Bethe-Bloch Equation different materials

[stutiger99]

I am trying to use the Bethe-Bloch equation to investigate how a Muon loses its energy as it penetrates through different materials in the muon lifetime experiment. In order to do so I need 2 determine both the total energy and kinetic energy.
For kinetic energy maximum I have been using

Tmx = (2*m*p^2*c^4) / (m^2*c^4 + M^2*c^4 + 2*m*c^2*(p^2*c^2 + M^2*c^4)^0.5 )

Typical values for Tmax I should be getting I have been informed are around 645MeV however I keep getting values in the region of 45MeV so I have obviously went wrong somewhere. Am I correct in saying:
m = electron mass (0.511MeV)
p = momentum of muon (achieved via p= gamma*mass of muon*c)
M = mass of the muon (105.6584 MeV)

If it is correct is there a chanae I am wrong with units of c somewhere? Or is it even density realted?


Also the velocity of a muon, is it correct that it is 0.9901c (i.e Beta = v/c for special relativity) before striking the material?

Much help would be greatly appreciated.

 

[Meir Achuz]

I don't understand your equation, but
if all masses, energies, and momenta are in MeV, you don't need any factors of c.

 

[stutiger99]

An equivalent of the equation is on wikipedia. It is slightly different but it can be simplified to what I have above for Tmax (N.B. this Tmax is the one used in the bethe-bloch equation.

 

[hamster143]

To get 645 MeV with either equation, you need gamma ~ 25. The velocity you're quoting (0.9901c) corresponds to gamma ~ 7. That could be the problem. I don't know what the correct velocity is, it depends on your specific experiment.

 

[Vanadium 50]

Also the velocity of a muon, is it correct that it is 0.9901c (i.e Beta = v/c for special relativity) before striking the material?

There aren't many muon sources with the velocity known to 4 digits of accuracy.

 

[hamster143]

There aren't many muon sources with the velocity known to 4 digits of accuracy.

Many muons produced in LHC collisions have energies greater than ~8 GeV and therefore have velocities of 0.9999c (to 4 digits of accuracy, rounded down).

 

[Vanadium 50]

Many muons produced in LHC collisions have energies greater than ~8 GeV and therefore have velocities of 0.9999c (to 4 digits of accuracy, rounded down).

Fine. You got me there.

Nonetheless, there are very few monoenergetic muon sources at the energies under discussion, which leads me to suspect something may be wrong here.

 

[malawi_glenn]

Many muons produced in LHC collisions have energies greater than ~8 GeV and therefore have velocities of 0.9999c (to 4 digits of accuracy, rounded down).

Have LHC started to colliding yet?

...trying to save Vanadium 50 ;-) ...

 

[Vanadium 50]

The LHC collided 450 GeV protons on steel. That gives a maximum muon energy of 30 GeV and an average muon energy of ~1.5 GeV.

 

[malawi_glenn]

The LHC collided 450 GeV protons on steel. That gives a maximum muon energy of 30 GeV and an average muon energy of ~1.5 GeV.

Ahh yeah that is true, during detector performance tests