half-lifeneutronsnuclear-physicsparticle-physicsweak-interaction
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?
Spontaneous processes such as neutron decay require that the final state is lower in energy than the initial state. In (stable) nuclei, this is not the case, because the energy you gain from the neutron decay is lower than the energy it costs you to have an additional proton in the core.
For neutron decay in the nuclei to be energetically favorable, the energy gained by the decay must be larger than the energy cost of adding that proton. This generally happens in neutron-rich isotopes:
An example is the $\beta^-$-decay of Cesium: $$\phantom{Cs}^{137}_{55} \mathrm{Cs} \rightarrow \vphantom{Ba}^{137}_{56}\mathrm{Ba} + e^- + \bar{\nu}_e$$
For a first impression of the energies involved, you can consult the semi-empirical Bethe-Weizsäcker formula which lets you plug in the number of protons and neutrons and tells you the binding energy of the nucleus. By comparing the energies of two nuclei related via the $\beta^-$-decay you can tell whether or not this process should be possible.
Conservation of energy and the electron-degenerate pressure.
For the neutron to decay you must have $$ n \to p + e^- + \bar{\nu}$$ or $$ n + \nu \to p + e^- \quad. $$
In either case that electron is going to stay around, but in addition to the neutrons being in a degenerate gas, the few remaining electrons are also degenerate, which means that adding a new one requires giving it momentum above the Fermi surface and the energy is not available.
Best Answer
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 $M_{i,f}$ 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