Verify the isospin factor $\sqrt{2}$ in (12.19). Note that ${ }^{14} \mathrm{C}$, ${ }^{14} \mathrm{~N}^*,{ }^{14} \mathrm{O}$ form an isospin triplet, which can be viewed as $\mathrm{nn}, \mathrm{np}, \mathrm{pp}$, together with an isospin zero ${ }^{12} \mathrm{C}$ core (see Fig. 2.2). Keep in mind that for indistinguishable proton decays, we must add amplitudes, not probabilities.
The rate $d \Gamma$ for the " $\mathrm{p}$ " $\rightarrow$ " $\mathrm{n}^{\prime} \mathrm{e}^{+} \nu$ transition is related to $|9|^2$ by (4.36). We obtain
$$
\begin{aligned}
d \Gamma= & G^2 \sum_{\text {spins }}\left|\bar{u}\left(p_v\right) \gamma^0\left(1-\gamma^5\right) v\left(p_e\right)\right|^2 \frac{d^3 p_e}{(2 \pi)^3 2 E_e} \\
& \times \frac{d^3 p_\nu}{(2 \pi)^3 2 E_\nu} 2 \pi \delta\left(E_0-E_e-E_v\right),
\end{aligned}
$$
where $E_0$ is the energy released to the lepton pair. The normalization factor $\left(2 m_N\right)^2$ cancels with the equivalent $2 E_p 2 E_n$ factor in (4.36), as indeed it must. The summation over spins can be performed using the techniques we introduced in Chapter 6. Neglecting the mass of the electron, we have
$$
\begin{aligned}
\sum_{\text {spins }}\left|\bar{u} \gamma^0\left(1-\gamma^5\right) v\right|^2 & =\sum\left(\bar{u} \gamma^0\left(1-\gamma^5\right) v\right)\left(\bar{v}\left(1+\gamma^5\right) \gamma^0 u\right) \\
& =\operatorname{Tr}\left(p_\nu \gamma^0\left(1-\gamma^5\right) p_e\left(1+\gamma^5\right) \gamma^0\right) \\
& =2 \operatorname{Tr}\left(p_v \gamma^0 p_e\left(1+\gamma^5\right) \gamma^0\right) \\
& =8\left(E_e E_v+\mathbf{p}_e \cdot \mathbf{p}_v\right) \\
& =8 E_e E_v\left(1+v_e \cos \theta\right),
\end{aligned}
$$
where $\theta$ is the opening angle between the two leptons and where the electron velocity $v_e=1$ in our approximation. Here, we have used the trace theorems of Section 6.4 ; see also (12.25) and (12.26). Substituting (12.21) into (12.20), the transition rate becomes
$$
d \Gamma=\frac{2 G^2}{(2 \pi)^5}(1+\cos \theta)\left[\left(2 \pi d \cos \theta p_e^2 d p_e\right)\left(4 \pi E_\nu^2 d E_v\right)\right] \delta\left(E_0-E_e-E_v\right),
$$
where $d^3 p_e d^3 p_v$ has been replaced by the expression in the square brackets.
Many experiments focus attention on the energy spectrum of the emitted positron. From (12.22), we obtain
$$
\begin{aligned}
\frac{d \Gamma}{d p_e} & =\frac{4 G^2}{(2 \pi)^3} p_e^2\left(E_0-E_e\right)^2 \int d \cos \theta(1+\cos \theta) \\
& =\frac{G^2}{\pi^3} p_e^2\left(E_0-E_e\right)^2 .
\end{aligned}
$$
Thus, if from the observed positron spectrum we plot $p_e^{-1}\left(d \Gamma / d p_e\right)^{1 / 2}$ as a function of $E_e$, we should obtain a linear plot with end point $E_0$. This is called the Kurie plot. It can be used to check whether the neutrino mass is indeed zero. A nonvanishing neutrino mass destroys the linear behavior, particularly for $E_e$ near $E_0$. (In practice, of course, we must examine the approximations we have made, correct $E_e$ for the energy gained from the nuclear Coulomb field, and allow for the experimental energy resolution.)
Our immediate interest here is of a different nature. We wish to determine $G$ from the observed value $E_0$ and the measured lifetime $\tau$ of the nuclear state to $\beta$-decay. We therefore carry out the $d p_e$ integration of (12.23) over the interval $0, E_0$. Making the relativistic approximation $p_e \approx E_e$, we find
$$
\Gamma=\frac{1}{\tau}=\frac{G^2 E_0^5}{30 \pi^3} .
$$
Now, for ${ }^{14} \mathrm{O} \rightarrow{ }^{14} \mathrm{~N}^* \mathrm{e}^{+} \nu$, the nuclear energy difference $E_0$ is $1.81 \mathrm{MeV}$, and the measured half-life is $\tau \log 2=71 \mathrm{sec}$. Using this information, we find
$$
G \simeq 10^{-5} / m_N^2 \text {. }
$$
Recall that $G$ has dimension (mass) ${ }^{-2}$. We have chosen to quote the value with respect to the nucleon mass.