Dimensions And Dimensional Formula

Dimensions and dimensional formula

Dimensions are the power to which the fundamental quantity is raised to represent any physical quantity. The dimension and unit of seven fundamental quantities can be written as:

Fundamental QuantityUnitDimension
length meter (m)[L]
masskilogram (kg)[M]
timesecond (s)[T]
temperatureKelvin (K)[K]
currentAmpere (A)[A]
luminous intensityCandela (cd)[J]
amount of substanceMole (mol)[N]

Let’s take an example to clarify dimension:

The physical quantity volume is calculated by length \times length \times length = (length)^3. So volume has the dimension of 3 in length and 0 in other fundamental quantities.

One more example:

    \[ \text{velocity} = \frac{\text{displacement}}{\text{time}} \]

    \[ [\text{velocity}] = \frac{[L]}{[T]} \]

    \[ [\text{velocity}] = [LT^{-1}] `\]

So, velocity has dimensions 1 in length and -1 in time. Thus, velocity is dependent to two fundamental quantities length and time.

Here [LT^{-1}] is called the dimensional formula of velocity.

Dimensional formula

The dimensional formula of a physical quantity is an expression that shows exactly how and which fundamental quantities are involved to obtain that physical quantity. for.eg: dimensional formula of density is [ML^{-3}], dimensional formula of work done is [ML^2T^{-2}], etc.

To see dimensional formula of other physical quanitties, check out this page.

Classification of physical quantities based on dimensions

Based on the dimensional analysis, there are four types of physical quantities.

a) Dimensional variables

These types of physical quantities have dimensions and do not have fixed values or constant values. For e.g: speed, acceleration, power, etc.

b) Dimensionless variables

These types of physical quantities do not have any dimensions and fixed values. For e.g: angle, specific gravity, refractive index, coefficient of friction, etc.

c) Dimensional constants

These types of physical quantities have dimensions and they are constants i.e. they have fixed values. For e.g: Gravitational constant, Universal gas constant, Boltzmann constant, Planck’s constant, etc.

d) Dimensionless constants

The constant value numbers are dimensionless constants. for eg. \pi, e, real numbers, etc.

Principle of homogeneity

This principle states that the dimensions of every term of a correct physical relation must be the same. In other words, every term, of both sides must have the same dimension in order for the physical relation to be correct.

This principle is used to check the correctness of any expression which is one of the applications of dimensional formula. Suppose, a physical relation A = B + C is to be checked. This relation is correct only if:

    \[ \text{dimensions of } A = \text{dimensions of } B = \text{dimensions of } C \]

Applications of dimensional formula or equation

Writing the dimensional formula of each term of any physical equation gives the dimensional equation. And there are some great applications of dimensional equations. Some are listed below.

  1. To check the correctness of any equation.
  2. To find the dimensional formula of dimensional constants.
  3. To convert one system of unit to other system of unit.
  4. To derive the relationsship between physical quanitties.

1. To check the correctness of any equation.

We use the principle of homogeneity to check the correctness of a given physical relation. This principle states that the dimensions of each term in an equation must be the same for the physical relation to be correct.

Let’s check whether the equation s = ut + \frac{1}{2}at^2 is correct or not.

    \[ s = ut + \frac{1}{2}at^2 \]

Let’s write the dimensional formula of each term of both sides,

Dimensional formula on L.H.S = [L]

Dimensional formula on R.H.S:

    \begin{align*} &= [LT^{-1}][T] + [\frac{1}{2}] [lT^{-2}] [T]^2  \\  &= [LT^{-1}][T] + [lT^{-2}][T]^2 \qquad [\frac{1}{2} is diemonsionless constant] \\ &=[L] + [L] \\ &= 2[L] \\ &=[L] \qquad [ \text{2 is dimensionless constant}]\end{align*}

Here: dimensions of L.H.S = dimensions of R.H.S

Hence, the relation s = ut + \frac{1}{2} at^2 is correct.

b) To find the dimensional formula of constant in any physcial relation.

Let’s calculate the dimensional formula of specific heat capacity. We have a physical relation Q = msdt

From that relation, we can write:

    \[ s = \frac{Q}{m \hspace{1mm}dt} \]

Writing dimensions of heat energy Q, mass m and change in temeperature dt, we get,

    \begin{align*} s &= \frac{[ML^2T^{-2}]}{[M][K]} \\ s &= [M^{1-1}L^2T^{-2}K^{-1}]  \\  s &= [M^0L^2T^{-2}K^{-1}] \end{align*}

Hence, dimensions of speicfic heat capacity is 2 in length, -2 in time and -1 in temperature.

c) To convert the value from one unit to other.

For complete measurement, we should know the numerical value as well as unit of that physical quantity. for e.g A car mass is 80 kg. Here, 80 is numerical value and kg is its unit.

