Why does the Sun keep shining for billions of years?

A candle consumes wax, needs a supply of oxygen and eventually goes out. The Sun also releases energy into its surroundings every second, yet it has been shining for billions of years. Where does such a reserve come from, and why is it not used up all at once?

Nuclear fusion in the hot core powers the Sun: hydrogen turns into helium through a sequence of reactions that release energy. A huge supply of accessible hydrogen, a large energy yield per kilogram and a slow, self-regulated reaction rate allow this stage to last billions of years. The supply is nevertheless finite. NASA: the source of solar energy.

A candle changes bonds; the Sun changes nuclei

In chemical combustion, a substance reacts with oxygen: atoms form new compounds by changing bonds involving electrons. The atomic nuclei remain nuclei of the same elements. Describing the Sun as “burning” can mislead, because its main energy source does not need this reaction with oxygen. NASA/RHESSI: why the Sun is not a chemical flame.

An atom has a tiny nucleus at its center. The nucleus of ordinary hydrogen is a single positively charged proton. The nucleus of the most common kind of helium contains two protons and two neutrons, particles without electric charge. Making helium from hydrogen requires a rearrangement of the nuclei themselves. That is a nuclear transformation.

Matter in the solar core is a plasma: nuclei and electrons do not form familiar neutral atoms. The temperature near the center is about 15 million °C. Gravity holds the enormous mass together and creates the conditions for a hot, dense core. This is where most energy production occurs, rather than in the visible outer layer. NASA: the core and photosphere.

A chain of rare transformations leads to helium

Protons repel one another because they have the same electric charge. High temperature makes their motion more energetic, while high density provides many close encounters. Solar fusion cannot, however, be explained simply by classical acceleration “over the barrier.” Quantum tunneling is essential: particles have a probability of passing through an energy barrier even when they lack the energy to cross it classically. This is a property of the microscopic world, rather than a hole between particles. DOE: tunneling and proton-proton fusion.

The main energy-producing route in the Sun is called the proton-proton, or pp, chain. First, two protons form a deuterium nucleus, a kind of hydrogen with a proton and a neutron. One proton must change into a neutron through the weak interaction. A positron, the electron’s antiparticle, and a neutrino, a particle that interacts very weakly with matter, also emerge. DOE: the beginning of solar fusion.

Further transformations build helium nuclei. Read one branch of the chain from top to bottom in each card. The first two steps are needed twice: only then are there two helium-3 nuclei for the final step.

One route from hydrogen to helium: pp-I

Needed twice
Two protons form deuterium
¹H + ¹H↓²H + e⁺ + νₑ
One proton changes into a neutron; a positron and a neutrino emerge.
Needed twice
Deuterium captures another proton
²H + ¹H↓³He + γ
A gamma photon, a quantum of high-energy radiation, emerges.
Needed once
Two helium-3 nuclei form helium-4
³He + ³He↓⁴He + ¹H + ¹H
Two protons return to the plasma and can take part in further reactions.

¹H — proton; ²H — deuterium; ³He and ⁴He — two kinds of helium nuclei. e⁺ — positron; νₑ — electron neutrino; γ — gamma photon.

Balance for this branch: 6 protons enter and 2 return. The net result converts 4 protons into one helium-4 nucleus, with other particles and energy released.

A sequence of transformations, rather than a simultaneous collision of four protons. Other pp-chain branches, subsequent positron annihilation, and time and distance scales are not shown.

The phrase “four hydrogen nuclei become helium” therefore describes the net outcome of several reactions, rather than four protons colliding at one point. The pp-I route shown here does not cover every branch of solar fusion. Ohio State: the proton-proton chain.

A rest-mass difference becomes other forms of energy

After the complete conversion of hydrogen into helium, the remaining particles have a slightly smaller total rest mass than the initial particles. The difference corresponds to energy carried by particle motion and radiation. The helium nucleus is more tightly bound: forming this bound state allows energy to be released.

For the overall hydrogen → helium balance, including electrons and subsequent positron-electron annihilation, the yield corresponds to about 0.7% of the initial hydrogen mass. Most of the rest remains in helium. The Sun does not turn all the fuel’s mass into light, and energy does not appear from nothing. OpenStax: mass and energy in solar reactions.

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What converting one kilogram of hydrogen yields

The relation between the rest-mass difference $\Delta m$ and released energy $E$ is:

E=\Delta m\,c^2.

Here $c$ is the speed of light in vacuum, rounded to $3\times10^8$ m/s. For $m=1$ kg of hydrogen completely converted into helium, take a fraction $\eta\approx0.007$ . Then:

\begin{aligned} \Delta m&\approx0.007\ \text{kg},\\ E&\approx6.3\times10^{14}\ \text{J}. \end{aligned}

Even a small fraction of a kilogram gives a large yield because of the factor $c^2$ . This is the total reaction energy, including the share carried away by neutrinos. Somewhat less remains in the plasma. The approximate 0.7% belongs to this transformation, rather than being a universal fraction for any nuclear reaction.

