Dimensional correctness and scaling

Quick description

Physical quantities and models need to be dimensionally consistent. See also dimensional analysis.

Prerequisites

Calculus, basic algebra

Example 1

From Newton's second law and basic assumptions, we might have derived the equation for a spring acting under gravity:

$m\frac{d^2 y}{dt^2} = -mg - ky$

where $y(t)$ is the height of the mass, $m$ its mass, $k$ the spring constant of the spring, $g$ is the gravitational acceleration at the Earth's surface, and $t$ is (of course) time.

Only quantities with the same dimensional units can be added. The dimensional units of a product is the product of the dimensional units: $[pq]=[p]\,[q]$ .

General discussion

The quantity $g$ must have the units of acceleration. If we use $[q]$ to denote the physical units of a quantity $q$ , then $[t]=T$ ( $T$ representing the units of time), $[m]=M$ (representing the units of mass), and $[y]=L$ (representing the units of length). Thus the dimensional units of $g$ are $[g]=L T^{-2}$ .

The spring constant $k$ must have units $[k]=M T^{-2}$ .

If we are looking for the period of oscillation of the spring-mass system, then we need to look for something that has units $T$ . Since $y(t)$ is an a priori unknown function of $t$ , our period must depend only on the other quantities: $m$ , $k$ , $g$ . Now $[m^\alpha k^\beta g^\gamma]=T$ means that $M^\alpha(M T^{-2})^\beta(L T^{-2})^\gamma=T$ . Since $M$ , $L$ and $T$ are independent units, this comes down to the equations for $M$ : $\alpha+\beta=0$ , for $L$ : $\gamma=0$ , and for $T$ : $-2\beta-2\gamma=1$ . This has the solution $\alpha=1/2$ , $\beta=-1/2$ and $\gamma=0$ . Thus we should be looking for a dimensionless multiple of $\sqrt{m/k}$ . Indeed the period is $2\pi\sqrt{m/k}$ .