Question

What time interval between the moments of decay of two particles will be observed in the frame $ K $ ?

Answer 1

Hint :In order to solve this question, we are going to first use the Lorentz contraction and compute the time intervals $ {t_1} $ and $ {t_2} $ . Compute the value of $ \gamma $ from the given value of the velocity and the speed of light, after that we have the values of intervals, the difference is calculated.
According to Lorentz contraction,
 $ {t_1} = \dfrac{1}{{2\gamma }}\left( {\dfrac{1}{{c + v}}} \right) $
Where, $ \gamma = \dfrac{1}{{\sqrt {1 - {\beta ^2}} }} $ , $ \beta = \dfrac{v}{c} $
 $ {t_2} = \dfrac{1}{{2\gamma }}\left( {\dfrac{1}{{c - v}}} \right) $

Complete Step By Step Answer:
Let us solve the question by applying the concept of Lorentz contraction, where the time intervals change due to the velocity of the particles in the frame of reference $ K $ .The velocity of the particles in the frame $ K $ is $ 0.990c $ , distance between the particles $ l = 120m $ ,
Now according to the Lorentz contraction, the time interval for the first particle will be
 $ {t_1} = \dfrac{1}{{2\gamma }}\left( {\dfrac{1}{{c + v}}} \right) $
And that for the second particle will be
 $ {t_2} = \dfrac{1}{{2\gamma }}\left( {\dfrac{1}{{c - v}}} \right) $
Where, $ \gamma = \dfrac{1}{{\sqrt {1 - {\beta ^2}} }} $ , $ \beta = \dfrac{v}{c} $
Computing the value of $ \gamma $ by putting the velocity $ v = 0.990c $
$ \gamma = \dfrac{1}{{\sqrt {1 - \dfrac{{{v^2}}}{{{c^2}}}} }} = \dfrac{1}{{\sqrt {1 - \dfrac{{{{\left( {0.99c} \right)}^2}}}{{{c^2}}}} }} \\
   \Rightarrow \gamma = \dfrac{1}{{\sqrt {1 - \dfrac{{0.9801{c^2}}}{{{c^2}}}} }} = \dfrac{1}{{0.141}} \\
   \Rightarrow \gamma = 7.09 \\ $
Now, the time interval between the moments of the decay of the two particles is
 $ {t_2} - {t_1} = \dfrac{{lv}}{\gamma }\left( {\dfrac{1}{{{c^2} - {v^2}}}} \right) $
Putting the values in the relation
$ {t_2} - {t_1} = \dfrac{{\left( {120} \right)\left( {0.99c} \right)}}{{\left( {7.09} \right)}}\left( {\dfrac{1}{{0.0199{c^2}}}} \right) \\
   \Rightarrow {t_2} - {t_1} = \dfrac{{11.88c}}{{\left( {7.09} \right)}}\left( {\dfrac{1}{{0.0199{c^2}}}} \right) \\
   \Rightarrow {t_2} - {t_1} = \left( {\dfrac{{1.676}}{{0.0199 \times 2.998 \times {{10}^8}}}} \right) \\
   \Rightarrow {t_2} - {t_1} = 28.08 \times {10^{ - 8}} \\ $
Hence, the time interval between the moments of decay of the two particles will be $ 28.08 \times {10^{ - 8}} $ .

Note :
For the first particle, the time interval $ {t_1} $ , follows the Lorentz contraction while the second particle follows the time dilation, the time intervals are affected by the particle’s velocity in the frame of reference $ K $ , the difference in the two time intervals gives the time interval between the moments of decay.