## How does the Sun shine? (Final part: The search for the proton-proton neutrinos)

Let’s return to the original goal of this series of posts on the Sun’s source of energy. After the post on the proton-proton chain model, it became quite clear that it was very difficult to experimentally prove that this mechanism was really happening inside the Sun. Because observing the low energy neutrinos from this first branch of the fusion process is itself an experimental challenge due to the fact that at these energies, there are many other things which can interfere with the results. Attempts to measure pp neutrinos directly over the past 30 years have been hindered by the presence of radioactive backgrounds in this low-energy region (0 to 420 keV).

It is to overcome this problem that the Borexino detector was designed. Borexino is very good at observing neutrinos of all known flavors and at all energies. Located deep beneath Italy’s Apennine Mountains, Borexino is designed to minimize backgrounds from radioactive isotopes both within, and external to, the liquid scintillator (also called the target). In the liquid scintillator, neutrinos interact with the electrons of this ultra-pure organic liquid scintillator at the center of a large stainless steel sphere surrounded by tons of water. Its great depth and multiple onion-like protective layers keep the core free from most radiation.

The Borexino experiment detects solar neutrinos by measuring the energy deposited in the liquid scintillator target by recoiling electrons undergoing neutrino-electron elastic scattering. That is, a neutrino of one of the three known neutrino flavours, collides with a free electron of the target/scintillator and gives it some of its kinetic energy (for more info on that, I’ll have to learn QFT, and Z & W bosons…). The target/scintillator converts the kinetic energy of electrons into photons (i.e., re-emit the absorbed energy in the form of photons/light). This light is detected and converted into electronic signals by photon multipliers mounted on a concentric stainless steel sphere. Despite extreme precautions against external sources of radiation from the earth and the equipment used by the instrument, the Borexino has to cope with the internal Carbon-14 beta radioactivity present inside the liquid scintillator and the cosmic ray muons which interfere with the detection of low-energy ppI neutrinos. (One can read about the statistical methods used to account for these interferences in the source below).

The primary aim of the experiment is to make a precise measurement of the beryllium-7 neutrino flux from the sun (coming from another branch of the fusion process) and comparing it to the Standard solar model prediction. This will allow scientists to further understand the nuclear fusion processes taking place at the core of the Sun and will also help determine properties of neutrino oscillations, including the MSW effect (The Mikheyev-Smirnov-Wolfenstein effect often referred to as matter effect, is a particle physics process which can act to modify neutrino oscillations in matter). Other goals of the experiment are to detect boron-8, pp, pep and CNO solar neutrinos as well as anti-neutrinos from the Earth and nuclear power plants. The project may also be able to detect neutrinos from supernova within our galaxy. (NOTE: Neutrinos are also important as early indicators of exploding stars, because neutrinos emitted by the dying star arrive before the light/photons, neutrinos still do not travel faster than light, it is only because they are produced earlier, before the photons that we often see on fantastic supernova pictures).

Finally, Borexino is currently the only detector on Earth capable of observing the entire spectrum of solar neutrino simultaneously. As we’ve seen previously, neutrinos come in three types, or flavors. Those from the Sun’s core are of the electron-flavor, and as they travel away from their birthplace, they oscillate or change between two other flavors, muon- to tau-. Since 2007, Borexino has, strongly confirmed this behavior of the neutrinos, and has proved with increased precision, that 99 per cent of the power of the Sun, 3.84 × 10^33 ergs per second, is generated by the proton-proton fusion process (the first branch or ‘ppI’ branch of the proton-proton chain).

This post is the last one on this series on the Sun’s source of energy, but the Sun has still many more mysteries like for example: What is responsible for the 11 years long solar cycle? Why is the corona’s temperature higher than the Sun’s surface? Why is the solar cycle interrupted every 70 years by a Maunder minimum? Why is the heliosphere shrinking and the solar wind flow that inflates it decreasing? The heliosphere is a bubble of magnetism springing from the sun and inflated to colossal proportions by the solar wind. Every planet of the solar system is inside it. The heliosphere is our shield against galactic cosmic rays, high-energy particles from black holes and supernovas, entering the solar system, but most are deflected by the heliosphere’s magnetic fields.

