> For the complete documentation index, see [llms.txt](https://slowdiveptg.gitbook.io/notes/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://slowdiveptg.gitbook.io/notes/stellar-evolution/chapter-9.-nuclear-reactions.md).

# Chapter 9. Nuclear Reactions

## Basic Considerations

Normally in a sun-like star, $\ce{4^1H -> ^4He}$. Here in nuclear physics, people use atomic mass unit (amu).

$$
1\text{ amu}=\frac1{12}m\left(\ce{^{12}C}\right)
$$

$m\left(\ce{^{1}H}\right)\simeq1.0081$ amu, $m\left(\ce{^{4}He}\right)\simeq4.0039$ amu, so the mass deficiency $\Delta m$ in this reaction is $0.0285$ amu. It will be transported into energy

$$
\Delta E=\Delta mc^2\simeq4.3\times10^{-5}\text{ erg}\simeq 27\text{ MeV}\sim10^{10}\text{ K}
$$

* The rest mass (energy) for important particles
  * Electron: $\sim 0.5$ MeV
  * Pion: $\sim10^2$ MeV
  * Proton: $\sim1$ GeV

Generally speaking, the nuclear binding energy $E\_B$ of a nucleus is

$$
E\_B=\left\[(A-Z)m\_\text{n}+Zm\_\text{p}-M\_\text{nuc}\right]c^2
$$

Where $Z$ is the proton number, $A$ is the atomic number, $m*\text{u}$, $m*\text{p}$, and $M\_\text{nuc}$ are neutron mass, proton mass, and nuclear mass, respectively.

Let us define the **average binding energy per nucleon**

$$
\epsilon\_B\equiv\frac{E\_B}A
$$

![](https://1509032923-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MPMxe8Bu9WDT3p-DA_8%2Fsync%2F1cb6e588ba654261b8c2c45cef30271d60960f2b.png?generation=1608873193189170\&alt=media)

The $\epsilon\_B-A$ curve peaks at $\ce{^{56}Fe}$, where $\epsilon\_B=8.5$ MeV. For $A<56$, nuclear reactions gain energy. Fusion for heavier elements requires energy.

**Note**

* It is interesting that $\ce{^4He}$ has a $\epsilon\_B=6.6$ MeV, far above the smoothed fitting function. From $\ce{^1H}$ to $\ce{^{56}Fe}$, $\epsilon\_B=8.5$ MeV, but $6.6$ MeV is already gained in the generation of $\ce{^4He}$.
* Nuclear fusion in stats produce heavy elements up to $\ce{^{56}Fe}$. Elements with $A>56$ are produced in violent explosions such as SNe, compact stars mergers, etc.

## Two-Body Nuclear Reaction Processes

### Reaction Rate

The number of reaction between species $i$ and $j$ per unit time in unit volume can be approximated as

$$
R\_{ij}=n\_in\_j\sigma v\ \[\text{1/s/cm}^3]
$$

where $n\_i,n\_j$ correspond to number desities, $\sigma$ denotes the cross section, and $v$ is the relative velocity. If we assume all nuclei are in thermal equilibrium so that the velocity magnitude $v$ obeys a Maxwell-Boltzmann distribution. The density function is

$$
f(v)\text dv=4\pi v^2\left(\frac{m}{2\pi k\_BT}\right)^{3/2}\exp\left(-\frac{mv^2}{2k\_BT}\right)\text dv
$$

If we define $E=\frac12mv^2$, we have

$$
f(E)\text dE=\frac2{\sqrt{\pi}}E^{1/2}\left(k\_BT\right)^{-3/2}\exp\left\[-\frac E{k\_BT}\right]\text dE
$$

where $m=m\_im\_j/(m\_i+m\_j)$ is the reduced mass. So statistically we can calculate the average $\sigma v$

$$
\begin{align\*}
\left\langle\sigma u\right\rangle&=\int\_0^\infty\sigma vf(v)\text dv\\
&=\int\_0^\infty\sigma(E)\cdot\left(\frac{2E}m\right)^{1/2}\cdot f(E)\text dE\\
&=\left(\frac{8}{\pi m}\right)^{1/2}\left(k\_BT\right)^{-3/2}\int\_0^\infty \sigma(E)\cdot E\exp\left\[-\frac E{k\_BT}\right]\text dE
\end{align\*}
$$

Since nuclei have positive charge, we need to consider the long-distance, repulsive force - Columb force. The **Columb barrier** is

$$
V\_C(r)=\frac{Z\_1Z\_2 e^2}{r}
$$

On the other hand, nuclear matters interact with short-distance, strong attractive force when

$$
r\<r\_0\simeq1.44\times10^{-13} A^{1/3}\text{ cm}\sim\text{fm}
$$

![](https://1509032923-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MPMxe8Bu9WDT3p-DA_8%2Fsync%2F0d8194796547acccc0718fb0fe53f35b63d23f52.png?generation=1608873192001976\&alt=media)

In a classical scenario, when the kinetic energy of partcles reaches $E\_\text{max}=V\_C(r\_0)\sim Z\_1Z\_2$ MeV, they are able to overcome the Columb barrier. But in this way crtical temperature must be over $\sim10^{10}$ K, far above the central temeperature of sun-like stars ($\sim10^{10}$ K).

