Relativity Q / Characteristic decay times

[CAF123]

Homework Statement

1)The pion is an unstable subatomic particle whose characteristic decay time in its own frame is 26ns. A beam of such particles emerges from a source at point X in the laboratory traveling with speed $v = \sqrt{3}/2\, c$ in the laboratory frame. Determine i) the decay time of the particles in the laboratory frame and ii) the distance (as measured in the lab) from the point X to the point Y at which it is found that the concentration of mesons in the beam is reduced to 1/e its value at the source.

2) Given a clock which is accurate to 1ns over a period of 1 second, estimate the speed that the clock has to attain for relativistic effects to become noticable.

The Attempt at a Solution

i) If we denote $\tau_o$ as the char decay time in the meson frame, then $\tau$ will be the char decay time in the lab frame. Just so I am clear on the meanings of these symbols: $\tau_o$ is the time interval of 2 events with the same spatial coordinates and so this interval is measured in the rest frame of some clock. $\tau$ is the time interval of 2 events without the same spatial coords so is measured in a frame moving relative to the stationary clock. Do these make sense? So for the question: in the lab frame, the char decay time is just $\tau = \tau_o \cdot \gamma$.

ii)It is a well known fact that 1/e of the sample remaining occurs at the char decay time. So the answer to this is, given we are concerned with the lab frame, $d = \tau \cdot v$?

2) When it says 'relativistic effects become noticable', I presume they just mean that the tick of the clock becomes longer and so the clock becomes less accurate? So in it's rest frame it is accurate to 1 ± 10^-9s. I am not sure where to go next with this question. Any hints? Many thanks.

 

[CAF123]

Can anyone confirm what I did in the first question and help me with the second?
Thanks!

 

[CAF123]

For 2), should I assume the clock travels some distance , say 1 m, and do v = d/t?

 

[haruspex]

2) When it says 'relativistic effects become noticeable', I presume they just mean that the tick of the clock becomes longer and so the clock becomes less accurate?

It means that the shift in timing becomes large enough that it is detectable, despite the clock's only being accurate to 1 in 10⁹. I.e., at what speed is the rate shifted by that much?

 

[CAF123]

I wouldn't be sure how to compute that. Any hints?

 

[haruspex]

I wouldn't be sure how to compute that. Any hints?

Hey, you're the studying relativity. I just look it up as necessary on the net... where I see:
t = t0/(1-v²/c²)^1/2
So you want the value of v at which (1-v²/c²)^1/2 < 1 - 10^-9.