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Dimensional correctness and scaling

Quick description

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


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}.