9
$\begingroup$

The dominant channels in the GZK process are

$$p+\gamma_{\rm CMB}\to\Delta^+\to p+\pi^0,$$ $$p+\gamma_{\rm CMB}\to\Delta^+\to n+\pi^+.$$

According to the pdg, $\Delta\to N+\pi$ makes up essentially 100% of the branching ratio (BR). It doesn't, however, say which process is favored: the proton and neutral pion or neutron and charged pion. My instinct is that they should each contribute about 50%, but I am not sure. So my question is, what are the BRs for each of the processes described above?

$\endgroup$
  • $\begingroup$ Interesting question! Note that the final-final states, after the neutrons and pions have decayed, are $$\begin{align*}p + \gamma_\text{CMB} &\to p + 2\gamma \\ p + \gamma_\text{CMB} &\to p + (e^- + \bar\nu_e) + (e^+ + \nu_e + \nu_\mu),\end{align*}$$ so the neutron channel probably contributes much more to the softening of the proton cosmic ray spectrum. $\endgroup$ – rob May 18 '14 at 15:22
  • $\begingroup$ @rob Thanks, but I am interested more in the exact decay process not the total softening. That is, how often does which process occur? Google-fu has not turned up anything as everyone just seems to talk about nucleons and pions rather than their charges. $\endgroup$ – jazzwhiz May 18 '14 at 15:26
  • 4
    $\begingroup$ Some relevant calculations as comment; I'm not quite sure about the appropriate wording to formulate an entire answer: With suitable isospin CL coefficients$$p+\pi^0\equiv |1/2,1/2\rangle\,|1,0\rangle=\sqrt{2/3}~|3/2,1/2\rangle-\sqrt{1/3}~|1/2,1/2 \rangle,$$and$$n+\pi^+ \equiv|1/2,-1/2\rangle\,|1,1\rangle=\sqrt{1/3}~|3/2,1/2\rangle+\sqrt{2/3}~|1/2,1/2\rangle.$$The branching ratio is thereby $$\frac{\Delta^+\rightarrow p+\pi^0}{\Delta^+\rightarrow n+\pi^+}\approx (\sqrt{2/3}~/\sqrt{1/3})^2=2.$$ $\endgroup$ – user12262 Jul 16 '14 at 18:02
  • $\begingroup$ @user12262 Thanks. I am assuming that you are using the fact that the Delta is a $|3/2,1/2\rangle$ particle, correct? It is a long time since I have done Clebsch-Gordon so I will have to refresh myself on this. Seeing that the ratio is $\mathcal O(1)$ is encouraging. $\endgroup$ – jazzwhiz Jul 16 '14 at 18:04
  • $\begingroup$ jazzwhiz: "I am assuming that you are using the fact that the Delta is a $|3/2,1/2 \rangle$ particle" -- The $\Delta^+$, right. (Also, I didn't point that out explicitly in my above comment since I had been using up pretty much all of the permitted 600 characters already &). "It is a long time since I have done Clebsch-Gordon" -- Same here. ($\approx$ 20 years since I first and last saw the similar problem of calculating $\frac{ \sigma(p \pi^+ \rightarrow p \pi^+)}{\sigma(p \pi^- \rightarrow p \pi^-)}$ etc.). So I'm a bit hesitant to "argue away" the remaining $|1/2,1/2 \rangle$ parts ... $\endgroup$ – user12262 Jul 16 '14 at 18:18
9
+300
$\begingroup$

[...] $\Delta^+ \rightarrow p + \pi^0$, [...] $\Delta^+ \rightarrow n + \pi^+$,

which process is favored: the proton and neutral pion or neutron and charged pion [?]

Since the kinematics (and corresponding "phase space" factors) for the two final states are presumably as good as equal, the evaluation of the branching ratio

$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]}$$

simplifies to determining the ratio of "state constituent" transition probabilities

$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{\left\lvert \langle p; \pi^0 \mid \Delta^+ \rangle \right\rvert^2}{\left\lvert \langle n; \pi^+ \mid \Delta^+ \rangle \right\rvert^2}.$$

