Where radiative capture actually happens

R-matrix theory begins by drawing a sphere of radius $a$ around a pair of fragments and describing the two sides differently: inside, a few poles stand in for the whole many-body problem; outside, only Coulomb and centrifugal forces act and the wave functions are known analytically.

For elastic scattering that division is invisible in the answer. For radiative capture it is not, because a photon can be emitted from either side of the surface. The electromagnetic operator has support everywhere, so the capture amplitude splits into an internal piece — the resonant one, carried by the poles — and an external piece, in which the photon is emitted while the two fragments are still approaching, in the region where they are not yet a compound nucleus at all. The two add coherently, and at low energy the external piece is very often the larger.

Getting this right requires knowing precisely which pathways the reaction has, so it is worth building the decomposition first.

How a reaction decomposes#

Everything is organised around the compound nucleus, and around the fact that total angular momentum $J$ and parity are conserved.

A pair is a two-body partition of the system: $\alpha + {}^{12}\mathrm{C}$, $p + {}^{15}\mathrm{N}$, or — for capture — a photon plus the residual nucleus in one of its bound states, $\gamma + {}^{16}\mathrm{O}(\text{g.s.})$. The pair fixes the masses, charges, spins and the separation energy, and therefore the penetrabilities and the kinematics.

A decay is a specific final pair the compound nucleus can break up into. One entrance pair generally has several decays: elastic back to itself, inelastic to an excited state, capture to each bound state of the residual nucleus.

Channel spins. Within a decay, the entrance pair's intrinsic spins couple to a channel spin $s$, and the exit pair's to $s'$. Grouping pathways by $(s, s')$ is what makes the statistical spin weights come out right, and it is why the model parameters — the reduced widths — are labelled by channel rather than by individual spin projections. A capture transition to a $2^+$ final state, for instance, has more than one allowed $s'$, and the pathways for each must be collected separately before they are weighted and summed.

Pathways. Within a fixed $(s, s')$, one concrete route is a triple $(J, \text{entrance channel}, \text{exit channel})$: a definite total angular momentum, a definite entrance orbital angular momentum, and a definite exit one. This is the level at which the collision matrix is actually computed — each such route contributes one element, and the observable is a coherent sum over routes followed by an incoherent sum over the things that were not measured.

The hierarchy matters here because the external capture amplitude has to be enumerated over the same set of routes as the internal one. If a route is counted twice, the two amplitudes are added twice, coherently, and the capture cross section is inflated by an amount that depends on how much of the total that route carried — which is exactly the kind of error that a good $\chi^2$ will absorb into the fitted widths rather than reveal.

The two flavours of external capture#

Once the pathway list exists, external capture comes in two physically distinct forms.

Direct, or hard-sphere, capture#

The particle arrives in the entrance channel, does not form a compound state, and radiates directly into a final bound state. The amplitude is a radial integral of the electromagnetic multipole operator between the entrance-channel scattering wave function and the bound-state wave function, taken over the external region only, $r > a$, where both are known Coulomb and Whittaker functions.

Two features follow from that being an integral over the outside.

It is non-resonant: a smooth function of energy, with no poles, contributing a slowly varying background to the S-factor. And it is controlled by the asymptotic normalisation coefficient of the final state, not by an interior property: what the integral samples is the tail of the bound-state wave function, so a loosely bound final state — a large ANC — gives a large direct-capture amplitude. This is why the low-energy capture cross section on light nuclei can be dominated by a quantity that no compound-nucleus picture contains.

Channel capture#

The second form is subtler, and it is the one that couples the two regions.

The particle arrives, scatters resonantly through an internal level — that is, it forms the compound nucleus and comes back out into a particle channel — and then radiates from the external region into the final bound state. The amplitude is the same external radial integral as before, but multiplied by the collision-matrix element of that intermediate particle channel.

So channel capture carries the resonant energy dependence of the intermediate level while the photon is still emitted outside. It is neither purely internal nor purely external, and it interferes with both. Building it requires walking the intermediate particle decay's pathway list, which is where the decomposition of the previous section stops being bookkeeping and starts being physics: each intermediate route contributes its own collision-matrix element, and the set of routes must be exactly the physical one.

Which final states have which#

Not every transition has a channel-capture contribution. It exists only if the intermediate particle decay has pathways whose quantum numbers connect the entrance channel to the final state through an allowed multipole. In practice this is a strong filter: in a given evaluation some final states acquire channel capture and others do not, and the two groups can behave quite differently as the model is varied.

The example: carbon-12(α,γ)oxygen-16#

This reaction sets the carbon-to-oxygen ratio at the end of helium burning, and with it the entire subsequent evolution of a massive star. It is also the standard hard case for external capture, for a reason that is easy to state: the astrophysical energy is around 300 keV, the nearest levels of $^{16}\mathrm{O}$ that carry the strength are sub-threshold, and the ground-state transition is dominated by E1 and E2 amplitudes whose external parts are large.

The transitions to the low-lying states of $^{16}\mathrm{O}$ divide cleanly:

final state	$J^\pi$	channel-capture pathways?
ground state	$0^+$	yes
6.05 MeV	$0^+$	no
6.13 MeV	$3^-$	yes
6.92 MeV	$2^+$	yes
7.12 MeV	$1^-$	no

The ones with channel capture are sensitive to the intermediate resonant scattering; the ones without are pure direct capture plus the internal resonant amplitude. A model change that touches the intermediate pathway enumeration moves the first group and leaves the second untouched — and switching external capture off entirely makes all five agree again, because then no external amplitude is built at all.

The size of the effect is not marginal. For the ground-state transition, doubling the channel-capture contribution moves the cross section at $E_{\mathrm{c.m.}} \approx 4.9$ MeV from $2.61 \times 10^{-7}$ to $6.56 \times 10^{-7}$ barn — a factor 2.5 — and shifts the total-capture excitation function by up to 108 %. The dominant single term in that coherent sum has a collision-matrix element of order $10^{-4}$ against an external normalisation of about 14; it is the product that matters, and it is comparable to everything else combined.

The practical consequence is worth stating plainly. Because the internal and external amplitudes interfere, an error in one is absorbed by the fit into the reduced widths of the other. A fit performed with a mis-enumerated external amplitude will still describe the data — the widths will simply have moved to compensate — and the damage appears only on extrapolation, where the two amplitudes have different energy dependences and no longer cancel in the same way. This is the general hazard of coherent decompositions, and the reason a capture evaluation should be checked by turning external capture off and on and confirming that the transitions which should be insensitive to it really are.

Why this structure exists at all#

It is tempting to see external capture as a correction bolted onto a compound-nucleus calculation. It is better understood as the price of the channel radius.

The sphere at $r = a$ was introduced so the interior could be replaced by poles. A photon emitted at $r < a$ is described by those poles; a photon emitted at $r > a$ is not, and would simply be missing. Adding the external amplitude restores it. The division between "resonant" and "direct" capture is therefore not a physical distinction between two mechanisms so much as a consequence of where the bookkeeping surface was drawn — which is the same lesson as in When a reduced width stops being measurable: move the surface and the split between the two amplitudes moves with it, while their sum, if the model is complete, does not.