### Tunneling Effect

Thanks to the quantum physics, we know that as long as the size of the potential well is finite, there can be **tunneling effects** against the well. In the stellar core, $k&#x54;*\text{c}\ll E*\text{max}$. Even so, the quantum tunneling allows penetration under a small, yet finite, probability, which is given by

$$
P(E)\sim\exp\left\[-\int\_{r\_0}^{r\_c}\frac{\sqrt{2m\[V(r)-E]}}{\hbar}\text dr\right]\equiv\exp(-2\pi\eta)
$$

Here $r\_c$ is defined so that $V(r\_c)=E$. Here $\eta$ is given by

$$
\eta=\left(\frac m2\right)^{1/2}\frac{Z\_1Z\_2 e^2}{\hbar E^{1/2}}
$$

We can understand this integration result in an alternative way. Considering $V\_C(\lambda)$, where $\lambda$ is the de Broglie length,

$$
\lambda\equiv\frac hp
$$

Then

$$
\frac{V\_C(\lambda)}{E}=\frac{Z\_1Z\_2 e^2}{h}\frac pE=\frac{Z\_1Z\_2 e^2}{h}\sqrt{\frac {2m}E}=\frac\eta\pi
$$

Thus $\eta$ corresponds to the ratio between the Columb potential and the particle specific energy,

$$
P(E)\sim\exp\left(-2\pi^2\frac{V\_C(\lambda)}{E}\right)
$$

Increasing $E$ and decreasing $Z\_1Z\_2$ (lighter particles) help promote the tunneling probability.

### Cross Section

The cross section is given by

$$
\sigma(E)\simeq\pi\lambda^2P(E)\xi(E)
$$

where $\lambda$ is the de Broglie length, $P(E)$ is the tunneling probability, and $\xi(E)$ is the **resonance**. The form of $\xi(E)$ is rather complicated (sort of a Lorentz profile), but when $E$ is far away from any resonance, $\xi\to1$. Anyway, we define

$$
\sigma(E)\simeq S(E)\frac{\exp\left\[{-b/\sqrt{E}}\right]}{E}
$$

$S(E)$ is known as the **astrophysical S-factor**. It is measured by nuclear physicists. Luckily, it depends only weakly on $E$. So we can treat it as a constant of $E$.

### Reaction Rate

Finally we go back to the integral

$$
\begin{align\*}
\left\langle\sigma u\right\rangle
&=\left(\frac{8}{\pi m}\right)^{1/2}\left(k\_BT\right)^{-3/2}\int\_0^\infty \sigma(E)\cdot E\exp\left\[-\frac E{k\_BT}\right]\text dE\\
&\simeq\left(\frac{8}{\pi m}\right)^{1/2}\left(k\_BT\right)^{-3/2}S\_0\int\_0^\infty \exp\left\[-\frac E{k\_BT}-\frac{b}{\sqrt{E}}\right]\text dE\\
\end{align\*}
$$

The integrand $f(E)$ vanishes either at large or small $E$, and only significantly contributes to the integration around its maximum.

Define $E\_0$, so that

$$
f'(E)=\frac{\text d}{\text dE} \left\[-\frac E{k\_BT}-\frac{b}{\sqrt{E}}\right]=0
$$

We can solve $E\_0$

$$
\begin{align\*}
E\_0=\left(\frac b2k\_BT\right)^{2/3}=\left\[\left(\frac{m}{2}\right)^{1 / 2} \pi \frac{Z\_{i} Z\_{k} e^{2} k\_B T}{\hbar}\right]^{2/3}
\end{align\*}
$$

Around $E\_0$, we can expand $f(E)$ into Tyler's series

$$
\begin{align\*}
f(E)&=f(E\_0)+\frac12f''(E\_0)(E-E\_0)^2+\mathcal O(E^3)\\
&=-\frac{3E\_0}{k\_BT}-\frac{3E\_0}8E\_0^{-3/2}b\left(\frac E{E\_0}-1\right)^2+\mathcal O(E^3)\\
&=-\frac{3E\_0}{k\_BT}-\frac{3E\_0}{4k\_BT}\left(\frac E{E\_0}-1\right)^2+\mathcal O(E^3)\\
&\equiv-\tau-\frac\tau4\left(\frac E{E\_0}-1\right)^2+\mathcal O(E^3)
\end{align\*}
$$

Thus the integration is approximately

$$
\begin{align\*}
\left\langle\sigma u\right\rangle
&\simeq\left(\frac{8}{\pi m}\right)^{1/2}\left(k\_BT\right)^{-3/2}S\_0\int\_0^\infty\exp\left\[-\tau-\frac\tau4\left(\frac E{E\_0}-1\right)^2\right]\text dE\\
&\simeq\left(\frac{8}{\pi m}\right)^{1/2}\left(k\_BT\right)^{-3/2}S\_0\text e^{-\tau}\int\_0^\infty\exp\left\[-\frac\tau4\left(\frac E{E\_0}-1\right)^2\right]\text dE\\
&\equiv \left(\frac{8}{\pi m}\right)^{1/2}\left(k\_BT\right)^{-3/2}S\_0\cdot\frac{2}{3} k\_B T\cdot \tau^{1 / 2} \mathrm{e}^{-\tau} \int\_{-\sqrt{\tau }/2}^{\infty} \mathrm{e}^{-\xi^{2}} d \xi
\end{align\*}
$$

where

$$
\xi\equiv\frac{\sqrt\tau}2\left(\frac E{E\_0}-1\right)
$$

The main contribution to $\langle\sigma u\rangle$ comes from a range close to $\xi=0$, so that no large errors are introduces when extending the range of integration to 􏰠$-\infty$, the integral over the Gaussian becoming 􏰕$\sqrt\pi$.

Therefore, one can obtain

$$
\left\langle\sigma u\right\rangle=\frac{4}{3} \left(\frac{2}{ mk\_BT}\right)^{1/2}S\_0 \tau^{1 / 2} \mathrm{e}^{-\tau}
$$

Recall that

$$
\tau=\frac{3E\_0}{k\_BT}\simeq20\left(Z\_1^2Z\_2^2A\right)^{1/3}T\_7^{-1/3}\propto T^{-1/3}
$$

$$
\Rightarrow\left\langle\sigma u\right\rangle\propto T^{-2/3}\exp\left(-cT^{-1/3}\right)
$$

Thus for the solar case, $\tau\sim20$, and

$$
\frac{\partial\ln\langle\sigma u\rangle}{\partial\ln T}=\frac\tau3-\frac23\sim6-7
$$

### Energy Generation Rate

The energy generation rate (per unit mass) between species $i$ and $j$ is simply

$$
\varepsilon\_\text{nuc}=\frac{Q\_{ij}R\_{ij}}{\rho}\propto\rho\langle\sigma u\rangle
$$

where $Q\_{ij}$ is the binding energy to release.

Define

$$
\nu\equiv\frac{\partial\ln\varepsilon\_\text{nuc}}{\partial\ln T}=\frac\tau3-\frac23\sim6-7
$$

The temperature dependence is really strong!

For heavier elements, since $\tau\propto\left(Z\_1^2Z\_2^2A\right)^{1/3}$, $\nu$ can be extremely high.

## Nuclear Burning Mechanisms

### pp-chain (in the Sun)

**pp-chain I**

* $\ce{^1H + ^1H -> ^2H + e+ +\nu\_e}$

  This reaction is slow. Once there is neutrino generation, weak interaction is involved, and the efficiency is much lower than other reactions.

  **Notes**

  In Big Bang nucleosynthesis (BBN), it causes the **deutron ($\ce{^2H}$) bottleneck**. All other reactions wait for the generation of $\ce{^2H}$.
* $\ce{^2H + ^1H -> ^3He + \gamma}$
* $\ce{^3He + ^3He -> ^4He + 2^1H}$

As a result, a positron and a gamma-photon are ejected to heat the gas up. The neutrino just escapes without any interaction.

**pp-chain II** and **pp-chain III** are rather complicated, and to form $\ce{^4He}$. $\ce{Li, Be, B}$ are also involved. The overall energy generation rate is

$$
\varepsilon\_\text{pp}=4.4\times10^5\rho X^2T^{-2/3}\_7\exp\left(-\frac{15.7}{T^{1/3}\_7}\right)\text{ erg/s/g}
$$

where $X$ is the abundance of hydrogen.

![](https://1509032923-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MPMxe8Bu9WDT3p-DA_8%2Fsync%2F353557de2dce8432f95c47b4dbedbfcdbc77168d.gif?generation=1608873193319639\&alt=media)

### CNO Cycle ($M>2M\_\odot$)

The CNO cycle is composed of **CNO-I (CN cycle)** and **CNO-II (CNO cycle)**. The overall energy generation rate is

$$
\varepsilon\_\text{CNO}=1.7\times10^{27}\rho XX\_\text{CNO}T^{-2/3}\_7\exp\left(-\frac{70.5}{T^{1/3}\_7}\right)\text{ erg/s/g}
$$

where $X\_\text{CNO}$ is the sum of carbon, nitrogen, and oxygen abundance.

![](https://1509032923-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MPMxe8Bu9WDT3p-DA_8%2Fsync%2Fde4039eca4cb9cb81ba401e4a9d3a554fd268ed1.png?generation=1608873193346583\&alt=media)