Answer to Question #88927 in Mechanics | Relativity for Sridhar

Question

The half life of a particle in the laboratory is 4×10^-8 sec, when its speed is 0.8c . with half life will be measured in the laboratory if the speed of the particle is reduces 0.6c
1)2×10^-8 sec
2)3××10^-8 sec
3)1×10^-8 sec
4)0.75×10^-8 sec

Answer 1

According to the equation of time dilation in special theory of relativity, the half life period of a particle in the laboratory frame is connected with its half life period in a frame moving with the particle as follows:

$$\Delta t = \gamma \Delta \tau,$$

where \gamma is the relativistic factor

$$\gamma = \frac{1}{\sqrt{1-v^2/c^2}}$$

Hence, for the two measurements we have

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

Finally, we derive:

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

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: 2) 3 x 10^-8 s.