Main Sequence Stellar Mass Function?

On 27 okt, 23:37, wrote:
On Oct 25, 5:48 pm, Erik Max Francis wrote:

wrote:
The function does not match observed stellar counts exactly, but it's
close enough for government work.
What "counts" are you trying to reproduce? You ask about mass, but it
seems here you're talking about _distributions_ of masses.

Yes, distribution of masses. The Initial Mass Functions that I found
on the internet are solved for mass. They take mass and spit out the
size of the population. I needed a function that would take a random
number and spit out a mass. The above function does that. (Excepting I
found after some tests that base 10 works much better than base 2.35.)
Run the equation about ten million times, the distribution comes out
to well within an order of magnitude for most spectral classes which
was all I was trying for.

Really, it was so simple that I should have recognized it right away
but my math was so rusty that I didn't. The computer does the math, I
just supply the logic. Which unfortunately means that I need to grease
the rusty wheels in my head before the high school/college algebra
comes back to me on those rare occasions that I need it.

Note that ANY power law distribution of that shape will represent an
infinite amount of total mass. When the exponent is 2,35 or any other
number over 1, there is always infinite mass of stars smaller than any
given mass; and when the exponent is under 1, there is infinite mass
of stars larger than any given mas.

Seeing how you have picked an exponent over 1, how do you propose to
handle the low mass end?

Crown-Horned Snorkack wrote:

Note that ANY power law distribution of that shape will represent an
infinite amount of total mass. When the exponent is 2,35 or any other
number over 1, there is always infinite mass of stars smaller than any
given mass; and when the exponent is under 1, there is infinite mass
of stars larger than any given mas.

Seeing how you have picked an exponent over 1, how do you propose to
handle the low mass end?

That one's easy, at least for his purposes. The main sequence ends at
about 0.08 solar masses. To include brown dwarfs, or planets, he'd have
to use a different model, but we don't have good empirical data on their
distribution yet.

--
Erik Max Francis && && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
To be refutable is not the least charm of a theory.
-- Friedrich Nietzsche

On Oct 28, 1:12 pm, Erik Max Francis wrote:
Crown-Horned Snorkack wrote:
Note that ANY power law distribution of that shape will represent an
infinite amount of total mass. When the exponent is 2,35 or any other
number over 1, there is always infinite mass of stars smaller than any
given mass; and when the exponent is under 1, there is infinite mass
of stars larger than any given mas.

Seeing how you have picked an exponent over 1, how do you propose to
handle the low mass end?

That one's easy, at least for his purposes. The main sequence ends at
about 0.08 solar masses. To include brown dwarfs, or planets, he'd have
to use a different model, but we don't have good empirical data on their
distribution yet.

--
Erik Max Francis && &&http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
To be refutable is not the least charm of a theory.
-- Friedrich Nietzsche

8% solar mass sounds about right, whereas at that point unless there's
a reactor or some other weird fusion thing going on, the star output
is either dead or soon enough turning itself into a brown dwarf.

~ BG

BradGuth wrote:

8% solar mass sounds about right, whereas at that point unless there's
a reactor or some other weird fusion thing going on, the star output
is either dead or soon enough turning itself into a brown dwarf.

Stars don't "turn themselves" into brown dwarfs; they start as brown
dwarfs and never reach the main sequence.

--
Erik Max Francis && && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
A physicist is an atom's way of knowing about atoms.
-- George Wald

On Oct 28, 6:37 pm, Erik Max Francis wrote:
BradGuth wrote:
8% solar mass sounds about right, whereas at that point unless there's
a reactor or some other weird fusion thing going on, the star output
is either dead or soon enough turning itself into a brown dwarf.

Stars don't "turn themselves" into brown dwarfs; they start as brown
dwarfs and never reach the main sequence.

A white dwarf eventually becomes a brown dwarf, then a black dwarf.

The universe is at least ten fold older than you and most others
think.

~ BG

BradGuth wrote:

On Oct 28, 6:37 pm, Erik Max Francis wrote:
BradGuth wrote:
8% solar mass sounds about right, whereas at that point unless there's
a reactor or some other weird fusion thing going on, the star output
is either dead or soon enough turning itself into a brown dwarf.
Stars don't "turn themselves" into brown dwarfs; they start as brown
dwarfs and never reach the main sequence.

A white dwarf eventually becomes a brown dwarf, then a black dwarf.

Nope, that's not compatible with how astronomers use those terms. A
brown dwarf is not an intermediate form between white dwarfs and black
dwarfs; it's something else entirely.

The universe is at least ten fold older than you and most others
think.

And I and most others know well that you're completely loopy.

--
Erik Max Francis && && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
It was involuntary. They sank my boat.
-- John F. Kennedy (on how he became a war hero)

On 28 okt, 23:12, Erik Max Francis wrote:
Crown-Horned Snorkack wrote:
Note that ANY power law distribution of that shape will represent an
infinite amount of total mass. When the exponent is 2,35 or any other
number over 1, there is always infinite mass of stars smaller than any
given mass; and when the exponent is under 1, there is infinite mass
of stars larger than any given mas.

Seeing how you have picked an exponent over 1, how do you propose to
handle the low mass end?

That one's easy, at least for his purposes. *The main sequence ends at
about 0.08 solar masses. *To include brown dwarfs, or planets, he'd have
to use a different model, but we don't have good empirical data on their
distribution yet.

Fine. Let´s consider the implications, then.

