# 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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## 3 Answers

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

As you correctly state, the neutron decay is an decay due to the weak interaction, these are quite a bit slower than other decays due the mass of the intermediate W boson, 81GeV, which slows the reaction, additionally the neutron decay only liberates a small amount of energy, around 1 MeV, it is the ratio of the liberated energy to the mass of the W which sets the speed of the reaction which is thus much slower than other decays as all the other particle decay release much more energy.

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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
Well, @dmckee is correct: That was a very incomplete explanation I attempted, and only some approach that addresses the product phase spaces can explain the huge range of weak force decay times. dmckee, ball's in your court if you want to give it a shot... – Terry Bollinger Jul 8 '12 at 2:46
"dmckee, ball's in your court if you want to give it a shot..." @TerryBollinger I actually started to right after the question was posted only to realize that I'm unable to offer a full picture right now. I need to take some of my copious spare time and refresh myself on this matter. – dmckee Jul 8 '12 at 2:49

You can estimate the neutron lifetime using dimensional analysis. Beta decay is correctly described by the well known four-fermion Fermi theory, so the amplitude must be proportional to the coupling $G_F\approx10^{-5}\text{GeV}^{-2}$ (the Fermi constant). The decay rate is proportional to the squared amplitude:

$$\Gamma\propto G_F^2\thinspace.$$

$\Gamma$ has units of Mass while $G_F^2$ has units of $[\text{Mass}]^{-4}$, so to get the units wright we must have

$$\Gamma\propto G_F^2 \Delta^5$$

where $\Delta$ is some quantity that has units of mass. The relevant mass scale in neutron decay is the mass difference between neutron and proton, so $\Delta=m_n-m_p\approx10^{-3}\text{GeV}$.

To be a bitsy more accurate one can try guess the $\pi$ dependence of the decay rate. This comes from the phase space of a 3-body decay, which usually goes as

$$(2\pi)^4\times\Big[\thinspace(2\pi)^{-3}\times(2\pi)^{-3}\times(2\pi)^{-3}\Big]\times (\pi^2)\ \propto\ \pi^{-3}\thinspace.$$

The first factor comes from the four momenta conservation delta function, the three $(2\pi)^{-3}$ in the bracket come from the integration measure of the 4-momenta of each outgoing particle and the last factor comes from integrating the angle variables. One finally gets the following estimation

$$\Gamma\propto \frac{1}{\pi^3}G_F^2(m_n-m_p)^5$$

If one plugs in all the numbers, the estimated lifetime of the neutron reads

$$\tau_{\text{neutron}}\ =\ \Gamma^{-1}\ \approx\ \pi^3 \ \text{sec.}$$

This is a bit shorter than the real value which is at least one order of magnitude larger. But it explains why the neutron's lifetime is so large (inverse of the 5th power of the small mass difference) with respect to other weak decay processes.

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## protected by Qmechanic♦Apr 28 '15 at 14:20

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