dots $, where $ k $ is a dimensionless constant and $ x, y, z, \dots $ are the dependent quantities.
3. Write down the dimensional formula for $ Q $ and for each of $ x, y, z, \dots $.
4. Substitute these dimensions into the assumed formula and equate the powers of the fundamental quantities (L, M, T) on both sides.
5. Solve the resulting system of linear equations for the exponents $ a, b, c, \dots $.
6. Substitute the values of the exponents back into the assumed formula to get the relation.

Example: Derive the formula for the period of a simple pendulum.**

The period $ T $ of a simple pendulum is expected to depend on its length $ L $, the mass of the bob $ m $, and the acceleration due to gravity $ g $. We assume the relation is of the form:

$ T = k \cdot L^a m^b g^c $

Where $ k $ is a dimensionless constant.

Write dimensions:

• $ [T] = \text{T} $
• $ [L] = \text{L} $
• $ [m] = \text{M} $
• $ [g] = \text{L T}^{-2} $

Substitute into the formula:

$ \text{T} = [\text{L}]^a [\text{M}]^b [\text{L T}^{-2}]^c $

$ \text{T} = \text{L}^a \text{M}^b \text{L}^c \text{T}^{-2c} $

$ \text{M}^0 \text{L}^0 \text{T}^1 = \text{M}^b \text{L}^{a+c} \text{T}^{-2c} $

Equating powers of fundamental quantities:

• For M: $ 0 = b \implies b = 0 $
• For L: $ 0 = a+c \implies a = -c $
• For T: $ 1 = -2c \implies c = -1/2 $

From $ a = -c $, we get $ a = -(-1/2) = 1/2 $.

So, the formula is $ T = k \cdot L^{1/2} m^0 g^{-1/2} = k \sqrt{\frac{L}{g}} $.

The experimental value of $ k $ is $ 2\pi $. Thus, $ T = 2\pi \sqrt{\frac{L}{g}} $. This method cannot determine dimensionless constants like $ k $.

Limitations of Dimensional Analysis

1. It cannot determine dimensionless constants (like $ 2\pi $).
2. It cannot deduce formulas involving more than three physical quantities if the relation involves more than three fundamental dimensions.
3. It cannot distinguish between quantities with the same dimensions (e.g., work and torque both have dimensions $ \text{M L}^2 \text{T}^{-2} $).
4. It is only applicable if the relation is in the form of a product of powers. It cannot be used for relations involving sums or differences of quantities with different dimensions.