> [!key-idea] Equation > **Energy Generated (per unit volume):** > $ >\epsilon_{\rm v} = Q_{\rm nuc} \, n_{A} \, n_{B} \, \sigma v \left( \frac{1}{1 + \delta_{AB}} \right) \quad \propto \quad Q_{\rm nuc} \, n_{A} \, n_{B} \, S(E_{0}) \; T_{6}^{-2/3} \, e^{-B T_{6}^{-1/3}} > $ > **Energy Generated (per unit mass):** > - *used in the [[Stellar Structure#Conservation of Energy|Conservation of Energy]] Stellar Structure Equation* > $ > \epsilon_{\rm m} = \frac{\epsilon_{\rm v}}{\rho} \quad \propto \quad Q_{\rm nuc} \, \rho X_{A} \, X_{B} \, S(E_{0}) \; T_{6}^{-2/3} \, e^{-B T_{6}^{-1/3}} > \hspace{1cm} \text{where} \hspace{1cm} X_{i} \equiv \text{mass fraction} > $ > **where** $A \equiv 1$ and $B \equiv 2$ > $ > T_{6} = \frac{T}{10^{6} \; {\rm K}} \hspace{1.5cm} > \mathcal{A} = \frac{A_{1} A_{2}}{A_{1} + A_{2}} \hspace{1.5cm} > B = 42.6 \,\left( Z_{1}^{2} Z_{2}^{2} \,\mathcal{A} \right)^{1/3} \; {\rm keV} > $ *What is the energy generation rate (per unit volume) for a given nuclear reaction?* $A + B \longrightarrow C + D$ If we have two sets of particles ($A\equiv 1$ and $B \equiv 2$), we can create a system to determine the number of interactions these two sets will have that can generate particles $C$ and $D$. For both particles types, their relative velocity (with respect to each other), $v_{\rm rel} \equiv v$, will follow a [[Statistical Mechanics#Maxwell-Boltzmann|Maxwellian]] distribution, $f(v)$. Using the kinetic energy relationship $E = \frac{1}{2} \mu v^{2}$, we can express this velocity distribution in terms of energy. $f(v) = 4 \pi v^{2} \, \left(\frac{\mu}{2 \pi k_{\rm B} T}\right)^{3/2} \exp \left( - \frac{\mu v^{2}}{2 k_{\rm B} T} \right)$ $\Downarrow$ $f(E) = \left(\frac{4 \beta^{3} E}{\pi}\right)^{1/2} \exp \left( - \beta E \right) \hspace{1cm} \text{where} \hspace{1cm} \beta = \frac{1}{k_{\rm B} T}$ Taking $\sigma(E)$ as the interaction cross section between the two nuclei approaching at energy $E$... $ \sigma(E) = \underbrace{\frac{1}{E} \;\exp \left[ - \frac{b}{\sqrt{E}} \right]}_{\text{quantum mechanics}} \; \underbrace{S(E)}_{\text{nuclear physics}} \hspace{1cm} \text{where} \hspace{1cm} \begin{aligned}[t] &b = \frac{1}{2 \epsilon_{0} \hbar} \left(\frac{\mu}{2}\right)^{1/2} Z_{1} Z_{2} e^{2} \\ \\ &\textcolor{gray}{S(E) = \text{slow-varying function, empirical}} \end{aligned} $ ...we can integrate over the [[Statistical Mechanics#Maxwell-Boltzmann|Maxwell]] distribution for average of all relative velocities. $ \begin{align} \langle \sigma v \rangle &= \int_{0}^{\infty} \sigma(E) \, v(E) \, f(E) \; \mathrm{d} E \\ &= \int_{0}^{\infty} \left[ \frac{S(E)}{E} \; e^{- \frac{b}{\sqrt{E}}} \right] \, \left[ \frac{2 E}{\mu} \right]^{1/2} \, \left[ \left(\frac{4 \beta^{3} E}{\pi}\right)^{1/2} e^{- \beta E} \right] \; \mathrm{d} E \\ &= \sqrt{\frac{8}{\pi \mu (k_{\rm B} T)^{3}}} \int_{0}^{\infty} S(E) \; \underbrace{e^{-E/k_{\rm B} T}}_{\rm decaying} \, \underbrace{e^{- b/\sqrt{E}}}_{\rm growing} \; \mathrm{d} E \\ \end{align} $ In this integral, the only significant contribution is near the "Gamow Peak" due to the product of decaying and growing exponentials. ![[gamow_peak.png|align:center|300]] Therefore, the only negligible range is near $E=E_{0}$ ($E_{0} \gg k_{B} T$) such that we can take $S(E) \simeq S(E_{0})$ and pull it out of the integral. $ \begin{align} \langle \sigma v \rangle &= \sqrt{\frac{8}{\pi \mu (k_{\rm B} T)^{3}}} \int_{0}^{\infty} S(E) \; e^{-E/k_{\rm B} T} \, e^{- b/\sqrt{E}} \; \mathrm{d} E \\ &= \sqrt{\frac{8}{\pi \mu (k_{\rm B} T)^{3}}} \, S(E_{0}) \;\underbrace{\int_{0}^{\infty} e^{-E/k_{\rm B} T} \, e^{- b/\sqrt{E}} \; \mathrm{d} E}_{I \, \equiv \, \text{integrand}} \end{align} $ The maximum of this integrand occurs at the effective mean energy ($E_{0}$) for fusion reactions at temperature $T$. $ e^{E/k_{\rm B} T} = e^{-b/\sqrt{E}} \hspace{1cm} \Rightarrow \hspace{1cm} E_{0} = \left( \frac{1}{2} b k_{\rm B} T \right)^{2/3} \textcolor{gray}{= 1.22 \; \left( Z_{1}^{2} Z_{2}^{2} \,\mathcal{A}\, T_{6}^{2} \right)^{1/3} \; {\rm keV}} $ $\hspace{1cm} \text{where} \hspace{1cm} \mathcal{A} = \frac{A_{1} A_{2}}{A_{1} + A_{2}} \quad {\rm and} \quad T_{6} = \frac{T}{10^{6} \; {\rm K}}$ $\Downarrow$ $I_{\rm max} = I(E_{0}) = e^{-3 E_{0} / k_{\rm B} T} \textcolor{gray}{\; = e^{- B T_{6}^{-1/3}} \hspace{1cm} \text{where} \hspace{1cm} B = 42.6 \,\left( Z_{1}^{2} Z_{2}^{2} \,\mathcal{A} \right)^{1/3} \; {\rm keV}}$ Approximating this distribution as a Gaussian, we can solve this integral by... $ \begin{align} \int_{0}^{\infty} \left[ e^{-E/k_{\rm B} T} \, e^{- b/\sqrt{E}} \right] \; \mathrm{d} E \; &\simeq \int_{0}^{\infty} I_{\rm max} \exp \left[ - \left(\frac{E - E_{0}}{\Delta E_{0} / 2}\right)^{2} \right] \; \mathrm{d} E \\ &= I_{\rm max} \int_{0}^{\infty} \exp \left[ - \left(\frac{E - E_{0}}{\Delta E_{0} / 2}\right)^{2} \right] \; \mathrm{d} E \\ &= I_{\rm max} \left( \sqrt{\frac{\pi}{4}} \, \Delta E_{0} \right) \textcolor{gray}{\hspace{1cm} \text{where} \hspace{1cm} \Delta E_{0} = 4 \sqrt{\frac{E_{0} k_{\rm B} T}{3}}} \end{align} $ $\Downarrow$ $\langle \sigma v \rangle = \sqrt{\frac{2}{\mu}} \left(\frac{\Delta E_{0}}{ (k_{\rm B} T)^{3/2}}\right) \; S(E_{0}) \; \exp \left( - \frac{3 E_{0}}{k_{\rm B} T} \right)$ Applying this to the two number densities ($n_{A}$ and $n_{B}$), we can find the interaction rate *per volume* between $A$ and $B$. This is known as the **nuclear reaction rate**. $r_{AB} = \frac{\text{\# of interactions}}{\text{volume} \times \text{time}} = n_{A} \, n_{B} \, \langle \sigma v \rangle \hspace{1cm} \text{where} \hspace{1cm} (A \ne B)$ If $A$ and $B$ are the same species of particle, then we need to divide by 2 to avoid double counting the interactions. $r_{AB} = n_{A} \, n_{B} \, \langle \sigma v \rangle \, \left(\frac{1}{1 + \delta_{AB}}\right) \hspace{1cm} \text{where} \hspace{1cm} \delta_{AB} = \begin{cases} 1 &{\rm if} \; A = B \\ 0 &{\rm if} \; A \ne B \end{cases}$ Then, if we take the energy generated in per interaction to be... $Q_{\rm nuc} = \text{energy of RHS - energy of LHS} = \left( m_{\rm C} + m_{\rm D} \right) c^{2} - \left( m_{\rm A} + m_{\rm B} \right) c^{2}$ ...we can express the energy generated (per unit volume) as: $ \boxed{\quad \epsilon_{\rm v} = Q_{\rm nuc} \, n_{A} \, n_{B} \, \langle \sigma v \rangle \left( \frac{1}{1 + \delta_{AB}} \right) \quad \propto \quad Q_{\rm nuc} \, n_{A} \, n_{B} \, S(E_{0}) \; T_{6}^{-2/3} \, e^{-B T_{6}^{-1/3}}\quad} $ > [!sun] Comparison of Burning Processes > > Below is a comparison of the main $\ce{^{4}He}$ production processes... > - [[Proton-Proton Chain]] (PP-Chain) > - Energy generation Rate: $\varepsilon \; \propto \; T^{4}$ > - [[Carbon-Nitrogen-Oxygen Cycle]] (CNO Cycle) > - Energy generation Rate: $\varepsilon \; \propto \; T^{20}$ > - [[Triple-Alpha Process]] (Triple-$\alpha$ Process) > - Energy generation Rate: $\varepsilon \; \propto \; T^{40}$ > > ![[stellar_nucleosynthesis.svg.png|align:center|500]] > > Important Notes: > - The sun's core temperature is $T_{\rm c, \odot} \approx 1.5 \times 10^{7} \; {\rm K}$. > - In the sun, about $\sim 99 \%$ of the energy produced is through the [[Proton-Proton Chain|pp chain]] while the other $\sim 1 \%$ is produced through the [[Carbon-Nitrogen-Oxygen Cycle|CNO cycle]]. > - The cross over between the [[Proton-Proton Chain|pp chain]] and the [[Carbon-Nitrogen-Oxygen Cycle|CNO cycle]] is at $T \sim 2 \times 10^{7} \; {\rm K}$. > - [[Triple-Alpha Process|Triple-alpha]] is highly temperature sensitive and explosive. Ignition happens near $T\sim 10^{8}\;{\rm K}$ ^nuclearBurning-comparison