How close is Earth to the closest black hole?

[JMz]

Analyze the statistics a bit:

Almost all M and K dwarfs ever formed are still around - the world has not lasted long enough for any of them to burn out. The only possible fates have been stellar collision or ejection from Milky Way, both of them rare.
Oldest G dwarfs are now burning out.
And for F, A, B and O dwarfs, the dwarfs presently existing are a small fraction of stars that have existed and burnt out.
Furthermore, the star formation rate of Milky Way has not been constant - young Milky Way at some time held more young stars at one time than now.
Now, what becomes or these short lived dwarfs?
G, F and A dwarfs become white dwarfs.
Above a certain mass, stars commonly become neutron stars.
What becomes of them?
Neutron stars can be destroyed by merger.
Also many pulsars have high peculiar velocities.
It appears that the formation of neutron stars often (not always) gives them high peculiar velocities.
Which fraction of pulsars in Milky Way are bound thereto? Which fraction of neutron stars formed in Milky Way is left in Milky Way?

What is the minimum mass of star to form a black hole rather than neutron star?
Does formation of black holes tend to give holes initial peculiar velocities, like the formation of pulsar does? How does the distribution of black hole initial peculiar velocities compare against the distribution of pulsar initial peculiar velocities?
Which fraction of black holes formed in Milky Way are left in Milky Way?

All good questions, but all the ones that matter to the OP were presumably factored into the estimate that there are 100M BH's in the MW.

 

[Buzz Bloom]

Yup. Note that your assumption of the more-or-less-constant ratio leads directly to the estimate of a few dozen LY. (I used 1000, instead of 1800, so the cube root is particularly easy. ;-)

Hi JMz:

One source

http://www.atlasoftheuniverse.com/50lys.html​

gives 2000 stars closer than 50 ly. Using 2000 stars per BH instead of 1800 therefore says in a random sphere of radius 50 ly one would expect to find one BH. Id a 50 ly radius sphere is centered on the earth, a random point inside the sphere is on the average 3/4 of the radius from the center. There for the average expected distance to the nearest BH would be 37.5 ly away.

Regards,
Buzz

 

[JMz]

Yes, that's just right (with the assumptions we are both making). Note that, for these purposes, "3/4=1" (just back of the envelope, no careful count of the number of stars in the MW, "1800=2000", etc.). So 50 LY it is. Or 37, or 43! :-)

This thread has made me curious whether more careful modeling of the MW's structure and history would give a significantly different answer. I suspect not (by more than a factor of 2 or 3, anyway), but I will leave it to others to pursue that.

 

[stefan r]

All good questions, but all the ones that matter to the OP were presumably factored into the estimate that there are 100M BH's in the MW.

No not exactly. The original paper says

...For example,m_bh>30M black holes should have a local number density of 0.9-2 x 10¹⁴ Gpc^-3with the range reflecting variations between our fiducial Kroupa (2002) and metallicity dependent (Geha et al., 2013) IMFs. This corresponds to an
occupation rate of ∼1 per 1000 M of stars formed in galaxies with...

I think it is relevant that they are using cubic gigaparsecs. It is an error to assume population III stars formed inside the disc. There was no disc when pop III formed. A 30Kpc radius makes a sphere with 1.1 x 10-13Gpc volume. The black holes and the gas clouds around them were falling into the milky way (or other galaxy). Gas clouds slow down when passing through but black holes and large stars pass through and retain most of their momentum.

We know 1 black hole formed at the center of the milky way. The heat from the resulting quasar would have effected star formation in the inner milky way. The first star population does not need to precisely match the galaxy's density. The stars would have formed in places where it was both dense and cool.

https://en.wikipedia.org/wiki/Milky_Way
Total mass of stars : 100 billion sun-mass units

https://en.wikipedia.org/wiki/Stellar_classification
TYPE MASS FRACTION
O >=16 ~0.00003%
B 2.1 - 16 0.13%
A 1.4 - 1.8 0.6%
F 1.15 - 1.4 3%
G 0.8 - 1.4 7.6%
K 0.45 - 0.8 12.1%
M 0.48 - 0.45 76.45%
(My apologies for the chart being hard to read. I do not know how to format it better here.)
-> Average star mass ~0.56 sun-mass units
-> number of stars in Milky Way ~180 billion

Type B and O stars have short lifetimes. 0.13% is the fraction that we can see. If you are looking for black hole frequency you need to look at how many there were.