Sources: Wikipedia and “Neutrinos from the primary proton-proton fusion process in the Sun”, Borexino Collaboration (27 August 2014), Nature 512 (7515): 383-386.

## Neutrinos (Part 2: Neutrino oscillations — simplified)

In my previous post on the neutrinos and the solar neutrino problem, I ended by writing that the current explanation for why there were missing neutrinos at the detector is because the neutrinos have changed state while travelling towards earth and since the detectors were calibrated on detecting only some particular states, they were missing some of them.

Discrepancies between the expected number of neutrinos of any given flavor and observations were also noted when counting neutrinos originating from other sources than the sun: for example, from nuclear reactors and from cosmic rays. This problem affects all three neutrino flavors (electron-like, muon-like or tau-like). Muon neutrinos are produced in muon decays, which affect muons produced naturally in the atmosphere by cosmic rays. Muons created in particle accelerators also produce muon neutrinos when they decay. Experiments that detect only muon neutrinos also find discrepancies.

To understand neutrino oscillation, one must think of neutrinos as waves rather than particles and have some basic knowledge of quantum mechanics. (*1*) At the start of their journey, neutrinos know definitely that they are electron-like, muon-like or tau-like what is called the different neutrino “flavors”. Because neutrinos are born in weak force interactions alone or together with other flavors of neutrinos, there is no law of Nature that dictates a certain flavor of particle should have a certain mass, so the neutrino is born knowing exactly what type of flavor but not exactly what mass it has. That is how quantum mechanics works, because of the Heisenberg uncertainty principle, you can never know both the position and the momentum (including its mass) of a particle with certainty (because momentum is mass times speed).

One theory states that neutrino flavor is a superposition of mass (Hamiltonian) eigenstates.

In the 2-flavors case (a simplification of the real 3-flavors theory), the neutrino flavors that we know from weak interactions (the weak eigenstates) are linear combinations of underlying mass eigenstates, where the mass eigenstates (denoted as ν1 and ν2) are the free particle solutions to the Schrodinger wave equation. We cannot assume a priori that the mass eigenstates and flavor eigenstates are equivalent. If they are, as we shall see momentarily, oscillation cannot take place. If they are not, then these mass eigenstates cannot be directly observed in their pure form, only weak eigenstates (via the weak interaction). The weak eigenstates are related to the mass eigenstates by a simple unitary (length preserving) matrix:

$\displaystyle \begin{bmatrix}\mid\nu_e\rangle\\\mid\nu_{\mu}\rangle\end{bmatrix} = \begin{bmatrix}\cos \theta & \sin \theta \\-\sin \theta &\cos \theta \end{bmatrix} \begin{bmatrix}\mid\nu_1\rangle\\\mid\nu_2\rangle\end{bmatrix}$

In other words, the weak eigenstates, aka the neutrino flavors, are coherent linear superpositions of the mass eigenstates. Here, $\displaystyle \theta$ is referred to as the mixing angle.

We assume that the mass eigenstates for the free neutrino as a function of time are simply plane waves. Combining the superposition in the Schrodinger equation results in the equation describing that the pure flavor state at t=0 (explained in *1*) is a superposition of different flavors states at t>0. From that, it is possible to calculate the probability of flavor transition as a function of time in vacuum.

Suppose we have a weak decay producing an electron neutrino at time t=0. That neutrino expressed in terms of the mass eigenstates ν1 and ν2 would be:

$\displaystyle \mid\psi(0)\rangle = \mid\nu_e\rangle = \cos\theta\mid\nu_1\rangle + \sin\theta\mid\nu_2\rangle$

The two-flavor oscillation probability in vacuum as a function of distance L is:

$\displaystyle P\left(\nu_e \rightarrow \nu_{\mu}\right) = \sin^22\theta\sin^2\left( \frac{\Delta m_{21}^2Lc^3}{4\hbar E} \right)$

Several things can be noted from this formula. Immediately after the particle is created, at L=0, the probability is 0 that its flavor has changed (because of the second sin^2 factor). The probability varies with the distance that a neutrino travels, fluctuating periodically as L increases. It reaches a maximum every time L is such that the second factor is 1. The first factor determines what the maximum of the probability can be, and it will be nonzero if the mixing angle is nonzero.

