Units & Measurements
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Units & Measurements - 02

Dimensional Analysis

In physics, when we speak of the dimensions of a quantity, we are referring to the type of base units or base quantities that make it up.

Table 1: Dimensions of some Quantities
Quantity	Unit	Dimension
area `A`	m²	[`L²`]
volume `V`	m³	[`L³`]
density `ρ`	kg/m³	[`ML⁻³`]
speed `v`	m/s	[`LT⁻¹`]
acceleration `a`	m/s²	[`LT⁻²`]
force `F`	kg m/s² or N	[`MLT⁻²`]
pressure `p`	N/m²	[`ML⁻¹T⁻²`]
energy `E`	N m or J	[`ML²T⁻²`]
power `P`	J/s or W	[`ML²T⁻³`]

The dimensions of length, mass, time, temperature, current, amount of substance and luminous intensity are denoted as [L], [M], [T], [K], [A], [mol], and [cd] respectively.

A dimensional formula shows how a combination of base quantities and their powers represent the dimensions of a physical quantity. Dimensions of some derived quantities are displayed in Table .

The mathematical formula for a physical quantity may be different in different cases, but the dimensions remain the same. For example, the area of a triangle and that of a circle have different formulas, but the dimensions of area will be the same, viz. $[L^{- 2}]$ .

When you equate a physical quantity with its dimensional formula, you get what is called a dimensional equation. For e.g. $[Area] = [L^{2}]$ .

Dimensions can be used as a help in working out relationships, a procedure referred to as dimensional analysis. Note that we add or subtract quantities only if they have the same dimensions; and the quantities on each side of an equals sign must have the same dimensions. Also, dimensions cancel just like algebraic quantities, and pure numerical factors have no dimensions, so they can be ignored.

The three main uses of dimensional analysis are:

To convert a physical quantity from one system of units to another.
To check for the correctness of a given physical relation.
To derive a relationship between different physical quantities.

According to the principle of homogeneity, the dimensions of each term on both sides of an equation must be the same. This is used in the point (2) above.

Conversion between unit systems

A physical quantity $Q = n \cdot u$ , where n is a real number and u the unit of measurement. If $n_{1}$ is the numerical value corresponding to unit $u_{1}$ , and $n_{2}$ for unit $u_{2}$ , then

$Q = n_{1} u_{1} = n_{2} u_{2} = constant$

If $m_{u}$ , $l_{u}$ and $t_{u}$ generically represent the sizes of the fundamental units of mass, length and time in any system, then

$n_{1} [{(m_{1})}^{a} \cdot {(l_{1})}^{b} \cdot {(t_{1})}^{c}] = n_{2} [{(m_{2})}^{a} \cdot {(l_{2})}^{b} \cdot {(t_{2})}^{c}]$

Thus,

$n_{2} = n_{1} {(\frac{m_{1}}{m_{2}})}^{a} {(\frac{l_{1}}{l_{2}})}^{b} {(\frac{t_{1}}{t_{2}})}^{c}$

.......(1)

Note that the sizes should be in terms of the units of the desired system (system 2). For e.g, for conversion from SI to CGS, the sizes should be used in terms of CGS units.

Solved Examples

Test dimensionally if the formula

t = 2 π \sqrt{\frac{m}{F / x}}

may be correct, where t is time period, m is mass, F is force and x is distance. (Verma 2019, Physics vol 1, p. 5 #Eg1.2)

The LHS of the formula is time period, and hence the dimension is $[T]$

For the RHS, the quantity $2 π$ is dimensionless. The dimension of force is $[M L T^{- 2}]$ . Thus, the dimension of the right-hand side is:

$\sqrt{\frac{[M]}{[M L T^{- 2}] / [L]}} = \sqrt{\frac{1}{[T^{- 2}]}} = [T]$

Since both sides match dimensionally, the formula is dimensionally correct.

