= Solution
At fixed initial zero-metal composition, equating parts (i) and (ii) gives $R\propto M^{3/4}$ and $L\propto M^3$. Therefore $T_{\rm eff}\propto(L/R^2)^{1/4}\propto M^{3/8}$, and the fully radiative zero-age line on the <Hertzsprung-Russell diagram> has slope
$$
\boxed{\frac{d\log L}{d\log T_{\rm eff}}=8}.
$$
For the convective sequence, equating parts (i) and (iii) at fixed composition gives
$$
R\propto M^{63/86},
\qquad
L\propto M^{141/43},
\qquad
T_{\rm eff}\propto M^{39/86}.
$$
Its slope is consequently
$$
\boxed{\frac{d\log L}{d\log T_{\rm eff}}=\frac{94}{13}\simeq7.23}.
$$
For a fully ionized zero-metal hydrogen-helium mixture with $Y=1-X$, the <mean molecular weight> obeys
$$
\frac1\mu=2X+\frac34Y=\frac{3+5X}{4}.
$$
For a radiative star of fixed mass,
$$
L\propto\frac{\mu^4}{1+X},
\qquad
R^{16}\propto X(1+X)\mu^9.
$$
At the pure-hydrogen point $X=1$,
$$
\boxed{\frac{d\log L}{dX}
=-\frac1{1+X}-\frac{20}{3+5X}=-3},
$$
and
$$
\frac{d\log R}{dX}
=\frac1{16}\left(\frac1X+\frac1{1+X}
-\frac{45}{3+5X}\right)
=-\frac{33}{128}.
$$
Hence
$$
\boxed{\frac{d\log T_{\rm eff}}{dX}
=\frac14\left(\frac{d\log L}{dX}
-2\frac{d\log R}{dX}\right)
=-\frac{159}{256}}.
$$
As hydrogen burns, $X$ decreases, so both $L$ and $T_{\rm eff}$ increase. The track moves upward and toward higher temperature from the <zero-age main sequence>, with initial slope $d\log L/d\log T_{\rm eff}=256/53\simeq4.83$.
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