Did the OP want 1 million stars, or 10 millions?

Exponent 2,35...2,33 has the nice property that 2 to power of 2,35 is
close to 5.

If there are absolutely no stars with masses below 0,08 solar, then
stars with masses 0,16 to 0,32 solar are 5 times less numerous than
stars with masses between 0,08 and 0,16 solar.

Thus, 80 % of all stars have masses between 0,08 and 0,16 solar.

Out of 10 millions, 2 millions have masses over 0,16 solar... then 400
000 have over 0,32 solar, 80 000 have over 0,64 solar, 16 000 have
over 1,28 solar, 3200 have over 2,5 solar, 640 have over 5 solar, 128
have over 10 solar, 25 have over 20 solar, 5 have over 40 solar and 1
has over 80 solar.

Note how the total is dominated by red dwarfs.

On Oct 29, 5:20*am, Crown-Horned Snorkack
wrote:

Note how the total is dominated by red dwarfs.

To answer specifically the question that was posed, I do not set the
minimum mass of 0.08 as a starting point. About 16.83% of the masses
fall below 0.08m, I simply discard them completely and do not factor
them into anything. (Technically these would be Brown Dwarfs or Super
Jupiter class planets depending on their mass. Currently I am not
using them, but I probably ought to code up a drawing routine for them
and reduce the number of empty nexuses I have by exchanging them for
Brown Dwarf systems.)

Below is some more detail:

In most space games you might care to name the distribution of solar
masses is generally quite off. Generally such games reduce the number
of M and K class stars greatly and increase the number of O, B, and A
class stars greatly, so that the distribution seems to be even across
the spectral classes at about 10% to 15% each. I was trying for a more
natural distribution, though matching the actual observed numbers
exactly is not a priority, since for my purposes, I would prefer fewer
O, B, and M stars and more K and G class stars than actually occur in
our general neighborhood.

Wikipedia gives the following distributions for the local area:
O: 0.00003%
B: 0.13%
A: 0.60%
F: 3.00%
G: 7.60%
K: 12.10%
M: 76.45%

The equation (mass = Math.Log(Random.NextDouble(), constantValue) *
-1) (where values 0.08 and 200 are disregarded and constantValue
= 10) provides the following distribution over 10 million iterations:
O: 0.00%
B: 0.12%
A: 2.57%
F: 6.01%
G: 8.26%
K: 22.83%
M: 60.21%

The numbers are within an order of magnitude (with the exception of
class O, which for map display purposes I don’t really want anyway
(they are just too large)) and classes K, G, and F are weighted more
heavily while M is weighted less heavily (which is also fine by me, it
gives more chance of habitable real-estate). I’d rather that A class
be lower, but 2% isn’t much of a problem. For my purposes this is
acceptable and probably more useful than more accurate numbers would
be, simply due to the reduction in the numbers of M class stars.

In article ,
Crown-Horned Snorkack writes:
Exponent 2,35...2,33 has the nice property that 2 to power of 2,35 is
close to 5.

The real stellar mass function turns over towards the low-mass end,
i.e., there are fewer low-mass stars than the power law would
suggest. The exact shape and turnover mass is a subject of current
research. It's not a subject I follow, but I seem to recall from
some talks that the turnover is around 0.4-0.5 solar masses, and the
mass function flattens or even declines below that. If there's
somebody reading who knows the correct answer, please post!

Note how the total is dominated by red dwarfs.

Yes, this is still true even with the turnover.

Regardless of the details, if the OP's purpose is a game, he need not
be too exact.

--
Steve Willner Phone 617-495-7123
Cambridge, MA 02138 USA
(Please email your reply if you want to be sure I see it; include a
valid Reply-To address to receive an acknowledgement. Commercial
email may be sent to your ISP.)

Steve Willner wrote:

In article ,
Crown-Horned Snorkack writes:
Exponent 2,35...2,33 has the nice property that 2 to power of 2,35 is
close to 5.

The real stellar mass function turns over towards the low-mass end,
i.e., there are fewer low-mass stars than the power law would
suggest. The exact shape and turnover mass is a subject of current
research. It's not a subject I follow, but I seem to recall from
some talks that the turnover is around 0.4-0.5 solar masses, and the
mass function flattens or even declines below that. If there's
somebody reading who knows the correct answer, please post!

Note how the total is dominated by red dwarfs.

Yes, this is still true even with the turnover.

Regardless of the details, if the OP's purpose is a game, he need not
be too exact.

The details of the "turnover" are error-prone enough that it's not clear
whether there really is a turnover, at least in terms of number density.
The initial mass function xi(M) as given in _Allen's Astrophysical
Quantities_ (p. 488) is

0.035 M^{-1.3+-0.6}, 0.08 = M = 0.50,
0.019 M^-2.2, 0.50 M = 1.0,
0.019 M^-2.7, 1.00 M = 100,

where xi(M) dM is the number of stars in the mass interval M to M + dM
(all masses in solar units).

Integrating this from M_0 to M_1 (where M_0 .. M_1 is the desired mass
range) will give you the number of stars in that mass range, but then
again you have to take into account the (huge) error bars on that first
range, which is precisely the one you're talking about.

--
Erik Max Francis && && http://www.alcyone.com/max/
San Jose, CA, USA && 37 18 N 121 57 W && AIM, Y!M erikmaxfrancis
The work will teach you how to do it.
-- (an Estonian proverb)