Testing Equations with Dimensional Analysis

Hello again. In the previous lesson, you represented constant-velocity motion as a formula, table, and graph, and rearranged

$$x=x_0+vt.$$

You also saw a small but important clue: when calculating $vt$, the seconds in velocity and time cancel, leaving metres. This lesson turns that observation into a systematic error-checking method: dimensional analysis.

By the end, you will be able to replace physical quantities in an equation with their dimensions, simplify them algebraically, and decide whether the equation is dimensionally consistent. This is a necessary first check before trusting any equation you remember, derive, or rearrange.

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Dimensions are not the same as units

A unit is a chosen standard for measuring a quantity. For example, length can be expressed in metres, kilometres, feet, or miles.

A dimension describes the underlying physical kind of quantity. Regardless of whether you measure a distance in metres or miles, it has the dimension of length.

For much of introductory mechanics, three base dimensions are enough:

Physical base quantity	Dimension symbol	Typical SI unit
Length	$L$	metre, $\mathrm{m}$
Mass	$M$	kilogram, $\mathrm{kg}$
Time	$T$	second, $\mathrm{s}$

Other quantities are built from these. For example:

Quantity	Meaning	Dimensions	SI units
Area	length times length	$L^2$	${\mathrm{m}}^2$
Volume	length cubed	$L^3$	${\mathrm{m}}^3$
Velocity	displacement per time	$L{T}^{-1}$	$\mathrm{m}{\mathrm{s}}^{-1}$
Acceleration	change of velocity per time	$L{T}^{-2}$	$\mathrm{m}{\mathrm{s}}^{-2}$
Density	mass per volume	$M{L}^{-3}$	$\mathrm{kg}{\mathrm{m}}^{-3}$
Force	mass times acceleration	$ML{T}^{-2}$	$\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2}$

Square brackets mean “the dimensions of.” Thus,

$$[v]=L{T}^{-1}$$

means “velocity has dimensions length divided by time.”

A useful distinction is:

$$[v]=L{T}^{-1}$$

versus

$$v=12\mathrm{m}{\mathrm{s}}^{-1}.$$

The first describes velocity in general; the second gives one particular measured value.

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The rule of dimensional consistency

A physical equation must be dimensionally consistent, sometimes called dimensionally homogeneous. This means two things.

1. Both sides of an equals sign must have the same dimensions.
2. Terms being added or subtracted must all have the same dimensions.

For instance, the expression

$$3\mathrm{m}+2\mathrm{s}$$

is meaningless: it tries to add a length and a time. The numbers do not solve the problem. It is the dimensions that are incompatible.

By contrast,

$$2.0\mathrm{m}+30\mathrm{cm}$$

is meaningful because both quantities are lengths. You must still convert to common units before doing the numerical addition:

$$2.0\mathrm{m}+0.30\mathrm{m}=2.30\mathrm{m}.$$

This exposes an important limitation. Dimensional analysis tests physical type, not numerical unit conversion. It will tell you that metres and centimetres are compatible, but it will not perform the conversion for you.

A reliable checking routine

When you are given a proposed equation:

1. Write the dimension of the quantity on the left-hand side.
2. Replace every variable on the right-hand side by its dimensions.
3. Treat multiplication, division, and powers as ordinary algebra.
4. Simplify every separate term.
5. Compare dimensions across the equality sign and across any additions or subtractions.

Ignore pure numerical factors such as $2$, $1/2$, and $\pi$, as well as a plus or minus sign. They carry no dimensions.

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Checking a motion equation

Return to the relation from the last lesson:

$$x=x_0+vt.$$

Here, position $x$ and initial position $x_0$ are lengths, velocity $v$ has dimensions $L{T}^{-1}$, and time $t$ has dimension $T$.

Check each term:

$$[x]=L$$ $$[x_0]=L$$ $$[vt]=[v][t]$$ $$[vt]=\left(L{T}^{-1}\right)(T)=L.$$

Every term has dimension $L$, so the equation is dimensionally consistent:

$$L=L+L.$$

That does not merely mean that the units happen to look tidy. It means the equation is structurally capable of describing a position.

Now consider a proposed formula for displacement:

$$x=v+\frac{1}{2}at.$$

Check its right-hand side:

$$[v]=L{T}^{-1}$$

and

$$[at]=\left(L{T}^{-2}\right)(T)=L{T}^{-1}.$$

Both terms on the right have dimensions of velocity, not length. So the equation is internally consistent on the right, but it fails to match the left:

$$L\ne L{T}^{-1}.$$

It cannot be an equation for displacement.

Compare it with the correct constant-acceleration form when the initial position is chosen as zero:

$$x=vt+\frac{1}{2}a{t}^2.$$

The first term is

$$[vt]=\left(L{T}^{-1}\right)(T)=L.$$

The second term is

$$\left[\frac{1}{2}a{t}^2\right]=\left(L{T}^{-2}\right)\left(T^2\right)=L.$$

Thus,

$$L=L+L.$$

The equation passes the dimensional check.

