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 

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