KAHIBARO
Table of Contents
Introduction
Dimensional analysis is a method that uses the basic “types” of physical quantities to check equations, derive relationships, and understand how different quantities depend on each other. It does not care about the exact numerical values, but about what kind of quantity you are dealing with, for example length, time, or mass. In this chapter we focus on what dimensions are, how to write them, and how to use them to reason about physical formulas without detailed calculations.
Dimensions versus Units
A dimension tells you what physical kind of quantity you have. A unit is a specific choice for measuring that quantity.
For example, length has the same dimension whether you measure it in meters, centimeters, or kilometers. Time has the same dimension whether you measure it in seconds or hours. Dimensional analysis works at the level of “length,” “time,” “mass,” and so on, not at the level of “meters” or “seconds.”
We usually describe dimensions in terms of a small set of base dimensions. In mechanics, the most important ones are:
| Quantity | Symbol | Dimension symbol |
|---|---|---|
| Length | $L$ | $[L]$ |
| Mass | $M$ | $[M]$ |
| Time | $T$ | $[T]$ |
Here $[L]$ does not mean a unit, such as meter, but “the dimension of length” in an abstract way.
Representing Dimensions
Many physical quantities can be written as products and powers of the base dimensions $[M]$, $[L]$, and $[T]$. For example, if a quantity depends on length and time, its dimension might be written as $[L][T]^{-1}$.
We usually show dimensions using square brackets and exponents:
- Velocity $v$ is distance per time, so
$$
[v] = \frac{[L]}{[T]} = [L][T]^{-1}.
$$ - Acceleration $a$ is velocity per time, so
$$
[a] = \frac{[L][T]^{-1}}{[T]} = [L][T]^{-2}.
$$ - Area $A$ is length times length, so
$$
[A] = [L]^2.
$$
In general we can write a quantity $Q$ as
$$
[Q] = [M]^a [L]^b [T]^c,
$$
where $a$, $b$, and $c$ are numbers (often integers, but not always).
Important: Dimensions are not numbers. They do not add or subtract. They multiply and divide, and their exponents add or subtract just like in algebra.
Dimensions of Common Mechanical Quantities
It is very useful to know, or to be able to derive, the dimensions of frequently used quantities. You can always start from their defining formulas.
| Quantity | Symbol | Definition (symbolic) | Dimension |
|---|---|---|---|
| Length | $x$ | basic | $[L]$ |
| Time | $t$ | basic | $[T]$ |
| Mass | $m$ | basic | $[M]$ |
| Velocity | $v$ | $v = \dfrac{\Delta x}{\Delta t}$ | $[L][T]^{-1}$ |
| Acceleration | $a$ | $a = \dfrac{\Delta v}{\Delta t}$ | $[L][T]^{-2}$ |
| Area | $A$ | $A = l \cdot w$ | $[L]^2$ |
| Volume | $V$ | $V = l \cdot w \cdot h$ | $[L]^3$ |
| Frequency | $f$ | $f = \dfrac{1}{T_{\text{period}}}$ | $[T]^{-1}$ |
| Force | $F$ | $F = m a$ | $[M][L][T]^{-2}$ |
| Work, Energy | $W,E$ | $W = F s$ | $[M][L]^2[T]^{-2}$ |
| Power | $P$ | $P = \dfrac{W}{t}$ | $[M][L]^2[T]^{-3}$ |
| Pressure | $p$ | $p = \dfrac{F}{A}$ | $[M][L]^{-1}[T]^{-2}$ |
| Momentum | $p$ | $p = m v$ | $[M][L][T]^{-1}$ |
| Density | $\rho$ | $\rho = \dfrac{m}{V}$ | $[M][L]^{-3}$ |
You should be able to derive each entry by substituting the definitions step by step.
Dimensional Homogeneity
A physically meaningful equation must be dimensionally homogeneous. This means that every term that is added or equated must have the same dimensions.
For example, consider the kinematic equation in one dimension:
$$
x = x_0 + v_0 t + \tfrac{1}{2} a t^2.
$$
We check dimensions term by term.
