Checking equations with dimensions

Apples and seconds

You can add $3\ \text{m}$ to $2\ \text{m}$. What is $3\ \text{m}$ plus $2\ \text{s}$? It means nothing. Quantities can be added, subtracted or set equal only if they are of the same kind.

So in a correct equation, every term has the same dimensions. This is the principle of homogeneity, and it is a free check on every formula you write.

Checking an equation

Take $s = ut + \frac{1}{2}at^2$.

• $[s] = [\text{L}]$.
• $[ut] = [\text{L}\,\text{T}^{-1}][\text{T}] = [\text{L}]$.
• $[\frac{1}{2}at^2] = [\text{L}\,\text{T}^{-2}][\text{T}^2] = [\text{L}]$.

All three terms are lengths, so the equation may be right. Now try $s = ut + \frac{1}{2}at^3$: the last term is $[\text{L}\,\text{T}]$, not a length. It is certainly wrong.

Functions need a pure number

What is $\sin(3\ \text{m})$? Or $e^{2\ \text{s}}$? Nothing: a sine, a logarithm or a power of $e$ works on a pure number. So in

$\omega t$ and $t/\tau$ must have no dimensions. That tells us $[\omega] = [\text{T}^{-1}]$ and $[\tau] = [\text{T}]$.

You should find the curve always reaches $37\%$ at $t = \tau$, whatever $\tau$ is. That only works because $\tau$ is a time.

Finding the dimensions of a constant

Homogeneity also tells us what a constant must be. Suppose a speed follows $v = A + Bt^2$.

1. $A$ is added to a speed, so $[A] = [\text{L}\,\text{T}^{-1}]$.
2. $Bt^2$ must also be a speed: $[B][\text{T}^2] = [\text{L}\,\text{T}^{-1}]$, so $[B] = [\text{L}\,\text{T}^{-3}]$.

The limits of the check

A dimension check catches many wrong equations, but it cannot prove one right.

• It cannot see a pure number: $s = ut + at^2$ and $s = ut + \frac{1}{2}at^2$ pass the same check.
• It cannot see a wrong sign, or a missing term that has the right dimensions.

So it can only say "certainly wrong" or "possibly right".

A worked example

A student writes the time for a ball to fall from a height $h$ as $t = \dfrac{2h}{g}$. Is it possible?

1. Left side. $[t] = [\text{T}]$.
2. Right side. $\dfrac{[\text{L}]}{[\text{L}\,\text{T}^{-2}]} = [\text{T}^2]$.

$[\text{T}] \neq [\text{T}^2]$, so the formula is wrong. Taking a square root mends the dimensions: $t = \sqrt{\dfrac{2h}{g}}$, the true answer.

Think first. A dimensionally wrong formula gives a different answer when you change units, say from metres to centimetres. A correct one never does. That is another way to see why the check works.

Practise with new numbers

Try it in Python

The program writes each term of $s = ut + \frac{1}{2}at^2$ as $(\text{M}, \text{L}, \text{T})$ powers and checks that they agree.

Remember these

• Every term of a correct equation has the same dimensions.
• The argument of sin, cos, log or $e$ must be dimensionless.
• Homogeneity gives the dimensions of a constant.
• A dimension check can prove an equation wrong, never right: it misses pure numbers and terms of the right kind.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Is $v = u + at^2$ possible?

2. In $y = A\cos(kx)$, $x$ is a distance. What are the dimensions of $k$?

3. In $F = av + bv^2$ ($F$ a force, $v$ a speed), find the dimensions of $b$.

4. A formula passes the dimension check. Can it still be wrong? Give an example.