The luminosity of the Milky Way galaxy according to this is $5\times10^{36}$ Watts, but this number suggests that there are about 10 billion stars with Solar luminosities in the Milky Way, which doesn't sound right considering that the Milky Way contains 200-400 billion stars of different luminosities.

The same goes for Andromeda, it has a luminosity similar to the Milky Way with even more stars.

I know red dwarfs are the most abundant type of stars and their luminosities are much less than the Solar luminosity, but still, that would mean that both the Milky Way and Andromeda must contain much less than 10 billion stars with masses > 1 solar mass. Does that sound correct ?


2 Answers 2


It turns out that it is the distribution of birth stellar masses and most importantly, the lifetimes of stars as a function of mass that are responsible for your result.

Let's fix the number of stars at 200 billion. Then let's assume they follow the "Salpeter birth mass function" so that $n(M) \propto M^{-2.3}$ (where $M$ is in solar masses) for $M>0.1$ to much larger masses. There are more complicated mass function known now - Kroupa multiple power laws, Chabrier lognormal, which say there are fewer low mass stars than predicted by Salpeter, but they don't change the gist of the argument. Using the total number of stars in the Galaxy, we equate to the integral of $N(M)$ to get the constant of proportionality: thus $$n(M) = 1.3\times10^{10} M^{-2.3}.$$

Now let's assume most stars are on the main sequence and that the luminosity scales roughly as $L = M^{3.5}$ ($L$ is also in solar units), thus $dL/dM = 3.5 M^{2.5}$.

We now say $n(L) = n(M)\times dM/dL$ and obtain $$ n(L) = 3.7\times10^{9} M^{-4.8} = 3.7\times10^{9} L^{-1.37}.$$

The total luminosity of a collection of star between two luminosity intervals is $$ L_{\rm galaxy} = \int^{L_2}_{L_1} n(L) L \ dL = 5.9\times 10^{9} \left[L^{0.63} \right]^{L_{2}}_{L_1}$$ This equation shows that although there are far more low-mass stars than high mass stars in the Galaxy, it is the higher mass stars that dominate the luminosity.

If we take $L_1=0.1^{3.5}$ we can ask what is the upper limit $L_2$ that gives $L_{\rm galaxy} = 1.3\times 10^{10} L_{\odot}$ ($=5\times10^{36}$ W)?

The answer is only $3.5L_{\odot}$. But we see many stars in the Galaxy that are way brighter than this, so surely the Galaxy ought to be much brighter?

The flaw in the above chain of reasoning is that the Salpeter mass function represents the birth mass function, and not the present-day mass function. Most of the stars present in the Galaxy were born about 10-12 billion years ago. The lifetime of a star on the main sequence is roughly $10^{10} M/L = 10^{10} M^{-2.5}$ years. So most of the high mass stars in the calculation I did above have vanished long ago, so the mass function effectively begins to be truncated above about $0.9M_{\odot}$. But that also then means that because the luminosity is dominated by the most luminous stars, the luminosity of the galaxy is effectively the number of $\sim 1M_{\odot}$ stars times a solar luminosity.

My Salpeter mass function above coincidentally does give that there are $\sim 10^{10}$ star with $M>1M_{\odot}$ in the Galaxy. However you should think of this as there have been $\sim 10^{10}$ stars with $M>1 M_{\odot}$ born in our Galaxy. A large fraction of these are not around today, and that is actually the lesson one learns from the integrated luminosity number you quote!

EDIT: A postscript on some of the assumptions made. The Galaxy is much more complicated than this. "Most of the stars present in the Galaxy were born 10-12 billion years go". This is probably not quite correct, depending on where you look. The bulge of the Galaxy contains about 50 billion stars and was created in the first billion years or so. The halo also formed early and quickly, but probably only contains a few percent of the stellar mass. The moderately metal-poor thick disk contains perhaps another 10-20% and was formed in the first few billion years. The rest (50%) of the mass is in the disk and was formed quasi-continuously over abut 8-10 billion years. (Source - Wyse (2009)). None of this detail alters the main argument, but lowers the fraction of $>1M_{\odot}$ stars that have been born but already died.

A second point though is assuming that the luminosity of the Galaxy is dominated by main sequence stars. This is only true at ultraviolet and blue wavelengths. At red and infrared wavelengths evolved red giants are dominant. The way this alters the argument is that some fraction of the "dead" massive stars are actually red giants which typically survive for only a few percent of their main sequence lifetime, but are orders of magnitude more luminous during this period. This means the contribution of of the typical low-mass main sequence stars that dominate the stellar numbers is even less significant than the calculation above suggests.

  • $\begingroup$ "Most of the stars present in the Galaxy were born about 10-12 billion years ago". Are you sure this is correct ? $\endgroup$ Nov 29, 2014 at 23:17
  • $\begingroup$ Could you please explain what the IMF actually gives ? Like dose it give the distribution of mass among stars when the galaxy was forming ? Or does it give the distribution for all the stars that will be formed in the lifetime of the galaxy ? Any explanation would actually be very appreciated. $\endgroup$ Nov 30, 2014 at 14:23
  • $\begingroup$ @AbanobEbrahim The IMF is defined as the number of stars born per unit mass interval per unit volume per unit time. You then need to multiply by a star forming rate to get the mass spectrum of stars that are born. $\endgroup$
    – ProfRob
    Nov 30, 2014 at 14:27
  • $\begingroup$ Great, but how about brown dwarfs ? If they form by collapsing just like stars, then shouldn't we be expecting that their number is larger than the red dwarfs according to the IMF ? $\endgroup$ Nov 30, 2014 at 14:29
  • $\begingroup$ this chart:i.stack.imgur.com/sgrnw.jpg doesn't mention how much less than 0.25 solar masses this 41% represent, but would this mean that brown dwarfs are included in this < 0.25 percentage ? $\endgroup$ Nov 30, 2014 at 14:33

Yes, your qualitative argument is correct and the number of stars brighter than the Sun is almost certainly much smaller than 10 billion. The reason is that the luminosities are hugely variable. Due to the mass-luminosity relation, each doubling of the stellar mass corresponds to increasing the luminosity 10 times.

So many if not most of the "stars brighter than the Sun" have luminosities greater than 10 times the solar luminosity, so the number of these stars has to be 10 times lower than you think to fit the fixed total luminosity.

On the other hand, most of the stars in the Milky Way – or any galaxy – are much less luminous than the Sun, and they make up those 100-400 billion in the Milky Way. Despite their clear dominance over the total number, their total luminosity is a small fraction of the total luminosity of the galaxy – again because most of these stars have luminosities smaller than the solar one by orders of magnitude.


This pie shows that 12% of stars are heavier than the Sun but for the luminosity, the percentage could be smaller (or much smaller) because the relationship only holds for a given sequence, like the main sequence, of stars – and the Sun is a member of the most luminous sequence.

  • $\begingroup$ I once thought that the luminosity of the Milky Way was estimated by multiplying the number of stars by the luminosity of the Sun, but obviously this is not correct. I wonder how this luminosity for the MW was estimated. $\endgroup$ Nov 26, 2014 at 14:00
  • $\begingroup$ @AbanobEbrahim Almost certainly by using a "present day" mass function and a mass-luminosity relationship combined with some estimate for how many stars there are in total in the Galaxy. See below. $\endgroup$
    – ProfRob
    Nov 26, 2014 at 14:16

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