research32 min studyLesson 592 in Physics

Proton–proton chains and solar neutrinos

Track proton recycling, branch-dependent neutrinos, and the distinction between nuclear release and deposited heat.

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What you’ll understand

  • Balance the net conversion of hydrogen into helium including weak-process leptons.
  • Count neutrinos for different completed pp-chain branches.
  • Convert a specified helium-production rate into heating and source flux under explicit assumptions.

Hydrogen fusion requires more than overcoming a Coulomb barrier. Some protons must become neutrons, so weak interactions enter the chain and produce electron neutrinos. The proton–proton chains assemble helium through light intermediates, with several possible endings. Their common net nuclear conversion does not make their neutrino spectra identical. Careful accounting distinguishes released rest-mass energy, energy carried away by escaping neutrinos, and heat available to support a stellar luminosity.

A slow weak gateway feeds fast captures

The ordinary pp gateway produces a deuteron, a positron, and an electron neutrino from two protons. Its weak conversion makes it slow compared with many subsequent electromagnetic captures. Deuterium can capture another proton to form helium three. A competing pep gateway consumes an electron with two protons and supplies a different neutrino line. The relative rates depend on the plasma conditions; assigning every hydrogen-burning neutrino to the ordinary pp reaction would miss this distinction.

In the pp-I termination, two helium-three nuclei fuse into one helium-four nucleus and return two protons. Preparing the two intermediates initially used six protons, but two are recycled, so only four are consumed net. Including positron annihilation with surrounding electrons, the net bookkeeping consumes two electrons and emits two neutrinos per newly made helium-four nucleus. Charge, baryon number, and lepton number all balance. The associated atomic-mass energy release must not then have annihilation energy added a second time.

4p+2e−⟶4He+2νe+Q,N˙ν,source=2N˙α,new4p+2e^-\longrightarrow{}^4\mathrm{He}+2\nu_e+Q,\qquad \dot N_{\nu,\rm source}=2\dot N_{\alpha,\rm new}
This is net bookkeeping after positron annihilation for completed hydrogen-to-helium cycles. Q includes the total rest-mass release, neutrinos carry part of it away, and the count excludes temporary storage in intermediates.↔ Scroll wide equations horizontally.

Branches alter neutrino labels, not the net conversion

The pp-II route uses helium three and an existing helium-four nucleus to make beryllium seven. Electron capture then makes lithium seven and a neutrino, and a proton reaction returns two helium-four nuclei. One of those replaces the original helium-four participant; only one is newly produced net. In the pp-III route, proton capture on beryllium seven produces boron eight, whose beta decay supplies a higher-energy neutrino spectrum before the light products complete the route.

A completed pp-I cycle normally has two gateway neutrinos. A pp-II or pp-III cycle normally has one gateway neutrino and one from its later weak step, with pep substitutions or small additional branches requiring separate accounting. A high-energy neutrino detector therefore does not directly count all helium production. Intermediate abundances and competing destruction rates determine branch fractions. The short lifetime of one intermediate does not identify the slowest rate controlling the complete conversion throughput.

Lheat=N˙α,new(Q−∑a∈cycle⟨Eν,a⟩)L_{\rm heat}=\dot N_{\alpha,\rm new}\left(Q-\sum_{a\in\rm cycle}\langle E_{\nu,a}\rangle\right)
For a stationary completed-cycle population with trapped electromagnetic products and freely escaping neutrinos, the heat per cycle is the total release minus mean neutrino energies. Energies must share units, and storage or other losses are omitted.↔ Scroll wide equations horizontally.

Source luminosity and detected flavor are separate

At a distance from an isotropic source, a neutrino production rate divided by sphere area gives a number flux, provided losses or direction-dependent emission are neglected. Neutrino flavor evolution changes the electron-flavor fraction arriving at a detector, so an electron-neutrino event rate also requires survival probabilities, cross sections, efficiencies, and the source spectrum. A source neutrino count is therefore neither an observed count rate nor a direct measurement of a particular fusion cross section without a model connecting those quantities.