The average of this probability across distances and energies is only a function of the mixing angle.

$\displaystyle \langle P\left(\nu_e \rightarrow \nu_{\mu}\right) \rangle = \frac{1}{2} \sin^22\theta$

Finally, let’s go back to the particle/wave analogy, the net result is that different mass states propagate at different speeds, and the mixing matrix (containing several mixing components) transfers this effect to flavor states. Interference between waves (representing mass states) for different flavor states is the basic cause of neutrino oscillation. So the probabilities associated with observation of specific flavor states depend in a complicated way on the (squared) mass differences between the specific mass states, as well as a particle’s energy and the distance it has traveled since creation.

Additionally, so far I’ve talked about oscillations in vacuum, but because ordinary matter is made from electrons (among other things) but not from muons and taus, electron-neutrinos have different interactions with ordinary matter than muon- and tau-neutrinos do. These interactions, which occur through the weak nuclear force, are tiny. But if a neutrino passes through a great deal of matter, such as the outer layers of the sun and then, the earth, these small effects can add up and have a big impact on the oscillations. Experiments have to take this into account to produce accurate results.

It is important to stress and maybe repeat myself again on this point. The equations presented before all use the squared mass difference and not the neutrino absolute mass. For neutrino oscillations to occur, at least one of the mass states must be non-zero. This simple statement has huge implications, for oscillations to happen, the neutrino must have mass. Furthermore the masses of the mass states must be different, else delta m is 0 and P is 0. In conclusion, careful measurements of neutrino oscillations are in fact the fastest way to learn about the properties of neutrinos and maybe determining the (absolute) neutrino masses once and for all, which has much deeper consequences and raises even more questions. Some of these questions will be addressed in the next part of this series on neutrinos.

Neutrino flavor-oscillations for dummies

Sources:

Fundamentals of Neutrino Physics and Astrophysics, C. Giunti and C.W. Kim. Oxford University Press (2007)

## How does the Sun shine? (Part 4: The Solar Neutrino Problem)

This post is based on my last two posts explaining the proton-proton chain and
the neutrino hypothesis and discovery.

As explained in the proton-proton chain post, solar neutrinos which are
also electron neutrinos, provide the only direct experimental evidence of the nuclear
fusion happening inside the core of the Sun. Because, unlike photons, neutrinos
produced by the ppI chain and other less probable branches are supposed
to go through the layers of the Sun almost undisturbed.

Solar neutrinos were studied in experiments undertaken by American physicist Raymond
Davis in 1968. He observed collisions between solar neutrinos and chlorine by
using underground tanks of chlorine.

$\large {\nu}_\textrm{e} + {}^{37}\textrm{Cl} \rightarrow {}^{37}\textrm{Ar} + \textrm{e}^{-}$

This pioneering experiment, called the Homestake experiment, on solar neutrinos started by Raymond Davis and collaborators (among them, John Bahcall, who made the predictions based on the Standard Solar Model) is based on the inverse beta decay process: Chlorine-37 absorbs the neutrino to yield Argon-37 and an electron. A tank containing 615 tons of a fluid rich in chlorine called tetrachloroethylene, a colorless liquid widely used for dry cleaning of fabrics, was placed in a gold mine in South Dakota (USA). The fluid was periodically purged with helium gas to remove the argon atoms which were then counted by means of their radioactivity. Due to the very weak interaction between neutrinos and the liquid, the experiment must have been like finding a particular grain of sand in the whole of the Sahara desert.

How many neutrinos were expected to show up during an experiment?

A simple calculation provides a good estimate of the expected flux of solar neutrinos produced by the PPI chain at the earth (and everywhere around the sun at a distance of 1 astronomical unit or AU). By first approximation, the PPI chain provides most of the
solar power we observe on earth, because it happens 85% of the time. As explained in a previous post, every iteration of this sequence results in 2 neutrinos for each 28 MeV of kinetic energy, which we observe eventually as Solar luminosity (Lsun).

