Question 15

A famous relation in physics relates 'moving mass' m to the 'rest mass' m0 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  =  m0  /  (1-v2)½

Answer

Given the relation,

m  =  m0  /  (1-v2)½

Dimension of m = M1 L0 T0

Dimension of = M1 L0 T0

Dimension of v = M0 L1 T-1

Dimension of v2 = M0 L2 T-2

Dimension of c = M0 L1 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-v2)½ is dimensionless i.e., (1 - v2) is dimensionless. This is only possible if v2 is divided by c2. Hence, the correct relation is

m  =  m0  /  (1-v2/c2)½ .

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