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 $$M_B^A<M_B^B$$ Given four numbers with relation $$x-y<a-b\not\Rightarrow x<a$$

Then how we can say so??

  • 1
    $\begingroup$ Could you please specify what all these variables are? It could be confusing to users who are new to this topic. $\endgroup$ Apr 28, 2021 at 1:22
  • 1
    $\begingroup$ You are neglecting reddening then? $\endgroup$
    – ProfRob
    Apr 29, 2021 at 11:36

1 Answer 1


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 $M_B^A<M_B^B$.

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, hotter objects are bluer and more luminous. This is a corollary of Planck's 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), 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).

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) 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


Stars are rather close to being blackbodies. However, a hot star could be reddened, e.g. by dust which preferentially absorbs blue wavelengths. Less hot stars may be blanketed 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.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Connor Garcia
    Apr 28, 2021 at 15:05

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