Analyzing (or defining) the initial state $\Delta^+$ and the two distinct final states in terms of isospin leads to the expressions

$$ \lvert \Delta^+ \rangle \equiv \big\lvert \left(3/2, 1/2\right)_i \big\rangle, $$

where the first value represents the magnitude of $\mathbf I$, and the second value represents the magnitude of $I_3$, along with

$$ \lvert p; \pi^0 \rangle \equiv \big\lvert (1/2, 1/2)_f; (1, 0)_f \big\rangle \equiv \sqrt{ \frac{2}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle - \sqrt{ \frac{1}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$ and

$$ \lvert n; \pi^+ \rangle \equiv \big\lvert (1/2, -1/2)_f; (1, 1)_f \big\rangle \equiv \sqrt{ \frac{1}{3} }~\big\lvert (3/2, 1/2)_t \big\rangle + \sqrt{ \frac{2}{3} }~\big\lvert (1/2, 1/2)_t \big\rangle, $$

where

  • the coefficients of the linear combinations on the right-hand sides are Clebsch-Gordan coefficients (specificly those values listed in table "$1/2 \otimes 1$"),

  • all states are normalized, and

  • the indices $f$ and $t$ are to distinguish final states and "state representations to evaluate transition probabilities"; such that in particular the states $(1/2, 1/2)_f$ and $(1/2, 1/2)_t$ are (meant to be) distinct; and both are distinct, and indeed disjoint, from the initial state $\lvert \Delta^+ \rangle \equiv \lvert (3/2, 1/2)_i \rangle$.

Now identifying

$$\big\lvert (3/2, 1/2)_t \big\rangle \equiv \big\lvert (3/2, 1/2)_i \big\rangle $$

we can evaluate

\begin{align} \langle p; \pi^0 \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t - \sqrt{ \frac{1}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \bigg\langle \sqrt{ \frac{2}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \sqrt{ \frac{2}{3} } \end{align}

and

\begin{align} \langle n; \pi^+ \mid \Delta^+ \rangle & \equiv \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t + \sqrt{ \frac{2}{3} }~ (1/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \bigg\langle \sqrt{ \frac{1}{3} }~ (3/2, 1/2)_t \bigg\vert (3/2, 1/2)_t \bigg\rangle \\ & = \sqrt{ \frac{1}{3} } \end{align}

obtaining the sought branching ratio value as

$$\text{BR} := \frac{\Gamma[ \Delta^+\rightarrow p+\pi^0 ]}{\Gamma[ \Delta^+\rightarrow n+\pi^+ ]} \simeq \frac{ (\sqrt{ 2/3 })^2 }{ (\sqrt{ 1/3 })^2} = 2.$$

$\endgroup$
  • 1
    $\begingroup$ The initial version of this answer has been formulated based on the comments of user12262 to the OP question above; and in consideration of remarks stated there (meta). $\endgroup$ – user12262 Jul 16 '14 at 21:57
  • $\begingroup$ That specific branching ratio is also discussed in Chapter 8, eq. (8.11) of these 't Hooft's lecture notes. The pdf file is available here. $\endgroup$ – Qmechanic Jul 16 '14 at 22:34
  • $\begingroup$ @user12262 I also threw in some formatting. If I clobbered it too much, just roll back the edit. $\endgroup$ – user10851 Jul 16 '14 at 22:34
  • $\begingroup$ @Qmechanic: "That specific branching ratio is also discussed in Chapter 8, eq. (8.11) of these 't Hooft's lecture notes. [ www.phys.uu.nl/~thooft/lectures/lieg07.pdf (p. 46) ]" -- With the same result (eq. 8.11:$$ \Gamma[\Delta^+\rightarrow n~\pi^+] : \Gamma[\Delta^+\rightarrow p~\pi^0] = 1 : 2$$), thank goodness. A bit above (on that same page) 't Hooft refers to two different irreducible representations of the isospin group in default of which the present version of this answer had coined the phrase "state representations to evaluate transition probabilities". Close enough, I suppose ... $\endgroup$ – user12262 Jul 17 '14 at 5:19
  • $\begingroup$ @Chris White: "I also threw in some formatting. [...]" -- Exemplary, IMHO; I like it a lot, FWIW. (Indeed: honestly nothing but formatting thrown in, too...) I recall my frustration when logging in here for the very first time (or was it even right before that) that I was not (yet) permitted to read the source "code" of answers (nor comments); so I felt unable to match some essential ways of formatting. (I'll check if at least community wiki source content is readable without login. (How to search for community wiki, btw. ?)) $\endgroup$ – user12262 Jul 17 '14 at 5:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.