$\large \phi = \frac{\textrm{n}_{\nu}}{S}$

$\large n_{reac} \times \frac{28MeV}{reac} = {\textrm{L}}_{sun}$

$\large n_{reac} = \frac{{n}_{\nu}}{2}$

$\large n_{\nu} = \frac{\textrm{L}_{sun} \times 2}{28 MeV}$

(*)$\large \phi = \frac{\textrm{L}_{sun} \times 2}{28 MeV} \times \frac{1}{4 \pi R^2}$

= units of particles per second per cm2
$\large \phi \approx 6.10^{10} / (cm^2 . s)$

At a radius R equal to 1 Astronomical Unit: R = 1 AU = 1.5*10^13cm,

Area = 4*pi*R^2 = 2.83*10^27cm^2

The problem with the Davis experiment was that the detection threshold in Davis’s experiment was 0.8 MeV. So, only neutrinos with energies above this threshold could be detected, and therefore only the less probable, and high-energy Boron-8 neutrinos (the ppIII chain) were detected. So they were not measuring the full spectrum of neutrinos from the Sun, they actually miss all of the most probable solar neutrinos and only measure the rarer types.

So, is there a solar neutrino problem?

The numbers that are being compared are the following: the theoretical neutrino flux in solar neutrino units (SNU) and the measured neutrino flux also in SNU. One SNU is equal to the neutrino flux producing 10^−36 captures per target atom per second. It is convenient given the very low event rates in radiochemical experiments. With typical neutrino flux of (*) 10^10 cm^−2 s^−1 and a typical interaction cross section of about 10^−45 cm^2, about 10^30 target atoms are required to produce one event per day.
Taking into account that 1 mole is equal to 6.022 10^23 atoms, this number corresponds to kilo tons of the target substances, whereas present neutrino detectors operate at much lower quantities.

So the expected neutrino flux of the 8B producing branch, is approximately 6 to 7.5 SNU (*), while Davis first experiment has observed only 2.5 SNU. Many similar experiments like Davis have been made at various locations, various neutrinos source have been used to calibrate these neutrino detectors but the results have been the same, only 1/3 of the expected neutrinos were observed. So, there seem to be a common problem of missing neutrinos everywhere the Solar neutrinos experiments took place.

In this picture, the estimated neutrino flux is given on the vertical axis in units cm^−2 s^−1. Different methods of detection, using Gallium and Chlorine, have various minimum neutrino’s energy threshold in order to be able to detect neutrinos. Note that the neutrinos with energies in the range 0.1-0.25 MeV does not have any associated detector. Also note that the 0 to 0.42 MeV range of the neutrinos from the PPI chain are only partly detected by Gallium-based detectors. The highest flux corresponds to the PPI chain neutrinos that we grossly estimated to be in the 10^10 cm−2 s−1 range, is the first curve with label “pp”. So, as explained previously, the Davis experiment detects neutrinos in the Chlorine detection range and misses the pp neutrinos.

The reason why only 1/3 of the expected neutrinos were detected in Davis experiment, remained a mystery until it was suggested that the electron-neutrinos produced in the sun may transform into other flavors of neutrino, which would not be detected, as they travel towards the Earth.

Sources:
Solar Neutrinos. John N. Bahcall. (http://arxiv.org/abs/hep-ph/0605186)

Solar Models: An Historical Overview. John N. Bahcall. (http://arxiv.org/abs/astro-ph/0209080)

## Five years in space for NASA’s Solar Dynamics Observatory

In honor of Solar Dynamics Observatory’s (SDO) 5th anniversary, NASA has released a video showcasing highlights from the last five years of sun watching. Watch the movie to see giant clouds of solar material hurled out into space, the dance of giant loops hovering in the corona, and huge sunspots growing and shrinking on the sun’s surface:

Here are some more pictures from the Solar Dynamics Observatory:

SDO Gallery

## Neutrinos (Part 1: Hypothesis and Discovery)

1. Wolfgang Pauli’s hypothesis

Before explaining how Pauli’s came to state his famous hypothesis, let’s do a quick recap on notation and beta decays.

