<p>Yes it would.</p>
<p>It is that way because the effective temperature is <em>defined</em> to be <span class="math-container">$(L/4\pi \sigma R^2)^{0.25}$</span>.</p>
<p>The radius of a neutron star is about 10 km <span class="math-container">$(1.4\times 10^{-5}R_\odot)$</span>. They are born with surface temperatures of around <span class="math-container">$10^8$</span> K. The coldest white dwarfs have effective temperatures of about 3000 K.</p>
<p>The luminosity ratio is
<span class="math-container">$$ \frac{L_{\rm NS}}{L_{\rm WD}}  = \left(\frac{10^{8}}{3000}\right)^4 \left(\frac{1.4\times 10^{-5}}{0.01}\right)^2 = 2.4\times 10^{12}\ .$$</span>
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 <span class="math-container">$\sim 10^5$</span>K, which could take as long as <span class="math-container">$10^7$</span> years (<a href="https://science.sciencemag.org/content/304/5670/536/tab-figures-data" rel="nofollow noreferrer">Lattimer &amp; Prakash 2004</a>).</p>

CC BY-SA 4.0
2021-04-27T01:15:37.257Z
Could you please clarify for $(L/4\pi \sigma R^2)^{0.25}$ as to where the division stops?
WarpPrime
CC BY-SA 4.0
2021-04-26T20:41:51.080Z
2021-04-26T23:36:20.543Z
2021-04-26T23:36:20.543Z
ProfRob
<p>A few million years after a white dwarf forms, its surface temperature reaches <span class="math-container">$100000\text{K}$</span>, while its radius is <span class="math-container">$0.01R_\odot$</span>. Would this mean that its luminosity is <span class="math-container">$\Big(\dfrac{100000}{5778}\Big)^4 \cdot 0.01^2 = 8.972 L_\odot$</span>? 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?</p>

<star><neutron-star><luminosity><white-dwarf><stefan-boltzmann-law>
Is something wrong with my luminosity calculation?
WarpPrime
# Is something wrong with my luminosity calculation?

(Asked by: username_0 on 2021-04-26)

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?

**Tags:** star, neutron-star, luminosity, white-dwarf, stefan-boltzmann-law

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## 1 answers

### Answer by username_1 on 2021-04-26. Score: 2

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](https://science.sciencemag.org/content/304/5670/536/tab-figures-data)).

> **Comments:**
> **username_0**: Could you please clarify for $(L/4\pi \sigma R^2)^{0.25}$ as to where the division stops?  

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