When we convert one unit to another unit, the numerical value changes as well as the unit. e.g That same car mass becomes 80,000 g. Here, 80 value is changed to 80,000, and kg is changed to g.

But, the overall quantity remains the same.

i.e

    \[ 80 \times 1 kilogram = 80,000 \times 1 gram \]

Hence,

(1)   \begin{equation*} \boxed{n_1 u_1 = n_2 u_2}   \end{equation*}

Where, n_1 is the numerical value in u_1 unit and n_2 is the numerical value in other unit u_2.

Let’s derive the formula for conversion from one unit to another using 1.

Suppose x, y, and z are the dimensions of mass, length, and time respectively. M_1, L_1 and T_1 are the units in first unit system and M_2, L_2 and T_2 are the units in another unit system. Then we can write:

    \[ n_1[{M_1}^x{L_1}^y{T_1}^z] = n_2[{M_2}^x{L_2}^y{T_2}^z] \]


(2)   \begin{equation*}\boxed{n_2 = n_1\left[\frac{M_1}{M_2}\right]^x\left[\frac{L_1}{L_2}\right]^y\left[\frac{T_1}{T_2}\right]^z} \end{equation*}

Q. Convert 1 N into dyne using dimensional analysis.

Both Newton and Dyne are the units of Force in the MKS and CGS systems respectively.

The dimensional formula of Force is [MLT^{-2}]. Force has dimensions 1 in mass, 1 in length, and -2 in time. Hence, x = 1, y = 1 and z = -2.

In MKS system, M_1 = 1 kg, L_1 = 1m, T_1 = 1s and n_1 = 1 because of 1N.

In CGS system, M_2 = 1gm, L_2 = 1cm, T_2 = 1s and n_2 = ?

Let’s use the formula derived in 2:

    \begin{align*} n_2 &= n_1\left[\frac{M_1}{M_2}\right]^x\left[\frac{L_1}{L_2}\right]^y\left[\frac{T_1}{T_2}\right]^z \\ n_2 &= 1 \times \left[\frac{1 kg}{1 gm}\right]^1\left[\frac{1m}{1 cm}\right]^1\left[\frac{1s}{1s}\right]^{-2} \\ n_2 &= 1 \times \left[\frac{1000gm}{1 gm}\right]^1\left[\frac{100cm}{1 cm}\right]^1\left[\frac{1s}{1s}\right]^{-2} \\ n_2 &= 1 \times[1000]^1 [100]^1 [1]^{-2} \\ n_2 &= 1 \times 1000 \times 100 \times 1 \\ n_2 &= 10^5 \end{align*}

Hence,

    \[ \boxed{1N = 10^5 dyne}\]

d) To derive the physical relation

Using dimensional analysis, we can easily drive physical relations.

Q. Derive a relation for the time period of a simple pendulum.

pendulum time period

We know the time period of a simple pendulum depends upon:

  • mass of the bob, T \propto m^a
  • length of the pendulum T \propto l^b
  • acceleration due to gravity T \propto g^c

Combining all these relations, we get:

    \begin{align*} T &\propto m^al^bg^c \\ T &= k \hspace{1mm}m^al^bg^c \end{align*}

where k is the proportionality constant and its value cannot be determined using dimensional analysis which is one of the limitations of dimensional analysis. From the experiment, the value of k is found to be 2\pi.

On writing dimensions, we get:

    \begin{align*} [T] &= [M]^a[L]^b[LT^{-2}]^c  \quad [2\pi \text{ is dimensionless constant}] \\  [T] &= [M]^a[L]^{b+c}[T^{-2c}]  \\ [T] &= [M^aL^{b+c}T^{-2c}] \end{align*}

Comparing the dimensions of mass, length and time on both sides, we get

    \begin{gather*} a = 0 \\b + c = 0 \\-2c = 1 \end{gather*}

Solving above equations, we get: a = 0, b = \frac{1}{2} and c = -\frac{1}{2}. And putting a, b, and c value, we get relation:

    \begin{gather*} T = k m^al^bg^c \\ T = 2\pi m^0l^{\frac{1}{2}}g^{-\frac{1}{2}} \\ T = 2 \pi \sqrt{\frac{l}{g}} \end{gather*}

Hence:

    \[ \boxed{T = 2\pi \sqrt{\frac{l}{g}}} \]

Limitations of dimensional analysis

Though it has many applications, it has some limitations too. Some of them are:

  • Dimensionless constants value cannot be determined using dimensional analysis.
  • It doesnot give information whether the given quantity is vector or scalar.
  • It doesn’t give any analysis on other functions such as trigonometric, logarithmic, exonential etc.

Some FAQ’s (frequently asked quetions)

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