If you assumed (wrongly) that the mass distribution was constant you would need to multiply the the 0.13% B stars by ~400.

The size of stars formed is heavily influenced by the metalicity of the gas could it forms in. A cloud of helium and hydrogen resists compression/collapse better than a cloud with water and methane ice. Metals in the stars core help to ignite fusion. An early start to fusion will blow material away from the forming star. High metalicity clouds form lots of smaller stars. So the mass fraction of stars in the early milky way is very different from the current mass fraction.

In order to form a black hole the progenitor star needs to have more mass than the resulting black hole. The ratio of progenitor size to black hole size is dependent on metalicity. The paper points out that at Z/Z_Θ -1.5 you can form a 30M_Θ black hole with a 33M_Θ progenitor. At Z/Z_Θ -0.5 you need a 90M_Θ progenitor. At solar metalicity you will not have many 30M_Θ black holes forming. see figure 1 and second to last paragraph of introduction. .

If a black hole happens to be in the disc it is likely a coincidence or the black hole is just passing through.

 

[JMz]

No not exactly. The original paper says

I think it is relevant that they are using cubic gigaparsecs. It is an error to assume population III stars formed inside the disc. There was no disc when pop III formed. A 30Kpc radius makes a sphere with 1.1 x 10-13Gpc volume. The black holes and the gas clouds around them were falling into the milky way (or other galaxy). Gas clouds slow down when passing through but black holes and large stars pass through and retain most of their momentum.
...
If a black hole happens to be in the disc it is likely a coincidence or the black hole is just passing through.

I think we are well past back-of-the-envelope answers here.
As for Pop III BH's, they presumably formed before the MW itself formed. However, whether they have experienced enough dynamical friction (with the disk) to live in the disk, instead of the halo, is hard for me to even guess: I have seen no numerical simulations that could give an answer, in light of the fact that many (most?) may have "joined" the MW after the disk was formed, in the course of merging with smaller galaxies.

 

[Buzz Bloom]

Note that, for these purposes, "3/4=1" (just back of the envelope, no careful count of the number of stars in the MW, "1800=2000", etc.). So 50 LY it is. Or 37, or 43! :-)

Hi JMz:

I get that the original numbers of 100 billion and 100 million are likely to be an order of magnitude estimate, and that such numbers limit the precision of any calculations based on them to be at best order of magnitude as well. It has become a "bad" habit of mine to use one or two digits as a result of a calculation even when the actual precision is not close to that. I find that this tends to help me remember some of the factors that might become relevant if future results improve the precision of the input numbers. I will try to remember to make clear in the future what the actual estimate of precision is when I post a number.

I am guessing that the original numbers have a confidence level which is something like 70% that the actual numbers are between 30 and 300 million and billion. With this in mind, I think the confidence level is 70% that nearest black hole to Earth is between 8 and 200 ly away.

8 = 40 * 0.2 and 200 = 40 * 5​

This range is based the fact that the result is a multiple of the ratio between the two input values. Therefore the standard deviation of increases from 0.5 for one logarithmic variable to 0.7 for two logarithmic. If you have what you think is a better guess, I would appreciate your sharing that with me.

Regards,
Buzz

 

[JMz]

Hi JMz:

I get that the original numbers of 100 billion and 100 million are likely to be an order of magnitude estimate, and that such numbers limit the precision of any calculations based on them to be at best order of magnitude as well. It has become a "bad" habit of mine to use one or two digits as a result of a calculation even when the actual precision is not close to that. I find that this tends to help me remember some of the factors that might become relevant if future results improve the precision of the input numbers. I will try to remember to make clear in the future what the actual estimate of precision is when I post a number.

I am guessing that the original numbers have a confidence level which is something like 70% that the actual numbers are between 30 and 300 million and billion. With this in mind, I think the confidence level is 70% that nearest black hole to Earth is between 8 and 200 ly away.

