Along the pp-chain homologous sequence, $R$ is constant and therefore $T_c\propto M$. Normalizing to the Sun gives the transition mass
$$
\boxed{M_{\rm tr}\simeq M_\odot
\frac{2.0\times10^7}{1.5\times10^7}
=\frac43M_\odot}.
$$

For the <CNO cycle> law $\epsilon\propto\rho T^{23/2}$, the nuclear scaling becomes
$$
L_{\rm nuc}\propto M\rho_cT_c^{23/2}
\propto M^{27/2}R^{-29/2}.
$$
The opacity law is unchanged, so $L_{\rm rad}\propto M^{11/2}R^{-1/2}$. Equating them gives
$$
M^8R^{-14}=\text{constant},
\qquad
\boxed{R\propto M^{4/7}},
$$
and then
$$
\boxed{L\propto M^{11/2}R^{-1/2}
\propto M^{73/14}}.
$$

The <effective temperature> follows from the <Stefan–Boltzmann law> $L=4\pi R^2\sigma T_e^4$. Thus the pp branch has $T_e\propto M^{11/8}$ and $L\propto T_e^4$, whereas the CNO branch has
$$
T_e\propto M^{57/56},
\qquad
L\propto T_e^{292/57}.
$$
On a <Hertzsprung-Russell diagram>, both branches rise toward larger luminosity and, conventionally, leftward toward larger temperature. They join near $4M_\odot/3$; the CNO branch has the steeper $L$--$T_e$ logarithmic slope and lies to the cooler side of the extrapolated constant-radius branch at fixed luminosity.

If $X$ is the <hydrogen mass fraction> and $X_{\rm CNO}$ the combined carbon-nitrogen-oxygen mass fraction, the reaction-pair and catalyst abundances give approximately
$$
\boxed{\epsilon_{pp}\propto X^2\rho T^{7/2}},
\qquad
\boxed{\epsilon_{\rm CNO}\propto XX_{\rm CNO}\rho T^{23/2}}.
$$
The CNO nuclei are catalysts, so their abundance multiplies the CNO-cycle rate rather than being consumed by the completed cycle.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-317/1/ii/solution