Calculating Fermi Coupling Constant (muon)

Click For Summary

Homework Help Overview

The discussion revolves around calculating the Fermi Coupling Constant, specifically using a formula that involves the mass of the muon, the speed of light, and the reduced Planck constant. Participants are exploring the relationship between these variables and the expected value of the constant.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to plug in values into the formula for the Fermi Coupling Constant and are questioning the correctness of their calculations. There is a focus on unit consistency and dimensional analysis.

Discussion Status

Some participants have identified potential issues with unit conversions and the application of the formula. There is ongoing exploration of the calculations, with requests for clarity on the steps taken and the units used. One participant has indicated a realization of an error in their initial calculation.

Contextual Notes

Participants are discussing the implications of expressing the mass of the muon in energy units and the necessary conversions involved in the calculations. There is a mention of homework constraints that may affect the approach to solving the problem.

tryingtolearn1
Messages
58
Reaction score
5
Homework Statement
Fermi Coupling Constant (muon)
Relevant Equations
##\Rightarrow \frac{G_F}{(\hbar c)^3}=\sqrt{\frac{\hbar}{\tau_\mu}\cdot\frac{192\pi^3}{(m_\mu c^2)^5}}##
I am trying to determine the Fermi Coupling Constant which is measured to be ##1.1663787 *10^{-5}\text{Ge}V^{-2}##. The formula for Fermi is ##\frac{G_F}{(\hbar c)^3}=\sqrt{\frac{\hbar}{\tau_\mu}\cdot\frac{192\pi^3}{(m_\mu c^2)^5}},## where ##m_\mu## is the mass of a muon which is ##\approx 0.105##GeV, ##c## is the speed of light ##\approx 3*10^8##m/s, ##\hbar## is reduced plank constant which is ##\hbar\approx 6.582\cdot 10^{-25}\,\rm{GeV}\cdot\rm{s}## and ##\tau_\mu## is the mean decay time of a muon which is ##\approx 2.2*10^{-6} s##. Now plugging all these values into the formula gives $$\frac{G_F}{(\hbar c)^3}=\sqrt{\frac{\hbar}{\tau_\mu}\cdot\frac{192\pi^3}{(m_\mu c^2)^5}}$$
$$=\sqrt{\frac{6.582*10^{-25}*192*\pi^3}{2.2*10^{-6}*(0.105*(3*10^8)^2)^5}}\approx 4.83157 × 10^{-48}$$ which is totally off from the expected value of ##1.1663787 *10^{-5}\text{Ge}V^{-2}##. Not sure what I am doing wrong?
 
Physics news on Phys.org
tryingtolearn1 said:
##m_\mu## is the mass of a muon which is ##\approx 0.105##GeV
Consider why the mass of the muon is expressed in units of energy.
 
TSny said:
Consider why the mass of the muon is expressed in units of energy.
Ops I meant the mass of the muon is ##\approx 0.105GeV/c^2## but even with that unit correction I still get a value that is way off.
 
tryingtolearn1 said:
Ops I meant the mass of the muon is ##\approx 0.105GeV/c^2## but even with that unit correction I still get a value that is way off.
Did you change anything in your first attempt shown below?
tryingtolearn1 said:
$$\frac{G_F}{(\hbar c)^3}=\sqrt{\frac{6.582*10^{-25}*192*\pi^3}{2.2*10^{-6}*(0.105*(3*10^8)^2)^5}}\approx 4.83157 × 10^{-48}$$
If so, can you show your new calculation? It would help if you show all units for all the numbers in your calculation. See if the units cancel to give the desired units for the answer.
 
tryingtolearn1 said:
the mass of the muon is ##\approx 0.105GeV/c^2##
Using this value for m, what is the value of mc2 in units of GeV?
 
TSny said:
Did you change anything in your first attempt shown below?

If so, can you show your new calculation? It would help if you show all units for all the numbers in your calculation. See if the units cancel to give the desired units for the answer.

My new calculation using only dimensional analysis is $$\sqrt{\frac{GeV\cdot s}{s}\cdot\frac{1}{[(\frac{GeV\cdot s}{c^2}) \cdot c^2]^5}}=GeV^{-2}\cdot s^{-5/2}$$ which has an extra ##s^{-5/2}##.
 
TSny said:
Using this value for m, what is the value of mc2 in units of GeV?

Actually I figured out my issue. I was include an extra factor of ##c^5##. Inside the square root it should have been ##.105^5## instead of ##(.105*(3*10^8)^2)^5##. Ty
 

Similar threads

Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 19 ·
Replies
19
Views
2K
Replies
15
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 9 ·
Replies
9
Views
13K
  • · Replies 0 ·
Replies
0
Views
3K