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The SI and CGS units of energy are joule and erg respectively. How many ergs are equal to one joule? (Verma 2019, Physics vol 1, p. 7 #3)

Dimensionally, energy = [ML²T⁻²]. Now,

1 joule = (1 kg) (1 m)² (1 s)⁻² and 1 erg = (1 g) (1 cm)² (1 s)⁻²

Thus,

$\begin{array}{l} \frac{1 joule}{1 erg} = (\frac{1 kg}{1 g}) {(\frac{1 m}{1 cm})}^{2} {(\frac{1 s}{1 s})}^{- 2} \\ = (\frac{1000 g}{1 g}) {(\frac{100 cm}{1 cm})}^{2} = 10^{7} \end{array}$

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A famous relation in physics relates 'moving mass' m to the 'rest mass' m₀ of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes :

m = \frac{m_{0}}{{(1 - v^{2})}^{1 / 2}}

What should be the position of the missing c? (NCERT 2019, Physics-I XI, p. 36 #2.15)

The velocity of light, c, has dimensions [LT⁻¹]. For the formula to be dimensionally consistent, the dimensions of both the LHS and RHS should be the same. Now, the dimension of the LHS is [M], which then should be the dimension of the RHS. Since the numerator of the RHS already has the dimension [M], it implies that the denominator is dimensionless. This is possible only if the expression within the parentheses, $(1 - v^{2})$ , is dimensionless. Since the expression involves subtraction, and the first term, i.e. 1, is without dimensions, the second term should also be so. However, the second term is a square of velocity, with dimensions [LT⁻¹]², and so it can be made dimensionless only when it is divided by the square of another velocity, in this case, the square of c. Hence, the correct formula is:

m = \frac{m_{0}}{{(1 - \frac{v^{2}}{c^{2}})}^{1 / 2}}

▼(show solution)

If the centripetal force is of the form

m^{a} v^{b} r^{c}

find the values of a, b and c. (Verma 2019, Physics vol 1, p. 8 #8)

Since the dimensions of force is [MLT⁻²], then the dimensional equation for the centrifugal force will be:

$[M L T^{- 2}] = {[M]}^{a} {[L T^{- 1}]}^{b} {[L]}^{c} Proceeding to simplify by equating coefficients with same bases, \begin{array}{l} [M^{1} L^{1} T^{- 2}] = [M^{a}] [L^{b+ c}] [T^{- b}] \\ \Rightarrow a = 1, b + c = 1, - b = - 2 \\ \Rightarrow a = 1, b = 2, c = - 1 \end{array}$

Thus, $F = m v^{2} r^{- 1} = \frac{m v^{2}}{r}$

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If velocity, time and force were chosen as basic quantities, find the dimensions of mass. (Verma 2019, Physics vol 1, p. 7 #5)

Dimensionally,

$force = mass \times acceleration = mass \times \frac{velocity}{time}$

Thus,

$mass = \frac{force \times time}{velocity} \Rightarrow [mass] = [F T V^{- 1}]$

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List of References

NCERT, Physics Part I – Textbook for Class XI, New Delhi, India: Publication Division, 2019.

Verma, HC, Concepts of Physics, vol. 1, Patna, India: Bharati Bhawan, 2019.

Bibliography

Arora, SL, New Simplified Physics – A Reference Book for Class XI, 8th edn, New Delhi, India: Dhanpat Rai & Co., 2019.

Cutnell, JD & Johnson, KW, Physics, 9th edn, USA: John Wiley & Sons, 2012.

Giancoli, DC, Physics – Principles with Applications, 7th edn, NJ, USA: Pearson, 2014.

Serway, RA & Jewett, JW, Principles of Physics – A Calculus-based Text, 5th edn, Boston, USA: Brooks/Cole, 2013.

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Units & Measurements

Dimensional Analysis

Units & Measurements - 02

Dimensional Analysis

Conversion between unit systems

Solved Examples

Feedback Form▼(open form)

List of References

Bibliography

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