Question

5 A famous relation in physics relates moving mass' m to the rest mass' m of
particle in terms of its speed vand the speed of light, C. (This relation ist arose
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 :
mo
m =-
Guess where to put the missing c.​

Answer 1

Given the relation,

m = m0 / (1-v2)1/2

Dimension of m = M1 L0 T0

Dimension of m0 = 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)1/2 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)1/2

Answer 2

${\mathfrak{\blue{\underline{\underline{Solution:-}}}}}$

Given the relation,

$\sf{m=\dfrac{m_{0}}{\Big(1-v^{2}\Big)^{\dfrac{1}{2}}}}$

$\sf{Dimension\;of\;m=M^{1}\;L^{0}\;T^{0}}$

$\sf{Dimension\;of\;m_{0}=M^{1}\;L^{0}\;T^{0}}$

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

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

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

The given formula will be dimensionally correct only when the dimension of L.H.S is the same as that R.H.S. This is only possible when the factor, $\sf{\Big(1-v^{2}\Big)^{\dfrac{1}{2}}}$ is dimensionless i.e., $\sf{(1-v^{2})}$ is dimensionless. This is only possible if v² is divided by c². Hence, the correct relation is :

$\sf{m=\dfrac{m_{0}}{\bigg(1-\dfrac{v^{2}}{c^{2}}\bigg)^{\dfrac{1}{2}}}}$