Physics of hydrogen burning in stellar cores

[TeleMad]

Opened this up today for the first time in a long while and worked on it some. Thought some might find it interesting/helpful, and it might also spark some discussion.

 

 

 

 

All main-sequence stars derive their energy from the thermonuclear process of “burning” hydrogen to helium. This involves three distinct combinations, that is, binding events:

 

(1) Two protons combine to form a deuteron

(2) A proton combines with the deuteron to form a helium-3 nucleus

(3) Two helium-3 nuclei (each produced from a separate occurrence of steps 1 and 2) combine to form a single helium-4 nucleus.

 

The following gives the details of these three binding events.

 

 

1) TWO PROTONS COMBINE TO FORM A DEUTERON

A deuteron consists of a single proton and a single neutron bound together (this arrangement is not unique to stellar interiors: it is found right here on Earth as the nucleus of deuterium, a heavy isotope of hydrogen). But if a deuteron is a proton and a neutron bound together, then how can two protons binding together form it? Doesn’t there have to be more than a single process involved in this binding event? Yes, so step 1 is broken down into two processes: here labeled as 1a and 1b. Loosely speaking, two protons must somehow “physically encounter each other” – despite the strong repulsion between them - and one of them must be transformed into a neutron to prevent them from then immediately flying apart.

 

 

1a) TWO PROTONS “PHYSICALLY ENCOUNTER EACH OTHER”

Every proton carries a +1 electric charge. So according to the first law of electrostatics (trivially, “opposites attract; likes repel”), two protons will always repel each other. In fact, the repulsion between two like electric charges increases by a factor of four for each halving of the separating distance (this follows the inverse square law), so it gets exponentially harder and harder to bring them closer together the closer together they get. Consequently, under ordinary circumstances, protons do not come close enough together to physically encounter each other. But what about under non-ordinary circumstances such as those found in the high-pressure, high-temperature cores of stars? After all, the temperatures there are so immense (for the center of our Sun, around twenty million Kelvins) that the average velocity of the protons is enormous (for the center of our Sun, an average of about 500 km/s). Wouldn’t the extremely high velocity of protons in the core of a star supply sufficient kinetic energy to overcome the strong electrical repulsion between protons? No, even that velocity falls way short (the required temperature is about 100 million Kelvins). How then do two protons approach one another closely enough to interact as needed if there is an insurmountable energy barrier that separates them? Quantum tunneling.

 

Quantum Tunneling

One simple way to picture quantum tunneling is the following. First, consider a proton to be a solid, spherical particle. As a macroscopic analogy, we’ll use a marble. What happens if you throw the marble at a barrier it cannot penetrate, say a foot-thick plate of glass? It simply bounces back, so if there is another marble on the other side of the glass the two cannot physically interact. Now instead of a marble-like particle, consider a proton to be a wave, like a pulse of light emitted from a flashlight that is quickly turned on and then off. What happens when the light encounters the glass? With a light wave, some of it gets reflected and some of it gets transmitted, ending up on the other side of the glass. So if a proton could be thought of as behaving like a wave, then we could imagine how it might ‘tunnel’ through the insurmountable barrier. Sounds like a good start, but can we actually consider a particle like a proton to be a wave? Yes, because of wave-particle duality. Perhaps this new concept itself needs an explanation.

 

Wave-Particle Duality

In 1905, Einstein began a revolution in thought when he stated that light, long known to be a wave, acted as if it were a particle in certain cases. Skepticism remained high for many years but experiments (such as one performed years later by Compton using X-rays) and the usefulness of the hypothesis eventually caused physicists to come to accept that light – a wave – did indeed act as discrete particles as well (these particles are now called photons). Another physicist, Louis de Broglie, suggested that the situation should be symmetrical: that since waves have particle aspects, then particles should have wave properties. Several years later experiments using electrons confirmed this: electrons fired through atomic-size slits in a crystal produced an interference pattern – a distinctly wave phenomenon. In fact, even large particles, such as atoms and even buckyballs, have been fired through double slits and found to produce interference patterns. That photons and electrons behave both like waves and like particles came to be called wave-particle duality. Just like electrons, protons too have a wave aspect.

