= Solution
The <standard thin-disk dissipation flux>, summed over both faces, follows by substituting the <Keplerian accretion disk> profile into the viscous heating rate:
$$
\mathcal H=\frac94\bar\nu\Sigma\Omega^2=\frac{3GM_*\dot M}{4\pi r^3}\left(1-\sqrt{\frac{r_*}{r}}\right).
$$
The <Stefan–Boltzmann law> gives $\mathcal C=2\sigma T^4$ because there are two emitting faces. Local <radiative equilibrium> consequently gives \b[the <effective-temperature profile of a zero-torque disk>]
$$
\boxed{T(r)=T_{\rm in}\left(\frac{r_*}{r}\right)^{3/4}\left(1-\sqrt{\frac{r_*}{r}}\right)^{1/4},\qquad T_{\rm in}^4=\frac{3GM_*\dot M}{8\pi\sigma r_*^3}.}
$$
Here $\sigma$ is the <Stefan-Boltzmann constant>. The scale $T_{\rm in}$ is not the <temperature> exactly at the inner boundary: the <zero-torque inner boundary condition> makes that formal <temperature> zero. Maximizing $T^4$ shows that the <maximum effective temperature of a zero-torque disk> occurs at $r=49r_*/36$, with $T_{\max}\simeq0.488T_{\rm in}$.
At equal central mass and <accretion rate>, characteristic <effective temperatures> scale as $r_*^{-3/4}$. Thus, comparing corresponding values of $r/r_*$,
$$
\boxed{\frac{T_{\rm NS}}{T_{\rm WD}}\simeq\left(\frac{10^4\,\mathrm{km}}{10\,\mathrm{km}}\right)^{3/4}=10^{9/4}\simeq178.}
$$
\b[The neutron-star disk is about 180 times hotter.] The <Planck law> and <Wien displacement law> move its characteristic emission to about 180 times higher <frequency>, or 180 times shorter <wavelength>. A <white dwarf> disk commonly emits in optical and ultraviolet bands, while the hotter <neutron star> disk can emit in X-rays. Absolute bands require an actual <accretion rate>; the relative shift follows directly from the stated scaling.
Back to article page