8 = 40 * 0.2 and 200 = 40 * 5​

This range is based the fact that the result is a multiple of the ratio between the two input values. Therefore the standard deviation of increases from 0.5 for one logarithmic variable to 0.7 for two logarithmic. If you have what you think is a better guess, I would appreciate your sharing that with me.

Regards,
Buzz

A small refinement: The numbers may be uncertain by these factors, but the distance scale (over which the requisite stars and estimated BH's are found) is probably good. So the 5x factors you are using should be cube-rooted, right?

 

[Buzz Bloom]

certain by these factors, but the distance scale (over which the requisite stars and estimated BH's are found) is probably good. So the 5x factors you are using should be cube-rooted, right?

Hi JMz:

You raise a good point, and I will have to think about it some. What comes to mind is that

y = a x^1/3​

where

y is the radius of a sphere of interest,
x a number of stars, and
a is a value which I need to think about how it might be related to x .​

If a is not dependent on x, then (If I remember my probability theory correctly) the standard deviation of the y variable is related to the standard deviation of the x variable in a manner that depends on

dy/dx = (1/3) a x^-(2/3).​

So I don't think the answer is the cube root of 5, but it may be 5^2/3 ~=3. If that is the case, then the 70% radius range would be about 10 to 100.

Regards,
Buzz

 

[JMz]

Hi JMz:

You raise a good point, and I will have to think about it some. What comes to mind is that

y = a x^1/3​

where

y is the radius of a sphere of interest,
x a number of stars, and
a is a value which I need to think about how it might be related to x .​

If a is not dependent on x, then (If I remember my probability theory correctly) the standard deviation of the y variable is related to the standard deviation of the x variable in a manner that depends on

dy/dx = (1/3) a x^-(2/3).​

So I don't think the answer is the cube root of 5, but it may be 5^2/3 ~=3. If that is the case, then the 70% radius range would be about 10 to 100.

Regards,
Buzz

I haven't thought about that closely, but my gut response is that the (2/3) applies to the variance, so it's still (1/3) for the std.dev. But an easier way might be to just ask for the distribution's 15th & 85th percentiles, which might be calculated easily. (Even if the distribution isn't straightforward, it might be estimated quickly with Monte Carlo.)

 

[Buzz Bloom]

Even if the distribution isn't straightforward, it might be estimated quickly with Monte Carlo.

Hi JMz:

Here is the result of the Monte Carlo calculation.

I generated 400 =rand() numbers, and for each group of four (a,b,c,d) I calculated a+b-c-d. This gave me 100 approximately Gaussian distributed numbers, A(1)..A(100), with a zero mean.​

I then calculated
B(1)..b(100), B(i) = A(i)².
I calculated the standard deviation as the square-root of the average of the squares of these 100 B(i) numbers. The result:

stdev(B) = 0.52418742.​

I also calculated the cube roots of each of the 100 B numbers to get C(i) = B(i)^1/3, and then:

stdev(C) = 0.72593341, and then five iterations to get five values for
stdev(C)/stdev(B) = 1.383 +/- 0.004.​

I do not know what the significance of 1.383 is, but the cube root of 3 is ~1.442, so maybe the difference is related to not having a good enough approximation for a Gaussian distribution. I will try again with more terms in the 100 numbers and see what happens.

Regards,
Buzz

 

[Chronos]

Rob Jeffriew, an astrophysicist at Keele University [UK] has offered the best estimate I 've found - that being 18 parsecs to the nearest black hole. He also calculated the probable distance to the nearest white dwarf and neutron star - those being 5 and 11 parsecs, respectively. Since white dwarfs are easily detectable over such distances, the 5 pc estimate offers an easy humor check- and the nearest known white dwarf [Sirius B], happens to be 2.3 parsecs from earth. So, Dr Jeffries calculations look rather convincing. For further details, see https://astronomy.stackexchange.com...-nearest-compact-star-remnant-likely-to-be..A current listing of known black holes can be found at http://www.astro.puc.cl/BlackCAT/

 

[Buzz Bloom]

Rob Jeffriew, an astrophysicist at Keele University [UK] has offered the most best estimate I 've found - that being 18 parsecs to the nearest black hole.

Hi Cronos:

Thank you for your post. I am very pleased to see the 18 pc expected closest distance which is very close to the 40 ly distance I calculated. It seems that he used approximately the same input values and a similar method of calculation to my effort.