The notation we will use is the following: the decay of a parent nucleus, P, to a daughter nucleus, D, generally involves the emission of other decay products:

P -> D + other products + Q

The quantity Q represents the energy required to balance this process. Its value can be determined by applying the conservation of (relativistic) energy to the process. That is, Q can be found by requiring that the total energy immediately before the decay is equal to the total energy immediately after the decay, where the term TOTAL energy is taken to include the energy equivalent of any rest masses involved (as given by Einstein’s mass-energy equation, E = mc2) as well as the relativistic kinetic energy of the particles. The energy represented by Q appears as the kinetic energy shared between the daughter nucleus, D, and other products. Beta decays, like other radioactive decays not discussed here, result in positive Q values.

In a two-body process, the lighter emitted particle, here, the electron (also called a BETA particle in the case of a beta decay), would be expected to carry away most of the released energy, which would have a unique value. Why?

Because the masses of the electron and the daughter nucleus are fixed, the available energy (or Q-value) would split in a definite proportion between the two and only one definite kinetic energy for the electron would be possible. ONLY if there is an additional third particle involved can the energy be split between the three in an infinite number of ways, limited only by the total energy available. All kinetic energies for the electron, up to some maximum (total energy available), become possible.

However, experiments show that the electrons are emitted with a continuous spectrum of energies. In fact, the observed differential distribution in the number of emitted electrons as a function of their energy has the shape given by this picture:

The endpoint corresponds to the maximum energy of any emitted electron. That is, the experiment shows that the electrons have a spectrum of energies, with most values lying well below that predicted by energy conservation in two-body decays.

When this was first observed, it appeared to be in contradiction with the conservation of energy and as we will see, the conservation of momentum too!

The questions that are at the origin of Pauli’s hypothesis are the following:

First, why beta particles have a range of energies? There must be some process which randomly selected energies. Why did it choose one energy and not another?

Second, when a nucleus emitted a beta particle it didn’t recoil backwards in a straight line but normally at an angle. This seemed to violate the principle of conservation of momentum. Why an angle? Is there a third particle possibly carrying some of the momentum?

Third, there is an apparent violation of the law of conservation of angular momentum. The total spin angular momentum of the initial particle before beta decay occurs is not equal the total angular momentum of all particles produced afterwards.

The beta decays experiment in the 1930s could not see the third particle but it looked like there was one “ghost” particle. This “ghost” particle was needed in order to satisfy the conservation of mass and energy but it had no charge and interacted very weakly with ordinary matter. The Italian nuclear physicist Enrico Fermi took up Pauli’s idea and on its basis developed a theory of beta decay. Fermi also coined the term “neutrino”, after Pauli had spoken of “neutron”, but the latter designation was reserved for the heavy component of the atomic nucleus discovered in 1932. Pauli’s hypothesis was presented in 1933. It then took a further 23 years (until 1956) before the experimental proof of the existence of the neutrino succeeded.

Wolfgang Pauli proposed in 1933 that the third particle, one that was difficult to detect, was emitted in beta-decay. Conservation of electric charge required this particle to be electrically neutral, just like the neutron and the photon. In fact, this would explain why it was so hard to detect this particle. We know now that this neutral particle, the neutrino, does not interact readily with matter, and this is the main reason why it is so difficult to observe. Because the maximum energies for electrons emitted in beta-decay corresponded to the disintegration energy of the nucleus, it meant that this new particle had
to be essentially massless. Furthermore, if the postulated neutrino were to restore the conservation of angular momentum, then it would have to be a fermion with spin angular momentum 1/2.

2. Discovery: The Cowan and Reines experiment of 1956

In short, the famous Cowan and Reines experiment is the first experiment which detects neutrinos indirectly by observing the effects of neutrino collisions with matter. After a collision, there is creation of an antiparticle, the annihilation of this antiparticle creates high energy photons or gamma rays, which are amplified by a liquid scintillator, which gives of flashes of visible light in response to invisible gamma rays. These flashes of visible light have to be detected by photomultiplier-tubes some microseconds after the gamma rays were emitted. The problem with this method is it’s sensibility to neutrinos created by cosmic rays and by the sun which are considered background noise for this experiment.

