Find total E of muon using time dilation

[oddjobmj]

Homework Statement

An Ω^- particle has rest energy 1672 MeV and mean lifetime 8.2*10¹¹ 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?

Homework Equations

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

t=t₀$\sqrt{1-(\frac{v}{c})^2}$

Rest E=mc²

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*10¹¹ seconds)$\sqrt{1-(\frac{v}{c})^2}$

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

Thank you!

 

[BvU]

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

 

[Doc Al]

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

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.

 

[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...

 

[Doc Al]

Also, I have a rather old particle data book. It claims these $\Omega^-$ live only $8.22 \times 10^{-11}$ seconds.

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

 

[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.

 

[Doc Al]

I'm just getting mixed up on which t is which and how to solve for t overall.

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.)

 

[oddjobmj]

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

 

[Doc Al]

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

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

 

[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?

 

[oddjobmj]

Given your suggestion:

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?

 

[Doc Al]

Moving clocks run slower

Right.

so the time in the particle's proper frame will measure longer than the time in the 'station' frame.

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.

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?

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

 

[Doc Al]

Given your suggestion:

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

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

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

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

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