advanced34 min studyLesson 590 in Physics

Fusion barriers and astrophysical S-factors

Separate kinematic Q-values, Coulomb penetration, intrinsic nuclear strength, and thermal reaction averaging.

Before you begin

On this page

The question for today

What you’ll understand

Interpret the astrophysical S-factor with units and assumptions.
Derive the leading Gamow-peak energy from competing exponentials.
Distinguish a monoenergetic cross section from a thermally averaged reaction coefficient.

Charged nuclei can react below their classical Coulomb barrier through tunneling. In a stellar plasma, the most important collision energies balance rapidly increasing penetration against the declining high-energy tail of the thermal distribution. The astrophysical S-factor separates a conventional barrier exponential from the remaining cross section. It is useful precisely when its residual energy dependence is mild, not because every fusion reaction becomes simple after that factor is removed.

A barrier suppresses an allowed channel

A positive fusion Q-value says the final rest energies can be lower than the initial ones. It does not guarantee a substantial rate at low incident energy. Two positively charged nuclei must approach through Coulomb repulsion, and angular momentum can raise the effective barrier. Quantum tunneling gives a finite sub-barrier amplitude. The classical barrier height is therefore not an absolute threshold; conservation-law thresholds and penetration suppression describe different restrictions on the reaction.

At low nonrelativistic relative energy, the long-range Coulomb penetration factor contains an exponential in inverse square root energy. Factoring this and an inverse-energy wavelength term out of the cross section defines S. Its dimensions are energy times area. Resonances, thresholds, subthreshold poles, changing partial waves, and nuclear-structure effects can still make S vary strongly. A smooth extrapolation from measured energies needs physical justification and uncertainties rather than being guaranteed by the definition.

\sigma(E)=\frac{S(E)}{E}e^{-2\pi\eta},\qquad\eta=\frac{Z_1Z_2e^2}{4\pi\epsilon_0\hbar v},\qquad 2\pi\eta=\frac{b}{\sqrt E},\quad E=\frac{\mu v^2}{2}

E is centre-of-mass relative energy, σ is an area, S has energy–area units, η is dimensionless, and b has units square root energy. The nonrelativistic Coulomb convention assumes positive charge numbers and relative speed v; strong screening and resonant energy dependence require additional modeling.↔ Scroll wide equations horizontally.

Thermal averaging weights collision rates, not just populations

For dilute nondegenerate nuclei in thermal equilibrium, relative velocities follow a Maxwell distribution. Reactions are weighted by relative speed as well as cross section, yielding an integral over energy with an extra energy factor. Substituting the S-factor representation cancels that explicit factor against the inverse energy in the cross section. If S changes slowly, the main exponential competition is then a Boltzmann factor decreasing with energy and a penetration factor increasing with energy.

The maximum of this product comes from minimizing the sum of energy over thermal energy and the Coulomb constant over square root energy. Differentiation gives the leading Gamow-peak energy, generally above the thermal-energy scale for a significant charged-particle barrier. This is a saddle-point estimate, not a statement that every reacting pair has exactly that energy. Broad windows, nearby resonances, screening, and nonthermal particle distributions can change which energies dominate the measured or astrophysical rate.

\langle\sigma v\rangle=\sqrt{\frac{8}{\pi\mu}}(k_BT)^{-3/2}\int_0^\infty\sigma(E)E e^{-E/(k_BT)}dE,\qquad E_0=\left(\frac{bk_BT}{2}\right)^{2/3}

With SI energies and μ in kg, the rate coefficient has units m³/s. E0 is the leading Gamow-peak energy under slowly varying S(E) and Maxwellian, nonrelativistic, unscreened assumptions. b and kBT must use consistent energy units.↔ Scroll wide equations horizontally.

A rate coefficient becomes a volume rate only after counting pairs

For two distinguishable species, multiply the thermal coefficient by both number densities to obtain a reaction rate per volume. For identical reactants, divide by two so the same pair is not counted twice. This pair-counting factor is separate from quantum symmetry effects already incorporated in the cross section. Energy production then multiplies the reaction rate by the appropriate deposited energy, which may differ from a total Q-value when neutrinos or escaping radiation carry energy away.

Laboratory beams usually measure cross sections at specified energies and target conditions, not directly a stellar thermal coefficient. Electron screening, thermally excited target states, and experimental energy loss must be treated when translating those data. An isolated narrow resonance can dominate the integral and requires its resonance strength and thermal weighting rather than an arbitrary constant-S fit. The useful lesson is to preserve the chain from microscopic amplitude to cross section to averaged coefficient to pair-counted rate, with each approximation named.

