How Far Does a Pion Travel in the Lab Frame?

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Homework Help Overview

The discussion revolves around calculating the distance a pion travels in the laboratory frame, given its mean lifetime and speed. The problem involves concepts from special relativity and particle physics.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between speed, time, and distance, questioning the appropriateness of a formula provided by the original poster's professor. There is discussion about the correct time duration to use for calculations and the implications of using the mean lifetime in different frames.

Discussion Status

Participants are actively engaging with the problem, offering hints and corrections regarding the calculations. Some suggest re-evaluating the time used in the distance calculation, while others emphasize the need for accuracy in the values employed.

Contextual Notes

There is mention of the pion's mean lifetime in its rest frame and the speed at which it travels in the lab frame. Participants are navigating discrepancies in their calculations and the interpretation of the problem's requirements.

kavamo
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Homework Statement



I have correctly figured out part a but am stuck on part b. Thanks for your help in advance.

An unstable particle called the pion has a mean lifetime of 25 ns in its own rest frame. A beam of pions travels through the laboratory at a speed of 0.587c.
(a) What is the mean lifetime of the pions as measured in the laboratory frame?
30.75 ns

(b) How far does a pion travel (as measured by laboratory observers) during this time?
m


Homework Equations



[ Delta t x c (speed of light) ] / 2

The Attempt at a Solution



using the answer from part a and converting to meters:

[(3.075E-8)(3E8)] / 2 = 4.6125 I've checked for errors and always come up with this number--which according to the web program is incorrect.

What am I not doing/understanding?
 
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kavamo said:

Homework Equations



[ Delta t x c (speed of light) ] / 2
:confused: Where in the world did you get this formula?

Hint: How fast is it moving? How long is it moving for? How do you calculate distance traveled?
 
I got the formula from my professor.
As far as I can tell it is moving at 0.587c = 176100000 m/s for 25 ns.

Distance = rate x time.

So then I suppose I would do the following:

176100000 m/s x (2.5 x 10^-10) = 0.044025 meters?
 
kavamo said:
I got the formula from my professor.
It must be meant for some specific application--but not this one.
As far as I can tell it is moving at 0.587c = 176100000 m/s for 25 ns.
Why 25 ns? Stick to the lab frame.

Distance = rate x time.
That's all you need.
 
o.k. then,

176100000m/s x (3.075 x 10^-10) = 0.05415075 ~ 0.0542 and convert to meters = 5.4 m
 
Last edited:
kavamo said:
o.k. then,

176100000m/s x (3.075 x 10^-10) = 0.05415075 ~ 0.0542 and convert to meters = 5.4 m
Two things, the first being the most important:
- check your exponents
- use a more accurate value for the time in part a (redo it); only round off at the end.
 
thank you I should have typed:

176100000m/s x (3x 10^8)x 30.75 (the answer from part a) = 5415075000

then convert from nano to meters (9 decimal spaces to the left) = 5.41 meters

which is the correct answer. Thank you for your guidance. K
 

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