Regards,
Buzz

 

[Chronos]

Buzz, as Dr. Jeffries has noted, this estimate is sensitive to a number of unknowns - lncluding historical supernova rates and stellar mass distributions, so getting a decent estimate based on current observational data is limited. My best guess is our current empirical data is at no better than the 1 sigma confidence level, so the relative accuracy of the one easily verifiable prediction [nearest white dwarf] is fairly impressive. Fortunately, this is great news for aspiring astrophysicists

 

[Buzz Bloom]

Hi @JMz:

I have decided that I need to redo my Monte Carlo calculations. First, I have found buggy behavior in the spread sheet package I am using on my Windows 7, so I am going to switch to a very old Excel on my very old XP PC that seems to still be a reliable tool, although more limited. Then I need to take into account that the random variable for the number of stars within a specified radius sphere is 10^R where R is a random number from an approximate Gaussian distribution with a mean of 3 and a standard deviation of 0.5, since the calculation is for a sphere containing 1000 stars with a random range between about 300 and 3000. I will post the results when I complete the new spread sheet.

RESULTS
I performed 20 runs of 100 trials each. Each trial generated a random number of stars and of BHs. This was done by generating the log based 10 of the number from an approximate Gaussian distribution with a standard deviation of 0.5, an respective means of 8 and 11. (I assumed that estimates of the 100 million BHs and 100 billion stars were independent.) For each run I calculated the expected value of the ratio of stars to BHs, as well as the corresponding value of 10^(mn+dv) and 10^(mn-sd). The results of the 20 runs showed a mean of approximately 1000 with 65% of the trials in the range 200 to 5000.

The average distance of a BH from the center of a sphere (at a random point in the sphere) with 1000 stars is approximately 40 ly. Taking into account that the size of a sphere corresponding to the star/BH ratio is proportional to the cube root of this ratio, the range of expected distance with a 65% confidence level would therefore be between
40 / 5^1/3 ~= 24 ly and 40 × 5^1/3 ~= 680 ly.

Regards,
Buzz

 

[JMz]

Hi JMz:

Here is the result of the Monte Carlo calculation.

I generated 400 =rand() numbers, and for each group of four (a,b,c,d) I calculated a+b-c-d. This gave me 100 approximately Gaussian distributed numbers, A(1)..A(100), with a zero mean.​

I then calculated
B(1)..b(100), B(i) = A(i)².
I calculated the standard deviation as the square-root of the average of the squares of these 100 B(i) numbers. The result:

stdev(B) = 0.52418742.​

I also calculated the cube roots of each of the 100 B numbers to get C(i) = B(i)^1/3, and then:

stdev(C) = 0.72593341, and then five iterations to get five values for
stdev(C)/stdev(B) = 1.383 +/- 0.004.​

I do not know what the significance of 1.383 is, but the cube root of 3 is ~1.442, so maybe the difference is related to not having a good enough approximation for a Gaussian distribution. I will try again with more terms in the 100 numbers and see what happens.

Regards,
Buzz

Purely technical question: In what language are you working? (Hence, what is the RAND function?) Matlab? XL? something else?

 

[JMz]

Buzz, as Dr. Jeffries has noted, this estimate is sensitive to a number of unknowns - lncluding historical supernova rates and stellar mass distributions, so getting a decent estimate based on current observational data is limited. My best guess is our current empirical data is at no better than the 1 sigma confidence level, so the relative accuracy of the one easily verifiable prediction [nearest white dwarf] is fairly impressive. Fortunately, this is great news for aspiring astrophysicists

Ahh, back to good old days, when "astrophysical accuracy" meant "within a factor of 2" (i.e., the error was in the exponent). ;-)

 

[JMz]

Purely technical question: In what language are you working? (Hence, what is the RAND function?) Matlab? XL? something else?

(I didn't see your statement that you are using XL until after I posted this.)