 

Schrondinger’s Wave Equation

Physicists knew they needed a mathematical equation that described the wave aspect of matter. Two key figures in quantum mechanics were up to the task: Werner Heinsenberg and Erwin Schrodinger. They created two different, but provably equivalent, mathematical methods that did the trick. However, Schrodinger’s wave equation used a more familiar math and so won out over Heinsenberg’s matrix-based method. Schrodinger’s wave equation produces a probability distribution, which is now widely interpreted to represent a probability wave. Suppose a physicist wants to know where an electron is. He plugs the relevant numbers into the wave equation and obtains the results, which show varying amplitudes (heights) of a wave. Where the amplitude is largest the physicist is most likely to find the particle if he does a measurement; where the amplitude is smallest he is least likely to find the particle. Now let’s return to the topic of a proton quantum tunneling through the insurmountable energy barrier. Considering the proton as a wave, Schrodingers wave equation tells us that there is a small - but non-zero - amplitude of the probability wave existing on the other side of the barrier. So although the probability is much greater for the proton to end up on the same side that it started from, there is still a slight probability that it will end up on the other side. If it does, then the proton is said to have quantum tunneled through the energy barrier.

 

Proton interaction: electrical repulsion and quantum tunneling

Now lets’ bring these two main ideas (electrical repulsion and quantum tunneling) together. The kinetic energies associated with protons in the core of a star, though enormous, are insufficient to overcome the energy barrier generated by the electrical repulsion that keeps two protons from approaching each other very closely. But a given proton can possibly quantum tunnel through the insurmountable energy barrier and find itself on the other side, in close enough proximity to another proton to react with it.

 

Climbing the energy barrier

If quantum tunneling allows protons to approach each other, then why is it that they do so in the center of stars but not here on Earth? Basically, because of probability differences. Think of the energy barrier that exists between two protons as a steeply sloping two-sided hill that must be climbed: if a particle makes it all the way up and over the hill, it has made it to the other side of the barrier … without tunneling. Protons in the core of a star, as well as those here on Earth, are incapable of climbing all the way up and over the hill. But protons in a star’s core, due toe their large velocities, can make it up the hill a good distance (unlike protons here are Earth) before falling back down. Here’s the key. The farther up the hill a particle finds itself, the narrower is the distance through barrier and therefore, the greater is the fraction of a proton’s wave that gets transmitted through. From the earlier discussion, it follows that the farther up the hill a particle climbs, the greater is the probability that it will tunnel through and end up on the other side.

 

At this point two protons have managed to overcome their electrostatic repulsions via quantum tunneling and are (essentially) in direct physical contact with one another.

 

 

1b) ONE PROTON TRANFORMED INTO A NEUTRON

Two protons in close proximity have a very large repulsive force between them. This strong repulsion does not disappear just because one proton managed to tunnel through the energy barrier that normally keeps them apart. Therefore, if nothing further occurs, the two protons will quickly fly apart. But if one of the two protons were converted into a neutron, which is an electrically neutral particle, then there would be no electrical repulsion between them to fling them apart. But can a proton be transformed into a neutron? Yes, the process, which is mediated by the weak nuclear force, involves a proton beta decaying into a neutron, an anti-electron (also called a positron), and an electron neutrino. But there is a catch!

 

Weak interaction is slow-acting

Though (1) the conversion of a proton to a neutron must occur very rapidly (or else the two protons will fly apart), (2) the force responsible for the conversion - the weak nuclear force (or the weak interaction) - is a very slow-acting interaction. The reason the weak interaction is so sluggish is because its force carriers, the so called W+, W-, and Zo bosons, are so massive that the probability of a virtual one arising is extremely small: again, probability ruling in the quantum world. As one might expect from combining facts (1) and (2) from above, in the vast majority of two-proton encounters that occur as a result of quantum tunneling the two protons repel each other and fly apart: only in an extraordinarily small fraction of two-proton encounters does the weak force manage to convert one of the protons into a neutron in time. Thus there is yet another probabilistic factor in this first combining step of hydrogen burning in main sequence stars (quantum tunneling being the other).

