Question

Famous relation in physics relates ${}^{\prime }{\text{moving mass}}^{\prime }m$ to the ${}^{\prime }{\text{rest mass}}^{\prime }{m}_{\circ }$ of a particle in terms of its speed $v$ and the speed of light, $c$. (This relation first arose as a consequence of special theory of relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant $c$. He writes:

$m=\frac{m_{\circ }}{(1-v^2{)}^{\frac{1}{2}}}$

Guess Where to put the missing $c$.

Answer 1

Given: $m=\frac{m_{\circ }}{(1-v^2{)}^{\frac{1}{2}}}$

Dimension of $m=M^1{L}^0{T}^0$

Dimension of $m_{\circ }=M^1{L}^0{T}^0$

Dimension of $v=M^0{L}^1{T}^{-1}$

Dimension of $v^2=M^1{L}^2{T}^{-2}$

Dimension of $c=M^0{L}^1{T}^{-1}$

The given formula will be dimensionally coreect only when the dimension of $L.H.S$ is the same as that of $R.H.S$. This is only possible when the factor, $(1-v^2{)}^{\frac{1}{2}}$ is dimensionless i.e., $(1-v^2)$ is dimensionless. This is only possible if $v^2$ is divided by $c^2$. Hence, the correct relation is $m=\frac{m_0}{{(1-\frac{v^2}{c^2})}^{\frac{1}{2}}}$.