Q 2.15: A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m_{o} of a particle in terms of its speed v and the speed of light, 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 c. He writes : m = \frac{m_{o}}{(1-v^2)^{1/2}}

Guess where to put the missing c.

Answers (1)

The relation given is m = \frac{m_{o}}{(1-v^2)^{1/2}}

Divide both sides by m_{o};  \therefore L.H.S becomes m / m_{o} which is dimensionless.

Hence, R.H.S must be dimensionless too. (After Dividing by m_{o} !)

\frac{1}{(1-v^2)^{1/2}} can be dimensionless only when v \rightarrow (v/c)

Therefore, the dimensional equation is m = \frac{m_{o}}{(1-(\frac{v}{c})^2)^{1/2}}

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