= Solution
In <stellar homology>, the dimensionless radial profiles are the same after scaling radius, <enclosed mass>, <pressure>, <temperature> and <luminosity>. At corresponding radii let $r'=\xi r$, $m_r'=m m_r$, $T'=tT$, $P'=pP$ and $L_r'=lL_r$. The printed $R'=\xi r$ is understood as this local radial scaling, with surface radius $R'=\xi R$.
Let $q=\rho'/\rho$ and keep composition, <opacity> coefficient and nuclear coefficient fixed first. <Mass conservation> gives $q=m/\xi^3$, and <hydrostatic equilibrium> gives $p=mq/\xi=m^2/\xi^4$. The <ideal gas> law then gives $t=p/q=m/\xi$. Scaling the <stellar energy-generation rate> equation and the <stellar radiative temperature gradient> gives respectively
$$
l=\xi^3q^2t^\eta=m^{\eta+2}\xi^{-(\eta+3)},\qquad
l=\xi t^{4-\nu}q^{-(\lambda+1)}
=m^{3-\lambda-\nu}\xi^{3\lambda+\nu}.
$$
The second follows from $dT/dr\propto-\kappa\rho L_r/(r^2T^3)$, not from energy production. Equating the two <luminosity> scalings gives
$$
\boxed{R\propto M^X,\quad
X=\frac{\eta+\lambda+\nu-1}{\eta+3\lambda+\nu+3},\qquad
L\propto M^Y,\quad Y=\eta+2-(\eta+3)X.}
$$
This assumes the denominator is nonzero and a consistent homologous family exists. If the <mean molecular weight> differs by $u=\mu'/\mu$, then $t=um/\xi$ and
$$
\xi^{\eta+3\lambda+\nu+3}
=u^{\eta+\nu-4}m^{\eta+\lambda+\nu-1}.
$$
Ratios of the <opacity> and energy-generation coefficients multiply the right-hand side. Thus fixed composition is a real restriction, not an automatic property of every stellar sequence.
Use the <Stefan–Boltzmann law> $L=4\pi R^2\sigma T_{\rm eff}^4$ to locate the family on a <Hertzsprung-Russell diagram>. It implies $T_{\rm eff}\propto M^{(Y-2X)/4}$ and
$$
\boxed{\frac{d\log L}{d\log T_{\rm eff}}=\frac{4Y}{Y-2X}.}
$$
For <proton–proton chain> burning take the usual local approximation $\eta=4$, with <Kramers opacity law> $\lambda=1$, $\nu=-7/2$. Then
$$
\boxed{R\propto M^{1/13},\quad L\propto M^{71/13},\quad
T_{\rm eff}\propto M^{69/52},\quad
\frac{d\log L}{d\log T_{\rm eff}}=\frac{284}{69}\simeq4.12.}
$$
For the <CNO cycle> with <electron-scattering opacity>, $\lambda=\nu=0$, so
$$
\boxed{R\propto M^{(\eta-1)/(\eta+3)},\quad L\propto M^3,\quad
\frac{d\log L}{d\log T_{\rm eff}}=\frac{12(\eta+3)}{\eta+11}.}
$$
A conventional local choice $\eta=16$ gives $R\propto M^{15/19}$ and slope \b[$76/9\simeq8.44$]. Choosing $\eta=17$ instead gives slope $60/7\simeq8.57$. The source specifies no numerical nuclear exponents, and the effective exponent changes with <temperature>; the general expression is the unambiguous answer.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-55-hr-homology.png]
{title=Idealized radiative homology branches on a Hertzsprung-Russell diagram, with pp exponent 4 and CNO exponent 16}
{height=520}
The <Hertzsprung-Russell diagram> places hotter stars to the left. Its branches rise toward higher <luminosity> and mass, and the <CNO cycle>/<electron-scattering opacity> branch has the steeper logarithmic slope. Their illustrative joining point and normalization are arbitrary because the proportional <opacity> and reaction laws do not specify absolute stellar scales. \b[These are the fully radiative ideal-gas homology predictions], not exact observed <main sequence> relations: <convection> and increasing radiation support limit those assumptions in real stars.
Back to article page
