1Nucleosynthesis of the Heavy ElementsThree basic processes can be identified by which heavy nuclei can bebuilt by the continuous addition of protons or neutrons:• p-process (proton)• s-process (slow neutron)• r-process (rapid neutron)Capture of protons on light nuclei tend to produce only proton-rich nu-clei. Capture of neutrons on light nuclei produce neutron-rich nuclei, butwhich nuclei are produced depends upon the rate at which neutrons areadded. Slow capture produces nuclei near the valley of beta stability, whilerapid capture (i.e., rapid compared to typical beta-decay timescales) ini-tially produces very neutron-rich radioactive nuclei that eventually beta-decay towards the valley of beta stability. Some nuclei can be built bymore than one process, as illustrated in Fig. 1.Figure 1: A small portion of the chart of the nuclides, illus-trating isotopes built by the three basic processes. Only stableisotopes are shown, and letters indicate the processes that cancontribute to these isotopes.2Neutron CaptureThe difference between the s-process and r-process nucleosynthesis iscontrolled by beta decay rates of nuclei. The s-process path lies in the valleyof beta stability, but the r-process path is shifted by many units towardneutron-richness. Fig. 2 shows the beta-decay timescales of radioactiveneutron-rich nuclei, the positions of stable nuclei, neutron and proton magicnumbers and the r-process path.Figure 2: Chart of the nuclides illustrating beta decay timescales(color), stable isotopes (black squares), magic numbers (hori-zontal and vertical black lines), the limit of known nuclei (di-agonal jagged black line), and the r-process path (purple line).Overlaid are the relative abundances of r-process nuclei, to il-lustrate the relation between magic numbers and abundances.For small neutron velocities v, i.e., energies up to a tens of keV, neutroncapture cross sections vary as σ ∝ v−1. Thus the product σv is roughly3constant for thermal energies. Characteristic cross sections are about 100-1000 millibarns (10−25− 10−24cm2). For temperatures around 30 keV,thermal velocities of neutrons are about 3 × 108cm s−3. Then, the lifetimeof a nucleus against neutron capture isτn∼1nnσv∼3 × 1016− 3 × 1017nns.s-process timescales are around 104yr, while r-process timescales are aroundmicroseconds, indicating neutron densities of 105− 106cm−3and 1023−1024cm−3, respectively. Neutron densities intermediate to these might beexpected to yield a nucleosynthesis pattern intermediate to the s- and r-processes. However, comparison to solar-system abundances indicates thatthis has not occurred.Neutron capture cross sections are generally smoothly varying with Abecause a captured neutron has an excitation energy of about 8 MeV (µn'µp≈ −8 MeV) where the density of levels if very high. However, for lightnuclei or magic number nuclei (N or Z ∈ [2, 8, 20, 28, 50, 82, 126]) this is notthe case and the neutron capture cross section are very small, comparatively,being from 1 to 10 mb.s-processA quantitative analysis can be performed with the basic assumption that,except for magic number nuclei, beta decay timescales are all very rapidcompared to the neutron cature rate. The nuclear chain thus indicatedis unique, because one proceeds to heavier nuclei as indicated in Fig. 2,stepping horizontally until an unstable nucleus is reached, then movingon a diagonal path increasing the proton number by 1 and decreasing theneutron number by 1. One can let NArepresent the abundance of nucleusA with out reference to Z, and we can ignore abundances of radioactivenuclei. The differential equation for abundances is then˙NA= − < σv >Ann(t) NA(t) + < σv >A−1nn(t) NA−1(t) ,where < σv >Ais thermally averaged for each nucleus A. Under the as-sumption that σ ∝ v−1, one can write < σv >A≈ vTσAwhere vTis thethermal velocity which is a constant. Defining a neutron exposure τ bydτ = vTnn(t) dt,4we findN0A= −σANA+ σA−1NA−1.This equation can only be solved with a suitable boundary condition. Be-cause neutron capture cross sections on light nuclei are so small, we caneffectively assume that NA(0) = 0 for A < 56 and NA(0) = N56(0) forA > 56. On the other hand, for A = 209 the chain cerminates becauseBi209is the most massive stable nucleus. Nuclei with mass 210 decay by αdecay to mass 206. We will ignore this complication.The above equation has the property that N0A< 0 if NA> (σA−1/σA)NA−1and vice-versa. Thus it is self-regulating: NAdecreases if it is too large withrespect to NA−1and increases if it is too small. If the process operates longenough, it will come to equilibrium: NA0 = 0, in which caseσANA≈ σA−1NA−1.In-between the magic numbers, the cross sections are large compared totheir differences, so locally this is a good approximation to the solution.Over a large range of A, however, this result is a poor approximation,because of the small cross sections associated with magic nuclei. A propergeneral soution is a little tricky, but can be analyzed by using the LaplacetransformˆNk(s) =Z∞0Nk(t) e−stdtusing k = A − 55. Using the fact that the Laplace transform of a derivativeisZ∞0N0k(t) e−stdt = sˆNk(s) − Nk(0) ,the transformed equations and boundary condition are:sˆN1(s) = − σ1ˆN1(s) + N1(0) ,sˆNk(s) = − σkˆNk(s) + σk−1ˆNk−1(s) .The solution of these equations, by substitution, isˆNk(s) =N1(0)σkYkσks + σk.Here the notation Πkindicates successive multiplications of arguments fromsubscript k to 1. Since the product σN is expected to be smoothly varying,5we choose to use that product as the dependent variable. We defineˆψk(s) ≡σkˆNk(s)N1(0)=Yk(1 + s/σk)−1.The solution is then the inverse Laplace transform:ψk(τ ) ≡ σNk(τ ) =12πiZi∞−i∞ˆψk(s) esτds=kXi=1e−σiτσiYk6=iσkσk− σi.Figure 3: The distribution function ψk(τ) using σ = 100.Left: Shown as a function of k for τ = 0.5012. Right: Shownas a function of τ for k = 50.This solution encounters severe numerical difficulties at large atomicweights. The sum must be taken over about 150 terms, which differ greatlyin order of magnitude but still must be included because of their near can-cellations. It is therefore necessary to find an approximate solution. Toillustrate the behavior of the exact solution, consider the idealized case inwhich all the cross sections are equal, σk= σ. The solution in that case,starting fromˆψk(s) =