Question

A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle In terms of its speed 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)^1/2 Guess where to put the missing c.

Answer 1

We need the dimension of the RHS to be equal to M¹. Since m₀ has the dimension of M¹, the denominator must be dimensionless. 

⇒ (1 - v²)^1/2 must be corrected to be dimensionless.

1 is dimensionless, v² has the dimensions [L¹T^-1]² = L²T^-2

Using c, we need to make 1 – v² dimensionless. Since c has the same dimensions as v, c/v and v/c are both dimensionless.

∴ The corrected formula must be

m = m₀/(1 - v²/c²)^1/2 or m = m₀/(1 - c²/v²)^1/2

Since m₀ is the rest mass, m = m₀ when v = o. (body at rest)

For the first formula, when v = 0, m = undefined, which is wrong.

∴ Corrected formula is m = m₀/(1 - v²/c²)^1/2