2.9 Dimensional formulae and dimensional equations

DIMENSIONS FORMULAE AND DIMENSIONAL EQUATIONS

The expression which shows how and which of the base quantities represent the dimensions of a physical quantity is called the dimensional formula of the given physical quantity. For example, the dimensional formula of the volume is \(\left[\mathrm{M}^{\circ} \mathrm{L}^{3} \mathrm{~T}^{\circ}\right]\), and that of speed or velocity is \(\left[\mathrm{M}^{\circ} \mathrm{LT}^{-1}\right]\). Similarly, \(\left[\mathrm{M}^{\circ} \mathrm{LT}^{-2}\right]\) is the dimensional formula of acceleration and \(\left[\mathrm{M} \mathrm{L}^{-3} \mathrm{~T}^{\circ}\right]\) that of mass density.

An equation obtained by equating a physical quantity with its dimensional formula is called the dimensional equation of the physical quantity. Thus, the dimensional equations are the equations, which represent the dimensions of a physical quantity in terms of the base quantities. For example, the dimensional equations of volume \([V]\), speed \([v]\), force \([F]\) and mass density \([\rho]\) may be expressed as
\(
\begin{aligned}
&{[V]=\left[\mathrm{M}^{0} \mathrm{~L}^{3} \mathrm{~T}^{0}\right]} \\
&{[v]=\left[\mathrm{M}^{0} \mathrm{~L} \mathrm{~T}^{-1}\right]} \\
&{[F]=\left[\mathrm{M} \mathrm{L} \mathrm{T}^{-2}\right]} \\
&{[\rho]=\left[\mathrm{ML}^{-3} \mathrm{~T}^{0}\right]}
\end{aligned}
\)

Example 1: Calculate the dimensional formula of energy from the equation \(E=\frac{1}{2} m v^2\).

Solution: Dimensionally, \(E=\) mass \(\times\) (velocity \(^2\), since \(\frac{1}{2}\) is a number and has no dimension.
\(
\therefore \quad[E]=\mathrm{M} \times\left(\frac{\mathrm{L}}{\mathrm{T}}\right)^2=\mathrm{ML}^2 \mathrm{~T}^{-2}
\)

Example 2: Test dimensionally if the formula \(t=2 \pi \sqrt{\frac{m}{F / x}}\) may be correct, where \(t\) is time period, \(m\) is mass, \(F\) is force and \(x\) is distance.

Solution: The dimension of force is \(\mathrm{MLT}^{-2}\). Thus, the dimension of the right-hand side is
\(
\sqrt{\frac{\mathrm{M}}{\mathrm{MLT}^{-2} / \mathrm{L}}}=\sqrt{\frac{1}{\mathrm{~T}^{-2}}}=\mathrm{T} \text {. }
\)
The left-hand side is time period and hence the dimension is \(\mathrm{T}\). The dimensions of both sides are equal and hence the formula may be correct.