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The temperature and pressure everywhere inside the Sun reach the critical point to start nuclear reactions - there is no reason for it to take such a long time to complete the reaction process.

Just like a nuclear bomb will complete all the reaction within $10^{-6}$seconds.

Why does most of the hydrogen of the Sun still not react even though it reaches the critical point, and why take stars billions of years to run out of fuel?

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There are four factors involved: (1) velocity distribution of the nuclei; (2) small geometrical cross-section for head-on collisions of nuclei; (3) quantum-mechanical tunneling probability; (4) for the p-p reaction, a weak-force effect is required. Dmckee's answer discusses 1 and 2. John Rennie's answer discusses 4. Maybe these could be edited into a single answer. –  Ben Crowell Aug 9 at 18:26
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In a way, the Sun is faster than a nuclear bomb. It emits the $E=mc^2$ equivalent of four million tons of matter per second, whereas a nuclear bomb converts only a tiny fraction of its mass (when we talk about a four megaton bomb, this refers to the chemical energy contained in 4 million tons of TNT). –  Hagen von Eitzen Aug 10 at 10:21
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@HagenvonEitzen I don't understand your point. 4 million tons is a minuscule fraction of the mass of the sun (about $2\times 10^{27}$ tonnes). –  David Richerby Aug 10 at 13:28
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@DavidRicherby I think the point is that the answer to the question "why doesn't the sun explode as fast as a nuclear bomb" is "it kinda does, but it's so huge that even at that rate it takes a long time to run out". –  anaximander Aug 11 at 11:36
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@anaximander Perhaps. But Hagen's use of "however" suggests that the sun is different from a nuclear bomb because the bomb "converts only a tiny fraction of its mass" whereas the sun implicitly doesn't. Except that, over the course of a second (a million times longer than the question states the reaction in the bomb runs for), the sun also only converts a tiny fraction of its mass. –  David Richerby Aug 11 at 16:54

8 Answers 8

The bottleneck in Solar fusion is getting two hydrogen nuclei, i.e. two protons, to fuse together.

Protons collide all the time in the Sun's core, but there is no bound state of two protons because there aren't any neutrons to hold them together. Protons can only fuse if one of them undergoes beta plus decay to become a neutron at the moment of the collision. The neutron and the remaining proton fuse to form a deuterium nucleus, and this can react with another proton to form $^{3}\text{He}$. The beta plus decay is mediated by the weak force so it's relatively slow process anyway, and the probability of the beta plus decay happening at just the right time is extremely low, which is why proton fusion is relatively slow in the Sun. It takes gazillions of proton-proton collisions to form a single deuterium nucleus.

Nuclear fusion weapons bombs fuse fast because they use a mixture of deuterium and tritium. They don't attempt to fuse $^{1}\text{H}$ so they don't have the bottleneck that the Sun has to deal with.

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This is an incomplete answer for the reasons explained in my comment at the top. –  Ben Crowell Aug 9 at 18:29
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@BenCrowell: True, but I echo dmckee's point: neither of us have the time to sit down and write an essay on Solar fusion. I've pinpointed what I consider be the key factor i.e. (1) there are plenty of protons with enough velocity (2) there are plenty of head on collisions (3) the probabilities are high enough but (4) the weak process is too slow. That's why I chose to discuss (4). If you want to copy and paste all our answers into a comprehensive answer please do so and I'll gladly upvote. Right now it's a Saturday night and I'm about to go drinking! :-) –  John Rennie Aug 9 at 18:46
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+1 for the usage of gazillions –  user1596244 Aug 11 at 14:12
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The answer inspired me to check if there is such a thing as helium-2. One way of explaining this, I guess, is to say the helium-2 is created all the time, but almost always decays by emitting one of the protons again. Only in extremely rare cases will helium-2 decay by the weak beta plus decay into deuterium (hydrogen-2). –  Jeppe Stig Nielsen Aug 12 at 10:15
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I disagree. The premise of the question seems wrong to me. The cross section of of D-T/T-T fusion is "only" 1 or 2 orders of magnitude greater than that of H-H fusion (via the inverse beta-decay chain, ie ignoring CNO). The only reason that nuclear weapons "use up" their fuel so quickly is because the inertial confinement is only capable of maintaining confinement for a fraction of a second. –  Aron Aug 12 at 11:11

This is an answer that I made, as suggested by John Rennie, by cutting and pasting his answer and dmckee's and adding a little more material.


There are four factors involved:

  1. Velocity distribution of the nuclei
  2. Small geometrical cross-section for head-on collisions of nuclei
  3. Quantum-mechanical tunneling probability
  4. For the p-p reaction, a weak-force effect is required

Velocity distribution of the nuclei

The interior of a star is a hot ionized gas at high pressure and temperature.

High temperature means a high average kinetic energy per particle, so all the nuclei of the atoms are whizzing around very fast (though for relatively short distance between collisions because the gas is so dense).

The thing is that they are not all whizzing around at the same speed, by random chance some are going fast and some are going slow. It's like the normal curve for grades of IQ or whatnot. The vast bulk of the atoms have very average speeds and just a very few are going either much faster or much slower than average.

What it means for a star to be "hot enough" is that if two of the very, very fast nuclei ram into each other head on, then there can be a fusion event.

Small geometrical cross-section

Not only are those very fast particles rare, but they have to hit head-on. This doesn't happen often.

Tunneling

Even the fastest particles do not have enough energy to overcome the electrical repulsion. Therefore fusion can occur only through quantum-mechanical tunneling, which is a low-probability process.

Weak interaction required

A further bottleneck is getting two hydrogen nuclei, i.e. two protons, to fuse together.

Protons collide all the time in the Sun's core, but there is no bound state of two protons because there aren't any neutrons to hold them together. Protons can only fuse if one of them undergoes beta plus decay to become a neutron at the moment of the collision. The neutron and the remaining proton fuse to form a deuterium nucleus, and this can react with another proton to form $^3$He. The beta plus decay is mediated by the weak force so it's relatively slow process anyway, and the probability of the beta plus decay happening at just the right time is extremely low, which is why proton fusion is relatively slow in the Sun. It takes gazillions of proton-proton collisions to form a single deuterium nucleus.

Nuclear fusion weapons bombs fuse fast because they use a mixture of deuterium and tritium. They don't attempt to fuse $^1$H so they don't have the bottleneck that the Sun has to deal with.

Stable equilibrium

The above factors explain why, given the prevailing conditions of temperature and pressure at the sun's core, we get such a slow reaction rate. MariusMatutiae's answer explains how this particular set of conditions comes about. The sun is in a stable equilibrium and acts as a thermostat.

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+1 Nice effort from all of you for providing these answers, and thanks Ben for completing and taking the time to sum everything up in one coherent answer. –  Phonon Aug 9 at 23:01
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Since this is an amalgamation of multiple answers, any point in making this a community wiki? –  Tyzoid Aug 12 at 17:09
    
@Tyzoid: How does one go about doing that? –  Ben Crowell Aug 13 at 21:49
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Also, the conditions for fusion are only present in the core of the sun, which is only a very small part of the sun. –  Jerry Schirmer Aug 13 at 21:52

The conditions at the core of the Sun are very different from those in a thermonuclear bomb. The first thermonuclear bomb used deuterium as the secondary. The Sun has to create deuterium before getting to this stage. It's the creation of deuterium that's the bottleneck in the fusion that occurs inside the Sun. Later bombs used lithium deuteride, which is even easier to fuse than is deuterium.

Fusion inside our Sun is incredibly slow process. Our Sun is not hot and bright because it produces a huge amount of energy per unit volume. A warm compost pile produces more energy per unit volume than does the core of the Sun. Our Sun is hot and bright because of the large volume over which that smallish amount of energy per unit volume is produced.

Fusion is slow in our Sun because it takes a good deal of energy to make two protons fuse to form deuterium. Once two protons have successfully fused to form deuterium, the rest of the proton-proton chain (p-p chain) that eventually produces helium-4 proceeds apace.

So why is proton-proton fusion in our Sun so slow? Two protons have to come within about 10-15 meters in order for the short range strong nuclear force to take control and make those two protons fuse to form deuterium. The amount of energy needed to overcome the Coulomb repulsion between two protons is immense. The root mean square velocity of protons at 15.6 million Kelvins is about 600 kilometers/second. That's not near enough energy to overcome that Coulomb repulsion. Only those protons from the very, very top end of the Maxwell–Boltzmann distribution at 15.6 million Kelvin have enough energy to overcome that Coulomb repulsion and bring two colliding protons close enough together to enable the strong force can take control.

