<h3>Color is a difference, not an absolute value</h3>
<p>Being &quot;blue&quot; doesn't (necessarily) mean that a light source has a large flux in the <span class="math-container">$B$</span> band. A color is not an absolute value; it is the <em>ratio</em> between two fluxes or, equivalently, the <em>difference</em> between two magnitudes.</p>
<p>Being blue means &quot;More flux in some short wavelength band (e.g. <span class="math-container">$B$</span>) than in a longer-wavelength band (e.g. <span class="math-container">$V$</span>)&quot;. Because of the stupid magnitude system, this then means a <em>lower</em> value in <span class="math-container">$B$</span> than in <span class="math-container">$V$</span>, i.e. <span class="math-container">$B-V&lt;0$</span>.</p>
<p>Hence, the smaller <span class="math-container">$B-V$</span> value a star has, the bluer it is.</p>
<h3>Intuitive example</h3>
<p>So when you write</p>
<blockquote>
<p>Star A is bluer than Star B means <span class="math-container">$M_B^A&lt;M_B^B$</span>.</p>
</blockquote>
<p>it's actually not true. This mathematical expression only means that Star A is brighter (in the <span class="math-container">$B$</span> band) than star B; it says nothing about the color.</p>
<p>For instance, if you compare the flux of the Sun with the flux of a blue laser pointer, the Sun most definitely emits more blue light than the laser pointer, i.e. <span class="math-container">$B_\odot &lt; B_\mathrm{laser}$</span>. But that doesn't mean that the Sun is blue, because it also emits much more red light.</p>
<h3>Astronomers are astronoming</h3>
<p>In the above, I've followed the sloppy astronomer notation of using <span class="math-container">$B$</span> and <span class="math-container">$V$</span> both for the filters/bands/bandpasses, and for the absolute magnitudes. More clearly, we can also write the magnitudes as <span class="math-container">$M_B$</span> and <span class="math-container">$M_V$</span>.</p>
<p>Not also that, in astronomy, the term &quot;bluer&quot; is often used interchangeably with &quot;shorter wavelength&quot;, irregardless of the actual wavelength. Similarly, the term &quot;redder&quot; is used for longer wavelengths.</p>
<p>For instance, we could say that a red photon is bluer than an infrared photons. Similarly, a UV photon can be redshifted as it travels through the expanding Universe. If it doesn't travel too long, it might be observed as &quot;blue&quot;, in the sense that its wavelength is around 4400 Å, but although it's not <em>red</em>, we still says that it has been redshifted.</p>
<h3>Relation of color to temperature</h3>
<p>For <a href="https://en.wikipedia.org/wiki/Black_body" rel="nofollow noreferrer">blackbodies</a>, hotter objects are bluer and more luminous. This is a corollary of <a href="https://en.wikipedia.org/wiki/Planck%27s_law" rel="nofollow noreferrer">Planck's law</a>, which relates the brightness <span class="math-container">$B$</span> per wavelength (&quot;spectral density&quot;) of a blackbody to its temperature <span class="math-container">$T$</span>.</p>
<p>If you integrate Planck's law you can show that the total power <span class="math-container">$P$</span> increases tremendously with temperature (<span class="math-container">$P\propto T^4$</span>; <a href="https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law" rel="nofollow noreferrer">Stefan-Boltzmann's law</a>), while if you differentiate it you can show that the peak wavelength <span class="math-container">$\lambda_\mathrm{peak}$</span> is inversely proportional to the temperature (<span class="math-container">$\lambda_\mathrm{peak}\propto1/T$</span>; <a href="https://en.wikipedia.org/wiki/Wien%27s_displacement_law" rel="nofollow noreferrer">Wien's displacement law</a>).</p>
<p>In other words, 1) hotter objects emit more flux (i.e. are brighter), and 2) hotter objects emit a larger fraction of their flux at short wavelengths (i.e. are bluer).</p>
<p>The figure below (an annotated version of <a href="https://en.wikipedia.org/wiki/Planck%27s_law#/media/File:Black_body.svg" rel="nofollow noreferrer">this</a>) shows an example of the colors of a <span class="math-container">$5000\,\mathrm{K}$</span> (blue solid line) and a <span class="math-container">$4000\,\mathrm{K}$</span> (dark green solid line), respectively. The <span class="math-container">$B$</span> and <span class="math-container">$V$</span> fluxes are measured at <span class="math-container">$4400\,\mathrm{Å}$</span> (purple dashed line) and <span class="math-container">$5500\,\mathrm{Å}$</span> (light green dashed line), respectively.</p>
<p>The <em>flux ratio</em> between the <span class="math-container">$B$</span> and <span class="math-container">$V$</span> wavelengths are larger for the <span class="math-container">$5000\,\mathrm{K}$</span> object, so its <em>magnitude difference</em> is smaller.</p>
<p>In other words, it's bluer.</p>
<p>(I've omitted a factor 2.5)</p>
<p><a href="https://i.sstatic.net/CZ660.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CZ660.jpg" alt="enter image description here" /></a></p>
<h3>Reddening</h3>
<p>Stars are rather close to being blackbodies. However, a hot star could be <a href="https://en.wikipedia.org/wiki/Extinction_(astronomy)" rel="nofollow noreferrer">reddened</a>, e.g. by dust which preferentially absorbs blue wavelengths. Less hot stars may be  <a href="https://en.wikipedia.org/wiki/Blanketing_effect" rel="nofollow noreferrer">blanketed</a> by metals absorbing light and re-emitting at lower wavelengths, effectively also reddening the star. Or an object could simply not be a blackbody, e.g. a blue laser pointer.</p>
<h3>Comparison to velocities</h3>
<p>If, for some reason, you don't know the value of <span class="math-container">$B-V$</span>, but you know that &quot;object A has a smaller <span class="math-container">$B-V$</span> value than object B&quot;, then you can't really say anything about their &quot;true&quot; color; you can just say that &quot;A is bluer than B&quot;, although neither of these objects need be blue.</p>
<p>You can compare this to velocities. All velocities are relative, so &quot;<span class="math-container">$v = 10\,\mathrm{km/h}$</span>&quot; doesn't mean anything, unless you specify a reference frame. But even given a reference frame, the statement <span class="math-container">$v_\mathrm{A} &gt; v_\mathrm{B}$</span> doesn't tell you how fast A and B are; it just tells you that A is faster than B.</p>

