<blockquote>
<p>What is the formula</p>
</blockquote>
<p><span class="math-container">$M = m - 5 \log(\frac{d}{10})$</span></p>
<blockquote>
<p>What do those symbols mean?</p>
</blockquote>
<p>M is the absolute magnitude of the star, m is the apparent magnitude of the star, and d is the distance to the star in parsecs.</p>
<blockquote>
<p>How to do what you want to do</p>
</blockquote>
<p>The absolute magnitude of the Sun is about +4.83. So, to express the absolute magnitude of a star as a multiple of the Sun's absolute magnitude, you can divide the star's absolute magnitude by +4.83.</p>
<p>For example, if a star has an absolute magnitude of +2.0, then the star is about 2.7 times brighter than the Sun (since 2.0 ÷ 4.83 = 0.414, and 1 ÷ 0.414 ≈ 2.7).</p>
<p>Note that this calculation assumes that the star is at the same distance as the Sun, which is not usually the case. Therefore, the actual value may differ from the calculated value if the star is at a different distance.</p>
<blockquote>
<p>That's annoying, how can I fix this?</p>
</blockquote>
<p>Not totally sure, but I believe that this follows the inverse square law. Therefore take the actual distance to the star. Then divide by the suns distance (use an units you wish). Now square this value and this is the factor it is off by from the real value, which you can now correct for.</p>

CC BY-SA 4.0
2023-03-31T00:06:59.437Z
2023-05-16T20:28:39.383Z
2023-05-16T20:28:39.383Z
Starship
<p>It is profoundly silly to multiply a magnitude value.</p>
<p>Contrast this with (for example) length.  If I have stick that is 1 metre long, and an other stick which is 2m long, then I can say that the second stick is double the length of the first. And if I have a third stick that is 4m long, then the ratio of the lengths of the first two sticks is similar to the ratio of the lengths of the second and third.  There is a common relationship of &quot;double the length&quot;.</p>
<p>But magnitude is a logarithmic scale which includes 0 and negative values.</p>
<p>So consider a star of mag 1, and compare it to a star of mag 2  Is this second star &quot;twice as dim&quot;  What about a star of mag 4, is that twice as dim again? What about a star of mag 0.1, is a star of mag 0.2  twice as dim (you couldn't easily distinguish the brightness of these stars by eye).</p>
<p>It becomes even stranger if you consider a star of mag 0, because twice zero is zero,  so this star is twice as dim as itself... and what about a star of magnitude -1, the the star that is twice as dim, is actually brighter (at magnitude -2)  All this goes to show that you can't treat a logarithmic scale with proportion.  It makes no sense to consider 2×the magnitude of the sun.</p>
<p>On the other hand, luminosity is measured on a linear scale, starting from 0, so it makes perfect sense to double the value to get the amount of light emitted by a star that is twice as bright</p>

CC BY-SA 4.0
2023-03-31T23:12:33.200Z
2023-03-31T23:12:33.200Z
James K
<p>This is a homework-related question.</p>
<p>I am trying to find the spectral class and luminosity (relative to the Sun) of a star using its absolute magnitude. The absolute magnitude is expressed as a fraction of the Sun's absolute magnitude, such as 0.3*M☉.</p>
<p>Despite what has felt like extensive research through Google and other Astronomy Stack Exchange posts, I've been unable to find a clear equation for calculating the hypothetical star's magnitude from this presentation.</p>
<p>EDIT: I have conferred with the professor on this topic, and the homework in question was poorly worded. Instead of M☉ meaning the Sun's absolute magnitude, it was meant to signify its mass. Using the mass-luminosity relation, I was able to find the spectral class.</p>
<p>However, I am still curious whether there is any meaningful way to multiply a magnitude value by either a fraction or any positive number, or if even trying to do that would be self-defeating and silly. Would converting into luminosity first make this possible?</p>

CC BY-SA 4.0
2023-03-31T00:40:10.770Z
Luminosity and magnitude are different things. Luminosity is in watts while luminosity is a dimensionless inverse logarithmic scale for brightness. You seem to use the terms interchangeable which they are not. And that might explain your confusion
planetmaker
<absolute-magnitude>
Finding the absolute magnitude of a star as a fraction/multiple of the Sun's absolute magnitude
ROPROPE
# Finding the absolute magnitude of a star as a fraction/multiple of the Sun's absolute magnitude

(Asked by: username_0 on 2023-03-30)

This is a homework-related question.

