Muon decay rate

  • Thread starter kelly0303
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  • #1
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Homework Statement


I need to calculate the muon decay rate, ignoring the mass of the outgoing particles.

Homework Equations


##d\Gamma = \frac{1}{2E_1}|M|^2d\Pi_{LIPS}##

The Attempt at a Solution


I am actually having problem with the math at a point. I reached this $$d\Gamma=\frac{mG^2|\vec{k_3}|^2}{8\pi^4}(m-2|\vec{k_3}|)\frac{sin(\theta)d|\vec{k_3}|d\theta d^3 k_4}{(|\vec{k_3}|^2+|\vec{k_4}|^2+2|\vec{k_3}||\vec{k_4}|cos(\theta))|\vec{k_4}|}\delta(m-|\vec{k_3}+\vec{k_4}|-|\vec{k_3}|-|\vec{k_4}|)$$ And I didn't know what to do. I looked online and I found something switching variables $$u^2 = |\vec{k_3}|^2+|\vec{k_4}|^2+2|\vec{k_3}||\vec{k_4}|cos(\theta)$$ $$2udu=-2|\vec{k_3}||\vec{k_4}|sin(\theta)d\theta$$ And upon replacement they get $$d\Gamma=\frac{mG^2|\vec{k_3}|}{8\pi^4}(m-2|\vec{k_3}|) \frac{dud|\vec{k_3}|d^3 k_4}{|\vec{k_4}|^2}\delta(m-u^2-|\vec{k_3}|-|\vec{k_4}|)$$ However I get something different: $$d\Gamma=\frac{mG^2|\vec{k_3}|}{8\pi^4}(m-2|\vec{k_3}|) \frac{dud|\vec{k_3}|d^3 k_4}{-u|\vec{k_4}|^2}\delta(m-u-|\vec{k_3}|-|\vec{k_4}|)$$ The step before this one, me and the solution I found have the same expression. I followed through the solution I found and in the end they reach the answer I need, so what they do is right but I am not sure what I am doing wrong. Basically they are getting $$\delta(m-u^2-|\vec{k_3}|-|\vec{k_4}|)$$ and I am getting $$\frac{1}{-u}\delta(m-u-|\vec{k_3}|-|\vec{k_4}|)$$ Is there a delta function identity to make the 2 expression equal? Or am I doing something wrong? Thank you!
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
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Homework Statement


I need to calculate the muon decay rate, ignoring the mass of the outgoing particles.

Homework Equations


##d\Gamma = \frac{1}{2E_1}|M|^2d\Pi_{LIPS}##

The Attempt at a Solution


I am actually having problem with the math at a point. I reached this $$d\Gamma=\frac{mG^2|\vec{k_3}|^2}{8\pi^4}(m-2|\vec{k_3}|)\frac{sin(\theta)d|\vec{k_3}|d\theta d^3 k_4}{(|\vec{k_3}|^2+|\vec{k_4}|^2+2|\vec{k_3}||\vec{k_4}|cos(\theta))|\vec{k_4}|}\delta(m-|\vec{k_3}+\vec{k_4}|-|\vec{k_3}|-|\vec{k_4}|)$$ And I didn't know what to do. I looked online and I found something switching variables $$u^2 = |\vec{k_3}|^2+|\vec{k_4}|^2+2|\vec{k_3}||\vec{k_4}|cos(\theta)$$ $$2udu=-2|\vec{k_3}||\vec{k_4}|sin(\theta)d\theta$$ And upon replacement they get $$d\Gamma=\frac{mG^2|\vec{k_3}|}{8\pi^4}(m-2|\vec{k_3}|) \frac{dud|\vec{k_3}|d^3 k_4}{|\vec{k_4}|^2}\delta(m-u^2-|\vec{k_3}|-|\vec{k_4}|)$$
This last expression cannot be right, since the dimensions in the delta function do not match (##u^2## has the dimensions of an energy squared). So there is something wrong with that. Can you show the rest of your steps, or at least the final expression?
 

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