What is the *fundamental* reason, at particle level, of the radioactivity?

Question

When giving fundamental reason for radioactivity, books or references most often state that this is due to the fact that nucleus is not in a stable enough state thus it goes back to the stability.

This is not a fundamental enough reason to me. Fundamental would need to involve the level of elementarly particle physics.

What is the fundamental reason, at particle level, of the radioactivity ?

Is it just due to the fact that Standard Model elementary particles couple with other particles, from Standard Model Feynman rules, and thus the radioactivity is only these reactions that occur, with the rates that just follow the cross-sections of the relevant particle level Feynman diagrams ?

If so, why books never explain it in this way, which looks much more "natural" than the explanations with "stability".

Answer 1

Radioactivity is an exothermic reaction: the decay products have kinetic energy. An unstable state is such that the "left-hand side" has more energy than the right-hand side, so there is enough kinetic energy for the products to fly apart.

You may think of "stability" as the low elevation in an energy graph, and "instability" a relative high elevation. If you are at the top of the hill, you are guaranteed to roll/slide downhill to a low point, which you reach with extra kinetic energy. But the converse does not happen, unless you get a kick by a bump that imparts the requisite kinetic energy to you.

How fast your decay will go will depend on 1) the absolute square of the decay matrix element, a fundamental, quantum quantity, as WP details; but also, 2) the phase space: the kinematic distribution of momenta and energies (relativistically invariant, of course) which you might (very loosely) analogize to the steepness off the hill.

The fundamental physics (computed, e.g., in QFT) is in the matrix element, and contributes to how fast you roll down (decay rate). But the same matrix element is totally powerless in the face of unfavorable phase-space: stability.

So, the weak interaction involved in β-decay, say, of a free neutron, involves virtually the same weak matrix element as the frustrated decay of a bound neutron in a nucleus; it does not decay, because its energy level is lowered below threshold by (strong) nuclear binding: it has been stabilized, analogous to the top of the hill sinking to a small sinkhole.

(Actually, a line-reversed matrix element is involved in inverse β-decay a scattering reaction involving the same particles.)

So, stability, is the first line of defense against decay, namely, unavailability of excess energy. However, if you do have the excess energy to decay to some products, as in the sought-after baryon decay, you will never decay if the fundamental theory does not support a non-vanishing matrix element. If that element is small, as in GUT nucleon decay, the decay will be seriously slow. If it vanishes, the decay won't go.

Answer 2

You're asking for the fundamental reason in terms of particles. That's Wong's equation. Under Wong's equation, under the influence of a non-Abelian gauge field that is turned on, the corresponding gauge charge will precess.

For the Weak Force, where the charge is (weak) isospin, the weak charges have the additional attribute that different charge states are different particles. For leptons, the left-handed electron is the isospin-down version of the left-handed (electron's) neutrino, which is isospin-up. The right-handed positron is the isospin-up version of the right-handed (electron's) anti-neutrino. A similar relation exists between the left-handed (down, up) quarks - separately for each color red, green and blue; and for the right-handed (anti-down, anti-up) anti-quarks separately for each color anti-red, anti-green and anti-blue (or cyan, magenta and amber, as I prefer to call them).

So, when the weak force is turned on, the weak charge can precess ... which means it also switches identity. A neutron, which has (up, down, down) quarks of separate colors (red, green, blue) can become a proton, which has (up, up, down) quarks of separate colors (red, green, blue), when one of the down's precesses to an isospin-up state and becomes an up quark.

This spits out a quantum of the weak force - a W⁻ particle - which then decays into an electron and anti-neutrino. That's the genesis of a radiation event.

Curiously, there aren't that many references on Wong's equation; not even a Wikipedia entry yet, as far as I can see. This one Canonical formulations of a classical particle in a Yang-Mills field and Wong's equations seems to be covering all of the bases, but it's in Letters in Mathematical Physics, which is not open access. Part of the sparsity of references is because you have to make full use of the geometry of principal bundles to provide any Lagrangian formulation for Wong's equation. There's no easy formulation of Wong's equation based on the action principle.

The simplest way to arrive at Wong's equation is to first treat the gauge field as just a Maxwell field that happens to have one set of potentials $\left(φ^a,𝐀^a\right)$ and field strengths $\left(𝐁^a, 𝐄^a\right)$ for each basis element $Y_a$ in the underlying Lie algebra. If the Lie algebra is non-Abelian, then the Lie brackets of the basis elements are non-zero and of the form $\left[Y_a, Y_b\right] = f^c_{ab} Y_c$ (with the summation convention used here and below), where the structure coefficients $f^c_{ab}$ are (generally) non-zero. The corresponding relation between the field strengths and potentials is, then, non-linear: $$𝐁^c = ∇×𝐀^c + \frac12 f^c_{ab} 𝐀^a×𝐀^b,\quad 𝐄^c = -∇φ^c - \frac{∂𝐀^c}{∂t} + f^c_{ab} φ^a 𝐀^b.$$

