Question

A famous relation in physics relates ‘moving mass’ $\text m$ to the ‘rest mass’ $\text m_0$ of a particle in terms of its speed v and the speed of light, $\text c$ . (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 $\text c$. He writes : $\text m$ = $\frac{\text m_0}{(1-v^2)^{\frac{1}{2}}}$ . Guess where to put the missing $\text c$.

Answer 1

Given the relation,
$\text m$ = $\frac{\text m_0}{(1-v^2)^{\frac{1}{2}}}$

Dimension of $\text m$ = $\text M^1 \; \text L^0\; \text T^0$

Dimension of $\text m_0$ = $\text M^1 \; \text L^0\; \text T^0$

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

Dimension of $\text v^2$ = $\text M^0 \; \text L^2\; \text T^{-2}$

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

The given formula will be dimensionally correct 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 $\text v^2$ is divided by $\text c^2$ . Hence, the correct relation is

$\text m = \frac{\text m_0}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}$.