Dimensional Analysis and Its Applications — JEE Advanced Physics

Checking Correctness of Equations

The Dimensional 'Lie Detector'

The first and most common use of dimensional analysis is as a 'lie detector' for equations. Based on the Principle of Homogeneity, we can quickly test if a formula is dimensionally plausible. If it fails the test, it is guaranteed to be incorrect. If it passes, it is dimensionally consistent, which is a good sign, but not a complete proof of its correctness.

How to Check an Equation

We test the formula for the period of a pendulum: $T = 2\pi \sqrt{l/g}$ , where T is the time period, l is the length, and g is the acceleration due to gravity.

Step 1: Isolate and Analyze the Left-Hand Side (LHS)

The LHS is just the time period, T.
[LHS] = [T]

Step 2: Isolate and Analyze the Right-Hand Side (RHS)

The RHS is $2\pi \sqrt{l/g}$ .
First, ignore any dimensionless constants (like $2\pi$ ).
Now, substitute the dimensions for the remaining variables:
[RHS] = $\sqrt{[l]/[g]} = \sqrt{\frac{[L]}{[LT^{-2}]}}$

Step 3: Simplify the RHS and Compare

Simplify the expression for the RHS dimensions using algebra.
[RHS] = $\sqrt{\frac{[L]}{[L][T^{-2}]}} = \sqrt{\frac{1}{[T^{-2}]}} = \sqrt{[T^2]} = [T]$
Now, compare: [LHS] = [T] and [RHS] = [T].

Step 4: Conclude

Since [LHS] = [RHS], the equation is dimensionally consistent.

The consistency check: write the dimensional formula of every term, compare the dimensions of all terms, and conclude — all equal means the equation is consistent; any difference means it is wrong.

You are checking the equation $F = ma$ . You've found the dimension for the LHS (Force) is [MLT⁻²]. What is the next step?

While checking $T = 2\pi\sqrt{l/g}$ , how should the factor $2\pi$ be treated?

Deriving Relations Between Quantities

Building a Formula from Scratch

Dimensional analysis can be used to deduce the form of a physical equation if we have a good idea of which physical quantities are involved. This technique allows us to see how a quantity must depend on others based solely on their dimensions.

The limitation: This method cannot determine any dimensionless constants in the formula (like $2\pi$ or $1/2$ ). Those must be found through experiment or a more complete theory.

How to Derive a Formula

We derive the formula for the time period (T) of a simple pendulum. Our physical intuition tells us that the period might depend on:

The mass of the bob (m)
The length of the string (l)
The acceleration due to gravity (g)

We can follow a clear, step-by-step process to find the relationship.

Deriving the pendulum formula: assume T ∝ mᵃlᵇgᶜ, write the dimensions, equate the powers of M, L and T to get a = 0, b = 1/2, c = −1/2, and conclude T = k√(l/g).

Result for the Pendulum

By following the steps, we equate the exponents for each dimension (M, L, and T) on both sides of our proposed equation. This gives us a system of equations that we can solve for the unknown powers.

For the pendulum, this process reveals that the period $T$ must be proportional to $m^0 l^{1/2} g^{-1/2}$ . This gives us the final relationship:

$T = k \sqrt{\frac{l}{g}}$

Dimensional analysis alone tells us that the period of a pendulum does not depend on its mass. The dimensionless constant, $k$ , is later found to be $2\pi$ through more advanced mechanics.

More Practice with the Method

The pendulum example shows the pattern; the following worked examples apply the exact same method — propose the dependent variables, write the equation with unknown powers, and equate exponents of each dimension — to a range of other physical situations.

Worked Example: Viscous Drag on a Sphere

Problem: A small sphere falling through a viscous fluid experiences a retarding (drag) force $F$ . Physical intuition (and experiment) suggests this force depends on the fluid's coefficient of viscosity $\eta$ , the sphere's radius $r$ , and its velocity $v$ . Derive the form of $F$ using dimensional analysis.

