3. Dimensional analysis

Learning outcomes

I can use dimensional analysis to verify equations.
I can determine whether equations are dimensionally consistent.
I can solve problems using conversion factors.
I can identify errors using dimensional reasoning.
I can explain why dimensional analysis is useful in physics.

What Is Dimensional Analysis?

Dimensional analysis is a method of examining the units or dimensions of physical quantities to check equations, perform conversions, and identify possible errors.

Physics equations describe relationships between quantities. For an equation to be physically meaningful, the quantities on both sides must have compatible dimensions.

For example:

v = d/t

Distance has the dimension of length and time has the dimension of time.

Therefore:

velocity = length/time

In SI units:

m/s

Dimensional analysis gives us a powerful way to check whether equations and calculations make physical sense.

Units and Dimensions Are Related but Different

A unit tells us how a quantity is measured.

Examples:

metre (m)
second (s)
kilogram (kg)
newton (N)

A dimension describes the physical type of the quantity.

Three important fundamental dimensions are:

Quantity	Dimension	Common SI Unit
Length	L	m
Mass	M	kg
Time	T	s

For example, a distance of 20 m has:

unit → m

dimension → L

Dimension Symbols

Dimensions are commonly written using square brackets.

For example:

[length] = L

[mass] = M

[time] = T

We can then describe more complicated quantities using combinations of these fundamental dimensions.

Dimensions of Speed

Speed is:

speed = distance/time

Distance has dimension:

Time has dimension:

Therefore:

[speed] = L/T

This can also be written:

[speed] = LT⁻¹

Its SI unit is:

m/s

Dimensions of Acceleration

Acceleration is:

acceleration = change in velocity/time

Velocity has dimensions:

LT⁻¹

Divide by time:

[acceleration] = LT⁻¹/T

Therefore:

[acceleration] = LT⁻²

Its SI unit is:

m/s²

Dimensions of Force

Newton's second law states:

F = ma

Mass has dimension:

Acceleration has dimension:

LT⁻²

Therefore:

[F] = MLT⁻²

The SI unit is the newton:

N = kg·m/s²

This shows that a newton is a derived unit.

Derived Quantities

Many physics quantities are combinations of fundamental dimensions.

Quantity	SI Unit	Dimensions
Area	m²	L²
Volume	m³	L³
Speed	m/s	LT⁻¹
Acceleration	m/s²	LT⁻²
Density	kg/m³	ML⁻³
Force	N	MLT⁻²
Pressure	Pa	ML⁻¹T⁻²
Energy	J	ML²T⁻²
Power	W	ML²T⁻³

Understanding these relationships makes complicated equations much easier to check.

Dimensional Consistency

An equation is dimensionally consistent when the dimensions on one side of the equation match the dimensions on the other side.

Consider:

d = vt

Left side:

[d] = L

Right side:

[v][t]

Velocity:

[v] = LT⁻¹

Time:

[t] = T

Therefore:

[vt] = LT⁻¹ × T

The T terms cancel:

[vt] = L

So:

L = L

The equation is dimensionally consistent.

Checking an Equation

Suppose someone proposes:

d = v/t

Is this dimensionally consistent?

Left side:

[d] = L

Right side:

[v/t] = LT⁻¹/T

Therefore:

[v/t] = LT⁻²

So we obtain:

L ≠ LT⁻²

The dimensions do not match.

Therefore:

d = v/t

cannot be a dimensionally correct equation for distance.

A Dimensional Check Does Not Prove an Equation Is Correct

This is an important limitation.

Suppose someone proposes:

d = 5vt

Dimensions of the right side:

[5vt] = LT⁻¹ × T

= L

So the equation is dimensionally consistent.

However, dimensional analysis cannot tell us whether the numerical factor 5 is correct.

Therefore:

dimensionally consistent does not necessarily mean physically correct

But:

dimensionally inconsistent means the equation cannot be correct as written

Dimensionless Numbers

Ordinary numerical constants have no physical dimensions.

Examples include:

For example:

Eₖ = ½mv²

The factor ½ does not affect the dimensional analysis.

We only need to examine:

mv²

Checking the Kinetic Energy Equation

Consider:

Eₖ = ½mv²

Ignore the dimensionless factor ½.

Mass:

[m] = M

Velocity:

[v] = LT⁻¹

Therefore:

[v²] = L²T⁻²

Multiply by mass:

[mv²] = ML²T⁻²

Energy has dimensions:

[E] = ML²T⁻²

Therefore:

ML²T⁻² = ML²T⁻²

The equation is dimensionally consistent.