Most energy is transferred to the surrounding matter. It then moves outward through radiation and, in the outer part of the interior, through plasma motion, or convection, as well. Absorption, scattering and fresh emission prevent us from picturing a single unchanged photon flying straight from the core to the surface.

Visible sunlight emerges mainly from the photosphere, at a temperature of about 5500 °C. Hot matter in this layer radiates; nuclear reactions in the much hotter core sustain the energy flow. NASA: energy transport and the visible surface.

Reactions are slow, and the star changes its own conditions

The Sun contains an enormous number of protons, but a successful initial pp transformation is extremely unlikely. Electric repulsion and the involvement of the weak interaction make this step slow. The hydrogen supply is therefore not all processed in a single flash. Yet there are so many protons that the combined rare reactions provide enormous power. DOE: why the initial step is slow.

There is a second reason for stability: negative feedback. Imagine a small increase in energy production in the present solar core. Heating raises pressure, and the core expands against gravity. Expansion cools it, reducing the fusion rate. The change triggers a response that weakens it.

If energy production becomes insufficient to cover losses, the core contracts under gravity. Contraction heats the matter and increases fusion. This is often called a stellar “thermostat,” although there is no sensor or switch inside. Gravity, pressure, energy transport and the temperature dependence of reactions do the work. Ohio State: hydrostatic self-regulation.

This explanation applies to the Sun’s long stable stage. It does not mean that every star in every state always returns to equilibrium, or that the Sun’s luminosity remains unchanged forever.

Billions of years come from the reserve divided by its use rate

The Sun has plenty of hydrogen, but the accessible hydrogen in the hot core matters for its current stage. Hydrogen in cooler outer layers cannot simply be added to the fuel currently sustaining central fusion. The core’s composition changes over time, with more helium and less hydrogen.

Combining a large accessible supply with the high nuclear energy yield gives a duration of billions of years. The Sun formed about 4.6 billion years ago. When the central hydrogen supply becomes insufficient, the star will change its structure and enter other evolutionary stages; it will not go out in an instant like a candle. NASA: the Sun’s formation and evolution.

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A timescale estimate, rather than a countdown

Set a simplified budget for the entire stage of central hydrogen fusion. Let the effective mass of hydrogen available for conversion, $m_f$ , be about one tenth of the solar mass. This is a chosen rough guide, rather than a measured current remainder or a claim that the entire core is pure hydrogen:

m_f\approx2\times10^{29}\ \text{kg}.

Take $\eta=0.007$ , $c\approx3\times10^8$ m/s and a constant luminosity $L\approx4\times10^{26}$ W, the energy the Sun radiates each second. Then:

\begin{aligned} E_f&\approx\eta m_f c^2,\\ E_f&\approx1.26\times10^{44}\ \text{J},\\ t&\approx\frac{E_f}{L},\\ t&\approx3.15\times10^{17}\ \text{s}. \end{aligned}

Using a rounded $3.16\times10^7$ seconds per year gives about 10 billion years. The units agree: joules divided by joules per second give seconds. A larger reserve at the same luminosity would give more time; greater luminosity for the same reserve would give less.

We have neglected a separate correction for neutrino energy and changes in luminosity, composition and fuel accessibility. The result therefore gives an order of magnitude, rather than the Sun’s precise age or the time remaining from today. Dividing the Sun’s entire rest energy $M_\odot c^2$ by its luminosity would be a mistake: neither is all its mass available as fuel, nor is all the fuel mass released as energy. University of Oklahoma: a nuclear-power lifetime exercise, p. 6.

Neutrinos bring a testable trace from the core

The calculation shows that fusion can provide the required energy. A calculation alone does not establish that this is what happens in the Sun, however. The model also predicts a flow of neutrinos escaping from the interior with almost no interaction with matter.

In 2014, the Borexino collaboration reported detecting neutrinos from the initial proton-proton process. An underground detector recorded rare interactions; researchers analyzed the signal’s energy spectrum and background contributions. This is an observable consequence of core reactions that supports the nuclear explanation of solar energy. The detector did not photograph an individual fusion event in the Sun. Borexino: the original result and open data.

Return to the candle. Would cutting off oxygen stop solar fusion? No: the pp chain needs hydrogen nuclei and the right conditions, while oxygen is not one of its reactants. If energy were used faster for the same available nuclear reserve, the stage would be shorter. These are two different questions: what powers the source, and how fast it uses its reserve.

The Sun shines for billions of years thanks to its nuclear reserve, rare initial reactions and the hot star’s self-regulation. It is finite, but its timescale greatly exceeds a human one. To follow this energy further, return to the difference between the Sun’s own light and the Moon’s reflected light. The next course topic examines atoms and how nuclear structure determines an element.