Solution: We propose $F = k\,\eta^a r^b v^c$ , where $k$ is a dimensionless constant. Writing the dimension of each quantity — $[F] = [MLT^{-2}]$ , $[\eta] = [ML^{-1}T^{-1}]$ , $[r] = [L]$ , $[v] = [LT^{-1}]$ — the dimensional equation becomes:

$[MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c = M^{a}L^{(-a+b+c)}T^{(-a-c)}$

Equating powers of M, L, and T on both sides gives three equations:

$M:\ a = 1 \qquad T:\ -a-c=-2 \qquad L:\ -a+b+c=1$

From the first, $a=1$ . Substituting into the T-equation gives $c=1$ , and then the L-equation gives $b=1$ . So the drag force must have the form:

$F = k\,\eta\, r\, v$

This is exactly the structure of Stokes' law, $F = 6\pi\eta r v$ , where the dimensionless constant $k = 6\pi$ comes from a full fluid-dynamics calculation that dimensional analysis alone cannot supply.

Worked Example: Dimensions of the van der Waals Constants

Problem: The van der Waals equation of state for a real gas is $\left(P + \dfrac{a}{V^2}\right)(V - b) = RT$ , where $P$ is pressure and $V$ is volume. Find the dimensional formulas of the constants $a$ and $b$ .

Solution: By the principle of homogeneity, every term added to or subtracted from another must share its dimension. Since $a/V^2$ is added to $P$ , it must itself have the dimension of pressure:

$\left[\frac{a}{V^2}\right] = [P] \quad\Rightarrow\quad [a] = [P][V]^2 = [ML^{-1}T^{-2}][L^3]^2 = [ML^5T^{-2}]$

Similarly, since $b$ is subtracted from $V$ , it must have the same dimension as volume:

$[b] = [V] = [M^0L^3T^0]$

This is a common exam trap: dimensional homogeneity lets you find the dimensions of an unfamiliar constant just by looking at what it is added to or subtracted from, without knowing the physical origin of the term.

Worked Example: Time Period of an LC Oscillator

Problem: An inductor of inductance $L$ and a capacitor of capacitance $C$ form an oscillating circuit whose natural time period is $T$ . Using $[L] = [ML^2T^{-2}A^{-2}]$ and $[C] = [M^{-1}L^{-2}T^4A^2]$ , derive how $T$ depends on $L$ and $C$ .

Solution: Assume $T = k\,L^{\,a}C^{\,b}$ . Substituting dimensions:

$[T] = [ML^2T^{-2}A^{-2}]^{a}[M^{-1}L^{-2}T^4A^2]^{b}$

Equating powers of M, L, T, and A on both sides:

$M:\ a-b=0 \qquad L:\ 2a-2b=0 \qquad T:\ -2a+4b=1 \qquad A:\ -2a+2b=0$

The M, L, and A equations all reduce to $a=b$ . Substituting into the T-equation: $-2a+4a=1 \Rightarrow a=\tfrac12$ , so $b=\tfrac12$ as well. Therefore:

$T = k\sqrt{LC}$

This matches the well-known result for an LC circuit, $T = 2\pi\sqrt{LC}$ , with the dimensionless constant $k = 2\pi$ supplied by the full circuit analysis.

Worked Example: Speed of a Water Wave

Problem: The speed $v$ of a gravity wave on deep water is believed to depend on its wavelength $\lambda$ and the acceleration due to gravity $g$ (density does not appear, since gravity waves at a free surface are not a function of the fluid's density). Derive the form of $v$ .

Solution: Assume $v = k\,\lambda^{a} g^{b}$ . Substituting dimensions $[v]=[LT^{-1}]$ , $[\lambda]=[L]$ , $[g]=[LT^{-2}]$ :

$[LT^{-1}] = [L]^{a}[LT^{-2}]^{b} = L^{(a+b)}T^{-2b}$

Equating powers of T: $-2b=-1 \Rightarrow b=\tfrac12$ . Equating powers of L: $a+b=1 \Rightarrow a=\tfrac12$ . So:

$v = k\sqrt{\lambda g}$

Once again, dimensional analysis pins down the functional form — wave speed grows with the square root of wavelength — while leaving the exact dimensionless constant $k$ (found from full wave theory to be $1/\sqrt{2\pi}$ for deep-water waves) outside its reach.

