<p>The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star.</p>
<p>Therefore, the second relation for the two fluxes is about the apparent magnitudes (which describe the brightness of an astronomical object observed from Earth),
<span class="math-container">$$m-m_\odot = -2.5 \log F/F_\odot$$</span></p>
<p>The first relation is alright. The absolute magnitudes are related with the luminosity of the star (the overall energy flux emitted by the star) and they are not dependent on the distance to the observer.</p>

CC BY-SA 4.0
2021-05-12T08:30:32.320Z
Here you can find an image for a better understanding, https://astronomy.swin.edu.au/cosmos/F/Flux
prody
CC BY-SA 4.0
2021-05-12T08:25:06.423Z
2021-05-12T08:25:06.423Z
prody
<p>The difference of 2 bolometric magnitudes is given by:</p>
<p><span class="math-container">$$M_{bol, ★} - M_{bol, ☉} = -2.5 \cdot \log \left( \frac{L_★}{L_☉} \right)$$</span></p>
<p>But <a href="http://www.vikdhillon.staff.shef.ac.uk/teaching/phy217/instruments/phy217_inst_mags.html" rel="nofollow noreferrer">Pogson's equation</a> is:</p>
<p><span class="math-container">$$M_{bol, ★} - M_{bol, ☉} = -2.5 \cdot \log \left( \frac{F_★}{F_☉} \right)$$</span></p>
<p>where <span class="math-container">$F_★=\frac{L_★}{4\pi R^2}$</span>, so how come the first equation isn't dependent on the radius?</p>

CC BY-SA 4.0
2021-05-12T07:09:27.640Z
Welcome to astronomy SE! Would you mind adding a link to Pogson's equation, just as a service for those who do want to learn (more) about it, please?
B--rian
CC BY-SA 4.0
2021-05-12T07:23:10.993Z
There ya go: http://www.vikdhillon.staff.shef.ac.uk/teaching/phy217/instruments/phy217_inst_mags.html
Pedro Nogueira
<star><luminosity><magnitude><absolute-magnitude><flux>
How is the difference of bolometric magnitudes not dependent on the stars' radii?
Pedro Nogueira
# How is the difference of bolometric magnitudes not dependent on the stars' radii?

(Asked by: username_0 on 2021-05-12)

The difference of 2 bolometric magnitudes is given by:

$$M\_{bol, ★} - M\_{bol, ☉} = -2.5 \cdot \log \left( \frac{L\_★}{L\_☉} \right)$$

But [Pogson's equation](http://www.vikdhillon.staff.shef.ac.uk/teaching/phy217/instruments/phy217_inst_mags.html) is:

$$M\_{bol, ★} - M\_{bol, ☉} = -2.5 \cdot \log \left( \frac{F\_★}{F\_☉} \right)$$

where $F\_★=\frac{L\_★}{4\pi R^2}$, so how come the first equation isn't dependent on the radius?

**Tags:** star, luminosity, magnitude, absolute-magnitude, flux
> **Comments:**
> **username_1**: Welcome to astronomy SE! Would you mind adding a link to Pogson's equation, just as a service for those who do want to learn (more) about it, please?  
> **username_0**: There ya go: http://www.vikdhillon.staff.shef.ac.uk/teaching/phy217/instruments/phy217_inst_mags.html  

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## 1 answers

### Answer by username_2 on 2021-05-12. Score: 5 | Accepted answer

The R in that equation is the distance from the star to observer, not the star radius. The light emitted from the star is distributed uniformly on a sphere of radius R, and when the light arrives to the Earth, that sphere will have a radius equal to the distance Earth-star.

Therefore, the second relation for the two fluxes is about the apparent magnitudes (which describe the brightness of an astronomical object observed from Earth),
$$m-m\_\odot = -2.5 \log F/F\_\odot$$

The first relation is alright. The absolute magnitudes are related with the luminosity of the star (the overall energy flux emitted by the star) and they are not dependent on the distance to the observer.

> **Comments:**
> **username_2**: Here you can find an image for a better understanding, https://astronomy.swin.edu.au/cosmos/F/Flux  

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