Dimensional correctness and scaling

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Dimensional correctness and scaling

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[QUICK DESCRIPTION]
Physical quantities and models need to be dimensionally consistent.  See also [[dimensional analysis]].

[PREREQUISITES]
Calculus, basic algebra

[EXAMPLE]
From Newton's second law and basic assumptions, we might have derived the equation for a spring acting under gravity:
[maths] m\frac{d^2 y}{dt^2} = -mg - ky [/maths]
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 $2\pi\sqrt{m/k}$.

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Dimensional correctness and scaling

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