1.1 Dimensional analysis  (Page 4/7)

We can break up the ratio as :

$$\Rightarrow \frac{e^2}{{\epsilon }_{0}hc}=\frac{e^2}{{\epsilon }_0}X\frac{1}{hc}$$

It is advantageous to use the expression of potential energy of two electrons system to evaluate the first term. The potential energy of two electrons system is :

$$U=\frac{e^2}{4\pi {\epsilon }_{0}r}$$

$$\Rightarrow \frac{\left[e^2\right]}{\left[{\epsilon }_0\right]}=\left[4\pi Ur\right]$$

Neglecting dimensionless constants,

$$\Rightarrow \frac{\left[e^2\right]}{\left[{\epsilon }_0\right]}=\left[Ur\right]=\left[M{L}^2{T}^{-2}L\right]=\left[M{L}^3{T}^{-2}\right]$$

The expression “hc” is related to energy of a photon as :

$$E=\frac{hc}{\lambda }$$

$$\Rightarrow \left[hc\right]=\left[E\right]\left[L\right]=\left[M{L}^2{T}^{-2}\right]\left[L\right]=\left[M{L}^3{T}^{-2}\right]$$

Putting dimensions as calculated in the original expression,

$$\Rightarrow \frac{\left[e^2\right]}{\left[{\epsilon }_{0}hc\right]}=\frac{\left[M{L}^3{T}^{-2}\right]}{\left[M{L}^3{T}^{-2}\right]}=\left[M^0{L}^0{T}^0\right]$$

Thus, given expression is dimensionless.

We can find the dimension of the constant “a” by applying principle of homogeneity, appearing in the numerator of the given expression. Hence,

$$\left[a\right]=\left[K^2\right]$$

The dimension of the sum of quantities in numerator is equal to the dimension of individual term. It means that :

$$\Rightarrow [a+t^2]=\left[K^2\right]$$

Now,

$$\left[P\right]=\frac{[a+t^2]}{\left[bh\right]}$$

$$\left[b\right]=\frac{[a+t^2]}{\left[Ph\right]}$$

$$\left[b\right]=\frac{\left[K^2\right]}{\left[\frac{F}{A}h\right]}=\frac{\left[K^2\right]}{\left[\frac{ML{T}^{-2}}{L^2}XL\right]}$$

$$\Rightarrow \left[b\right]=\left[M^{-1}{T}^2{K}^2\right]$$

The dimension of the ratio “ $\frac{a}{b}$ ” is :

$$\Rightarrow \left[\frac{a}{b}\right]=\frac{\left[K^2\right]}{\left[M^{-1}{T}^2{K}^2\right]}=\left[M{T}^{-2}\right]$$

Here, the basic quantities of the system are assumed to be different than that of SI system. We see here that mass can be linked to force, using equation of motion,

$$\mathrm{Force}=\text{mass X acceleration}=\frac{\text{mass X velocity}}{\text{time}}$$

$$\Rightarrow \text{mass}=\frac{\text{force X time}}{\text{velocity}}$$

Each of the quantities on the right hand is the basic quantity of new system. Let us represent new basic quantities by first letter of the quantities in capital form as :

$$\Rightarrow \left[\mathrm{mass}\right]=\left[V^{-1}TF\right]$$

Similarly, we can dimensionally link length to basic quantities, using definition of velocity.

$$\text{velocity}=\frac{\text{displacement}}{\text{time}}=\frac{\text{length}}{\text{time}}$$

$$\Rightarrow \text{length}=\text{velocity X time}$$

$$\Rightarrow \left[\text{length}\right]=\left[VT\right]$$

Now, work is given as :

$$\text{work}=\text{force X displacement}$$

Dimensionally,

$$\Rightarrow \text{work}=\text{force X length}$$

$$\Rightarrow \left[\text{work}\right]=\left[F\right]\left[VT\right]=\left[VTF\right]$$

We are required to find the units of mass and length in terms of the newly defined units.

We need to first find dimensional expression of the quantity as required in terms of units defined in the question. We see that dimensional formula of physical quantities as basic units are :

$$\left[F\right]=\left[ML{T}^{-2}\right]$$

$$\left[T\right]=\left[T\right]$$

$$\left[P\right]=\left[M{L}^2{T}^{-3}\right]$$

Dividing dimensional formula of “P” by “F”, we have :

$$\frac{\left[P\right]}{\left[F\right]}=\frac{\left[M{L}^2{T}^{-3}\right]}{\left[ML{T}^{-2}\right]}=\left[\frac{L}{T}\right]$$

As we are required to know the units of length, its dimension in new units are :

$$\Rightarrow \left[L\right]=\frac{\left[PT\right]}{\left[F\right]}$$

In order to find the unit of “L” in terms of new system, we have :

$$n_2=\frac{\left[P_1\right]\left[T_1\right]\left[F_2\right]}{\left[P_2\right]\left[T_2\right]\left[F_1\right]}$$

We should note here that subscript “1” denotes new system of units, which comprises of force, time and power as basic dimensions. Hence,

$$\Rightarrow n_2=\frac{\left[{10}^{3}W\right]\left[s\right]\left[1N\right]}{\left[1W\right]\left[1s\right]\left[{10}^{3}N\right]}={10}^{-3}$$

Hence, length has the unit of length,

$$\Rightarrow L={10}^{-3}m$$

From the dimensional formula of force, we have :

$$\left[F\right]=\left[ML{T}^{-2}\right]$$

$$\Rightarrow \left[M\right]=\frac{\left[F\right]}{\left[L{T}^{-2}\right]}=\frac{\left[{10}^{3}N\right]}{\left[\left({10}^{-3}m\right){\left({10}^{-3}\right)}^{-2}\right]}=\left[1kg\right]$$

Hence, unit of mass is "1 kg".