Calculating Mean Life of Mesons at Rest

[fluidistic]

Homework Statement

The half life time of mesons $$\mu$$ with velocity $$0.95 c$$ has been obtained experimentally as $$6\times 10^{-6}s$$.
Calculate the half life of those mesons in a system in which they are at rest.

Note: It shouldn't be "half life" but something like "mean life" or something like that, I don't know how to translate. Anyway it's not relevant to the problem.

Homework Equations

Lorentz transformations.

The Attempt at a Solution

Say I have a reference frame O and O' where O is the system at rest and O' the one moving at 0.95 c from the mesons.
They give me, I believe, $$t_B'-t_A'=6 \times 10^{-6}s$$. They ask me $$T_B-T_A$$.
Using Lorentz transformations, I get that $$T_B-T_A=\frac{vx_B}{c^2}-\frac{vx_A}{c^2}$$. So my problem is to get rid of a distance relationship. They don't say anything about distance... or should I calculate the distance the mesons travel in $$6 \times 10^{-6}s$$ which would give me $$x_B'-x_A'$$? And then I could calculate $$x_A-x_B$$ with Lorentz transformations and I think the problem would be solved... does this sound correct?

 

[vela]

or should I calculate the distance the mesons travel in $$6 \times 10^{-6}s$$ which would give me $$x_B'-x_A'$$?

You have two events: the muon's creation and the muon's decay. What you know about the events is the temporal difference between them in the lab frame and the spatial difference in the muon's rest frame. That's enough info to calculate the time elapsed in the muon's rest frame. Alternately, you could also calculate the spatial difference in the lab frame, which then allows you to calculate the time elapsed in the muon frame. In either case, you'll use a Lorentz transformation equation, but you use different ones depending on which method you choose.

 

[fluidistic]

Ok so I just followed my thoughts which is basically what you said. I reach as a final answer: $$5.415001691 \times 10 ^{-6}s$$ which is roughly 10% less time of life than mesons with 0.95c speed. Don't know if it's correct.

 

[vela]

Hmm, maybe you should show us your work. I get about 2 μs.

 

[fluidistic]

Hmm, maybe you should show us your work. I get about 2 μs.

Sure.
Wow, when I was writing my work here, I saw my error... Nevermind, I restart it all. I will try to see if my future result will match yours.

 

[fluidistic]

I now get (with less algebra!) $$1.8734994 \times 10^{-6}s$$. If you don't get this, let me know and I show my work.

 

[vela]

Yup, that matches what I got.

 

[fluidistic]

Yup, that matches what I got.

Thanks for the confirmation, problem solved.