Checking Gravitational Potential Energy

Consider:

Eₚ = mgh

Mass:

[m] = M

Acceleration:

[g] = LT⁻²

Height:

[h] = L

Therefore:

[mgh] = M × LT⁻² × L

= ML²T⁻²

This matches the dimensions of energy.

Therefore:

Eₚ = mgh

is dimensionally consistent.

Checking Momentum

Momentum is:

p = mv

Mass:

Velocity:

LT⁻¹

Therefore:

[p] = MLT⁻¹

The SI unit is:

kg·m/s

Checking Pressure

Pressure is:

p = F/A

Force:

[F] = MLT⁻²

Area:

[A] = L²

Therefore:

[p] = MLT⁻²/L²

Simplify:

[p] = ML⁻¹T⁻²

The SI unit is:

kg/(m·s²)

This derived unit is called the:

pascal (Pa)

Checking Density

Density is:

ρ = m/V

Mass:

[m] = M

Volume:

[V] = L³

Therefore:

[ρ] = M/L³

or:

[ρ] = ML⁻³

Its SI unit is:

kg/m³

Addition and Subtraction Require Matching Dimensions

Quantities can only be meaningfully added or subtracted if they have compatible dimensions.

For example:

5 m + 3 m

makes sense.

But:

5 m + 3 s

does not.

Length and time represent different physical quantities.

The same rule applies inside equations.

Checking a Motion Equation

Consider:

v = u + at

Check each term.

Velocity:

[v] = LT⁻¹

Initial velocity:

[u] = LT⁻¹

Acceleration × time:

[at] = LT⁻² × T

= LT⁻¹

Therefore:

LT⁻¹ = LT⁻¹ + LT⁻¹

Every term has the dimensions of velocity.

The equation is dimensionally consistent.

Checking Another Motion Equation

Consider:

s = ut + ½at²

Left side:

[s] = L

First term:

[ut] = LT⁻¹ × T

= L

Second term:

[at²] = LT⁻² × T²

= L

Therefore:

L = L + L

The equation is dimensionally consistent.

Spotting an Error

Suppose someone writes:

s = ut + ½at

Check the second term:

[at] = LT⁻² × T

= LT⁻¹

The equation would therefore attempt to add:

length + velocity

That is not dimensionally valid.

So the equation contains an error.

The missing factor is related to time:

s = ut + ½at²

Dimensions Can Reveal Missing Powers

Suppose an equation contains:

distance ∝ timeⁿ

For motion from rest under constant acceleration, suppose we know acceleration must also appear:

d ∝ atⁿ

Determine n using dimensions.

Distance:

Acceleration:

LT⁻²

Time:

Tⁿ

Therefore:

L = LT⁻² × Tⁿ

L = LTⁿ⁻²

For the time dimensions to cancel:

n − 2 = 0

Therefore:

n = 2

So:

d ∝ at²

Dimensional reasoning has revealed the required power of time.

Conversion Factors

Dimensional analysis is also extremely useful for unit conversions.

A conversion factor is a ratio of equivalent quantities.

For example:

1 km = 1000 m

So either of these ratios equals 1:

1000 m / 1 km

or:

1 km / 1000 m

We choose the version that causes the unwanted unit to cancel.

Example: Kilometres to Metres

Convert:

6.2 km → m

Use:

6.2 km × (1000 m / 1 km)

The kilometres cancel:

6.2 × 1000 m

Therefore:

6.2 km = 6200 m

The cancellation of units provides a built-in check.

Example: Centimetres to Metres

Convert:

45 cm → m

Use:

45 cm × (1 m / 100 cm)

Cancel cm:

45/100 m

Therefore:

45 cm = 0.45 m

Multi-Step Dimensional Analysis

Suppose we want to convert:

90 km/h → m/s

Use two conversion factors:

90 km/h × (1000 m / 1 km) × (1 h / 3600 s)

Cancel:

and:

This leaves:

m/s

Calculate:

90 × 1000 / 3600

= 25

Therefore:

90 km/h = 25 m/s

The units tell us exactly how the conversion factors should be arranged.

Why Unit Cancellation Is Powerful

Suppose you accidentally write:

90 km/h × (1 km / 1000 m)

The kilometres do not cancel correctly.

You would end up with an inappropriate combination of units.

That immediately warns you that the conversion factor has been arranged incorrectly.

