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Q2.15:
A famous relation in physics relates ‘moving mass’ $m$ to the ‘rest mass’ $m_0$ 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 : $m = \frac{m_0}{(1 - v^2)^{1/2}}$. Guess where to put the missing $c$.

Solution :

Step 1: Dimensional Analysis of the Given Expression

To determine the placement of the constant $c$, we must invoke the Principle of Homogeneity of Dimensions [Per the Principle of Dimensional Homogeneity, an equation is physically valid only if the dimensions of all terms on both sides of the equation are identical].

The given formula is: $m = \frac{m_0}{(1 - v^2)^{1/2}}$

  • The dimension of the LHS (mass $m$) is $[M^1 L^0 T^0]$.
  • The dimension of the numerator $m_0$ is $[M^1 L^0 T^0]$.
  • For the equation to be dimensionally consistent, the denominator $(1 - v^2)^{1/2}$ must be dimensionless (i.e., $[M^0 L^0 T^0]$).

Step 2: Analyzing the Denominator

In the term $(1 - v^2)$, we encounter a physical impossibility: a dimensionless constant ($1$) is being subtracted by a term ($v^2$) with dimensions $[L^2 T^{-2}]$. [Per the Law of Dimensional Addition, only quantities with identical dimensions can be added or subtracted].

To make $v^2$ dimensionless, it must be divided by another quantity that possesses the dimensions of velocity squared ($[L^2 T^{-2}]$). Since $v$ is the speed of the particle and $c$ is the speed of light, $c$ is the only physical constant present with the required dimensions of speed ($[L^1 T^{-1}]$).

Step 3: Correcting the Expression

To ensure $(v^2 / c^2)$ is dimensionless, we rewrite the denominator as $(1 - \frac{v^2}{c^2})$.

Verification of dimensions for $\frac{v^2}{c^2}$:

$\frac{[L^1 T^{-1}]^2}{[L^1 T^{-1}]^2} = \frac{[L^2 T^{-2}]}{[L^2 T^{-2}]} = [M^0 L^0 T^0]$ (Dimensionless)

Thus, the expression becomes:

$m = \frac{m_0}{(1 - \frac{v^2}{c^2})^{1/2}}$

Step 4: Visualizing the Relativistic Factor

Velocity (v/c) Mass (m) m = m₀ / √(1 - v²/c²)

Final Solution: The correct relation is $m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}$. The constant $c$ must be placed in the denominator such that $v$ is divided by $c$ before being squared.


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