Deducing Relations among Physical Quantities

You have already seen how dimensional analysis can check whether an equation makes sense. But the technique can do more than just verify. Under the right conditions, it can actually help you discover a formula from scratch, even before you work through the detailed physics. The idea is surprisingly simple: if you know which quantities a result depends on, the dimensions themselves can pin down how those quantities must combine.

The Core Idea

Suppose you want to find how some physical quantity $Q$ depends on a set of other quantities. The method of dimensions lets you figure out the mathematical form of that relationship, provided two conditions are met:

You already know which quantities matter. You need to identify (or guess, based on physical reasoning) the variables that $Q$ depends on. The method cannot tell you what the relevant variables are; that comes from your understanding of the physics.
The dependence is a product of powers. You assume the relationship takes the form

$Q = k \, a^x \, b^y \, c^z$

where $k$ is a dimensionless constant and $x$ , $y$ , $z$ are unknown exponents that you need to find.

Once you write this product-type expression, you take the dimensions of both sides and match the exponents of each base dimension ( $M$ , $L$ , $T$ ). That gives you a set of simple equations, one per base dimension, which you solve for $x$ , $y$ , and $z$ .

There is a practical ceiling on how many variables this approach can handle. Since mechanics involves three base dimensions ( $M$ , $L$ , $T$ ), matching exponents gives you at most three independent equations. That means you can solve for at most three unknown exponents. If your quantity depends on more than three independent variables, the system becomes underdetermined and the method cannot pin down all the exponents.

Worked Example: Time Period of a Simple Pendulum

This classic example shows the method in action. Picture a bob hanging from a string, swinging back and forth under gravity. You want to find a formula for the time period $T$ (the time for one complete oscillation).

Step 1: Identify the relevant quantities

From physical intuition, the period might depend on:

$l$ : the length of the string
$m$ : the mass of the bob
$g$ : the acceleration due to gravity

Step 2: Write the product-type expression

Assume the dependence takes the form:

$T = k \, l^x \, g^y \, m^z$

Here $k$ is a dimensionless constant and $x$ , $y$ , $z$ are the exponents to be found.

Step 3: Write the dimensions of each quantity

Quantity	Dimensional formula
$T$ (time period)	$[M^0 \, L^0 \, T^1]$
$l$ (length)	$[L^1]$
$g$ (acceleration)	$[L^1 \, T^{-2}]$
$m$ (mass)	$[M^1]$

Step 4: Substitute dimensions into the equation

Replace each quantity with its dimensional formula on both sides:

$[M^0 \, L^0 \, T^1] = [L^1]^x \, [L^1 \, T^{-2}]^y \, [M^1]^z$

The constant $k$ is dimensionless, so it drops out. Now expand the right side by applying the exponent to each base dimension:

$[M^0 \, L^0 \, T^1] = [L^x] \, [L^y \, T^{-2y}] \, [M^z]$

Combine like bases:

$[M^0 \, L^0 \, T^1] = [M^z \, L^{x+y} \, T^{-2y}]$

Step 5: Equate the exponents of each base dimension

For this equation to hold, the exponent of each base dimension must be the same on both sides:

For $M$ :

$z = 0$

For $L$ :

$x + y = 0$

For $T$ :

$-2y = 1$

Step 6: Solve the system

From the $T$ equation:

$y = -\frac{1}{2}$

Substitute into the $L$ equation:

$x + \left(-\frac{1}{2}\right) = 0 \implies x = \frac{1}{2}$

And the $M$ equation already gives:

$z = 0$

Step 7: Write the final relation

Put the exponents back into the original expression:

$T = k \, l^{1/2} \, g^{-1/2} \, m^0$

Since $m^0 = 1$ , the mass drops out entirely. The time period does not depend on the mass of the bob. Rewriting the fractional exponents as a square root:

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

This tells you that the period grows with the square root of the string length and shrinks with the square root of the gravitational acceleration.

What about the constant?

Dimensional analysis gives you the structure of the formula but leaves $k$ undetermined. You cannot find its value by matching dimensions because $k$ is a pure number and has no dimensions to match. The actual value, obtained from a full derivation using Newton’s laws, turns out to be $k = 2\pi$ , giving the well-known result:

$T = 2\pi\sqrt{\frac{l}{g}}$

Whether $k$ is $1$ or $2\pi$ or any other number makes no difference to the dimensions of the right-hand side. The constant is invisible to dimensional analysis.

Limitations You Must Know

Dimensional analysis is a powerful shortcut, but it has clear boundaries. Three limitations stand out:

It cannot find dimensionless constants. As the pendulum example showed, the method gives you $T = k\sqrt{l/g}$ but not $k = 2\pi$ . Any pure number could multiply the right side without changing the dimensions. The constant must come from a complete theoretical derivation or from experiment.
It can only test dimensional validity, not the exact relationship. Passing a dimensional check does not guarantee an equation is correct. Two different formulas can share the same dimensional form but differ in their dimensionless factors. Dimensional analysis will accept both equally.
It cannot tell apart quantities that share the same dimensions. Several physically distinct quantities can have identical dimensional formulae. For example, work, energy, and torque all carry dimensions $[M \, L^2 \, T^{-2}]$ . If your formula involves one of these, dimensional analysis has no way to confirm you picked the right one rather than a different quantity that happens to look the same dimensionally.

Despite these limits, the method remains extremely useful. It gives you the functional form of a relation quickly, narrows down the possibilities, and serves as a strong consistency check on results obtained by other means.