A few million years after a white dwarf forms, its surface temperature reaches $100000\text{K}$, while its radius is $0.01R_\odot$. Would this mean that its luminosity is $\Big(\dfrac{100000}{5778}\Big)^4 \cdot 0.01^2 = 8.972 L_\odot$? If so, why is it that way (I do understand the Stefan-Boltzmann law), and does that mean that all neutron stars are less luminous than white dwarfs?


Yes it would.

It is that way because the effective temperature is defined to be $(L/4\pi \sigma R^2)^{0.25}$.

The radius of a neutron star is about 10 km $(1.4\times 10^{-5}R_\odot)$. They are born with surface temperatures of around $10^8$ K. The coldest white dwarfs have effective temperatures of about 3000 K.

The luminosity ratio is $$ \frac{L_{\rm NS}}{L_{\rm WD}} = \left(\frac{10^{8}}{3000}\right)^4 \left(\frac{1.4\times 10^{-5}}{0.01}\right)^2 = 2.4\times 10^{12}\ .$$ Thus the most luminous neutron stars are (briefly) much more luminous than the coldest white dwarfs. They would remain more luminous until they cooled to $\sim 10^5$K, which could take as long as $10^7$ years (Lattimer & Prakash 2004).

  • $\begingroup$ Could you please clarify for $(L/4\pi \sigma R^2)^{0.25}$ as to where the division stops? $\endgroup$ – fasterthanlight Apr 27 at 1:15

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