= Solution
For a homologous star, <hydrostatic equilibrium>, <mass conservation>, and the <ideal gas> equation give the central scalings
$$
\rho_c\propto\frac M{R^3},
\qquad
P_c\propto\frac{GM^2}{R^4},
\qquad
T_c\propto\frac{\mu M}{R},
$$
where $\mu$ is the <mean molecular weight>. Integrating the nuclear energy-generation law over a fixed homologous profile gives
$$
L_{\rm nuc}\propto X\rho_cT_c^{13}M
\propto X\mu^{13}\frac{M^{15}}{R^{16}}.
$$
<Radiative stellar structure> gives independently
$$
L_{\rm rad}\propto\frac{R T_c^4}{\kappa\rho_c}
\propto\frac{\mu^4M^3}{\kappa}.
$$
Equating the two luminosities yields
$$
R^{16}\propto\kappa X\mu^9M^{12}.
$$
At fixed zero-age composition, the stars are therefore homologous with
$$
\boxed{R\propto M^{3/4}},
\qquad
\boxed{L\propto M^3}.
$$
The <effective temperature> satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The <zero-age main sequence> consequently has
$$
\boxed{\frac{d\log L}{d\log T_e}=8}.
$$
It is a steep line rising toward high luminosity and high temperature on a <Hertzsprung-Russell diagram>.
For fully ionized hydrogen and helium with $Y=1-X$,
$$
\frac1\mu=2X+\frac34Y=\frac{3+5X}{4},
$$
while $\kappa\propto1+X$. At fixed mass,
$$
\boxed{L\propto(1+X)^{-1}(3+5X)^{-4}},
$$
and the corresponding radius relation is
$$
\boxed{R\propto[X(1+X)]^{1/16}(3+5X)^{-9/16}}.
$$
At the pure-hydrogen zero-age point $X=1$,
$$
\frac{d\log L}{dX}=-\frac1{1+X}-\frac{20}{3+5X}=-3,
$$
whereas
$$
\frac{d\log R}{dX}=\frac1{16}\left(\frac1X+\frac1{1+X}-\frac{45}{3+5X}\right)=-\frac{33}{128}.
$$
Hence
$$
\frac{d\log T_e}{dX}
=\frac14\left(\frac{d\log L}{dX}-2\frac{d\log R}{dX}\right)
=-\frac{159}{256},
$$
and
$$
\boxed{\frac{d\log L}{d\log T_e}=\frac{256}{53}\simeq4.83}.
$$
As hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.
Back to article page
