Question

The average lifetime of mu-mesons with a speed of 0.95c is measured to be 6 times 10^-6 s. Compute the average lifetime of mu-mesons in a system in which they are at rest. Pions have a half-life of 1.8 times 10^-8 s. A pion beam leaves a particle accelerator at a speed of 0.8c. What is the expected distance over which half of the pions should decay classically? Relativistically? Rocket A travels to the right and rocket B travels to the left, with speeds of 0.8c and 0.6c, respectively relative to the Earth. What is the velocity of rocket A measured from rocket B? Consider a radioactive nucleus that moves with a constant speed of 0.5c relative to the laboratory. The nucleus decays and emits an electron with a speed of 0.8c relative to the nucleus along the direction of motion. Find the velocity of the electron in the laboratory frame. Suppose the Doppler shift in wavelength of the sodium D_2 line (589.0 nm) is 10 nm when the light is observed from a distant star, Determine the star's velocity of recession.

Answer 1

6) we use the following equation

    $\Delta t=\gamma \Delta t_{o}---(1)$

where $\Delta t$ is the average time in the rest frame, $\Delta t_{o}$ is the proper time for muon moving with velocity 0.95c

$\Rightarrow \Delta t=\frac{\Delta t_{0}}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$

$\Rightarrow \Delta t=\frac{6*10^{-6}}{\sqrt{1-{0.95^{2}}}}=1.92*10^{-5}sec$

7) Classically:

    $d=v*t=(0.8c)*(1.8*10^{-8})=4.32m$

   Relativistically:

   $d=v*t'=v*\gamma t=v*\frac{t}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=0.8c*\frac{1.8*10^{-8}}{\sqrt{1-(0.8)^{2}}}=7.2m$

8) the velocity of A relative to the earth ($u$)=0.8c

    the velocity of B relative to earth ($v$)= -0.6c

    the velocity of A measured from B is

   ${u}'=\frac{u-v}{1-\frac{uv}{c^{2}}}=\frac{0.8c-(-0.6c)}{1-\frac{0.8c*(-0.6c)}{c^{2}}}=0.96c$

9) the speed of nucleus relative to lab ($v$)=0.5c

   the speed of electron relative to nucleus (${u}'$)=0.8c

   the velocity of electron in lab frame ($u$) is given by

$u=\frac{{u}'+v}{1+\frac{{u}'v}{c^{2}}}=\frac{0.8c+0.5c}{1+(0.8*0.5)}=0.93c$

10) the observed wavelength is $\lambda _{obs}=589+10=599nm$

      $\lambda _{source}=589nm$

    we use the doppler relation

     $\lambda _{obs}=\lambda _{source}\sqrt{\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$

$\Rightarrow \left [ \frac{\lambda _{obs}}{\lambda _{sou}} \right ]^{2}={\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$

$\Rightarrow \left [ \frac{599}{589} \right ]^{2}={\frac{1+\frac{v}{c}}{1-\frac{v}{c}}}$

Simplyfying the above equation we get

$\frac{v}{c}=0.017$