This is one reason dimensional analysis is more reliable than simply memorizing whether to multiply or divide.

Multi-Step Example

Convert:

72 km/h → cm/s

Start with:

72 km/h

Use:

72 km/h × (1000 m / 1 km) × (100 cm / 1 m) × (1 h / 3600 s)

Cancel:

then:

The remaining unit is:

cm/s

Calculate:

72 × 1000 × 100 / 3600

= 2000

Therefore:

72 km/h = 2000 cm/s

Squared Units

Dimensional analysis becomes especially useful when converting areas.

Remember:

1 m = 100 cm

For area:

1 m² = (100 cm)²

Therefore:

1 m² = 10,000 cm²

The conversion factor must be squared because the unit itself is squared.

Example: Area Conversion

Convert:

3.5 m² → cm²

Use:

3.5 m² × (100 cm / 1 m)²

Square the conversion factor:

3.5 × 10,000 cm²

Therefore:

3.5 m² = 35,000 cm²

Cubed Units

Volume requires the conversion factor to be cubed.

Since:

1 m = 100 cm

then:

1 m³ = (100 cm)³

Therefore:

1 m³ = 1,000,000 cm³

or:

1 m³ = 10⁶ cm³

Example: Volume Conversion

Convert:

250 cm³ → m³

Use:

250 cm³ × (1 m / 100 cm)³

Therefore:

250 × 10⁻⁶ m³

= 2.50 × 10⁻⁴ m³

The cubic units cancel correctly.

Dimensional Analysis in Physics Problems

Dimensional analysis can be used before, during, and after solving a physics problem.

Before

Check that the equation is appropriate.

During

Check that units are compatible.

After

Check that the final units match the quantity being calculated.

This can catch many mistakes before they become final answers.

Worked Example: Force

A student calculates force using:

F = ma

Suppose:

m = 4.0 kg

a = 3.0 m/s²

Look at the units:

kg × m/s²

This gives:

kg·m/s²

which is:

Now calculate:

F = 4.0 × 3.0

F = 12 N

The dimensional check confirms that the final unit is appropriate for force.

Worked Example: Speed

A student travels:

600 m

in:

50 s

Use:

v = d/t

Units:

m/s

Calculation:

v = 600/50

v = 12 m/s

The units match the expected dimensions of speed:

LT⁻¹

Worked Example: Energy

Suppose:

m = 5.0 kg

v = 4.0 m/s

Use:

Eₖ = ½mv²

Check units:

kg × (m/s)²

= kg·m²/s²

This is equivalent to:

Now calculate:

Eₖ = ½(5.0)(4.0)²

Eₖ = 40 J

Identifying an Incorrect Answer Unit

Suppose a student calculates acceleration and reports:

a = 7.5 m/s

Something is wrong.

Acceleration has dimensions:

LT⁻²

Its SI unit should therefore be:

m/s²

The reported unit:

m/s

belongs to velocity.

Even before checking the arithmetic, dimensional reasoning reveals a problem.

Identifying an Incorrect Equation

Suppose a student proposes:

F = mv

Check dimensions.

Right side:

[mv] = M × LT⁻¹

= MLT⁻¹

But force has dimensions:

MLT⁻²

Therefore:

MLT⁻¹ ≠ MLT⁻²

So:

F = mv

is not dimensionally consistent.

The quantity mv represents momentum, not force.

Rearranging Equations and Dimensions

Dimensional analysis can also check rearranged equations.

Suppose:

F = ma

Rearrange for mass:

m = F/a

Check dimensions:

[m] = MLT⁻² / LT⁻²

Cancel:

[m] = M

The result has the dimensions of mass, so the rearrangement is dimensionally consistent.

Example: Rearranging Density

Start with:

ρ = m/V

Rearrange:

m = ρV

Check:

Density:

ML⁻³

Volume:

L³

Therefore:

[ρV] = ML⁻³ × L³

= M

The result has the dimensions of mass.

Dimensional Analysis and Constants

Some physical constants have dimensions.

For example, Newton's law of gravitation is:

F = Gm₁m₂/r²

The gravitational constant G must have dimensions that make the entire equation produce force.

We can determine them.

