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.