 

Neutron more massive than proton

An observant reader might see that there seems to be a slight mistake here. Fusion, such as occurs in the core of a star, is supposed to be an energy-yielding process, but conversion of a proton into a neutron seems to go in the opposite direction. Why? A neutron is more massive than a proton, so the conversion from a proton to a neutron increases the mass of the two-particle system. According to Einstein’s energy-mass equivalence formula, E = mc2, an increase in mass is equivalent to an increase in energy. Since another law of physics mandates that total mass-energy be conserved, if the two-particle system gains mass-energy, then the surroundings must lose an equal amount of mass-energy to the two-particle system (otherwise, the “books” would become unbalanced). So the formation of a deuteron, which requires the conversion of a proton into a neutron, consumes energy instead of liberating it … doesn’t it? No, but only because another factor comes into play: binding energy.

 

Binding energy

Binding energy is the amount of energy it takes to separate two bound particles such that they are free, or, the amount of energy released when two free particles bind: the magnitude is the same in both cases, but the signs differ. Since it requires an input of energy to separate bound particles the value for doing so is considered to be positive (the system requires the addition of energy). On the other hand, energy is released when two particles bind and so for that process the value is considered to be negative.

 

Formation of deuteron: increase or decrease in mass?

But wait a minute…wasn’t it just stated above that energy and mass are equivalent? Yes, it was. So if the two-particle system loses energy due to binding, doesn’t it lose mass too? Yes, it does. But how can that be since it was already stated that there is an increase in mass of the two-particle system when the proton is converted into a neutron? A paradox? No, because those are separate events. The conversion of a free proton to a free neutron does involve an increase in mass, and the binding together of two free particles does result in a decrease in mass. Overall, the decrease in mass associated with binding of the two nucleons exceeds the slight increase in mass associated with the proton-to-neutron conversion, so the net process results in a decrease in mass. The “missing mass” is converted to energy so, just as one would expect for a solar process, step 1 yields energy.

 

Binding energy greater than proton-to-neutron mass difference

One might wonder, “why is the binding energy so large that it exceeds the mass difference between a proton and a neutron?” Two reasons: the strength of the force responsible for binding nucleons - protons and neutrons - and the small mass difference between those two nucleons. Many types of bound systems exist, such as a solar system consisting of planets gravitationally bound to a star. But since the gravitational force is by far the weakest of the four fundamental forces of nature, the gravitational binding energy between two objects as minute as nucleons is for all practical purposes zero. So what force is responsible for the large change in mass-energy in deuteron formation? When considering the binding together of, or separation of, nucleons, the binding energy is dependent upon the strength of the very intense strong nuclear force. This force is by far the strongest of the four fundamental forces of nature: it is about 137 times as strong as the electromagnetic force that causes repulsion between like electric charges (such as two protons), and is ten million trillion trillion trillion times as strong as the gravitational force that binds planets to stars. Consequently, the effects of the strong force are associated with large or enormous magnitudes. Finally, the difference between a proton’s mass and a neutron’s mass is quite small, so the mass lost due to binding does not actually have to be all that great to exceed the proton-neutron mass difference.

 

Why don’t two protons bind together?

