<p>You just use the Luminosity:</p>
<p><span class="math-container">$M = -2.5  \log_{10}(L / L_0)$</span></p>
<p>Where <span class="math-container">$L_0$</span> is the luminosity of a magnitude 0 star <span class="math-container">$L_0=3\times10^{28}\mathrm{W}=79L_\odot$</span></p>
<p>You don't need the other variables. You would need the distance if you wanted to calculate the apparent magnitude.  You could also use the radius and mass to estimate the luminosity, if you weren't given it.</p>
<p>This gives the bolometric magnitude, which includes the brightness of the star at infrared and ultraviolet wavelengths (which, of course, we can't see) You could use the temperature to correct for this and get a visual magnitude, but the calculation then becomes more complex, you would need to estimate the temperature from the mass and radius, which is left as an exercise for the reader.</p>

CC BY-SA 4.0
2020-10-30T17:41:22.823Z
2020-10-30T17:43:27.667Z
2020-10-30T17:43:27.667Z
James K
<p>Yes. You derive the absolute bolometric magnitude by reference to the Sun. The Sun has a bolometric magnitude (i.e. integrated over all wavelengths) of about 4.75. Thus the difference in bolometric magnitudes is given by the log of the ratio of luminosities like so
<span class="math-container">$$ M_{\rm bol} = 4.75 -2.5 \log_{10} \left(\frac{L_*}{L_{\odot}}\right)\ ,$$</span>
where <span class="math-container">$L_{\odot} = 3.8\times 10^{26}$</span> W is the luminosity of the Sun and <span class="math-container">$L_*$</span> is the luminosity of the star.</p>
<p>To get the absolute visual magnitude then you need to subtract a <a href="https://en.wikipedia.org/wiki/Bolometric_correction" rel="noreferrer">bolometric correction</a>, which to first order, depends on the temperature of the star (it also depends to second order on the composition and gravity).
<span class="math-container">$$ M_V = M_{\rm bol} - BC_V\ .$$</span></p>
<p>A table of bolometric corrections vs temperature can be found in the 6th column of <a href="https://www.pas.rochester.edu/%7Eemamajek/EEM_dwarf_UBVIJHK_colors_Teff.txt" rel="noreferrer">this authoritative table</a>, which is in common use by astronomers/astrophysicists.</p>

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2020-10-30T18:57:44.303Z
2020-10-30T18:57:44.303Z
ProfRob
<p>I have a star that I need to calculate the absolute magnitude of. I am given the temperature, luminosity, radius, mass, and distance in light-years. So I am wondering, what is the formula to compute this, using <strong>only</strong> these terms?</p>

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2020-10-30T20:37:58.883Z
A general point on SE is that users are encouraged to wait about 24 hours before accepting an answer to give people (globally !) a chance to provide an answer.  It is also not uncommon that better answers turn up later (and users sometimes like to do more detailed fact checking on answers before posting).  Note you can vote up any answer anytime, so you're still free to reward users promptly.
StephenG - Help Ukraine
<star><absolute-magnitude>
Is there a formula for absolute magnitude that does not contain an apparent magnitude term?
WarpPrime
# Is there a formula for absolute magnitude that does not contain an apparent magnitude term?

(Asked by: username_0 on 2020-10-30)

I have a star that I need to calculate the absolute magnitude of. I am given the temperature, luminosity, radius, mass, and distance in light-years. So I am wondering, what is the formula to compute this, using **only** these terms?

**Tags:** star, absolute-magnitude

> **Comments:**
> **username_1**: A general point on SE is that users are encouraged to wait about 24 hours before accepting an answer to give people (globally !) a chance to provide an answer.  It is also not uncommon that better answers turn up later (and users sometimes like to do more detailed fact checking on answers before posting).  Note you can vote up any answer anytime, so you're still free to reward users promptly.  

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## 2 answers

### Answer by username_3 on 2020-10-30. Score: 11 | Accepted answer

Yes. You derive the absolute bolometric magnitude by reference to the Sun. The Sun has a bolometric magnitude (i.e. integrated over all wavelengths) of about 4.75. Thus the difference in bolometric magnitudes is given by the log of the ratio of luminosities like so
$$ M\_{\rm bol} = 4.75 -2.5 \log\_{10} \left(\frac{L\_\*}{L\_{\odot}}\right)\ ,$$
where $L\_{\odot} = 3.8\times 10^{26}$ W is the luminosity of the Sun and $L\_\*$ is the luminosity of the star.

To get the absolute visual magnitude then you need to subtract a [bolometric correction](https://en.wikipedia.org/wiki/Bolometric_correction), which to first order, depends on the temperature of the star (it also depends to second order on the composition and gravity).
$$ M\_V = M\_{\rm bol} - BC\_V\ .$$

A table of bolometric corrections vs temperature can be found in the 6th column of [this authoritative table](https://www.pas.rochester.edu/%7Eemamajek/EEM_dwarf_UBVIJHK_colors_Teff.txt), which is in common use by astronomers/astrophysicists.


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### Answer by username_2 on 2020-10-30. Score: 7

You just use the Luminosity:

$M = -2.5 \log\_{10}(L / L\_0)$

Where $L\_0$ is the luminosity of a magnitude 0 star $L\_0=3\times10^{28}\mathrm{W}=79L\_\odot$

You don't need the other variables. You would need the distance if you wanted to calculate the apparent magnitude. You could also use the radius and mass to estimate the luminosity, if you weren't given it.

This gives the bolometric magnitude, which includes the brightness of the star at infrared and ultraviolet wavelengths (which, of course, we can't see) You could use the temperature to correct for this and get a visual magnitude, but the calculation then becomes more complex, you would need to estimate the temperature from the mass and radius, which is left as an exercise for the reader.


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