In our Sun, the probability of fusion per collision is only $2\times10^{-31}$. The pressure and temperature are greater in slightly larger stars, increasing the probability that two colliding protons will fuse. A different method for producing helium from hydrogen occurs in even larger stars, the CNO cycle. This process is even more temperature sensitive than is the p-p chain. A tiny amount of the helium produced in our Sun results from the CNO cycle. Most of it results from the p-p chain.

In large stars, those with a mass greater than 1.3 solar masses, the CNO cycle dominates over the p-p chain because proton-proton fusion remains rather difficult to achieve even in those large stars. The bottleneck in the CNO cycle becomes much less of a bottleneck in large stars. The CNO cycle is by far the dominant method of helium production in very large stars. Those very large stars don't live very long compared to our Sun, but they still live a whole lot longer than does a thermonuclear device.

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So, if an alien civilization wanted to destroy earth, all they have to do would be to figure out a way to send a lots of deuterium into the sun to let the sun burn out in a matter of minutes and kill us all ? –  Wildling Aug 14 at 10:18
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@Wildling That sounds like a LOT of deuterium. They'd probably have an easier time bombarding us with H-bombs directly. –  Brilliand Aug 18 at 4:31

The premise that the sun has the same conditions all throughout is incorrect. For the most part the conditions (Temperature and Pressure) necessary for nuclear fusion to occur are only found within a small region in the core.

For example, when hydrogen fusion occurs and creates helium, since that helium is heavier it tend to coalesce as the core. In regular stars this helium won't reach fusion conditions for billions of years (until the star becomes a red giant) since the conditions for helium fusion are far more intense than hydrogen fusion. Around this ball of helium, you will have a fusion front, a region where hydrogen fusion will occur, and the helium product will (for the most part) deposit on the forming core. For larger stars that can have multiple stages of fusion occurring there can actually be multiple layers of fusion occurring. fusion occurring at multiple layers within the star

The stages shown here are only the case for the most massive stars, while for a red giant there will only be a helium shell to form generally around a carbon core.

A good thing to remember is that fusion is occurring in response to gravity's attempts to compress the mass of the star into a black hole. The countermeasure to this compression is fusion, and the star only pushes back (approximately) however hard it needs to prevent compression further. Huge amounts of (already radiated) energy would be required to expand the diameter of a star a few kilometers, and if a star were to expand that much fusion activity would decrease in the star until it reaches an equilibrium again that provide enough fusion to support the star. The barrier required to get the entire star fusing is generally not present during the life cycle of a star. However, there is one case where what you are expecting more or less occurs: a Supernova.

A little look at this table from Exploring the Universe (Cengage) shows us how long each fuel can sustain a star, but you'll note iron isn't listed there. enter image description here

Nuclear binding energy (net energetics of a nuclear reaction) caps out at Iron, and something interesting occurs, the iron-nickel core fails to support the star at all. In the last second of a stars' (pre-supernova) life, this iron-core expands to the size of the earth (with the mass of our sun).

enter image description here

Since iron fusion fails to provide any support to higher levels of the star they all start compressing uncontrollably. Fusion starts to occur throughout the entire star in the s-process and r-process as runaway fusion begins to occur throughout the star. The huge amounts of energy throughout the entire star begin creating every natural element we have encountered, and as the star rapidly releases energy the upper layers collide and rebound from the core, ripping the star apart in a massive release of energy. It's only in these conditions, where a star has finally lost it's fight against gravity, that a huge wave of fusion tears the star apart.

Summary:

Material must be very dense to fuse, and as they fuse they produce energy. This extra energy expands the star and the decrease in density slows the reaction. As long as the star fuses elements lighter than iron, nuclear binding energy shows us that that energy is added to the star. Expansion from the heat produced by fusion forms an equillibrium with the gravitational force compressing the star. Once iron fusion occurs, that fusion will quickly no longer supply heat, and since heat generation will no longer counterbalance the density increase due to gravity, the star will rapidly start compressing, and create the only (temporary) instance where there is star wide fusion occurring (s and r processes) as the star goes supernova.