 

[JMz]

Hi @JMz:

I have decided that I need to redo my Monte Carlo calculations. First, I have found buggy behavior in the spread sheet package I am using on my Windows 7, so I am going to switch to a very old Excel on my very old XP PC that seems to still be a reliable tool, although more limited. Then I need to take into account that the random variable for the number of stars within a specified radius sphere is 10^R where R is a random number from an approximate Gaussian distribution with a mean of 3 and a standard deviation of 0.5, since the calculation is for a sphere containing 1000 stars with a random range between about 300 and 3000. I will post the results when I complete the new spread sheet.

RESULTS
I performed 20 runs of 100 trials each. Each trial generated a random number of stars and of BHs. This was done by generating the log based 10 of the number from an approximate Gaussian distribution with a standard deviation of 0.5, an respective means of 8 and 11. (I assumed that estimates of the 100 million BHs and 100 billion stars were independent.) For each run I calculated the expected value of the ratio of stars to BHs, as well as the corresponding value of 10^(mn+dv) and 10^(mn-sd). The results of the 20 runs showed a mean of approximately 1000 with 65% of the trials in the range 200 to 5000.

The average distance of a BH from the center of a sphere (at a random point in the sphere) with 1000 stars is approximately 40 ly. Taking into account that the size of a sphere corresponding to the star/BH ratio is proportional to the cube root of this ratio, the range of expected distance with a 65% confidence level would therefore be between
40 / 5^1/3 ~= 24 ly and 40 × 5^1/3 ~= 680 ly.

Regards,
Buzz

Typo in that last number: 40*5^(1/3) ~ 68. This all seems quite believable.

 

[Buzz Bloom]

Purely technical question: In what language are you working? (Hence, what is the RAND function?) Matlab? XL? something else?

Hi JMz:

RAND() is a function provided by Excel. It generates a (pseudo) random number with a flat distribution between zero and one.

Regards,
Buzz

 

[JMz]

Hi JMz:

RAND() is a function provided by Excel. It generates a (pseudo) random number with a flat distribution between zero and one.

Regards,
Buzz

OK. Note that there are simple methods to quickly generate exact standard Normal samples from Uniform ones: e.g., https://en.wikipedia.org/wiki/Box–Muller_transform. The polar version (Bell, Knop) needs about 20% more U's than can be used, but both methods otherwise produce ~ 1 Normal for each Uniform, rather than needing several U's for a single N.

 

[hilbert2]

Could there be some "anthropic" reason why a black hole can't exist near a planet that has life on it? If a significant amount of mass per second is spiralling into a BH, how far away from it do you have to be to not get fried by X-rays and gamma rays? I don't even know how to do an order of magnitude estimate for this, but I remember seeing a claim that a supernova explosion taking place too close would destroy everything living on Earth.

 

[graybass]

So we agree the estimated (but not known) distance to be approx. 3-400 LY?

 

[stefan r]

Could there be some "anthropic" reason why a black hole can't exist near a planet that has life on it? If a significant amount of mass per second is spiralling into a BH, how far away from it do you have to be to not get fried by X-rays and gamma rays? I don't even know how to do an order of magnitude estimate for this, but I remember seeing a claim that a supernova explosion taking place too close would destroy everything living on Earth.

No, Jupiter gives of x-rays for example. We also get lots of x-rays from sun storms. Doctors on Earth use molybdenum targets in x-ray generators because the k-alpha radiation is a higher energy. Copper targets are better because they cool faster. Copper is also cheaper and mechanically better. Copper k-alpha radiation is used in research x-ray machines. No high energy radiation is really safe but the medical/dental x-rays are much less damaging to living tissue. The atmosphere blocks most of the x-rays that are most harmful.

The level of x-rays that are life threatening and the level that is detectable has many orders of magnitude difference. So there would be plenty of sources for Chandra to find.

Particles falling in radiate energy.

This process of accretion is one of the most efficient energy-producing processes known; up to 40% of the rest mass of the accreted material can be emitted as radiation.

Someone with a better background with plasmas might be able to estimate the rate interstellar gas falls in. The density of the interstellar medium varies a lot.

 

[Buzz Bloom]

So we agree the estimated (but not known) distance to be approx. 3-400 LY?

Hi graybass:

I can't figure out where your numbers come from, but the statement I would make based on my Monte Carlo calculation with @JMz 's correction, is:
the probability is 65% that the distance to the nearest black hole is between 30 and 85 ly.