The experiment used a nuclear reactor as the source of a neutrino flux of 5×10^13 neutrinos per second per square centimeter. It was far higher than any attainable flux from other radioactive sources. Negative beta decay in the nuclear reactor, which is only one type of radioactive decay, produces an electron antineutrino, the electron antineutrinos interacted with protons in two tanks of water, creating neutrons and positrons:

(1) $\bar\nu_{e} + p \rightarrow n + e^{+}$

The positron quickly finds an electron, and they annihilate each other, so that each positron created a pair of gamma rays when it annihilated. The gamma rays were detected by sandwiching the water tanks between tanks filled with liquid scintillator. The scintillator material gives off flashes of light in response to the gamma rays, and these light flashes are detected by photomultiplier tubes.

This experiment was not conclusive enough, so they devised a second layer of certainty. They detected the neutrons by placing cadmium chloride in the tank. Cadmium is a highly effective neutron absorber and gives off a gamma ray when it absorbs a neutron. The gamma ray from the cadmium capturing one neutron produced by reaction (1) would be detected 5 microseconds after the gamma ray from the positron annihilation, if it were truly produced by a neutrino.

The experiment was moved several times in order to decrease the background noise, and get better results by shielding the experiment against cosmic rays. Because cosmic rays are also producing neutrinos in the upper atmosphere. This shielded location was far away from the reactor and several meters underground, as depicted on this picture:

After months of data collection, they had accumulated data on about three neutrinos per hour in their detector! To be absolutely sure that they were seeing neutrino events from the detection scheme described above, they shutdown the reactor to show that there was a difference in the number of detected events.

Sources: Wikipedia and Das & Ferbel, Introduction to Nuclear and Particle Physics (2003)

## The Gamow Window

Stars are composed of hot gases in which the atoms and molecules are almost completely ionized in the interior (the state of matter called a plasma). The question of whether fusion reactions can occur in this plasma is primarily one of the density/pressure(*) and the temperature of the gas, because the density controls the number of collisions and the temperature controls their average energy. (*)Please note that for a fixed volume and temperature, the density and pressure are directly related, we may also use the pressure instead of the density as a variable.

The higher the electric charges of the interacting nuclei, the greater the Coulomb repulsive force, hence the higher the kinetic energy and temperature required before reactions occur. Highly charged nuclei are the more massive, so reactions between light elements occur at lower temperature than reactions between heavy elements. But, in other words, classical mechanics prevents the two protons from fusing because they will never have enough energy to overcome the Coulomb repulsion.

The energy of the Coulomb barrier is given by:

$\textrm{E}_{Coulomb} \simeq \frac{Z_1Z_2e^2}{r_0}$

Where Z is the atomic number for each particle, in this case, Z=1 for 2 protons. r0 is the radius at which nuclear attraction overcomes Coulomb repulsion, r0 ~ 10^-15m. This energy is thus approximately equal to 1MeV for 2 protons.

In the kinetic theory of gases, all gas particles have kinetic energy due to their motions as they bounce around. Temperature is a measure of this microscopic kinetic energy. The kinetic energy of any moving particle is 1/2mv2, where m is the particle mass and v is its velocity. Maxwell and Boltzmann (M&B) deduced that the mean kinetic energy is proportional to T. This statement is usually written E = 1/2 mv2 = 3/2 kT

In this equation, k is a fundamental constant called the Boltzmann constant, which has the tiny value of 1.38 × 10^-23 Joules per Kelvin. The temperature is proportional to the square of the average velocity and it is proportional to the mass of the particle.

However, in the solar core, the temperature is Tc ∼ 1.5×10^7 K. Then, with the M&B equation: E(= 3 / 2 kTc ) ~ 1keV, or ∼ 10^−3 Ec. That is, despite that the core is very hot, it is not hot enough to bring the protons close enough to trigger fusion, but only brings protons to within ∼ 10^3 r0 of each other. In other words, the protons are not close enough by a factor of 10^3!

George Gamow calculated that protons with 3 to 10 keV of energy (which there are plenty of in the Sun’s core) can overcome the Coulomb barrier (of 1 MeV) through a process of quantum tunneling or barrier penetration. It is (1) the low probability of this quantum tunneling, along with (2) the need for a weak interaction in order to fuse two protons into a deuterium nucleus, which are the two factors that make stars have lifetimes billions of years long!