CC BY-SA 4.0
2021-04-28T15:05:04.227Z
Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/123549/discussion-on-answer-by-pela-given-that-a-color-index-b-v-of-star-a-is-lower).
Connor Garcia
CC BY-SA 4.0
2021-04-27T12:41:44.093Z
2021-05-10T09:29:01.003Z
2021-05-10T09:29:01.003Z
pela
<p>The color index <span class="math-container">$B-V$</span> of the star is defined as
<span class="math-container">$$B-V\equiv M_B-M_V$$</span>
Let us suppose it's given
<span class="math-container">$$(B-V)^{(A)}&lt;(B-V)^{(B)}\Rightarrow M_B^A-M_V^A&gt;M^B_B-M^B_V$$</span>
The notations are a bit confusing, but I think you would pardon me.</p>
<p>Now we know that Stellar magnitudes decrease with increasing brightness which means a star with absolute magnitude <span class="math-container">$-5$</span> will be brighter than a star with absolute magnitude <span class="math-container">$-2$</span>.</p>
<p>I'm reading <em>Introduction to Modern Astrophysics</em> By Carroll and Ostlie. The  author says,</p>
<blockquote>
<p>A star with a smaller <span class="math-container">$B-V$</span> color index is bluer than a star with a larger value of <span class="math-container">$B-V$</span>.</p>
</blockquote>
<p>I don't understand, How??
Star A is bluer than Star B means
<span class="math-container">$$M_B^A&lt;M_B^B$$</span>
Given four numbers with relation
<span class="math-container">$$x-y&lt;a-b\not\Rightarrow x&lt;a$$</span></p>
<p>Then how we can say so??</p>

CC BY-SA 4.0
2021-04-28T01:22:28.900Z
Could you please specify what all these variables are? It could be confusing to users who are new to this topic.
WarpPrime
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2021-04-29T11:36:26.980Z
You are neglecting reddening then?
ProfRob
<star><temperature><absolute-magnitude><color-index>
Given that a color index ($B-V$) of star A is lower than that of Star B, What can we say about the stars?
Himanshu
# Given that a color index ($B-V$) of star A is lower than that of Star B, What can we say about the stars?

(Asked by: username_0 on 2021-04-27)

The color index $B-V$ of the star is defined as
$$B-V\equiv M\_B-M\_V$$
Let us suppose it's given
$$(B-V)^{(A)}<(B-V)^{(B)}\Rightarrow M\_B^A-M\_V^A>M^B\_B-M^B\_V$$
The notations are a bit confusing, but I think you would pardon me.

Now we know that Stellar magnitudes decrease with increasing brightness which means a star with absolute magnitude $-5$ will be brighter than a star with absolute magnitude $-2$.

I'm reading *Introduction to Modern Astrophysics* By Carroll and Ostlie. The author says,

> A star with a smaller $B-V$ color index is bluer than a star with a larger value of $B-V$.

I don't understand, How??
Star A is bluer than Star B means
$$M\_B^A
Given four numbers with relation
$$x-y

Then how we can say so??

**Tags:** star, temperature, absolute-magnitude, color-index
> **Comments:**
> **username_1**: Could you please specify what all these variables are? It could be confusing to users who are new to this topic.  
> **username_2**: You are neglecting reddening then?  

