# Special relativity

1. Aug 15, 2010

### fluidistic

1. The problem statement, all variables and given/known data
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.
2. Relevant equations Lorentz transformations.

3. 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?

2. Aug 15, 2010

### vela

Staff Emeritus
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.

Last edited: Aug 15, 2010
3. Aug 15, 2010

### 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.

4. Aug 15, 2010

### vela

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

5. Aug 15, 2010

### fluidistic

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.

6. Aug 15, 2010

### 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.

7. Aug 15, 2010

### vela

Staff Emeritus
Yup, that matches what I got.

8. Aug 15, 2010

### fluidistic

Thanks for the confirmation, problem solved.