Regards,
Buzz

 

[JMz]

Hi graybass:

I can't figure out where your numbers come from, but the statement I would make based on my Monte Carlo calculation with @JMz 's correction, is:
the probability is 65% that the distance to the nearest black hole is between 30 and 85 ly.

Regards,
Buzz

Out of curiosity, Buzz, what are the median and quartile distances? These won't depend on any parameter estimates like the std.dev. They are robust to outliers (non-Normal variates).

 

[mfb]

We know 60 black holes in our galaxy today.

Gaia is expected to find a few thousand. It is basically guaranteed that many of them will be closer than the currently closest known.
As these are double systems, we might have to wait for the final data release, 2022-2024.

 

[stefan r]

... find a few thousand. ...

I the article

https://www.sciencenews.org/article/we-share-milky-way-100-million-black-holes​

says

The Milky Way teems with black holes — about 100 million of them.​

...

The arxiv article says 10⁵ black holes in the milky way. Did someone miss by x10³?

 

[mfb]

That is a good question. Both numbers are recent, and a factor 1000 is large even for astrophysical uncertainties.

Gaia will measure roughly 1% of the stars in our galaxy, measuring O(1%) of the black holes in binary systems doesn't look unreasonable.

 

[Buzz Bloom]

The arxiv article says 10⁵ black holes in the milky way. Did someone miss by x10³?

Hi stefen:

Here is a quote from the https://arxiv.org/pdf/1703.02551.pdf article.
... a galaxy like the Milky Way should host millions of ∼30M_sun black holes ...
I confess that the bulk of the article is over my head. However, Figure 2 seems to show the basis for estimating the MW BHs.
https://en.wikipedia.org/wiki/Milky_Way says the mass of the MW is ~10¹² solar masses.

As I read this chart, the black line gives the number of BHs of mass greater than 10 solar masses. The X-axis shows the mass of a galaxy. One has to extrapolate the black line since the max Y-axis value is 10⁸.

Regards,
Buzz

 

[Buzz Bloom]

Out of curiosity, Buzz, what are the median and quartile distances?

Hi JMz:

I will have to go back and explore the Monte Carlo data. It will try to post the answer later today or tomorrow.

Regards,
Buzz

 

[Buzz Bloom]

Gaia is expected to find a few thousand.

Hi mfb:

Thanks very much for the link. The previous abstract about GAIA failed to mention anything about finding black holes.

Regards,
Buzz

 

[snorkack]

That is a good question. Both numbers are recent, and a factor 1000 is large even for astrophysical uncertainties.

The article contains a bland statement that most black holes should be binaries with a luminous partner.

Absurd on its face.

Luminous stars are short lived. A great majority of black holes should not have a luminous partner now, even if they had one at some point of evolution.

 

[mfb]

The article contains a bland statement that most black holes should be binaries with a luminous partner.

Absurd on its face.

Luminous stars are short lived. A great majority of black holes should not have a luminous partner now, even if they had one at some point of evolution.

Why do you expect binary stars to have symmetric masses?

 

[snorkack]

Why do you expect binary stars to have symmetric masses?

I did not.

On closer examination, it turns out that the article does interpret "luminous" apparently to mean "giving off any, even small amount of light", and expressly considers binaries of black hole and low mass old main sequence star - although these are hard to detect.

So: binary systems of a black hole and a massive star should commonly form, but are short-lived.
Binary systems of a black hole and a low mass main sequence star may form less often, but then last longer.

 

[stefan r]

Hi mfb:

Thanks very much for the link. The previous abstract about GAIA failed to mention anything about finding black holes.

Regards,
Buzz

I believe all the black holes GAIA will find have luminous companions. Gaia is looking at momentum and position.

Why do you expect binary stars to have symmetric masses?

There is this paper

...massive stars preferentially have massive companions...

If they came from the same cloud at the same time my first guess would be somewhat similar stars form. They have nearly identical metalicity and were probably kick started by the same shockwave. Most of the poster image open clusters are all the same color. The small dim companions do not show up as well in the poster so that is not an accurate measurement. In pictures of stellar nurseries like Orion nebula you have a group of B and O stars blowing material away and smaller stars forming in the remnants. Again that is not statistical evidence but would be a good first guess.