The Coulomb barrier for charged particle reactions and the distribution of velocities implied by the kinetic theory of gases imply that there is a narrow range of energies where nuclear reactions involving charged nuclei occur in stars. This window is called the Gamow window. The peak is the product of the two curves decreasing in opposite directions: The probability for penetrating the Coulomb barrier decreases rapidly with decreasing energy (wave behavior and Energy = wavelength x frequency), but with M&B, at a given temperature the possibility of having a particle of high kinetic energy (and therefore high velocity for the particle behavior) decreases rapidly with increasing energy (because of E=mc^2 and mass resists to acceleration, that is, a change in velocity).

To conclude, the sum of these opposing effects produces an energy window for the nuclear reaction: only if the particles have energies approximately in this window can the reaction take place. This is a plot of the normalize probability (range between 0 and 1) on the Y-axis and the particle energies on the X-axis:

Source: Wikipedia Gamow window as starting point, but a lot of personal research, since this topic is often skipped in Astronomy text books.

## The proton-proton 1 chain

In a previous article, I have written about the binding energy. If we look again at the diagram of binding energies per nucleus, we see that the first big peak is at the $^{4}\textrm{He}$ isotope, so that is where we expect the fusion reaction to go, because, as I also wrote in that article, the system wants to reach a lower energy state, i.e., a state with higher binding energy between protons and neutrons in the nucleus.

When using classical physics in order to estimate the temperature of the Sun, it was discovered that the Sun’s temperature was considered too low to overcome the Coulomb barrier between protons. After the development of quantum mechanics, it was discovered that quantum tunneling of the wavefunctions of the protons through the repulsive Coulomb barrier allows for fusion at a lower temperature than the classical prediction. However, even with quantum tunneling, it was unclear how proton-proton fusion might proceed since creating $^{4}\textrm{He}$ starting from 2 hydrogen nuclei (or 2 protons), is, as we will see in the next paragraph not easy. It all starts with this reaction:

$\large _{1}^{1}\textrm{H} + _{1}^{1}\textrm{H} \rightarrow _{2}^{2}\textrm{He} + \gamma$

The result of the reaction above is a diproton (with Z=2 protons and N=0 neutrons in the nuclei). This nuclei is unstable and immediately dissociates back into a pair of protons.

As we have seen in the article on binding energies and the weak force, neutrons are required to have a stable nuclei, in order to counteract the effect of the electromagnetic repulsion between protons. And in order to reduce the disruptive energy of the electric force, the weak interaction allows the number of neutrons to exceed that of protons. Therefore, without any neutrons, $_{2}^{2}\textrm{He}$ almost immediately disintegrates into two protons, producing no net energy release, and that also explains why $_{2}^{2}\textrm{He}$ is not in the diagram of binding energies but only $^{3}\textrm{He}$ (Z=2 and N=1) and $^{4}\textrm{He}$ (Z=2 and N=2) are because they have N > 0.

Thus for a reaction to generate energy, one must find a way to bypass $_{2}^{2}\textrm{He}$ and jump to a stable state. One possible solution to this problem was discovered by Hans Bethe in 1939: during the very brief period that $_{2}^{2}\textrm{He}$ lives, a weak nuclear reaction (also called a beta-plus decay) can occur that converts one of the protons into a neutron plus a positron (the antiparticle equivalent to an electron) plus a neutrino, a nearly massless particle with a spectrum of energies between 0.0 and 0.42 MeV.

$\large _{2}^{2}\textrm{He} \rightarrow _{1}^{2}\textrm{D} + {e}^{+} + \nu_{e} (\small\textrm{0.0 .. 0.42MeV})$

That is because the only nucleus that can exist with 2 nucleons is the deuteron. So when 2 protons combine, one of the proton has to convert into a neutron through beta+ decay simultaneously that the fusion reaction happens. 2 processes occurring simultaneously one driven by strong forces between proton and proton in order to get into one nucleus in the 10^-20 seconds timescale. The other process, a proton converting into neutron is driven by weak interaction, with a timescale that is about similar to the lifetime of a free neutron 14 minutes or 840 seconds. But here we are demanding that this beta decay happens in the same timescale as 10^-20 seconds. Therefore, the probability that this process happens in this timescale is very very low. So that first step of the proton-proton chain is very very slow because of this demand that the beta decay occurs together with the fusing of the two protons. This is known as the deuteron bottleneck. The probability that two protons turn into a deuteron is very very very small. Let’s say you can wait 10^10 years to see 1 proton out of 10^18 fuse with another proton! But because the total number of protons in the sun is so large, 6×10^11kg of hydrogen, approx 3.7×10^38 protons fuse each second inside the core of the sun.