- $[x] = [L]$.
- $[x_0] = [L]$.
- $[v_0 t] = [v_0][t] = [L][T]^{-1}[T] = [L]$.
- $[a t^2] = [a][t]^2 = [L][T]^{-2}[T]^2 = [L]$.
Every term has dimension $[L]$, so the equation is dimensionally homogeneous.
If someone wrote instead
$$
x = x_0 + v_0 + a t^2,
$$
then $[v_0] = [L][T]^{-1}$ but $[x_0] = [L]$, so you would be adding a length and a velocity, which is dimensionally inconsistent.
Rule: In any valid physical equation, all terms that are added or compared (for example, on both sides of an equality) must have the same dimensions.
You can use this as a quick check on formulas you derive or that you see in textbooks or problems.
Dimensionless Quantities
Not all physical expressions have dimensions. Some are dimensionless, which means their overall dimension is $[M]^0 [L]^0 [T]^0$, often written simply as 1.
Some important examples are:
- Pure numbers such as $\pi$ or 2.5.
- Ratios of similar quantities, such as $\dfrac{x}{x_0}$, where both $x$ and $x_0$ are lengths, so their ratio is dimensionless.
- Trigonometric functions like $\sin \theta$ and $\cos \theta$, where the angle $\theta$ must be expressed in radians. A radian is not a separate dimension, it is a ratio of lengths, so it is dimensionless.
- Exponents in expressions like $\exp(y)$ or $e^{y}$, where $y$ must be dimensionless.
Rule: Arguments of functions like $\sin(\cdot)$, $\cos(\cdot)$, $\exp(\cdot)$, and logarithms must be dimensionless. If they are not, the expression is not physically meaningful.
Dimensionless combinations often appear as important physical parameters, such as Reynolds number in fluid mechanics, but those will be discussed in later sections of the course.
Checking Equations with Dimensional Analysis
Dimensional analysis cannot prove that an equation is correct, but it can quickly show if an equation is definitely wrong.
The procedure is:
- Identify the dimension of each symbol in the equation.
- Replace each symbol by its dimension inside $[\;]$.
- Manipulate the dimensions using algebra of exponents.
- Check that all terms on both sides have the same dimensions.
Consider an example involving a force $F$ applied to a spring with extension $x$ and spring constant $k$. Suppose you see the formula
$$
F = k x^2.
$$
From physics, the correct linear relation is $F = k x$, but imagine you do not remember it.
We know $[F] = [M][L][T]^{-2}$ and $[x] = [L]$. Suppose we do not know $[k]$ and we want to find it.
Assuming $F = k x^2$ is correct, then
$$
[F] = [k][x]^2.
$$
So,
$$
[M][L][T]^{-2} = [k][L]^2.
$$
Rearranging,
$$
[k] = [M][L]^{-1}[T]^{-2}.
$$
Now compare this with what we get from the standard linear formula $F = k x$:
$$
[F] = [k][x] \quad \Rightarrow \quad [M][L][T]^{-2} = [k][L],
$$
so
$$
[k] = [M][T]^{-2}.
$$
The two possible dimensions for $k$ disagree. Only one can be correct in nature. The simple linear law has extensive experimental support, so $[k] = [M][T]^{-2}$ is correct, and the formula $F = k x^2$ has incompatible dimensions and cannot represent Hooke’s law.
Deriving Forms of Relationships
Sometimes you can use dimensional analysis to guess the form of a physical law, up to a dimensionless constant. Suppose you suspect that the period $T$ of a simple pendulum depends on its length $L$, the mass $m$ at the end, and the gravitational acceleration $g$. You want to know how $T$ might depend on these quantities.
Assume
$$
T \propto L^{a} m^{b} g^{c},
$$
or more generally
$$
T = C L^{a} m^{b} g^{c},
$$
where $C$ is a dimensionless constant and $a$, $b$, $c$ are unknown exponents.
Write dimensions of each quantity:
- $[T] = [T]$.
- $[L] = [L]$.
