---
title: "Find the collisions that burn deuterium"
description: "Derive a thermal reaction rate, locate the Gamow window and carry common calibration uncertainty through the integral."
canonical_url: "https://sandboxphysics.com/stories/nuclear-bridge/thermal-rates/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/nuclear-bridge/thermal-rates/"
last_modified: "2026-10-27"
---

[Nuclear physics](https://sandboxphysics.com/stories/nuclear-bridge/)

L3 · Chapter 7

# Find the collisions that burn deuterium

Sandbox Physics · 2026-10-27

A nuclear reaction in a hot gas is usually carried by a particular range of collision energies. Slow nuclei are plentiful but seldom penetrate the repulsive electric barrier. Fast nuclei penetrate more easily but are scarce. Their competition selects a window, and that window tells an experimenter which energies matter.

## Before you begin

Collision cross sections, kinetic energy and probability densities. A definite integral is used as a weighted sum; the required energy distribution is introduced here.

[Review the preceding prerequisite](https://sandboxphysics.com/stories/nuclear-bridge/inverse-scattering/)

- Turn a measured cross section into a Maxwellian rate with explicit units.
- Estimate the burning window before integrating.
- Propagate one calibration shared by every measured point.

## 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 $E$ be their centre-of-mass collision energy and $\mu=m_pm_d/(m_p+m_d)$ 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.

$$
\begin{aligned} f_E(E)&=\frac{2\sqrt{E}}{\sqrt{\pi}(k_BT)^{3/2}}e^{-E/(k_BT)},\\ v(E)&=\sqrt{2E/\mu},\\ R&=N_A\int_0^\infty \sigma(E)v(E)f_E(E)\,dE.\end{aligned}
$$

The energy density integrates to one. Temperature is absolute, and every energy in the integral must use the same unit. The two species are distinct, so the volumetric rate needs no identical-pair factor.

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 $0.6714\,\mathrm{u}$. With these masses, the barrier energy is about $658\,\mathrm{keV}$. Inserting the S-factor definition into the thermal rate cancels the explicit powers of collision energy and leaves two competing exponentials.

$$
\begin{aligned}E_G&=2\mu c^2(\pi\alpha Z_pZ_d)^2,\\ \sigma(E)&=\frac{S(E)}{E}e^{-\sqrt{E_G/E}},\\ R&=N_A\sqrt{\frac8{\pi\mu}}(k_BT)^{-3/2}\int_0^\infty S(E)e^{-E/(k_BT)-\sqrt{E_G/E}}\,dE.\end{aligned}
$$

The fine structure constant is alpha. The first exponential suppresses the rare fast pairs; the second suppresses slow pairs that cannot easily tunnel. Bare nuclei and a nonrelativistic Maxwell distribution are assumed.

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 $T=100\,\mathrm{MK}$, the thermal energy is $k_BT=8.62\,\mathrm{keV}$. The formula gives $E_0=23.0\,\mathrm{keV}$, 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.

$$
\begin{aligned}\frac{d}{dE}\left(\frac{E}{k_BT}+\sqrt{\frac{E_G}{E}}\right)&=\frac1{k_BT}-\frac{\sqrt{E_G}}{2E^{3/2}}=0,\\ E_0&=\left[\frac{E_G(k_BT)^2}{4}\right]^{1/3}.\end{aligned}
$$

This estimate locates the maximum of the exponential part. The full window has finite width, and the actual contribution also includes the energy dependence of the fitted S factor.

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 $\boldsymbol a$. The integral is linear in those coefficients. Its response vector $\boldsymbol g$ 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 $(\delta_NR)^2$ 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.

$$
R=\boldsymbol g^{\!T}\boldsymbol a,\qquad \operatorname{Var}(R)=\boldsymbol g^{\!T}C\boldsymbol g+(\delta_NR)^2+\left(\frac{\partial R}{\partial\epsilon_E}\delta_E\right)^2
$$

C is the coefficient covariance from independent point errors. The final term shifts all assigned energies coherently by epsilon\_E, rederives the S factors at fixed recorded cross sections, and refits. The common scale, point errors and energy calibration are assumed independent.

## Try the reasoning in an experiment

### [Move the window across a fixed set of measurements](https://sandboxphysics.com/labs/stellar-reaction-rates/)

1. Predict which way the important energies move when the temperature rises. Compare A young star with A hotter early universe.
2. Choose Stop measuring too soon. Add the higher-energy measurements and watch the fitted curvature, rate and measured-span fraction together.
3. Keep temperature fixed and double Shared scale uncertainty. Explain why the central curve stays in place even though the rate interval widens.

## Exercises and explanations

1. The temperature increases from $100\,\mathrm{MK}$ to $800\,\mathrm{MK}$. By what factor does the Gamow estimate move?

   Read the explanation

   Its temperature power is two thirds. The temperature ratio is eight, so $E_{0,2}/E_{0,1}=8^{2/3}=4$. The estimate moves from about 23 to 92 keV; the full contribution can peak elsewhere because the S factor varies.

2. A common efficiency calibration has a 3% fractional uncertainty. How much does that component shrink when thirteen measured energies become fifty-two?

   Read the explanation

   It stays at 3%. All points share the same scale, so every trial calibration moves the whole fitted cross section and thermal integral together. Independent point fluctuations can shrink with more information; a common calibration requires a better calibration measurement.

[Previous: Why different nuclear potentials can fit the same scattering](https://sandboxphysics.com/stories/nuclear-bridge/inverse-scattering/) [Next: Why shells and tails change near the neutron drip line](https://sandboxphysics.com/stories/nuclear-bridge/shell-evolution-and-halos/)

## Sources and reproduced scope

Evaluated level energies and published reaction measurements retain their source versions and uncertainties. Model curves, schematic figures and synthetic scattering records are identified separately. The offline benchmark files state which calculations have been reproduced.

1. [V. Mossa et al., The baryon density of the Universe from an improved rate of deuterium burning, Nature 587, 210–213 (2020)](https://doi.org/10.1038/s41586-020-2878-4)

   Accepted manuscript, Eqs. 2–3 and Extended Data Tables 1–2: measured proton-deuteron S factors and reference thermal rates. The local subset fit and the published global fit are different calculations.

2. [R. Longland et al., Charged-particle thermonuclear reaction rates: I. Monte Carlo method and statistical distributions (2010), arXiv:1004.4136v1](https://arxiv.org/pdf/1004.4136v1)

   Thermal-rate probability distributions and the need to keep common normalization errors correlated. The course derives its own first-order covariance propagation for the chosen fit.