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This is true but doesn't answer the question. This would be better as a comment. –  Ben Crowell Aug 9 at 18:21
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I like this answer more than the other more technical ones because it focuses on the aspect of equilibrum. For example, the "velocity distribution of nuclei" is such that only the top speed nuclei can react because (simplified) a larger fraction of reactive nuclei would lead to more reaction, would lead expansion against the gravitational pressure, would lead to lower density, would lead to less reacton –  Hagen von Eitzen Aug 10 at 10:25
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@wallyk - The image has absolutely nothing to do with why fusion in our Sun is so slow. While a good image is worth a thousand words, the converse is true for a bad image. In this case, it's worth a minus one vote from me. I don't understand why this answer has so many upvotes. This answer is a bad answer. Note well: I'm not downvoting to promote my answer. To wit, I just gave Ben's answer a plus one vote. –  David Hammen Aug 10 at 20:29
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You guys seem to be missing the fact that this question also says other stars in the main question. I explained at the large scale the equilibrium present in ALL stars, and used that to say why instantaneous fusion never occurs within the entire star during the normal lifecycle. Then I proceeded to build up to a supernova since it explains the only time you will ever actually see star-wide fusion. –  Skyler Aug 11 at 5:16
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@Skyler: The question is (effectively) Why do stars last so long? Your answer does nothing to actually address this issue. Instead, it talks about (quite literally) the very last few hours of a massive stars life. –  Kyle Kanos Aug 11 at 14:44

The interior of a star is a hot ionized gas at high pressure and temperature.

High temperature means high average kinetic energy per particle, so all the nuclei of the atoms are whizzing around very fast (though for relatively short distance between collisions because the gas is so dense).

The thing is that they are not all whizzing around at the same speed, by random chance some are going fast and some are going slow. It's like the normal curve for grades or IQ or whatnot. The vast bulk of the atoms have very average speeds and just a very few are going either much faster or much slower than average.

What it means for a star to be "hot enough" is that if two of the very, very fast nuclei ram each other head on there can be a fusion event.

Because those very faster particles are rare and because they have to hit head-on this doesn't happen often, which means that the time scales over which the star can burn a significant amount of fuel are very long.

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This is an incomplete answer for the reasons explained in my comment at the top. –  Ben Crowell Aug 9 at 18:28
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Ben, I agree that a much more complete answer could be written, but I do not intend to do so today. Feel free if you have the time and inclination. –  dmckee Aug 9 at 18:33

Fusion, as it occurs within stars, is in fact very unlike what happens in a bomb.

An "H-bomb" is actually a mixture of fission and fusion. The fission part works on a chain reaction: when a fissile nucleus absorbs a neutron, it vibrates madly and then splits into several components, in particular two or three neutrons. These extra neutrons go on breaking other nuclei. When the "critical mass" is reached, an average of more than one of these neutrons triggers further fission, leading to an exponentially increasing reaction.

When you want to do fusion, you have to convince the positively charged nuclei to come close enough to each other for strong interaction to overcome electrostatic repulsion. In controlled fusion, as is sought after in ongoing experiments like ITER, heat is used: the high kinetic energy induced by the severe heat is enough to push the nuclei together. Magnetic confinement is used to prevent the hot plasma from expanding. This is also what happens within a star: gravitation maintains the pressure. All of this makes for slow fusion.

In an H-bomb, though there is indeed a lot of heat, this mechanism does not contribute in non-negligible amounts to the fusion. The whole explosion implies a fireball which expands way too fast; there is nothing to keep the nuclei close enough. Instead, the primary (the fission core) produces a lot of highly energetic photons (X-rays) which travel at the speed of light, i.e. much faster than the emitted neutrons, and even more than the shock wave. These photons, when they reach the deuterium-tritium fuel, induce fusion (they yield enough energy to the nuclei to make them dance like John Travolta and bump into their neighbours). The fusion energy adds to the resulting fireball, and, crucially, emits a lot of extra neutrons which induce a lot more fission in the secondary (which again uses fission).

Thus, H-bombs explode fast because they are not, in fact, heat/confinement engines. Instead, they use the fission-based chain reaction to get a lot of X-ray and neutrons in a very short time; the fusion reactions add to the weapon yield, but their main use is to produce extra neutrons for more fission to happen. In a modern H-bomb, fusion and fission contribute similar amounts of energy to the total yield. The common explanation of H-bombs as "an A-bomb which sparks a much stronger fusion-based reaction" is flawed.

The Wikipedia page on nuclear weapon designs is a good place to start reading on the subject; it includes nice schematics and many pointers.


Within a star, there is an equilibrium between pressure from gravity, and expansion from the heat. The star's core remains at exactly the right temperature where the heat from fusion reactions counteract the gravitation. If the star is bigger, there is more gravitation, hence more heat and more reactions, which is why bigger stars live less long (very big stars will have a lifetime of a few millions of years, instead of a few billions for smaller stars like our Sun).