- $[m] = [M]$.
- $[g] = [L][T]^{-2}$.
So
$$
[T] = [L]^a [M]^b ([L][T]^{-2})^c = [L]^{a+c} [M]^b [T]^{-2c}.
$$
Now match exponents of the base dimensions on both sides.
On the left side, $[T]^1$ but no $[M]$ or $[L]$. So:
- For $[M]$: exponent must be zero, so $b = 0$.
- For $[L]$: exponent must be zero, so $a + c = 0$.
- For $[T]$: exponent must be 1, so $-2c = 1$, hence $c = -\tfrac{1}{2}$.
Then $a = -c = \tfrac{1}{2}$. So we find
$$
T = C \, L^{1/2} g^{-1/2} = C \sqrt{\frac{L}{g}}.
$$
Experiments show that for small oscillations $C = 2\pi$, giving
$$
T = 2\pi \sqrt{\frac{L}{g}}.
$$
Dimensional analysis gave us the structure $\sqrt{L/g}$ and showed that the mass does not appear in the formula at all. It cannot predict the numerical factor $2\pi$.
Rule: Dimensional analysis can determine how a quantity scales with others (powers of $L$, $M$, $T$, etc.), but it cannot determine pure numerical constants like $2$, $\pi$, or $\sqrt{2}$.
Using Dimensional Analysis to Build New Quantities
Sometimes you want to construct a quantity that has a certain dimension, for example an energy-like quantity from other variables you already know. Dimensional analysis can guide you.
Imagine you know a mass $m$ and a speed $v$ and you want to construct a quantity with the dimensions of energy $[M][L]^2[T]^{-2}$ using only $m$ and $v$.
Try a general combination
$$
E \propto m^{a} v^{b}.
$$
Write dimensions:
- $[m] = [M]$.
- $[v] = [L][T]^{-1}$.
Then
$$
[E] = [M]^a ([L][T]^{-1})^b = [M]^a [L]^b [T]^{-b}.
$$
We want $[E] = [M]^1 [L]^2 [T]^{-2}$. So we match exponents:
- For $[M]$: $a = 1$.
- For $[L]$: $b = 2$.
- For $[T]$: $-b = -2$, which is consistent with $b = 2$.
Thus
$$
E \propto m^1 v^2 = m v^2,
$$
which suggests the kinetic energy has the form $E = C m v^2$. From experiments and more detailed theory, we learn the constant is $C = \tfrac{1}{2}$, giving $E = \tfrac{1}{2} m v^2$.
Limitations of Dimensional Analysis
Dimensional analysis is powerful but limited.
First, it cannot distinguish between different dimensionally correct expressions. For example, both
$$
s = v t \quad \text{and} \quad s = \tfrac{1}{2} a t^2
$$
have the correct dimension for a distance $s$ if you choose appropriate meanings for $v$ or $a$, but they represent different physical situations.
Second, it cannot give numerical factors or dimensionless constants. It can tell you $s \propto t^2$ but not whether $s = \tfrac{1}{2} a t^2$ or $s = 3 a t^2$.
Third, if more than one dimensionless combination can be built from the given variables, dimensional analysis alone may not narrow down the correct form. You then need physical reasoning or experimental data.
Despite these limitations, dimensional analysis is extremely useful as a quick check and as a way to get insight into how quantities must be related.
A Simple Diagram of Dimension Combinations
The following schematic shows how base dimensions combine to produce common physical quantities. It is not a graph in the numerical sense, but a conceptual map.
This drawing helps you visualize how more complex quantities are built from the base dimensions through multiplication and division.
Summary
Dimensional analysis treats physical quantities in terms of their basic dimensions, typically mass, length, and time in mechanics. By representing each quantity as a product of powers of these base dimensions, you can check whether equations are dimensionally homogeneous, detect likely errors, and often determine how one quantity must depend on others up to a dimensionless constant.
It cannot replace full physical theory, but it is a powerful tool for sanity checking formulas and for getting the right “shape” of relationships between physical quantities before you know all the details.