Answer to Question #89214 in Mechanics | Relativity for Sridhar

Question

The half life of a particle in the laboratory is 4×10^-8 sec,when it is speed is 0.8C what half life will be measured in the laboratory if the speed of particle is reduced 0.6C

Answer 1

Within the special theory of relativity, the half life period of a particle measured in the laboratory frame is connected with its half life period measured in a frame moving with the particle as follows:

$$\Delta t = \gamma \Delta \tau, \quad \gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

For both measurements we obtain:

$$\Delta \tau = \frac{\Delta t_1}{\gamma_1} = \frac{\Delta t_2}{\gamma_2}$$

$$\Rightarrow \, \Delta t_2 = \frac{\gamma_2}{\gamma_1} \Delta t_1$$

Substituting the numerical values, we obtain:

$$\Delta t_2 = \frac{\sqrt{1-0.8^2}}{\sqrt{1-0.6^2}} \cdot 4 \cdot 10^{-8} = \frac{0.6 \cdot 4}{0.8} \cdot 10^{-8} = 3 \cdot 10^{-8} \,s$$

Answer: 3 x 10^-8 s.