The minimum KE required for muons to travel a given distance

[Rahulrj]

Homework Statement

Muons of Kinetic energy 'E' are produced in collision with a target in a laboratory. The mass of a muon is ##106MeV/c^2## and its half life in it's rest frame is ## 1.4 * 10^{-6}## in its rest frame. what should be the minimum KE such that more than half of the muons createdwould travel a distance of 840 m to reach a detector from the target?

Homework Equations

## \Delta T_0= \gamma \Delta T'##
## \Delta x_0= \gamma (\Delta x'+v\Delta T')##
## E = (\gamma - 1)mc^2##

The Attempt at a Solution

I am bit confused since the mass is given in terms of energy/c^2. I do not know if the energy represented is Total energy or kinetic energy. If its KE, I can write the answer right away as 106 MeV and if its total energy then from ## E = (\gamma - 1)mc^2 ## ## \gamma mc^2 ## is the total energy
therefore 106 = E + mc^2 and from here E becomes 0 so both idea seems nonsensical to me.
So then it tells me I have to find ## \gamma ##. However I do not know how to get it from the info given.
Please help!

 

[PeroK]

I am bit confused since the mass is given in terms of energy/c^2. I do not know if the energy represented is Total energy or kinetic energy.
Please help!

##106MeV/c^2## is the rest mass of the muon. These units are useful because you can get the energy of the muon from:

##E = \gamma mc^2 = \gamma 106 MeV##

as the factors of ##c^2## cancel.

To solve this problem you need to do two things:

a) Find out the required speed of the muons.

b) Calculate the required KE from the speed.

 

[Rahulrj]

##106MeV/c^2## is the rest mass of the muon. These units are useful because you can get the energy of the muon from:

##E = \gamma mc^2 = \gamma 106 MeV##

as the factors of ##c^2## cancel.

To solve this problem you need to do two things:

a) Find out the required speed of the muons.

b) Calculate the required KE from the speed.

How can I find speed since I am unable to find ## \gamma## ?
the distance given is 840 m in lab frame and the time given is the half life in muons frame, So dividing them does not make sense and moreover v then becomes 2c. So can you give a hint on how to calculate speed?

I figured I can find ##\gamma ## from ##E^2 = (pc)^2+ (E_0)^2##
and it gives me ##\gamma = 2 ##. am I right?

 

[PeroK]

How can I find speed since I am unable to find ## \gamma## ?
the distance given is 840 m in lab frame and the time given is the half life in muons frame, So dividing them does not make sense and moreover v then becomes 2c. So can you give a hint on how to calculate speed?

This is a general problem in SR. You can look at it two ways.

1) In the lab frame, the faster the muon travels the longer it lives. The distance traveled in this time is a function of ##\gamma## and ##v##, but ##\gamma## is a function of ##v##, so you should be able to solve for ##v##.

2) In the muon's frame, the faster the lab is moving, the more the lab is length contracted. You should get the same equation for ##v## (and ##\gamma##) as you did above.

 

[Rahulrj]

This is a general problem in SR. You can look at it two ways.

1) In the lab frame, the faster the muon travels the longer it lives. The distance traveled in this time is a function of ##\gamma## and ##v##, but ##\gamma## is a function of ##v##, so you should be able to solve for ##v##.

2) In the muon's frame, the faster the lab is moving, the more the lab is length contracted. You should get the same equation for ##v## (and ##\gamma##) as you did above.

I do not see the need for calculating speed since I found that ##\gamma = 2## like I said in the above comment therefore ##E= (\gamma - 1)mc^2## which gives me 106 MeV as the answer. I do not understand why calculate KE from speed.

 

[PeroK]

I do not see the need for calculating speed since I found that ##\gamma = 2## like I said in the above comment therefore ##E= (\gamma - 1)mc^2## which gives me 106 MeV as the answer. I do not understand why calculate KE from speed.

How did you get ##\gamma = 2##?

 

[Rahulrj]

How did you get ##\gamma = 2##?

Oh I was wrong, I made a mistake in substitution using ##E^2 = (pc)^2+ (E_0)^2##.
So then like you said using the equation

Δx0=γ(Δx′+vΔT′)

I am only able to get ##v=2c/\gamma##
Don't see how this helps which was why I tried using the other equation to find ##\gamma##

 

[PeroK]

Oh I was wrong, I made a mistake in substitution using ##E^2 = (pc)^2+ (E_0)^2##.
So then like you said using the equation

I am only able to get ##v=2c/\gamma##

That's correct. Remember that ##1/\gamma^2 = 1- \frac{v^2}{c^2}##, so just square that equation and solve for ##v##, then get ##\gamma##.

 

[Rahulrj]

That's correct. Remember that ##1/\gamma^2 = 1- \frac{v^2}{c^2}##, so just square that equation and solve for ##v##, then get ##\gamma##.

Ah yes indeed, all long I forgot what \gamma was. So I get v = 0.894 c
and now how do I find KE directly from v? or were you referring to the equation ##E= (\gamma - 1)mc^2)## where I find \gamma by substituting for v?

 

[PeroK]

Ah yes indeed, all long I forgot what \gamma was. So I get v = 0.894 c
and now how do I find KE directly from v? or were you referring to the equation ##E= (\gamma - 1)mc^2)## where I find \gamma by substituting for v?

Once you have ##v## you can calculate ##\gamma##. Then, you have the kinetic energy ##T = (\gamma -1)mc^2##

 

[Rahulrj]

Once you have ##v## you can calculate ##\gamma##. Then, you have the kinetic energy ##T = (\gamma -1)mc^2##

Thank you