# The P-P Chain

This the second post in a series about the fundamentals equations for modeling astrophysics. I'm not sure how long this series will end up being, but you can see the whole list here.

The P-P chain is the system in which most main-sequence stars produce energy vis-a-vis hydrogen burning. This post mostly summarizes that process, while not covering more complex topics like Helium burning.

## 1 Background

Ooooh boy, if you're not familiar with quantum physics or the Coulomb barrier, strap in, you're in for a doozy. If you've ever watched any science documentary about stars, they'll tell you the star spends most of it's life burning hydrogen in the main sequence phase. What they don't tell you is that this burning actually occurs as pretty low temperatures, but is actually dependent on the wave-like properties of atoms.

Alright I sorta lied about how approachable this topic is, but if you're not familiar with the wave-particle duality, for the purposes of this post you just need to know that at very tiny scales, particles act very much like waves. Check out the Wikipedia post if you're interested in knowing more.

Let's take two atomic nuclei: particle A with charge $$Z_A$$ and mass $$m_A$$ and particle B with charge $$Z_B$$ and mass $$m_B$$. On classical scales, these nuclei are guided by the repulsive Coulomb potential, that is: $$\frac{Z_A Z_B e^2}{4 \pi \epsilon_0 r}$$. Effectively what this means is that they either repel or attract each other based on their charge, and the force of this is determined by the radius between them, a la gravity-style. However, if we dig down to a fermi (the measure of distance with 10-15 m), there becomes a strong nuclear potential on them, attracting each other. But the meddlesome parent of classical physics intervenes as well, creating a Coulomb Barrier that prevents the nuclei from fusing. This barrier not only prevents the nuclei from fusion, it also causes them to bounce back once the nuclei hit the barrier, which is described by the classical mechanics in which kinetic energy (energy of approach) is converted into potential energy (energy of fusion), until the nuclei hit the Coulomb barrier. Think of it as an "energy" hill, as the nuclei approach each other, the energy the nuclei create must surpass this hill in order to create fusion, shown below:

If the energy generated from the two particles attracting is surmounts the Coulomb barrier, the nuclei successfully enable fusion. However, it's important to note that the height is this barrier is far greater than the energies generated from the nuclei is stars. The equation for the energy required for fusion is given by:

$$\begin{gathered} E = \frac{Z_A Z_B e^2}{4 pi \epsilon_0 r} \approx \frac{1.4 Z_A Z_B}{(r \text{ in fermis)}} \text{MeV} \end{gathered}$$

But the energy generated from inside a star is on the magnitudes of keV, not MeV. Furthermore, only an extremely small subset of nuclei are even able to obtain energies on the magnitude of MeV at all. So how do stars do it?

The answer is quantum tunneling, which relies on the wave-particle duality. Let us assume the energy curve is simply a "barrier" that the particle must surpass, shown below:

Due to the wave-particle duality, we can model it as a sinusoidal wave (you can move the demo around too!):

There's three things going on:

1. First, the particle has a wave function, represented by the blue sinusoidal function at the right.
2. Second, as the particle passes through the grided area (the "barrier" of the energy curve), it decays.
3. Third, it exits the barrier with a significantly smaller wave function.

What's happening here is that even though it's forbidden in classical physics for the nuclei to pass through the barrier, in quantum physics, the particle can simply"tunnel" through the energy barrier, reaching the forbidden region previous unaccessible to it. There's a lot more going on than just that, but that's a post for another time.

So now we've gotten the general idea of how fusion works, we can start talking about the two types of fusion occurring in the main sequence phase of stars: the P-P chain (I'll talk about the CNO cycle at another time).

## 2 The Proton - Proton Chain

Despite the name P-P chain (go on, giggle, we've all done it), the P-P chain is the the fusion of protons through a weak nuclear reaction.

$$\begin{gathered} p + p \rightarrow d + e^+ + \nu_e \end{gathered}$$

The mechanism for this is two protons fusing to become a deuteron, a spare positron, and a electron neutrino. The underlying mechanism is that one the protons participating in the fusion undergoes inverse beta decay, $$p \rightarrow n + e^+ \nu_e$$, then the neutron produced is bound to the other proton to become a deuteron.

The key feature to the the P-P chain is that the process itself is very slow. However, despite the slowness of the reaction, it paves the way for 3 different reactions:

### 2.1 Branch 1

$$\begin{gathered} p + p \rightarrow d + e^+ + \nu_e \\ p + pd \rightarrow {}^{3}\text{He} + \gamma \\ {}^{3}\text{He} + {}^{3}\text{He} \rightarrow {}^{4}\text{He} + 2p \end{gathered}$$

This is the first, and simplest brach of the P-P chain. We have the initial proton-proton reaction, followed by another proton interacting with the deuteron just created, which in turn creates $${}^{3}\text{He}$$ and a photon. Two $${}^{3}\text{He}$$ fuse to form $${}^{4}\text{He}$$ and 2 spare protons, and the process begins again. This branch releases an effective energy (energy not carried away by neutrinos) of $$26.2 \text{MeV}$$, occupying $$85\%$$ of reactions in our sun.

### 2.2 Branch 2

$$\begin{gathered} p + p \rightarrow d + e^+ + \nu_e \\ p + pd \rightarrow {}^{3}\text{He} + \gamma \\ {}^3\text{He} + {}^4\text{He} \rightarrow {}^7\text{Be} + \gamma \\ e^- + {}^7\text{Be} \rightarrow {}^7\text{Li} + \nu_e \\ p + {}^7\text{Li} \rightarrow {}^4\text{He} + {}^4\text{He} \end{gathered}$$

Branch 2 has a little more going on. After the initial formation of $${}^3\text{He}$$, it fuses with an $${}^4\text{He}$$ created from branch 1, forming $${}^7\text{Be}$$ (Beryllium 7) and a photon. This Beryllium then interacts with a stray electron, creating $${}^7\text{Li}$$ (Lithium) and an electron neutrino. The Lithium proceeds further to react with a spare proton, splitting itself into two Helium isotopes. Branch 2 creates $$25.7$$ MeV, and accounts for $$15\%$$ of reactions in our sun.

### 2.3 Branch 3

$$\begin{gathered} p + p \rightarrow d + e^+ + \nu_e \\ p + pd \rightarrow {}^{3}\text{He} + \gamma \\ {}^3\text{He} + {}^4\text{He} \rightarrow {}^7\text{Be} + \gamma \\ p + {}^7\text{Be} \rightarrow {}^8\text{B} + \gamma \\ {}^8\text{B} \rightarrow {}^8\text{Be}^* + e^+ \nu_e \\ {}^8\text{Be}^* \rightarrow {}^4\text{He} + {}^4\text{He} \end{gathered}$$

Branch 3 is the black sheep of the P-P chain family, creating only $$19.1$$ MeV and accounting for only $$0.02 \%$$ of all reactions within our sun. It begins the same way as brach 2, except the Beryllium binds with another proton to form Boron, which then decays back into Beryllium 8. Through the alpha decay process, Beryllium 8 decays back into two stable Helium isotopes.

Because all the branches from the P-P chain start from the initial proton-proton fusion, the proton-proton fusion governs the rate of energy release in the chain as a whole.

With that, we've discovered the reactions of hydrogen burning that account for main sequence stars whose masses are comparable to our sun. The P-P chain fails to account for the reactions of much larger stars, and that's because larger stars are dominated by the CNO cycle, which I'll talk about in my next post.

Posted: 2016-09-06
Filed Under: Astrophysics, Fundamentals