I am trying to find the spectral class and luminosity (relative to the Sun) of a star using its absolute magnitude. The absolute magnitude is expressed as a fraction of the Sun's absolute magnitude, such as 0.3\*M☉.

Despite what has felt like extensive research through Google and other Astronomy Stack Exchange posts, I've been unable to find a clear equation for calculating the hypothetical star's magnitude from this presentation.

EDIT: I have conferred with the professor on this topic, and the homework in question was poorly worded. Instead of M☉ meaning the Sun's absolute magnitude, it was meant to signify its mass. Using the mass-luminosity relation, I was able to find the spectral class.

However, I am still curious whether there is any meaningful way to multiply a magnitude value by either a fraction or any positive number, or if even trying to do that would be self-defeating and silly. Would converting into luminosity first make this possible?

**Tags:** absolute-magnitude

> **Comments:**
> **username_1**: Luminosity and magnitude are different things. Luminosity is in watts while luminosity is a dimensionless inverse logarithmic scale for brightness. You seem to use the terms interchangeable which they are not. And that might explain your confusion  

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## 2 answers

### Answer by username_2 on 2023-03-31. Score: 3

> What is the formula

$M = m - 5 \log(\frac{d}{10})$

> What do those symbols mean?

M is the absolute magnitude of the star, m is the apparent magnitude of the star, and d is the distance to the star in parsecs.

> How to do what you want to do

The absolute magnitude of the Sun is about +4.83. So, to express the absolute magnitude of a star as a multiple of the Sun's absolute magnitude, you can divide the star's absolute magnitude by +4.83.

For example, if a star has an absolute magnitude of +2.0, then the star is about 2.7 times brighter than the Sun (since 2.0 ÷ 4.83 = 0.414, and 1 ÷ 0.414 ≈ 2.7).

Note that this calculation assumes that the star is at the same distance as the Sun, which is not usually the case. Therefore, the actual value may differ from the calculated value if the star is at a different distance.

> That's annoying, how can I fix this?

Not totally sure, but I believe that this follows the inverse square law. Therefore take the actual distance to the star. Then divide by the suns distance (use an units you wish). Now square this value and this is the factor it is off by from the real value, which you can now correct for.


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### Answer by username_3 on 2023-03-31. Score: 2

It is profoundly silly to multiply a magnitude value.

Contrast this with (for example) length. If I have stick that is 1 metre long, and an other stick which is 2m long, then I can say that the second stick is double the length of the first. And if I have a third stick that is 4m long, then the ratio of the lengths of the first two sticks is similar to the ratio of the lengths of the second and third. There is a common relationship of "double the length".

But magnitude is a logarithmic scale which includes 0 and negative values.

So consider a star of mag 1, and compare it to a star of mag 2 Is this second star "twice as dim" What about a star of mag 4, is that twice as dim again? What about a star of mag 0.1, is a star of mag 0.2 twice as dim (you couldn't easily distinguish the brightness of these stars by eye).

It becomes even stranger if you consider a star of mag 0, because twice zero is zero, so this star is twice as dim as itself... and what about a star of magnitude -1, the the star that is twice as dim, is actually brighter (at magnitude -2) All this goes to show that you can't treat a logarithmic scale with proportion. It makes no sense to consider 2×the magnitude of the sun.

On the other hand, luminosity is measured on a linear scale, starting from 0, so it makes perfect sense to double the value to get the amount of light emitted by a star that is twice as bright


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