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).

$m\left(\ce{^{1}H}\right)\simeq1.0081$ 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

$$\epsilon_B\equiv\frac{E_B}A$$

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

$$r<r_0\simeq1.44\times10^{-13} A^{1/3}\text{ cm}\sim\text{fm}$$

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, $kT\text{c}\ll E\text{max}$. 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.

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.