How Fast Must a Pion Travel to Avoid Decay Over 1.9 km?

[student34]

Homework Statement

After being produced in a collision between elementary particles, a positive pion must travel down a 1.90 km long tube to reach an experimental area. A positive pion has an average lifetime of t_o = 2.60*10^(-8)s; the pion we are considering has this lifetime. How fast must the pion travel if it is not to decay before it reaches the end of the tube?

(I put "basic" in the title because it is a chapter in my first year physics program that introduces relativity)

Homework Equations

L = L_o*(1 - u^2/c^2)^(1/2)

or maybe Δt = Δt_o/(1 - u^2/c^2)^(1/2)

where

L = 1900m, and L_o = ? (the length of the tube relative to the pion).

Δt_o = 2.60*10^(-8)s, and Δt = ?

The Attempt at a Solution

I don't even know where to start because these formulas require more information. Is it even possible to do this question with the information given? If so, can someone help please?

 

[Bandersnatch]

Hi, student34.

First of all, double check the meaning of symbols in the equations. You've got them mixed at one point.(which symbols represent the length and time in a stationary reference frame?)

As for the solution, consider:

When you observe the pion, its time is dilated, but the length it has to travel is the same.

When the pion observes the tube, its time is unchanged, but the tube is length contracted.

Choose one and simply use V=L/t.

 

[student34]

Hi, student34.

First of all, double check the meaning of symbols in the equations. You've got them mixed at one point.(which symbols represent the length and time in a stationary reference frame?)

I figured it out, but V = L/t did not work for V = (1.9km)/(2.60*10^(-8)s). Forgive me if that was not what you meant. And in case your interested, I will explain everything.

To answer your first question, I was thinking that Δt_o is the time relative to the pion (proper time) which is 2.60*10^(-8)s. And Δt would be the unknown "observed" time. We know the length relative to the observer L = 1.9km, but we don't know L_o (length relative to the pion).

As for the solution, consider:

When you observe the pion, its time is dilated, but the length it has to travel is the same.

When the pion observes the tube, its time is unchanged, but the tube is length contracted.

Choose one and simply use V=L/t.

My issue was that the equation that seemed to be needed for this question Δt = Δto/(1 - u^2/c^2)^(1/2) has two unknowns, u and Δt. But it finally dawned on me that I can put u in terms of L/Δt and then just isolate Δt. It gives the right answer.