It was stated earlier that two protons will not approach each other closely because there is a strong electrical repulsion between like charges. Yet just above it was stated that the strong force – which is the force that is responsible for binding two nucleons together - is about 137 times as strong as the electromagnetic force. So why doesn’t the strong force simply overpower the electromagnetic force and cause two protons to bind together? Limited range. The strong force is sometimes more accurately called the nuclear strong force, which gives an idea of how small its sphere of influence is. The primary responsibility of the strong force is the binding together of pairs or triplets of subatomic particles called quarks. Protons and neutrons are triplets of quarks bound together by the strong force, also called the color force. For example, a proton consists of two up quarks and one down quark, while a neutron consists of one up quark and two down quarks. Up quarks have a charge of +2/3 and down quarks have a charge of –1/3. Doing the math shows that protons end up with +1 electric charge and neutrons end up with 0. The effective range of the strong force extends only a minute distance past the boundary of an individual quark so it barely reaches out past a single proton or neutron (and in fact disappears completely after that, instead of trailing off according to the inverse square law). It is this slight residual affect of the color or strong force that binds nucleons together. Let us consider two protons once again. At a separating distance much greater than the sphere of influence of the residual strong force, the only force worth mentioning is the electromagnetic force which causes the two positive charges to repel each other. As the two particles are forced to come closer to one another, the electrical repulsion increases exponentially. This effectively prevents them from ever coming close enough to one another that the residual effects of the extremely short-range color force could be an overwhelming factor.

 

So these two processes – (1) one proton quantum tunneling through the energy barrier that is produced by the electrical repulsion between it and the other proton, and (2) one of the two protons transforming via the weak force into a neutron – allow the two particles to bind, thereby releasing energy.

 

 

2) A PROTON COMBINES WITH A DEUTERON TO PRODUCE A HELIUM-3 NUCLEUS

Since no particle transformations are involved (a proton is not converted into a neutron), the slowness of the weak interaction is not a problem here. However, since both the deuteron and the proton that will bind carry a net charge of +1, there is still the strong electrical repulsion between like charges that must be considered. The first fact to note is that the net +1 charge on the deuteron is not equally distributed: the proton carries all of that electric charge and the neutron is as always neutral overall. So in this binding step, the incoming proton does not have to approach and interact directly with the proton portion of the deuteron, but can instead approach and interact with the neutron portion. Since there is no electrical repulsion between a proton and a neutron, and the separating distance between the two positive electric charges is greater here than in step 1, the strength of repulsion is not as great (remember that the strength of repulsion between like charges falls off exponentially with distance, following the inverse square law). In fact, once the proton does bind to the deuteron, the resulting helium-3 nucleus is stable (that is, it will not immediately fly apart as the two protons in step 1 would do were one of them not immediately converted into a neutron). Note that this is a nucleon binding event (whose product’s mass does not exceed iron-56), and as such, it too releases energy.

 

 

3) TWO HELIUM-3 NUCLEI COMBINE TO FORM A SINGLE HELIUM-4 NUCLEUS

Obviously, if this step involves the combining of two helium-3 nuclei, then binding steps 1 and 2 above must each occur twice, in order to produce each of the individual helium-3 nuclei. In this final step of the “burning” of hydrogen to helium, the inputs (“reactants”) each consist of two protons and one neutron for a total of four protons and two neutrons. Since the output (“product”) consists of only two protons and two neutrons, it is clear that two of the protons are recycled – that is, they are returned to their free state to combine again with other protons at a later time. Thus, using a ‘type’ of chemical equation, with P representing protons and N representing neutrons, we have:

 

P2N + P2N -> P2N2 + 2P + energy

 

This binding step too liberates energy.

 

 

ALL DONE

At the end of this series of events (which involves both all four of the fundamental forces of nature, even though gravity was not explicitly mentioned here, as well as several key aspects of quantum mechanics), four free protons have been converted into a single bound helium nucleus consisting of two protons and two neutrons. The difference in mass between the starting free particles and the ending nucleus is about one percent. Hence, about one percent of the original mass has been converted into energy via E = mc2.

 

 

 

 

PS: There are several things I feel iffy on in my descriptions of quantum mechanics, and I am very unsure about the beta decay I mentioned for the conversion of a proton to a neutron (I am familiar with the other way around, in which a neutron beta decays into a proton, an electron/beta particle, and an electron antineutrino, but not with this opposite process ... and I don't feel like looking it up!)