I'm currently preparing for an examination of course in introductory (experimental) particle physics. One topic that we covered and that I'm currently revising is the universality in weak interactions.

However I don't really understand the point my professor wants to make here.

Let me show you his exposure to the topic:

We stark by looking at three different weak decays:

  • $\beta$-decay in a nuclei: $^{14}O \rightarrow ^{14}N^{*} + e^{+} + \nu_{e}$ (Lifetime $\tau$=103sec)
  • $\mu$-decay: $\mu^{+}\rightarrow e^{+} + \nu_{e} + \overline{\nu_{\mu}}$ ($\tau$=2.2$\mu$sec)
  • $\tau-decay$: $\tau^{+}\rightarrow$ $\mu^{+} + \nu_\mu + \overline{\nu_{\tau}}$ or $\tau^{+}\rightarrow$ $e^{+} + \nu_e + \overline{\nu_{\tau}}$ ($\tau = 2 \cdot 10^{13}$ sec).

Now he points out that the lifetimes are indeed quiet different. Never the less the way the reactions behave is the same? (Why? Okay we always have a lepton decaying into another leptons and neutrinos? But what's the deal?)

Then he writes down Fermi's golden rule given by: $W=\frac{2\pi}{\hbar} |M_{fi}|^{2} \rho(E')$.

Now he says that universality means that the matrix element $|M_{fi}|$ is the same in all interactions. The phase space however is the same (Why? First of all I have often read on the internet that universality means that that certain groups of particles carry the same "weak charge"? And secondly: What do particle physicists mean when they talk about greater phase-space? Three dimensional momentum space? But how do you see or measure that this space is bigger? And bigger in what respect? More momentum states?)

Now he says that the different phase spaces come from the different lifetimes. He calculates $\rho(E')=\frac{dn}{dE}=\int \frac{d^{3}p_{\nu} d^{3}p_{e}}{dE} = p_{max}^{5} \cdot \int \frac{d^{3}(p_{\nu}/p_{max}) d^{3}(p_{e}/p_{max})}{d(E/p_{max})}$.

Now the last integral is supposed to be identical for all decays. And $p_{max}$is suppose to be different in all decays . But why? And what is the definition of $p_{max}$?

Now he has $\tau = \frac{1}{M}$. So he gets $\tau = const \cdot \frac{1}{|M_{fi}|^{2} p_{max}^5}$. Hence he gets in the ln($\tau$)-ln($p_{max}$) diagram a line with slope -5.

Now he claims that this "proofs" that $|M_{fi}|$ is constant in all processes. Again why?

Can someone please give me some overview and explain to me why he is doing all that stuff? I din't really have much background when it comes to particle physics. So can someone explain it to me in a clear and easy way?


The phase-space represents the "number" of allowed final states (think of a discrete quantum system with degenerate final states, except that here we have a continuum).

More final states makes the transition more likely to happen and thus gives it a shorter lifetime.

Each of the reactions that you show has a three body final state and a single body initial state. Without loss of generality, we consider the system in the rest frame of the initial particle and write it's energy and momentum as $\mathbf{\mathcal{P}} = (\mathcal{M},\vec{0})$. For each of the products write $\mathbf{p}_i = (E_i,\vec{p}_i)$.

This leaves us 12 unknown parameters at this point, but each product particle has a definite mass, which puts three constraints on the system, and energy and momentum are conserved putting another four constrains on the system. Only some of these may be redundant (in particular one of the masses is fixed by the other concerns).

How do we visualize the parameter space?

  • Assume (again, without loss of generality) that the particle indexed 1 will have it's momentum parallel with $+\hat{z}$. So the first parameter is $p_1 = \vec{p}_1 \cdot \hat{z}$. This is the choice of a kinetic energy (1 parameter).
  • Each of the other two particles must have $\vec{p}_2 = (p_\perp, p_\|)_2$. This is equivalent to choosing a angle with respect to $\hat{z}$ and a kinetic energy (2 parameters).
  • The final particle's energy and momentum are fixed by the choices we have made so far.

We have a three parameter phase space $\langle p_{1,z}, p_{2,\perp}, p_{2,\|} \rangle$ (and a three fold ambiguity of orientation for the whole system) and these parameters are linked by the necessity of conserving energy and momentum (which you represent in the integrals as a bunch of Dirac $\delta$ functions), so they map out an allowed volume.

Now the important part: the size of that volume depends in large measure on the energy excess represented by $\mathcal{M} - \sum_i m_i$, and on the relative masses of the products.

I once wrote a Mote Carlo generator for a reaction with un-measured branching ratios using this kind of phase-space calculation to compute the relative likelihood of each final state. It got branching ratios in as good agreement with the data we eventually took as those based on a many-body calculation I found.

  • $\begingroup$ Hey. Thanks for your answer. I think it really explains quite well what the phase space is. However could you please also adress my other questions if possible? In particular the question concerning the integral manipulation, etc.? Thanks in advance! $\endgroup$ – fermi333 Jun 30 '11 at 7:31

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