# Energy powers in decay widths

1. Jul 25, 2013

### Dreak

In Muon decay, the decay width is:
$\Gamma$$\mu$ = hbar/$\tau$$\mu$ = GF² / (192π³(hbar.c)6) . (m$\mu$c²)5(1+$\epsilon$)

It's this (m$\mu$c²)5 that bothers me.. How do you decide the powers in other reactions? I know it has something to do with the degrees of freedom, but I don't know how.

Also: what would be the answer to following question?
p+ + p- -> top particles.
The decay width will go according to GF.(Mt)x
What is the value of x?

Last edited: Jul 25, 2013
2. Jul 25, 2013

### Bill_K

Doesn't it come from the phase space factors? If you have N particles in the final state there are N-1 factors of d3p to integrate over. Two factors in this case. Then Γ also has a 1/2E in front from the initial state, so a total of 2*2 -1 = 5 factors of mc2 in this case.

3. Jul 25, 2013

### clem

G_F~1/M_p^2, so m_\mu^5 is needed to give an energy.

4. Jul 25, 2013

### Bill_K

Isn't it the other way around? the (mc2)5 factor from phase space is what requires the dimensionality of GF to be what it is.

5. Jul 25, 2013

### Dreak

It has indeed something to do with phase space (I can remember something from that).
But I don't really understand why there are N-1 factors of d³p.
The d³p comes from the phase space of one particle, but why N-1 and not N?

I thought the (hc)^x was added to make the dimensionalty fit?

6. Jul 25, 2013

### fzero

I would say that $G_F$ is determined from the interaction vertices and propagator (in terms of whichever EW parameters make sense for the calculation), while the phase space factors are responsible for making sure that the whole thing has the dimensions of a decay rate.

7. Jul 25, 2013

### Bill_K

No, I would disagree. There is nothing fundamental about GF. It was not defined from any fundamental theory. It does not appear in any Lagrangian. Fermi wrote it down as an empirical "thing in front". Change the underlying theory from V-A to any other interaction, and you would still need to put a thing in front with exactly the same dimensions. Its dimensionality does not come from theory, it is dictated solely by the kinematics.

Last edited: Jul 25, 2013
8. Jul 25, 2013

### fzero

Yes, the factor is really $g^2/m_W^2$ (for an appropriate diagram, $m_Z$ might well appear in another), which is being called $G_F$ here because of historical convention. The factor appears because of two vertices and weak boson propagator. The remaining dimensionful quantities in the decay rate arise from power counting of the remaining kinematical factors.

9. Jul 25, 2013

### RGevo

I think bills point is that the vertex could have been anything.

I.e some higher dimensional operator.

It just happened to be that it was the w propagator and couplings in there.

10. Jul 25, 2013

### fzero

Let's consider a simplified theory where there is one dimensionless coupling $g$ and a collection of masses $m_i$. In general we can write

$$\Gamma(g,m_i) = \Gamma_0 (g,m_i)+ \Gamma_1 (g,m_i)+ \cdots ,$$

where $\Gamma_0$ is partial width for the lowest order process and the rest of the terms are an expansion in powers of $g$. Several diagrams might contribute to each $\Gamma_a$, but we don't need to be so refined. We can always express

$$\Gamma(g,m_i) = \Gamma_0 (g,m_i) (1 + F (g,x_i)) ,$$

where now $F(g,x_i)$ is a function of the dimensionless variables $x_i = m_i/m_k$, for some conveniently chosen reference mass $m_k$. The expression in the OP is of this type. The factor of $m_\mu$ out in front is determined from the leading-order contribution, while the other diagrams contribute to $F$. For any given diagram we can read off certain factors by counting vertices and propagators, but if the diagram is too complicated, we would probably need to actually do some work to figure out the leading dependence on the ratios $x_i$.

In any case, the lesson to learn is that the leading coefficient is tied to the lowest order process, which should be easy enough to work out for any process of interest.

11. Jul 25, 2013

### Staff: Mentor

You don't care about the motion of the center of mass. In the center of mass system, one momentum can be determined based on the other particles.

12. Jul 26, 2013

### bootstrap63

I recommend Commins and Bucksbaum "Weak Interactions Of Leptons And Quarks" (1983) Cambridge, particularly Chap. 3, where you will find not only the decay width of the muon, but also a very clear discussion of the parameters involved, including their experimental justification and analysis, all worked out in most satisfying detail. You will find your formula patiently arrived at on pp.98. of Commins Bucksbaum (1983).

13. Jul 26, 2013

### andrien

If fermi interaction is written as something like G(ψ-γμψ)(ψ-γμψ),Then on the basis of simple dimensional analysis as we know ψ has mass dimension 3/2(from the term like mψ-ψ)in 4-D space time.So GF comes out to be having mass dimension -2.

14. Jul 26, 2013

### Dreak

wait, so we got this equation:
p_muon = p_e + p_nu(muon) + p_anti-nu(e)

and we need to integrate over the momenta in phase space:

∫p_muon².p_e².p_nu(muon)².p_anti-nu(e)²d³p

in which we can change p_muon with the above momentum equation?

I'm really not good in setting up these equations and I'm currently unable to get my book.. :/

15. Jul 26, 2013

### clem

When two things are related, half of the people will say one is basic and the other is derived from it, while half will say v-v. The statistics in this thread verify that.