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?


1 Answer 1


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? $\endgroup$
    – WarpPrime
    Apr 27, 2021 at 1:15

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