Can someone explain this diagram?

The text is in Dutch, free translation: "You are given a graph: a histogram of the 10 000 most apparent bright stars. Explain 1, 2 and 3."

A histogram of the 10 000 stars with highest apparent magnitude

The fact that the number of counted stars of types O and M is small due to

  • O is bright but just very rare and,
  • M is common but quite dim.

The puzzling bit in this diagram is the sudden dip from a local maximum F (why is there a maximum) to G (why is G a local minimum). The drop from K to M is explained above - that being that M class stars are just very dim (is this all there is to say here?).

It seems that if there wasn't a dip at 3, and the graph would just continue, that the local max at K can be explained as being there just because there is a dip at G. If there wasn't a dip it wouldn't be a maximum and it'd just look like a bell-curve.


  • Why is there a dip at G?
  • Why does it rise so smooth from O to F? Is it just because B - F are more common?
  • Why is there a maximum at F?
  • 2
    $\begingroup$ Homework questions are not necessarily frowned upon, but it would be better if you provided any though process you may have tried when solving this. What exactly are you stuck on? What have you tried so far? Just posting your homework problem and asking someone to solve isn't how this site works. As a hint though, consider two things: (1) the quantity of stars in each stellar class, and (2) the absolute (as opposed to apparent) brightness of the stellar classes. $\endgroup$
    – zephyr
    Jun 16, 2017 at 12:50
  • $\begingroup$ What happened to N and S ? :-) $\endgroup$ Jun 16, 2017 at 13:23
  • 2
    $\begingroup$ @adrianmcmenamin There are not more K type stars than M. M is by far the most prevalent stellar type, being about $\sim76\%$ of all stars. The point of the question is to explain why the 10,000 most apparently bright stars have so few M stars, despite these stars being the most prevalent. $\endgroup$
    – zephyr
    Jun 16, 2017 at 13:49
  • $\begingroup$ The en.wikipedia.org/wiki/Hertzsprung%E2%80%93Russell_diagram tells us type O stars are the brightest, B second brightest, etc, but en.wikipedia.org/wiki/… tells us they are also fairly rare. The diagram you have above apparently combines these two pieces of information. $\endgroup$
    – user21
    Jun 16, 2017 at 14:05
  • 1
    $\begingroup$ @zephyr, Okay, I'll try to think of it next time. I've indeed spend a lot of time thinking about this and have come to some conclusion, but just wasn't up for it to write it all down, but I'll add something more specific to the question. $\endgroup$ Jun 16, 2017 at 15:40

2 Answers 2


It's all to do with the relationships between mass, spectral-type and luminosity and the initial mass function of stars.

I think your explanation of points 1 and 2 are completely correct. O and B stars are rarely born and short-lived; so even though they have enormous luminosities relatively few make it into a list of stars ordered by apparent brightness. A further important reason is that although they are immensely luminous, the more distant examples that might have made it into the list, cannot make it into the list because they are in the Galactic plane (they are young and still close to where they were born) and thus severely affected by extinction.

M-dwarfs are indeed the most common type of star, but not common enough to overcome the fact that their luminosities are much, much lower than hotter stars.

The dip in G stars is actually a peak in the K-stars caused by giants. Most low(ish) mass stars will evolve towards more or less the same point in the HR diagram forming the red clump. These stars are luminous, older and not concentrated in the galactic plane, can therefore be seen to great distances and produce a peak in the number counts.

So you can see this using actual data rather than a cartoon, I producedthe HR diagram for the brightest 10,000 stars in the revised Hipparcos catalogue (about V<6.7). You can clearly see the red clump of giants. Now collapse this distribution onto the B-V axis and you can see there is a local minimum at $0.6<B-V<0.8$ corresponding to G-stars.

Absolute colour-magnitude diagram for the brightest 10,000 stars Hipparcos HR diagram

The corresponding B-V distribution for these stars. G-stars have $0.6<B-V<0.8$ Histogram of B-V

  • $\begingroup$ So if there wasn't this red clump the curve would have a bell-shape with a peak at F? And the drop in G stars isn't a drop, but just a normal count that we would expect anyway - and it looks like a drop because there's actually a peak at K due to the red clump as you explained? Just recapitulating because I may or may not get this as an exam question this monday - yet my course book strangely doesn't say anything about a red clump. $\endgroup$ Jun 17, 2017 at 9:10
  • $\begingroup$ @StijnD'hondt Well the thing you showed is a cartoon, not real data. You also have to think about how steep the relationship is between spectral type and luminosity - spectral class is hardly a linear unit. The conclusion about the red clump is obvious if you look a the HR diagram above, but I'm not sure the decline at (2) is as steep as your cartoon shows. I'll do the equivalent for B-V. $\endgroup$
    – ProfRob
    Jun 17, 2017 at 14:32
  • $\begingroup$ @StijnD'hondt see new pictures. $\endgroup$
    – ProfRob
    Jun 17, 2017 at 15:05

I'd say there are fewer M stars on list because they are least bright. they may outnumber other star types but only nearest ones would make list of brightest stars. Type O stars are rare hence the small number. This info is available from a Wikipedia page on 'stellar classifications' and refer mostly to the mail sequence stars on the Hertzsprung–Russell diagram. The upper peak around type K (your point 3) is maybe explained by the addition of Red Giants and Bright Giants, but this is just a guess really from the HR diagram.

  • 1
    $\begingroup$ Feel free to provide direct links as necessary. That saves the person reading your answer from trying to find the source you mentioned themselves (and potentially finding the wrong one in the process). $\endgroup$
    – zephyr
    Jun 16, 2017 at 15:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .