# Find total E of muon using time dilation

1. Feb 24, 2014

### oddjobmj

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

An Ω- particle has rest energy 1672 MeV and mean lifetime 8.2*1011 s. It is created and decays in a particle track detector and leaves a track 24 mm long. What is the total energy of the  particle?

2. Relevant equations

E=$\frac{mc^2}{\sqrt{1-(\frac{v}{c})^2}}$

t=t0$\sqrt{1-(\frac{v}{c})^2}$

Rest E=mc2

3. The attempt at a solution

Because Rest E=mc2 I know that:

E=$\frac{1672 MeV}{\sqrt{1-(\frac{v}{c})^2}}$

What I need to do now is solve for v using the given time and distance (.024 meters). The time is measured in the particle's proper frame so:

t=(8.2*1011 seconds)$\sqrt{1-(\frac{v}{c})^2}$

Of course, this contains v itself. What am I missing?

Thank you!

2. Feb 25, 2014

### BvU

I am missing what you do with the track length.....

3. Feb 25, 2014

### Staff: Mentor

Set up an equation for distance in terms of speed and time. All from the view of the lab frame. Then you can solve for v.

4. Feb 25, 2014

### BvU

Also, I have a rather old particle data book. It claims these $\Omega^-$ live only $8.22 \times 10^{-11}$ seconds. For our convenience, it also mentions $c\tau = 0.0246$ m....

5. Feb 25, 2014

### Staff: Mentor

LOL.... yeah, the OP contains a typo. :tongue:

6. Feb 25, 2014

### oddjobmj

Hah, yes! That would be a rather long time. Unfortunately it is too late to edit the original post.

Also, as Doc Al suggested, I need v to solve this problem. They give me the time and distance it travels in that time but not the velocity. I need to use time dilation to find the time in the rest (station?) frame and divide the distance by that time to find the velocity. I'm just getting mixed up on which t is which and how to solve for t overall.

7. Feb 25, 2014

### Staff: Mentor

The only time you are given is the proper time (in the muon's frame). What's the time in the lab frame? (Give an expression in terms of v, not a number.)

8. Feb 25, 2014

### oddjobmj

t=t0$\sqrt{1-(\frac{v}{c})^2}$, where t0 is given

9. Feb 25, 2014

### Staff: Mentor

Almost. Don't mix up t and t0. (Which must be larger?)

10. Feb 25, 2014

### oddjobmj

Moving clocks run slower so the time in the particle's proper frame will measure longer than the time in the 'station' frame. In that case t0 needs to be larger but you suggest t is given which flips my assumptions around. Is t0 not the time measured in the particle's proper frame?

11. Feb 25, 2014

### oddjobmj

t0=$\frac{t}{\sqrt{1-(\frac{v}{c})^2}}$=$\frac{t}{\sqrt{1-(\frac{x}{ct})^2}}$

Should I be using t or t0 inside the square root?

Once I find t0 I can use that in my E calculation to find v?

12. Feb 26, 2014

### Staff: Mentor

Right.

No, just the opposite. The proper time is the shortest time. Every one else observes the particle's 'clock' to run slow. So everyone else measures a greater time.

Yes, t0 is the time measured in the particle's frame.

13. Feb 26, 2014

### Staff: Mentor

That's not my suggestion. That's just rewriting what you wrote before.

Inside the square root there should only be v, not time. You're going to solve for v.

t0 is given. Once you have the correct expression for t (lab frame time), you'll use it to express distance = speed*time as seen in the lab frame. The only unknown will be v, which you will solve for.