Give every collision its own weight
Consider a dilute gas of protons and deuterons in thermal equilibrium at the same temperature. A deuteron contains one proton and one neutron. A proton can be captured by it to form helium-3 and emit a photon. We follow the relative motion of the incoming pair, so the mass in the kinetic energy is their reduced mass.
Let be their centre-of-mass collision energy and their reduced mass. The Maxwell energy density gives the fraction of pairs in a small energy interval. Multiply it by relative speed and cross section: a faster pair sweeps out a larger collision volume each second.
Adding these contributions gives the rate coefficient. Multiplying it by the two number densities would give reactions per volume per time. The molar coefficient displayed here includes Avogadro’s constant, a convenient convention for reaction networks; it still needs the composition of the gas before it becomes a burning rate.
Source: R. Longland et al.
Separate the barrier from the nuclear response
An experimental cross section rises steeply as the incident energy increases. Much of that slope comes from the chance of crossing the Coulomb barrier. Pulling out its strongest energy dependence leaves the astrophysical S factor, a smoother quantity that is easier to compare between nearby energies.
The proton and deuteron each have charge number one. Their reduced mass is about . With these masses, the barrier energy is about . Inserting the S-factor definition into the thermal rate cancels the explicit powers of collision energy and leaves two competing exponentials.
Source: V. Mossa et al. · R. Longland et al.
Estimate the window before calculating its area
For a first estimate, hold the S factor constant across the important energies. Maximizing the exponential weight is equivalent to minimizing its positive exponent. Differentiate once with respect to energy and solve for the stationary point. The result is called the Gamow peak.
At , the thermal energy is . The formula gives , substantially above the thermal energy. This is a worked example of why evaluating a cross section at the mean energy misses the relevant weighting.
The S factor for this reaction grows across the window. Including it moves the full contribution toward higher energies. In the default fit, the median contributing energy is about 31 keV, and only about 47% of the calculated rate lies between the lowest and highest selected LUNA energies. The contribution outside those endpoints is an extrapolation of the fitted quadratic.
Source: R. Longland et al.
Carry the common calibration through the sum
Imagine measuring at thirteen energies with the same efficiency calibration. A counting fluctuation at one energy can average against a fluctuation at another. A calibration that shifts all thirteen measurements upward moves the entire fitted curve. Its effect on the integral survives even if many more energies are measured.
Write the fitted S factor as a sum of three polynomial basis functions with coefficient vector . The integral is linear in those coefficients. Its response vector is obtained by integrating each basis function with the same thermal weight. The coefficient covariance then propagates by a quadratic form. A fractional common scale uncertainty contributes to the rate variance.
For the default LUNA-only quadratic at 100 MK, the calculated rate is 5.38 in cubic centimetres per mole per second. Independent point uncertainties contribute about 3.13%, the assumed common scale 2.40%, and the chosen coherent energy uncertainty about 1.54%. Combining these independent components gives 4.23%. Doubling the common scale component changes the uncertainty and leaves the central rate fixed.
That interval conditions on the fitted quadratic. It does not describe how far another plausible extrapolation could bend below the first measurement. The separate coverage fraction tells you when that missing model uncertainty matters most.