Units and Measurements · Lesson 5 of 7

Checking Equations with Dimensions

“Use dimensional homogeneity to reject inconsistent expressions and locate missing factors.”

Learning Objectives

• Apply dimensional homogeneity to every term. • Check equations without substituting numerical values. • Recognise dimensionless function arguments. • Explain why passing a check is not proof of correctness.

Every added term must have the same dimensions

An equation must represent compatible physical quantities on both sides. Within a sum or difference, every term must have the same dimensions. Adding a length to a time is meaningless until a physical relationship supplies the needed conversion. Multiplication and division are different: multiplying speed by time produces a length, and dimensions show why.

Dimensional analysis is useful before numerical substitution because a wrong formula can still yield an ordinary-looking number. Replace quantities by their dimensions, simplify powers, then compare all terms. Constants such as 1/2 and π have no dimensions and do not affect the check.

Checking a motion equation

Problem
Check $x = x_{0} + v_{0} t + \frac{1}{2} a t^{2}$ .

1.Position $x$ and initial position $x_{0}$ each have dimension $[L]$ .
2.The product $v_{0} t$ has dimensions $[L T^{- 1}] [T] = [L]$ .
3.The product $a t^{2}$ has dimensions $[L T^{- 2}] [T^{2}] = [L]$ .
4.Every term represents a length, so the equation passes the check. This alone does not establish the physical conditions under which the equation applies; constant acceleration is still required.

An energy comparison

Checking two energy expressions

Problem
Compare the dimensions of $\frac{1}{2} m v^{2}$ and $m g h$ .

1. $[m v^{2}] = [M] [L T^{- 1}]^{2} = [M L^{2} T^{- 2}]$ .
2. $[m g h] = [M] [L T^{- 2}] [L] = [M L^{2} T^{- 2}]$ .
3.They have equal dimensions, so an equation relating them can be dimensionally consistent. Whether their numerical values are equal depends on the actual physical situation.

Candidate for energy	Dimensional judgement
$m^{2} v^{3}$	Reject: the powers of mass, length and time do not match energy
$ma$	Reject: this has the dimensions of force
$\frac{1}{2} m v^{2}$	Passes the dimensional check
$\frac{3}{16} m v^{2}$	Also passes: dimensions cannot select the coefficient
$\frac{1}{2} m v^{2} + ma$	Reject: energy and force cannot be added

Arguments of functions

A pure number has no dimensions. The input of a logarithm, exponential or trigonometric function must likewise be dimensionless in a physical equation. For example, an exponential involving time can contain a time divided by another time. A sine representing an angle uses the angle expressed as a dimensionless radian measure. A bare velocity cannot be subtracted from the pure number 1.

Recovering a missing speed factor

Problem
A recalled historical moving-mass expression is $m = m_{0} / 1 - v^{2}$ . Where must the speed of light c appear?

1.The pure number 1 can be combined only with a dimensionless term. Since $[v^{2}] = [L^{2} T^{- 2}]$ , divide by another squared speed.
2.The ratio $v^{2} / c^{2}$ is dimensionless, giving $m = m_{0} / 1 - v^{2} / c^{2}$ .
3.This repairs the dimensional inconsistency using the specified variables. It is a dimensional exercise, not a derivation of relativity; modern treatments usually reserve mass for invariant rest mass.

What the test can establish

A dimensionally inconsistent equation is wrong as written. A dimensionally consistent equation is only a candidate. Dimensions cannot establish a numerical coefficient, a sign, an additive term of matching dimension, or the assumptions needed by a model. They also cannot distinguish two different quantities that happen to share a dimensional formula. Use dimensional checking together with physical reasoning and limiting cases.

Quick check

Two energy expressions differ only by the factors 1/2 and 3/16. What can dimensions tell us?

Key Takeaways

• Compare every added term as well as both sides of an equation. • Special-function arguments must be dimensionless. • Dimensional inconsistency rejects a formula. • Dimensional consistency cannot prove a formula or determine its numerical coefficient.

Previous · Lesson 4

Dimensions and Dimensional Formulae

Next · Lesson 6

Finding Relationships by Dimensional Analysis

Every added term must have the same dimensions

An energy comparison

Arguments of functions

What the test can establish

Quiz

Key Takeaways