Rearrange:

G = Fr²/(m₁m₂)

Force:

MLT⁻²

Distance squared:

L²

Mass × mass:

M²

Therefore:

[G] = (MLT⁻²)(L²)/M²

Simplify:

[G] = M⁻¹L³T⁻²

This corresponds to the SI units:

N·m²/kg²

Deriving the Dimensions of Energy

Work is:

W = Fd

Force:

MLT⁻²

Distance:

Therefore:

[W] = MLT⁻² × L

[W] = ML²T⁻²

Since work is an energy transfer, energy has the same dimensions:

[E] = ML²T⁻²

Deriving the Dimensions of Power

Power is:

P = E/t

Energy:

ML²T⁻²

Divide by time:

Therefore:

[P] = ML²T⁻³

Its SI unit is the watt:

W = J/s

Derived Units Can Be Expanded

A named SI unit can often be rewritten using base units.

For example:

Newton

N = kg·m/s²

Joule

J = N·m

Substitute the newton:

J = kg·m²/s²

Pascal

Pa = N/m²

Substitute the newton:

Pa = kg/(m·s²)

Watt

W = J/s

Therefore:

W = kg·m²/s³

Expanding derived units can make dimensional checking easier.

Comparing Two Sides of an Equation

A systematic method is:

Step 1: Write the dimensions of the left side.

Step 2: Write the dimensions of every quantity on the right side.

Step 3: Combine the dimensions using the mathematical operations.

Step 4: Simplify.

Step 5: Compare both sides.

If they match:

dimensionally consistent

If they do not:

dimensionally inconsistent

Worked Check

Is this equation dimensionally consistent?

v² = u² + 2as

Left side:

[v²] = (LT⁻¹)²

= L²T⁻²

First term on right:

[u²] = L²T⁻²

Second term:

[as] = LT⁻² × L

= L²T⁻²

The factor 2 has no dimensions.

Therefore:

L²T⁻² = L²T⁻² + L²T⁻²

The equation is dimensionally consistent.

Worked Check: Finding an Error

Consider:

v² = u² + 2at

Left side:

L²T⁻²

First right-hand term:

L²T⁻²

Second right-hand term:

[at] = LT⁻¹

The equation tries to add:

L²T⁻² + LT⁻¹

These dimensions are different.

Therefore, the equation is dimensionally inconsistent.

Dimensional Reasoning Without Knowing the Formula

Suppose you forget whether the period of a simple pendulum depends on:

L/g

or:

g/L

We can examine dimensions.

Length:

[L] = L

Acceleration:

[g] = LT⁻²

For:

L/g

the dimensions are:

L / (LT⁻²)

= T²

Taking the square root gives:

which is the dimension required for a period.

Therefore, dimensional reasoning suggests a relationship of the form:

period ∝ √(L/g)

Dimensional analysis cannot determine a dimensionless numerical factor, but it can reveal the required dimensional form.

Limits of Dimensional Analysis

Dimensional analysis is powerful, but it cannot tell us everything.

It cannot usually determine:

Numerical constants such as ½ or 2π.
Whether addition or subtraction should be used when terms have the same dimensions.
The exact physical mechanism involved.
Whether an equation is experimentally correct.
Whether two quantities with the same dimensions represent the same physical concept.

For example:

work and torque can both be expressed using:

N·m

but they represent different physical quantities.

Dimensionally Correct but Physically Wrong

Consider:

d = vt

This is dimensionally consistent.

Now consider:

d = 17vt

This is also dimensionally consistent because 17 has no dimensions.

But that does not mean both equations correctly describe the same situation.

Dimensional analysis tests consistency, not complete physical correctness.

Real-World Importance

Dimensional analysis is used throughout science and engineering.

It helps scientists and engineers:

Check equations.
Convert measurement units.
Detect calculation mistakes.
Design mathematical models.
Interpret unfamiliar formulas.
Compare different measurement systems.
Determine possible relationships between variables.
Check computer calculations and simulations.

A simple unit check can sometimes prevent a very large mistake.

The Mars Climate Orbiter

A famous example of the importance of consistent units occurred with NASA's Mars Climate Orbiter in 1999.

One engineering team supplied data using U.S. customary units while spacecraft navigation software expected metric units. The mismatch contributed to an incorrect spacecraft trajectory and the mission was lost.

The incident demonstrates an important lesson:

A calculation can be numerically performed correctly and still produce a disastrous result if the units are wrong.

A Practical Problem-Solving Routine

When solving a physics problem:

Identify

Write down the known quantities and their units.

Convert

Convert quantities into compatible units where necessary.

Select

Choose an appropriate equation.

Check Dimensions

Make sure the equation produces the dimensions of the quantity you are trying to calculate.

Substitute

Insert the numerical values with their units.

Calculate

Perform the arithmetic.

Check

Ask:

Are the final units correct?
Is the magnitude reasonable?
Does the sign make sense?
Does the answer fit the physical situation?

This routine catches many common errors.

Common Mistakes

Checking Only the Numbers

A calculator can perform arithmetic without understanding units.

Always check both:

number + unit

Assuming Matching Units Prove an Equation

Matching dimensions show that an equation could be valid.

They do not prove that it is physically correct.

Ignoring Powers on Units

If:

1 cm = 10⁻² m

then:

1 cm² = 10⁻⁴ m²

and:

1 cm³ = 10⁻⁶ m³

Adding Different Dimensions

You cannot meaningfully add:

5 m + 3 s

Similarly, terms added within a physics equation must have compatible dimensions.

Treating Numerical Constants as Physical Dimensions

Numbers such as:

2, ½ and π

are dimensionless.

They do not change the dimensional form of an equation.

Confusing Units and Dimensions

metre is a unit.

length is a dimension.

second is a unit.

time is a dimension.

Forgetting Compound Units

When converting something such as:

km/h → m/s

both the numerator and denominator must be considered.

Check Your Understanding

1. What is dimensional analysis?

2. Explain the difference between a unit and a dimension.

3. What are the dimensions of length, mass, and time?

4. Determine the dimensions of speed.

5. Determine the dimensions of acceleration.

6. Determine the dimensions of force.

7. Show that d = vt is dimensionally consistent.

8. Explain why d = v/t is dimensionally inconsistent.

9. Show that F = ma is dimensionally consistent.

10. Determine the dimensions of density.

11. Determine the dimensions of pressure.

12. Determine the dimensions of energy.

13. Determine the dimensions of power.

14. Check whether v = u + at is dimensionally consistent.

15. Check whether s = ut + ½at² is dimensionally consistent.

16. Explain why the equation s = ut + ½at cannot be correct.

17. Convert 4.5 km to metres using a conversion factor.

18. Convert 108 km/h to m/s using dimensional analysis.

19. Convert 2.5 m² to cm² using conversion factors.

20. Convert 500 cm³ to m³.

21. A student reports an acceleration of 8.0 m/s. Use dimensional reasoning to identify the problem.

22. Expand the unit N into SI base units.

23. Expand the unit J into SI base units.

24. Why can dimensional analysis identify an incorrect equation but not prove that a dimensionally consistent equation is physically correct?

25. Explain how dimensional analysis can help detect mistakes when solving physics problems.

Key Terms

Dimensional analysis – method of analyzing physical dimensions to check equations, conversions, and calculations.
Dimension – physical nature of a quantity, such as length, mass, or time.
Unit – agreed standard used to measure a physical quantity.
Fundamental dimension – basic dimension such as mass, length, or time.
Derived quantity – quantity constructed from combinations of fundamental quantities.
Dimensional consistency – condition in which the dimensions on both sides of an equation agree.
Conversion factor – ratio of equivalent measurements used to change units.
Dimensionless quantity – quantity with no physical dimensions.
Derived unit – unit constructed from combinations of base SI units.
Unit cancellation – cancellation of identical units in the numerator and denominator during a calculation.

Key Takeaways

Dimensional analysis uses the dimensions and units of physical quantities to examine equations and calculations.
Length, mass, and time are represented by L, M, and T.
Speed has dimensions LT⁻¹.
Acceleration has dimensions LT⁻².
Force has dimensions MLT⁻².
Energy has dimensions ML²T⁻².
Power has dimensions ML²T⁻³.
Density has dimensions ML⁻³.
Pressure has dimensions ML⁻¹T⁻².
Both sides of a valid physical equation must be dimensionally compatible.
Terms being added or subtracted must have compatible dimensions.
Dimensional analysis can reveal missing or incorrect powers in equations.
Conversion factors can be arranged so unwanted units cancel.
Squared units require squared conversion factors.
Cubed units require cubed conversion factors.
Dimensional analysis can detect many equation, calculation, and unit-conversion errors.
A dimensionally inconsistent equation cannot be physically correct as written.
A dimensionally consistent equation is not automatically physically correct.
Dimensionless constants cannot normally be determined through dimensional analysis alone.
Checking units is one of the fastest ways to catch mistakes in physics.

A useful habit for every physics calculation is:

Check the equation → follow the units → calculate → check the final dimensions.