---

## 1 answers

### Answer by username_3 on 2021-04-27. Score: 4 | Accepted answer

### Color is a difference, not an absolute value

Being "blue" doesn't (necessarily) mean that a light source has a large flux in the $B$ band. A color is not an absolute value; it is the *ratio* between two fluxes or, equivalently, the *difference* between two magnitudes.

Being blue means "More flux in some short wavelength band (e.g. $B$) than in a longer-wavelength band (e.g. $V$)". Because of the stupid magnitude system, this then means a *lower* value in $B$ than in $V$, i.e. $B-V<0$.

Hence, the smaller $B-V$ value a star has, the bluer it is.

### Intuitive example

So when you write

> Star A is bluer than Star B means $M\_B^A.

it's actually not true. This mathematical expression only means that Star A is brighter (in the $B$ band) than star B; it says nothing about the color.

For instance, if you compare the flux of the Sun with the flux of a blue laser pointer, the Sun most definitely emits more blue light than the laser pointer, i.e. $B\_\odot < B\_\mathrm{laser}$. But that doesn't mean that the Sun is blue, because it also emits much more red light.

### Astronomers are astronoming

In the above, I've followed the sloppy astronomer notation of using $B$ and $V$ both for the filters/bands/bandpasses, and for the absolute magnitudes. More clearly, we can also write the magnitudes as $M\_B$ and $M\_V$.

Not also that, in astronomy, the term "bluer" is often used interchangeably with "shorter wavelength", irregardless of the actual wavelength. Similarly, the term "redder" is used for longer wavelengths.

For instance, we could say that a red photon is bluer than an infrared photons. Similarly, a UV photon can be redshifted as it travels through the expanding Universe. If it doesn't travel too long, it might be observed as "blue", in the sense that its wavelength is around 4400 Å, but although it's not *red*, we still says that it has been redshifted.

### Relation of color to temperature

For [blackbodies](https://en.wikipedia.org/wiki/Black_body), hotter objects are bluer and more luminous. This is a corollary of [Planck's law](https://en.wikipedia.org/wiki/Planck%27s_law), which relates the brightness $B$ per wavelength ("spectral density") of a blackbody to its temperature $T$.

If you integrate Planck's law you can show that the total power $P$ increases tremendously with temperature ($P\propto T^4$; [Stefan-Boltzmann's law](https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law)), while if you differentiate it you can show that the peak wavelength $\lambda\_\mathrm{peak}$ is inversely proportional to the temperature ($\lambda\_\mathrm{peak}\propto1/T$; [Wien's displacement law](https://en.wikipedia.org/wiki/Wien%27s_displacement_law)).

In other words, 1) hotter objects emit more flux (i.e. are brighter), and 2) hotter objects emit a larger fraction of their flux at short wavelengths (i.e. are bluer).

The figure below (an annotated version of [this](https://en.wikipedia.org/wiki/Planck%27s_law#/media/File:Black_body.svg)) shows an example of the colors of a $5000\,\mathrm{K}$ (blue solid line) and a $4000\,\mathrm{K}$ (dark green solid line), respectively. The $B$ and $V$ fluxes are measured at $4400\,\mathrm{Å}$ (purple dashed line) and $5500\,\mathrm{Å}$ (light green dashed line), respectively.

The *flux ratio* between the $B$ and $V$ wavelengths are larger for the $5000\,\mathrm{K}$ object, so its *magnitude difference* is smaller.

In other words, it's bluer.

(I've omitted a factor 2.5)

[![enter image description here](https://i.sstatic.net/CZ660.jpg)](https://i.sstatic.net/CZ660.jpg)

### Reddening

Stars are rather close to being blackbodies. However, a hot star could be [reddened](https://en.wikipedia.org/wiki/Extinction_(astronomy)), e.g. by dust which preferentially absorbs blue wavelengths. Less hot stars may be [blanketed](https://en.wikipedia.org/wiki/Blanketing_effect) by metals absorbing light and re-emitting at lower wavelengths, effectively also reddening the star. Or an object could simply not be a blackbody, e.g. a blue laser pointer.

### Comparison to velocities

If, for some reason, you don't know the value of $B-V$, but you know that "object A has a smaller $B-V$ value than object B", then you can't really say anything about their "true" color; you can just say that "A is bluer than B", although neither of these objects need be blue.

You can compare this to velocities. All velocities are relative, so "$v = 10\,\mathrm{km/h}$" doesn't mean anything, unless you specify a reference frame. But even given a reference frame, the statement $v\_\mathrm{A} > v\_\mathrm{B}$ doesn't tell you how fast A and B are; it just tells you that A is faster than B.

> **Comments:**
> **username_4**: Comments are not for extended discussion; this conversation has been [moved to chat](https://chat.stackexchange.com/rooms/123549/discussion-on-answer-by-pela-given-that-a-color-index-b-v-of-star-a-is-lower).  

---