(1) $\large _{1}^{1}\textrm{H} + _{1}^{1}\textrm{H} \rightarrow _{1}^{2}\textrm{D} + {e}^{+} + \nu_{e} (\small\textrm{0.0 .. 0.42MeV})$

The positron and neutrino, which do not feel the strong nuclear force, immediately escape from the nucleus, leaving behind a proton plus a neutron. The proton plus neutron do constitute a stable nucleus: deuteron (Z=1 and N=1) ($_{1}^{2}\textrm{D}$ OR $^{2}\textrm{H}$). The net reaction is exothermic, and the excess energy mostly goes into the recoil of the deuteron and positron. This excess energy is then turned into heat when the nuclei collide with other particle. Additionally, the positron will rapidly be attracted to any nearby electron. Since these are anti particles to each other they annihilate each other producing a power gamma ray (a form of electromagnetic energy far more violent than x-rays).

As soon as the first deuteron is created, it combines with another proton to form 3-Helium, strong interaction only no weak interaction this time.

(2)$\large _{1}^{2}\textrm{D} + _{1}^{1}\textrm{H} \rightarrow _{2}^{3}\textrm{He} + \gamma (\small\textrm{5.49MeV})$

This 3-He cannot combine with one more proton because it’s 4-Li and it will not exist with 3 p and 1 n. Combine with a D is almost unlikely because as soon as D is produced it combines with one p. Combine with n is also not likely because there aren’t any free n around. D has to WAIT for another 3He to be produced in order to fuse with it and produce 4He.

(3)$\large _{1}^{2}\textrm{D} + _{1}^{2}\textrm{D} \rightarrow _{2}^{4}\textrm{He} + _{1}^{1}\textrm{H} + _{1}^{1}\textrm{H} + 12.86MeV$

85% of the time it will follow this path.

The other 15% of the time, we have deuteron combining with existing 4-Helium: D + 4He -> 7Be + gamma because in the core there are enough 4He already produced during the first 4.5 billion years to take this path in the core.

So, to summarize:

2 times reaction (1), 2 times reaction (2) and one time reaction (3) is the p-p chain which happens 85% of the time.

That means 4 protons into one 4-Helium 2 positrons and 2 neutrinos with a total energy output of 26.7 MeV: 4p -> 4He + 2positrons+ 2neutrinos (26.7MeV)

The other paths also called ppII and ppIII in the remaining 15% also produce 4-Helium but I’ll leave them for the moment. Here is the 3 complete pp chains, of which I mainly talked about ppI in this post:

The only direct experimental evidence of this process are the neutrinos produced, why?

Because the energy we are getting in the form of “sunlight” and that is observed with spectroscopy is not coming from the fusion in the core! Fusion takes place inside the core and from there the energy goes through the radiative zone and the convective zone and then to photosphere.

The photosphere transparent to electromagnetic waves. The photons from the photonsphere they cross the photosphere and come to the earth but the passing from the core to the phostosphere may take some million years. So what we are receiving is only remotely connected to the fusion inside the core.

In the radiative zone, photons are emitted, absorbed than reemitted, and so on with different energies and frequencies. so what reach the photoshpere is very different from what was produced inside the core. So the IR and UV light we receive is not only very different than what is issued from the core it is also very old. So how do I know that fusion is still going on in the core because if it stops today, it will take one million year before we notice… These neutrinos very weakly interact with matter, million and million of neutrinos keep going through our bodies every seconds and we do not notice anything. However, scientists have built huge neutrinos detectors and instead of getting answers, they were puzzled by the experimental results. It took until 2001 to actually get an answer on the missing solar neutrinos.