# Why is the (free) neutron lifetime so long?

A neutron outside the nucleus lives for about 15 minutes and decays mainly through weak decays (beta decay). Many other weakly decaying particles decay with lifetimes between $10^{-10}$ and $10^{-12}$ seconds, which is consistent with $\alpha_W \simeq 10^{-6}$.

Why does the neutron lives so much longer than the others?

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NB: I feel like this is a pretty half-assed job, and I apologize for that but having opened my mouth in the comments I guess I have to write something to back it up.

We start with Fermi's golden rule for all transitions. The probability of the transition is $$P_{i\to f} = \frac{2\pi}{\hbar} \left|M_{i,f}\right|^2 \rho$$ where $\rho$ is the density of final states which is proportional to $p^2$ for massive particles. To find the rate1 for all possible final states we sum over these probabilities incoherently. When the mass difference between the initial and final states is much less than the $W$ mass the matrix element depends only weakly (hah!) on the particular state and the sum is well approximated by a sum only over the density of states: $$P_\text{decay} \approx \frac{2\pi}{\hbar} \left|M_{}\right|^2 \int_\text{all outcomes} \rho .$$ This sum is collectively called the phase space available to the decay. In these cases the matrix element is also quite small for the reason that Dr BDO discusses.

The phase space computation can be quite complicated as it must be taken over all unconstrained momenta of the products. For decays to two body states it turns out to be easy, there is no freedom in the final states except the $4\pi$ angular distribution in the decay frame (their are eight degrees of freedom in two 4-vectors, but 2 masses and the conservation of four momentum account for all of them except the azimuthal and polar angles of one of the particles).

The decays that you have asked about are to three body states. That gives us twelve degrees of freedom less three constraints from masses, four from conservation of 4-momentum which leaves five. Three of these are the Euler angles describing the orientation of the decay (and a factor of $8\pi^2$ to $\rho$), so our sum is over two non-trival momenta. The integral looks something like $$\begin{array}\\ \rho \propto \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - E_1 - E_2-E_3 ) \\ &\delta(E_1^2 - m_1^2 - p_1^2) \\ &\delta(E_2^2 - m_2^2 - p_2^2) \\ &\delta(E_2^2 - m_2^2 - p_2^2) \\ &\delta(\vec{p}_1 + \vec{p}_2 + \vec{p}_3) \end{array}$$ which is easier to compute in Monte Carlo than by hand. (BTW--the reason for introducing the seemingly redundant integral over the angle $\theta$ between the momenta of particles 1 and 2 will become evident in a little while).

For beta decays the remnant nucleus is very heavy compared to the released energy, which simplifies the above in one limit.

In the case of muon decay, it is not unreasonable to treat all the products as ultra-relativistic, and the above reduces to $$\begin{array}\\ \rho \propto \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - E_1 - E_2 - E_3 ) \\ &\delta(E_1 - p_1) \\ &\delta(E_2 - p_2) \\ &\delta(E_3 - p_3) \\ &\delta(\vec{p}_1 + \vec{p}_2 + \vec{p}_3) \\ = \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - p_1 - p_2 - p_3 ) \\ &\delta(\vec{p}_1 + \vec{p}_2 + \vec{p}_3) \\ = \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta(m_0 - p_1 - p_2 - \left|\vec{p}_1 + \vec{p}_2\right| )\\ = \int p_1^2 \mathrm{d}p_1 \int p_2^2 \mathrm{d}p_2 \int \mathrm{d}(\cos\theta) &\delta\left(m_0 - p_1 - p_2 - \sqrt{p_1^2 + p_2^2 - p_1p_2\cos\theta} \right) \end{array}$$ The integral over the angle will evaluate to one in some regions and zero in others and as such is equivalent to correctly assigning the limits of the other two integrals, so writing $\delta m = m_0 - m_1 - m_2 - m_3$ we get $$\begin{array} \rho & \propto \int_0^{\delta m/2} p_1^2 \mathrm{d}p_1 \int_0^{\delta m-p_1} p_2^2 \mathrm{d}p_2 \\ & \propto \int_0^{\delta m/2} p_1^2 \mathrm{d}p_1 \left[ \frac{p_2^3}{3}\right]_{p_2=0}^{\delta m-p_1} \\ & \propto \int_0^{\delta m/2} p_1^2 \mathrm{d}p_1 \frac{(\delta m - p_1)^3}{3} \end{array}$$ which I am not going to bother finishing but shows that that phase space can vary as a high power of the mass difference (up to the sixth power in this case).

1 The lifetime of the state is inversely proportional to the probability

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dmckee, I'm much amused but highly appreciative: You made me realize how truly off-base my flip answer about the weakness of the weak force was for neutrons... and you put together an answer worth digging into here! So, thanks. –  Terry Bollinger Jul 10 '12 at 2:18
@Terry Well, you can find example of strong decays and weak decays with similar mass differences and final states and the difference in the time of those is all down to the mass of the weak bosons. It's just not the whole story. –  dmckee Jul 10 '12 at 2:32
Actually, that's a pretty good answer. I was anticipating something a lot more cryptic, but it was both readable and good math -- well, I assume it's good, I sure didn't cross-check it! Thanks. (Er... did we sort of forget the fellow who asked this originally?... :) –  Terry Bollinger Jul 10 '12 at 2:57
@Terry - maybe you did (forget me) but I didn´t forget the question ;) Great answer by the way! My thanks to everyone involved –  Forever_a_Newcomer Jul 19 '12 at 18:12
Dr BDO already nailed it, so just to give another angle on it: The neutron decays very slowly because it has one option available for decaying into the slightly lower mass proton. That option is weak decay, which requires that one of the neutrons inner parts, a down quark (charge -1/3) emit a very massive force particle called a $W^-$. The $W^-$ then promptly decays into a ordinary electron and an anti-neutrino. Quantum time-energy uncertainty allows $W^-$ particles to form, but only very rarely due to their super-high masses. –  Terry Bollinger Jul 7 '12 at 14:48
This is actually very incomplete, because other weak decays can proceed very much faster. Muon decay is a weak process and has a halflife of $10^{-6}$ seconds. The charged pion decay is a weak process with a halflife of $10^{-8}$ seconds and so on. Then there are weak decay processes which are much slower as in many long-lived beta active isotopes. The phase space available to the products plays a big part in the full answer. –  dmckee Jul 7 '12 at 16:14
I not completely satisfied with the answer. It seems to me that dr-bdo-adams and @terry-bollinger are explaining this based on the off-shellness of the W. But still the diference is too big, the neutron lifetime is $10^9$ times bigger than that of the Muon, but the "liberated energy" is only $10^2$ smaller. Is this going like $\left(\frac{E_L}{M_W}\right)^4$? ($E_L$ is the liberated energy). Can someone give me pointers of why we have such a high power in the dependence? –  Forever_a_Newcomer Jul 8 '12 at 0:54