Dimensions and consistencyAQA A-Level Further Maths: Revision notes

Every mechanical quantity is built from three base dimensions: mass $M$, length $L$ and time $T$. The dimensions of a quantity $x$ are written $[x]$. For example $[\text{mass}]=M$, $[\text{distance}]=L$, $[\text{time}]=T$. Dimensions describe the kind of quantity, not its units: speed has dimensions $LT^{-1}$ whether it is measured in m s⁻¹ or km h⁻¹. To find the dimensions of a new quantity, write down a defining equation, substitute the dimensions of each symbol, and combine the powers. When quantities are multiplied their powers of $M$, $L$ and $T$ add; when divided, they subtract.

Build these from definitions and learn the results:

• Speed or velocity: $LT^{-1}$; acceleration: $LT^{-2}$.
• Force: $F=ma$, so $[F]=MLT^{-2}$.
• Momentum $mv$ and impulse $Ft$: $MLT^{-1}$.
• Work, energy: force $\times$ distance, so $ML^{2}T^{-2}$; power: energy $\div$ time, so $ML^{2}T^{-3}$.
• Density: $ML^{-3}$; pressure (force per area): $ML^{-1}T^{-2}$.
• Frequency or angular speed: $T^{-1}$.
• Spring stiffness $k$ (from $F=kx$): $MT^{-2}$. A quantity with no dimensions (power zero for $M$, $L$ and $T$) is dimensionless. Pure numbers such as $\frac12$ or $2\pi$, angles in radians, ratios such as the coefficient of restitution $e$, and the argument of any $\sin$, $\ln$ or exponential function are all dimensionless.

An equation is dimensionally consistent if both sides have the same dimensions, and every term added or subtracted has the same dimensions. Only quantities with the same dimensions can be added, subtracted or equated. Method: find the dimensions of each term separately, then compare. Example: is $s=ut+\frac12at^2$ consistent? $[s]=L$, $[ut]=LT^{-1}\times T=L$ and $[\tfrac12at^2]=LT^{-2}\times T^2=L$. All three terms are $L$, so it is consistent. Example: $E=\frac12mv$ for kinetic energy: $[mv]=MLT^{-1}$, but $[E]=ML^{2}T^{-2}$, so it cannot be correct. Dimensional consistency is a necessary test but not a sufficient one: it cannot detect a wrong numerical factor. $s=ut+at^2$ is also consistent, yet incorrect.

If you are told which quantities a result depends on, you can find the powers by dimensional analysis. Write the quantity as a product $x=\lambda\,a^{p}b^{q}c^{r}$ where $\lambda$ is a dimensionless constant, substitute dimensions, and equate the power of each of $M$, $L$ and $T$ on both sides. Example: the period $\tau$ of a simple pendulum depends on its length $l$, mass $m$ and $g$. Then $T=L^{a}M^{c}\left(LT^{-2}\right)^{b}$. Equate: $M$: $c=0$; $T$: $1=-2b$ so $b=-\frac12$; $L$: $0=a+b$ so $a=\frac12$. Hence $\tau=\lambda\sqrt{\frac{l}{g}}$. The method gives the form of the result and the powers, but not the value of $\lambda$; that needs an experiment or a full calculation.

Rearrange the given formula to make the constant the subject, then substitute dimensions. Example: Newton's law of gravitation $F=\frac{Gm_1m_2}{r^2}$. Then $G=\frac{Fr^2}{m_1m_2}$, so $[G]=\frac{MLT^{-2}\times L^2}{M^2}=M^{-1}L^{3}T^{-2}$. Example: air resistance $R=kv^2$ gives $[k]=\frac{MLT^{-2}}{L^2T^{-2}}=ML^{-1}$. You can then use these dimensions to test other formulae. For terminal speed $V=\sqrt{\frac{mg}{k}}$: $\left[\frac{mg}{k}\right]=\frac{MLT^{-2}}{ML^{-1}}=L^2T^{-2}$, so $[V]=LT^{-1}$, as required.