Others have pointed out that the proton-proton chain at works in most stars includes a slow step: when two protons fuse, they usually don't stay there, and separate again, reabsorbing the fusion energy. For the protons to stick, one of them has to morph into a neutron (emitting a positron with the positive charge), a process which involves the weak interaction and has only a very small probability of occurring.

This particularity explains why massive stars explode into supernovae. During most of its life (millions of years), the star consumes its hydrogen with the proton-proton chain. When enough helium has been produced, the alpha and triple-alpha process begin to take over, and then other fusion mechanisms, which are substantially faster. Things then happen within a few hours, a very short time compared to the previous millions of years, but still quite a lot longer than the microseconds during which an H-bomb detonates.

Summary: stars last for millions or billions of years, instead of mere hours, because of the weak-interaction step in the proton-proton chain. H-bomb go boom within microseconds, instead of hours, because they rely on a fission-based chain reaction, which allows for exponential cascading.

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None of these answers seems to explain correctly why the Sun differs from a nuclear bomb.

The reason is that any star, including the Sun, acts as a thermostat. If the Sun were to produce more energy than it can radiate away, the energy thus freed would make it hotter; a hot gas expands, and simultaneously cools. Both factors (lower densities and lower temperatures) would then contribute to reducing the rate of nuclear energy generation.

Conversely, if the Sun were to produce less energy than it radiates away, it would then contract; in a contraction temperature increases, and both factors (larger densities and higher temperatures) lead to an increase in the production of nuclear energy, thusly restoring once again equilibrium.

This is exactly the behaviour of a thermostat. It is often said that the structure of a star is dictated not by the nuclear sources involved, but by the extent of its envelope. The reason for this has been described above: the rate of nuclear energy generation simply adapts itself to what is demanded by the stars' energy transfer processes.

On the other hand, matter inside a nuclear bomb cannot expand and cool if energy is produced in excess; actually, the exact opposite is true: fissionable matter is disposed in such a way that the initial fusion explosion heats and compresses the fusion material to make sure that fusion reaction can procede without impediments. And this is exactly the opposite of the interiors of a star.

This process is described just about anywhere, including Martin Schwarzschild's now obsolete book, Structure and evolution of the stars, and also online, see the paragraph entitled A stellar thermostat here-

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Good point, +1. I would say that this is orthogonal to the other answers, which start from assumed conditions of pressure and temperature at the sun's core. However, the fact that the sun is in a stable equilibrium isn't the only difference between the sun and an H-bomb. The other differences include the need for the weak interaction and quantum-mechanical tunnelling. –  Ben Crowell Aug 13 at 21:54
    
Might be worth pointing out that the energy production for the two most common reaction chains in stars (p-p chain and CNO cycle) go as $T^4$ and $T^{20}$ respectively, which leads to very stable equilibrium states indeed. –  Kyle Aug 13 at 21:55
    
@Kyle What you say is true, at least at Solar temperatures, but I fail to see how this is relevant. Stability is due to the fact that the dynamical timescale (about 10 minutes for the Sun) is much shorter than the energy removal timescale, aka Kelvin-Helmholtz timescale, about 30 million years for the Sun. This means: if excess energy is produced, it will be used to power an expansion before it manages to be carried away. Thus extra energy production is used exactly to quench the conditions that brought it about. That's all. –  MariusMatutiae Aug 14 at 10:06

Because stars are not confined.

As @MariusMatutiae says the fusion in a star is maintained at equilibrium by the thermostat of pressure versus gravity. An even more apt appliance for analogy is a nuclear power plant. In nuclear fission power, control rods or other mechanisms adjust concentration so as to prevent explosion. The fissionable material is positioned at a precise balance just shy of critical mass, avoiding the positive-feedback runaway-train that would consume the fuel quickly.

In a star, thermonuclear fusion rates are balanced against proceeding quickly by thermal expansion against the force of gravity. Instead of pulling control rods away from each other to reduce the reaction rate as in nuclear power, heat expands and diminishes the concentration of fusionable materials.

In a thermonuclear bomb, fuel is deliberately confined by very elaborate (and somewhat secret) processes, for the rapid use of fuel.

In a nova or supernova, equilibrium is upset and some fuel is confined (e.g. by the gravity of a dense companion star) and consumed rapidly.

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