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Special Relativity, calculating velocity of Kl0 meson with reference frames

1. The problem statement, all variables and given/known data

K mesons (“kaons”) are unstable particles composed of a quark and an antiquark. They can be produced copiously in energetic collisions between stable particles at accelerator laboratories. Soon after they are produced, kaons decay to lighter particles. One type of kaon, the KL0, has a lifetime of 5 × 10–8 seconds in its own rest frame. Now imagine that a beam of fast KL0 mesons is produced at a national laboratory. The average distance the kaons travel before decaying in flight is found to be 45 meters in the lab frame.
(a) What is the speed of the kaons in the lab frame? Use the value c = 3 × 108 m/sec in your calculation.
(b) What is the kaons’ gamma factor in the lab frame?
(c) Isaac Newton didn’t know about special relativity. If you told Newton that a kaon with a lifetime of 5 × 10–8 sec was zipping through your lab at the speed you calculated in part (a), how far would Newton expect the kaon to travel before decaying?


2. Relevant equations

So from my understanding, the time given is the t' (the time in the meson's reference frame), while the length given (L) is from the lab's reference frame. If this is true, when calculating velocity, don't you need either both t' and L', or t and L to solve for some velocity?

3. The attempt at a solution

From the given variables that was all I could do. If I threw out my conceptual understanding of the problem and just did v=L/t you get 3 times the speed of light. Kind of lost on where to go from there.
 
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You can convert t' to t; the conversion will depend on velocity, unknown so far. Then L/t = v, which you can treat as an equation for the velocity.
 
t' is indeed 5x10^-8.

We know the kaons travel 45m in time t in the lab frame.
So 45=vt

and t=gamma t'.

So 45=gamma x t' x v.

Now solve for v.
 
t' is indeed 5x10^-8.

We know the kaons travel 45m in time t in the lab frame.
So 45=vt

and t=gamma t'.

So 45=gamma x t' x v.

Now solve for v.
Hm, but gamma requires v too? I also find this confusing.
 
Thanks for the help voko & apelling. Kaldanis, I worked it through this way and it seemed to work...

t'=5x10^-8s
L=45m

So.. we know that t (time in the lab frame) must be longer than t' because the Kaon's clock ticks slower due to its high velocity, so..

t=[itex]\gamma[/itex]t'

v=L/t

v=L/([itex]\gamma[/itex]t')

[itex]\gamma[/itex]=1/√(1-v^2/c^2)

v=L√(1-v^2/c^2)/t'

v^2t'^2=L^2(1-v^2/c^2)

v^2t'^2=L^2-L^2v^2/c^2

v^2t'^2 + L^2v^2/c^2=L^2

v^2(t'^2 + L^2/c^2)=L^2

v=L/√(t'^2+L^2/c^2)

v=2.846 x 10^8 m/s

...
Thanks again for the help!



[itex]\gamma[/itex]
 

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