Work it through

Locate a toy Gamow maximum

Assume a smooth constant S-factor, $k_BT=2\,\mathrm{keV}$ , and a toy Coulomb constant $b=8\sqrt{\mathrm{keV}}$ . Find the Gamow-peak energy and the ratio of the exponential integrand at that energy to its value at 1 keV. Ignore nonexponential prefactors after the S-factor cancellation.

The leading peak is $E_0=(bk_BT/2)^{2/3}=(8\,\mathrm{keV}^{3/2})^{2/3}=4\,\mathrm{keV}$ . Units are consistent because the quantity inside the power has energy to the three-halves.
At 4 keV the suppressing exponent is $E/(k_BT)+b/\sqrt E=4/2+8/2=6$ . At 1 keV it is $1/2+8=8.5$ .
Taking the ratio of exponentials gives $e^{-6}/e^{-8.5}=e^{2.5}\approx12.18$ . This compares rate-integrand weights under the declared model, not cross sections alone or a fraction of all thermal particles.

The result

The toy peak lies at 4 keV, with exponential weight about 12.18 times the 1 keV weight. The peak differs from both the mean thermal energy and a classical barrier threshold.

The living notebook

Explore the model

Interactive lab

Interactive lab · root-finding

Resolve a square root used in penetration estimates

For the dimensionless energy ratio E/(1 keV) = 2, Reset and set both controls to c = 2 and U = 3. Predict the positive root lies between 1 and 2. After one Step the bracket is [0, 1.5], so its midpoint readout is 0.75 and error bound 0.75; after two Steps the midpoint is 1.125 and bound 0.375. This exact bisection instrument can estimate a square root used in a penetration factor; it has no fusion cross section, thermal distribution, or Gamow-peak optimizer.

Quadratic / midpointCurrent bracketPositive root

Midpoint after 0 refinements: 1.5
Bracket error bound: 1.5
Computed residual · x² − c: 0.25

Target c2Initial upper bound U3

t = 0 s

Bisection of x² − c on [0,U], capped at 50 iterations for floating-point resolution. The error bound is the maximum distance from the displayed midpoint to either bracket endpoint (half the width in exact arithmetic). Compensated square evaluation preserves the bracket when ordinary multiplication rounds a residual to zero. A truly exact endpoint or midpoint collapses the bracket. This is one monotone quadratic, not a general root solver. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNumerical substeps need their own error meaning even inside an approximate physical model.

Pause & reason

Check your understanding

Interactive checks need JavaScript. You can still read the explanation.

Read the checkpoint explanation

Low energies are suppressed by Coulomb penetration, while high energies are suppressed by the Maxwellian tail. Their competing exponentials select a window; smooth S is an approximation, and resonances or screening can change the result.

Your turn

Try it yourself

A toy measured cross section at 4 keV is $2\times10^{-6}\,\mathrm b$ and the Coulomb exponent there is $2\pi\eta=\ln100$ . Find S at that energy. Separately, for an identical-species plasma with density $10^6\,\mathrm{m}^{-3}$ and stipulated $\langle\sigma v\rangle=10^{-12}\,\mathrm{m}^3\mathrm{s}^{-1}$ , find the reaction rate per volume.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Multiply the cross section by energy and the inverse penetration factor. Count identical pairs only once.

Compare with the solution

The definition gives $S=\sigma E e^{2\pi\eta}=(2\times10^{-6})(4)(100)=8\times10^{-4}\,\mathrm{keV\,b}$ . This is a value at one energy, not proof of energy independence. The identical-pair volume rate is $n^2\langle\sigma v\rangle/2=(10^{12})(10^{-12})/2=0.5\,\mathrm{m}^{-3}\mathrm{s}^{-1}$ . Omitting the pair-counting factor would double count the same reacting pairs.

Keep the idea

Take it with you

Q-values and Coulomb penetration constrain different parts of a fusion prediction.
S has units energy times area and need not be slowly varying.
Thermal averaging and identical-pair counting are additional steps beyond a cross section.

Read further

Original sources and deeper explanations to follow at your own pace.

G. Colò, Nuclear Physics, §§6.2 and 7.1: S-factor, Gamow peak, thermal averaging, and fusion barriers, pp. 113–116 and 121–126

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