The usual luminosity comparison assumes the burning network is close to a stationary state. A changing reservoir of helium three or another intermediate can delay the relation between gateway reactions and newly completed helium nuclei. Photon transport also delays a star's surface response, whereas escaping neutrinos probe ongoing core reactions more directly. These timing distinctions matter when interpreting a hypothetical transient or comparing different epochs. The simple arithmetic below deliberately specifies a stationary ideal source rather than asserting a complete solar model.

Φν,all=N˙ν,source4πd2,Φνe(E)=Pee(E)Φνe,source(E)\Phi_{\nu,\rm all}=\frac{\dot N_{\nu,\rm source}}{4\pi d^2},\qquad \Phi_{\nu_e}(E)=P_{ee}(E)\Phi_{\nu_e,\rm source}(E)
d is distance in metres and number flux is in inverse square metres per second. The second expression is a simple source-to-electron-flavor relation with no initial non-electron component; a detector rate needs a further cross-section convolution.↔ Scroll wide equations horizontally.

An explicitly specified ideal hydrogen-burning source

An illustrative stationary source makes 3×10363\times10^{36} new helium nuclei per second. Use a stipulated total release Q=26 MeVQ=26\,\mathrm{MeV} and mean escaping neutrino energy sum 2 MeV2\,\mathrm{MeV} per cycle. With 1 MeV=1.6×10−13 J1\,\mathrm{MeV}=1.6\times10^{-13}\,\mathrm J and observing sphere area 1.2×1023 m21.2\times10^{23}\,\mathrm m^2, find the heat power and all-flavor neutrino flux.

  1. The completed-cycle count gives N˙ν=2(3×1036)=6×1036 s−1\dot N_\nu=2(3\times10^{36})=6\times10^{36}\,\mathrm{s}^{-1}. This counts source neutrinos before any flavor-dependent detector selection.

  2. The heat per cycle is 26−2=24 MeV=3.84×10−12 J26-2=24\,\mathrm{MeV}=3.84\times10^{-12}\,\mathrm J. Multiplying by the helium-production rate gives Lheat=1.152×1025 WL_{\rm heat}=1.152\times10^{25}\,\mathrm W.

  3. Divide the neutrino rate by the stipulated sphere area to obtain Φν=5×1013 m−2s−1\Phi_\nu=5\times10^{13}\,\mathrm m^{-2}\mathrm{s}^{-1}. No measured solar constants or branch fractions have been assumed; every numerical input describes this illustrative source.

The result

The model deposits 1.152 times ten to the twenty-fifth watts and supplies an all-flavor flux of five times ten to the thirteenth per square metre per second.

Explore the model

Interactive lab

Interactive lab · probability

Count an illustrative termination branch

For a purely illustrative independent branch label, Reset and set success probability to 0.25 and trials to 20. Predict mean five labeled branches and variance 3.75. Playback highlights exact binomial outcomes rather than measuring neutrinos. This instrument neither solves pp-chain abundances nor predicts solar branch fractions, flavor conversion, or detector efficiency.

Count an illustrative termination branchThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forA branch count is conditional on its assumed probability and is distinct from a reaction-network prediction.

Check your understanding

Why does a completed pp-I cycle consume four protons net even though two helium-three precursors initially require six?

Try it yourself

In an illustrative sample of 100 completed cycles, 75 follow pp-I and 25 follow pp-II. Neglect pep and all other branches. Count ordinary pp neutrinos and beryllium-electron-capture neutrinos, checking the total against helium production.

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

Give me a hint

A pp-I cycle supplies two ordinary gateway neutrinos; a pp-II cycle supplies one ordinary gateway neutrino and one later electron-capture neutrino.

Compare with the solution

The ordinary pp count is 2(75)+25=1752(75)+25=175, and the beryllium-capture count is 25. The sum is 200 neutrinos for 100 new helium nuclei, as required by the net weak conversion. The two neutrino classes have different energies, so equal total neutrino counts alone would not determine the deposited heat or a detector's event count.

Take it with you

  • Weak reactions provide the neutron conversions and neutrinos in hydrogen burning.
  • Proton recycling and pre-existing helium must be subtracted when computing net production.
  • Neutrino energy losses, flavor conversion, and detector response are separate parts of the prediction.

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