The gauge charge, then, has components which may be expressed as $e_a$. Wong's equation then arises by requiring that the two forms of the corresponding force law for the momentum $𝐩$ and power law for the energy $E$ of a gauge charge reconcile: $$ \frac{d\left(𝐩 + e_a 𝐀^a\right)}{dt} = -∇U,\quad \frac{d\left(E + e_a φ^a\right)}{dt} = \frac{∂U}{∂t}, $$ versus $$ \frac{d𝐩}{dt} = e_c \left(𝐄^c + 𝐯×𝐁^c\right),\quad \frac{dE}{dt} = e_c 𝐯·𝐄^c, $$ where $𝐯$ is the velocity of the charge, $𝐫,t$ indicates the dependence on the coordinates for position and time and $$U(𝐫,t,𝐯) = e_a (φ^a(𝐫,t) - 𝐯·𝐀^a(𝐫,t)).$$

The parts of the expressions on the right-hand side of the second set of equations that arise from the non-linear parts of the potential-to-field strength equations: $$ ⋯ = e_c \left(f^c_{ab} φ^a 𝐀^b + 𝐯×\left(\frac12 f^c_{ab} 𝐀^a×𝐀^b\right)\right),\quad ⋯ = e_c 𝐯·\left(f^c_{ab} φ^a 𝐀^b\right) $$ should be negatively offset by the parts of the first set of equations, on the left-hand sides, that arise from making the charge $e_a$ time-dependent, which yields the following subtractions: $$-\frac{de_a}{dt} 𝐀^a = ⋯,\quad -\frac{de_a}{dt} φ^a = ⋯.$$

Working out the vector algebra and using the anti-symmetry of the Lie bracket $\left[Y_a, Y_b\right] = -\left[Y_b, Y_a\right]$ and structure coefficients $f^c_{ab} = -f^c_{ba}$, we have: $$ f^c_{ab} 𝐀^a 𝐯·𝐀^b = -f^c_{ba} 𝐀^a 𝐯·𝐀^b = -f^c_{ba} 𝐯·𝐀^b 𝐀^a = -f^c_{ab} 𝐯·𝐀^a 𝐀^b $$ thus, $$ 𝐯×\frac12 f^c_{ab} 𝐀^a×𝐀^b = \frac12 f^c_{ab} \left(-𝐯·𝐀^a 𝐀^b + 𝐀^a 𝐯·𝐀^b\right) = f^c_{ab} 𝐀^a 𝐯·𝐀^b $$ and $$ f^c_{ab} φ^a 𝐀^b = -f^c_{ba} φ^a 𝐀^b = -f^c_{ba} 𝐀^b φ^a = -f^c_{ab} 𝐀^a φ^b $$ thus, in turn, $$ e_c \left(f^c_{ab} φ^a 𝐀^b + 𝐯×\left(\frac12 f^c_{ab} 𝐀^a×𝐀^b\right)\right) = -f^c_{ab} e_c 𝐀^a \left(φ^b - 𝐯·𝐀^b\right). $$

In addition, noting that $$ f^c_{ab} φ^a φ^b = -f^c_{ba} φ^a φ^b = -f^c_{ba} φ^b φ^a = -f^c_{ab} φ^a φ^b, $$ it following that $f^c_{ab} φ^a φ^b = 0$. Therefore, $$ e_c 𝐯·\left(f^c_{ab} φ^a 𝐀^b\right) = f^c_{ab} e_c φ^a \left(𝐯·𝐀^b - φ^b\right) = -f^c_{ab} e_c φ^a \left(φ^b - 𝐯·𝐀^b\right). $$

Thus, $$ -\frac{de_a}{dt} 𝐀^a = -f^c_{ab} e_c \left(φ^b - 𝐯·𝐀^b\right) 𝐀^a,\quad -\frac{de_a}{dt} φ^a = -f^c_{ab} e_c \left(φ^b - 𝐯·𝐀^b\right) φ^a. $$ In order for this to hold for arbitrary potentials $\left(φ^a, 𝐀^a\right)$, this requires that $$\frac{de_a}{dt} = f^c_{ab} e_c \left(φ^b - 𝐯·𝐀^b\right).$$ That's Wong's equation.

For the weak force, there are three Lie components and $f^c_{ab}$ is completely anti-symmetric with $f^1_{23} = 1$. So, one can arrange the Lie components of the charge and velocity-dependent potential into vectors: $$𝐞 = \left(e_1, e_2, e_3\right),\quad \overrightarrow{(φ - 𝐯·𝐀)} = \left(φ^1 - 𝐯·𝐀^1, φ^2 - 𝐯·𝐀^2, φ^3 - 𝐯·𝐀^3\right),$$ and write this as $$\frac{d𝐞}{dt} = -𝐞×\overrightarrow{(φ - 𝐯·𝐀)}.$$ So it's an equation that describes the precession of weak isospin.

Answer 3

The reason of radioactive decay is the second law of thermodynamics. Matter particles are connected to interaction particles, which latter have vacuum energy. The vacuum fluctuation forces matter to release energy and decay. The vacuum state is an infinitely large bath. According to the second law of thermodynamics, energy is distributed evenly among degrees of freedom. Interaction particles (like photon) tend to take up the whole energy of matter. This is the reason why matter tends to decay into the ground state, the lowest state of energy.

Answer 4

The object is in a superposition of "nucleus" and "reduced nucleus plus decay products" states. There is a potential barrier between the "nucleus" states and the "reduced nucleus plus decay products" states, so if the object starts measured in a " nucleus" state, the probability of finding it on the far side of the potential barrier is low. Iff the total configuration energy of the system in a decayed state $\le$ the energy in a non decayed state, the probability is nonzero, so if you observe for long enough you will eventually measure the system in a decayed state.