You want to derive the formula for centripetal force (F). You assume it depends on mass (m), velocity (v), and radius (r). What is the correct initial setup for the dimensional equation?

For the simple pendulum, the method gives $a = 0$ as the exponent of mass. What does this result mean?

Limitations of Dimensional Analysis

What Dimensional Analysis Cannot Tell You

Dimensional analysis is a powerful tool, but it has definite limits. To apply it correctly, we must understand what it cannot do. The main limitations are these:

1. It Cannot Find Dimensionless Constants

The biggest limitation is that the method has no way of finding numerical constants like $2\pi$ , $1/2$ , or any other 'pure number' in an equation. It can tell us that $KE \propto mv^2$ , but it cannot tell us the correct formula is $KE = \frac{1}{2}mv^2$ . These constants must be found by experiment or a full mathematical derivation.

2. It Cannot Handle Sums or Complex Functions

Dimensional analysis only works for relationships that can be expressed as a product of powers (like $F = k m^a v^b r^c$ ). It cannot derive equations that involve sums or differences (like $s = ut + \frac{1}{2}at^2$ ) or equations that use trigonometric, exponential, or logarithmic functions.

3. It Cannot Distinguish Between Similar Dimensions

Some very different physical quantities happen to have the same dimensions. For example, Work and Torque both have the dimensions $[ML^2T^{-2}]$ . Dimensional analysis cannot tell them apart; it only sees their fundamental recipe.

You use dimensional analysis to derive the formula for kinetic energy and arrive at $KE = k \cdot mv^2$ , which is dimensionally consistent. How would you find the value of the dimensionless constant 'k'?

Unit Conversion Using Dimensions

Same Quantity, Different Numbers

A third powerful use of dimensional analysis is to convert a quantity from one system of units to another (e.g., from SI to CGS). The core principle is simple: the physical quantity itself does not change. A certain amount of force is the same amount of force whether you measure it in Newtons or Dynes. Only the numerical value and the unit change.

$Q = n_1 u_1 = n_2 u_2$

Our goal is to find the new numerical value, $n_2$ , given the old value $n_1$ .

How to Convert Units

We convert 1 Newton (the SI unit of force) into dynes (the CGS unit of force).

Step 1: Find the Dimensional Formula

First, we need the 'recipe' for the quantity. For Force, the dimensional formula is $[M^1 L^1 T^{-2}]$ .

Step 2: Set Up the Conversion Equation

We start with the core principle: $n_1 u_1 = n_2 u_2$ . We want to find $n_2$ .
$n_2 = n_1 \frac{u_1}{u_2}$

Step 3: Substitute Dimensions for Units

Replace the units $u_1$ and $u_2$ with their base units from each system, raised to the powers from the dimensional formula.
$n_2 = n_1 \left( \frac{M_1}{M_2} \right)^1 \left( \frac{L_1}{L_2} \right)^1 \left( \frac{T_1}{T_2} \right)^{-2}$

Step 4: Plug in Conversion Factors and Solve

Now, substitute the known ratios of the base units.
$n_2 = 1 \times \left( \frac{1 \text{ kg}}{1 \text{ g}} \right) \times \left( \frac{1 \text{ m}}{1 \text{ cm}} \right) \times \left( \frac{1 \text{ s}}{1 \text{ s}} \right)^{-2}$
$n_2 = 1 \times (1000) \times (100) \times (1)^{-2} = 100,000 = 10^5$

Result: 1 Newton = $10^5$ dynes.

The dimensional formula for Energy is [ML²T⁻²]. You are converting from SI (Joules) to CGS (ergs). Which term in the conversion formula will contribute the most to the final numerical value?