2
$\begingroup$

a) We cannot detect all black holes and therefore don't know the percentage of the galaxy's mass they make up.

b) We do not know the mass of the Milky Way's central black hole.

c) According to our understanding of stellar evolution, black holes should make up a much lower percentage of the galaxy's mass.

d)According to our understanding of stellar evolution, black holes should make up a much higher percentage of the galaxy's mass.

In my own understanding, I feel like it should be a). Since the black holes are forming all the time, I think it is hard to detect all of them. But I am not quite sure, could someone give any insight?

$\endgroup$
  • $\begingroup$ We do know the mass of the Milky Way's central black hole. It's called Sagittarius A*, and it has a mass of ~4.1-4.5 million solar masses. $\endgroup$ – Phiteros Nov 20 '16 at 8:35
  • 1
    $\begingroup$ Is this some sort of homework question? Answering your homework is not the aim of this site. $\endgroup$ – Rob Jeffries Nov 20 '16 at 20:20
7
$\begingroup$

(a) We can't detect ALL of anything, but we can certainly makes estimates of their integral properties. For example, we estimate that there are about 100 billion stars in the galaxy, but we can observe only a tiny fraction of these. We have also observed a tiny fraction of the stellar black holes in our galaxy - those few examples that have revealed themselves through their binary companions.

(b) Yes we do. It is just over 4 million solar masses and thus a negligible fraction of the Galaxy's total mass.

(c) and (d) It is not just stellar evolution that is involved, it is the distribution of stellar birth masses (the initial mass function) that needs to be known, possibly as a function of time.

The stellar evolution part has a major uncertainty in that although it is reasonably certain that all stars above about 10 solar masses end their lives in supernovae, it is not certain what fraction will leave black hole remnants. This is probably a strong function of mass - higher masses favour a black hole and progenitor metallicity; low metallicity stars lose less mass during their lives and are more likely to directly collapse to a black hole - see Heger et al. (2003).

You can do a back of the envelope calculation. Suppose the IMF has the Saltpeter form $N(M)\propto M^{-2.35}$ between 0.1 and 100 solar masses. Further assume that: because a 1 solar mass star has a lifetime of 10 billion years, the age of the Galaxy is 10 billion years and the stellar lifetime is a very strong inverse function of mass $(\propto M^{-2.5})$, that the 100 billion stars that we deduce exist in our Galaxy (by counting up the local stars and measuring stellar densities elsewhere and extrapolating) are predominantly all the 0.1 to 1 solar mass stars that have ever been born. This allows us to then estimate how many stars have been born in any mass range - even if they have long-since died.

So making the above assumptions, I will make the further assumption that all stars from 20-100 solar masses die and leave a black hole remnant.Then if my maths is correct then there are about $10^{11}/1400 \sim 7\times 10^{7}$ black holes in our Galaxy. That is there is 1 black hole for every 1400 stars of 0.1-1 solar mass.

The remaining parts of the puzzle are what is the mass of a typical black hole and what is the mass of the Galaxy? A good number for the former is about 10 solar masses, since the black hole binaries appear to cluster at this value or a little lower (Ozel et al. 2010). The latter is still a topic of investigation. It appears to be an order of magnitude bigger than all the stars, gas and dust due to some form of dark matter. Let's use the number $7\times 10^{11}$ solar masses (Eadie & Harris 2016).

So my final number is that black holes form $7\times 10^{7}\times 10/7\times 10^{11} = 0.1$ percent of the Milky Way mass.

You could push this number up by a factor of a few by assuming that the IMF contained more higher mass stars in the past, or that the threshold mass for black hole production was lower in the past for low metallicity stars. Both seem theoretically possible and are active topics of investigation.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.