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A force example: checking Newton’s second law

Newton’s second law is

$$F=ma.$$

Force is measured in newtons. In base SI units,

$$1\mathrm{N}=1\mathrm{kg}\mathrm{m}{\mathrm{s}}^{-2},$$

so the dimensions of force are

$$[F]=ML{T}^{-2}.$$

The right-hand side has dimensions

$$[ma]=[m][a].$$

Since mass has dimension $M$ and acceleration has dimension $L{T}^{-2}$,

$$[ma]=M\left(L{T}^{-2}\right)=ML{T}^{-2}.$$

Therefore,

$$[F]=[ma]=ML{T}^{-2}.$$

Newton’s second law is dimensionally consistent.

Now examine a tempting but incorrect alternative:

$$F=mv.$$

The right-hand side has dimensions

$$[mv]=M\left(L{T}^{-1}\right)=ML{T}^{-1}.$$

But force requires $ML{T}^{-2}$, not $ML{T}^{-1}$. One factor of inverse time is missing:

$$ML{T}^{-2}\ne ML{T}^{-1}.$$

Dimensional analysis catches the error immediately, even before any values are substituted.

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Passing the test does not prove an equation is true

This point is essential: dimensional consistency is a necessary condition for a physical equation, but it is not a sufficient condition.

Consider these two expressions:

$$x=ut+\frac{1}{2}a{t}^2$$

and

$$x=7ut+4a{t}^2.$$

Both are dimensionally consistent, because $ut$ and $a{t}^2$ each have the dimension $L$, while $7$, $4$, and $1/2$ are dimensionless numbers.

But only the first expression is the standard constant-acceleration equation. Dimensional analysis cannot determine the numerical coefficients $1/2$, $7$, or $4$. It also cannot reveal whether an equation has omitted a relevant physical quantity if the remaining expression still has the right dimensions.

So use the test with the correct level of confidence:

• If an equation fails, it is definitely wrong, incomplete, or being used for the wrong quantity.
• If an equation passes, it is plausible, but it still needs justification from experiment, derivation, or known physical laws.

Think of dimensional analysis as a strict entry requirement, not a proof of admission.

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Dimensionless quantities and mathematical functions

A quantity is dimensionless if all its base-dimension powers cancel:

$$L^0{M}^0{T}^0=1.$$

Examples include a ratio of two lengths, such as $x/L$, or a numerical coefficient such as $2$.

There is a further rule that becomes increasingly useful in calculus, waves, and quantum physics: the input to functions such as sine, cosine, exponential, and logarithm must be dimensionless.

For example, an oscillation might be written as

$$x=A\sin (\omega t).$$

The sine function produces a dimensionless number between $-1$ and $1$. Therefore $\omega t$ must be dimensionless. Since time has dimension $T$, angular frequency must have dimensions

$$[\omega ]=T^{-1}.$$

Then the dimensions of the right-hand side are simply those of amplitude $A$:

$$[A\sin (\omega t)]=[A].$$

Thus, if $x$ is a position, $A$ must have dimension $L$.

You do not yet need to manipulate sine functions in detail. The key habit is to notice their arguments. An expression such as

$$\sin (t)$$

is not physically meaningful when $t$ is an ordinary time measured in seconds. A dimensionless combination such as $\omega t$ is needed instead.

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A compact dimensional-analysis template

For your physics notes, use this layout whenever you check an equation:

$$\text{Claim: }\text{physical quantity}=\text{expression}$$ $$[\text{left side}]=\text{dimensions}$$ $$[\text{first right-hand term}]=\text{dimensions}$$ $$[\text{second right-hand term}]=\text{dimensions}$$ $$\text{Verdict: consistent or inconsistent}.$$

For the kinematic equation,

$$x=ut+\frac{1}{2}a{t}^2,$$

that template becomes

$$[x]=L$$ $$[ut]=\left(L{T}^{-1}\right)(T)=L$$ $$[a{t}^2]=\left(L{T}^{-2}\right)\left(T^2\right)=L.$$

Verdict: all terms have dimensions of length, so the equation is dimensionally consistent.

Use this check after rearranging an equation as well. It is particularly good at detecting a missing power of time, an accidental multiplication in place of division, or an attempt to add incompatible quantities.

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Wrap-up

You now have a fast, general test for physical equations:

• Dimensions describe physical type, while units are specific measurement standards.
• In mechanics, many quantities can be expressed using length $L$, mass $M$, and time $T$.
• Every term in a sum or difference must share dimensions.
• The two sides of a physical equation must have matching dimensions.
• Arguments of sine, cosine, logarithmic, and exponential functions must be dimensionless.
• A failed dimensional check proves that an equation cannot be correct; a passed check only shows that it is possible in form.

Next, you will connect this unit-aware approach to graphs by interpreting slope and signed area physically. In particular, you will see why the units of a graph’s slope and the units of its area reveal what physical quantities they represent.