How bright is the Milky Way?
Abdul Ahad
across the whole night sky (what can only be done with some calculated
"guesswork" unless I'm mistaken), I wonder how much collective light
that would amount to?
Even at midnight in the most favourable latitudes on Earth, the sky is
never totally pitch black for meaningful measurements to be made which
exclude sky background brightness. I suppose from really *far*
orbiting satellites such measurements could be easily recorded with a
360-degree field, and manually edited to exclude some of the brighter
nearby stars that happen to lie in the galactic plane. Likewise some
adjustments could be made for obscuring areas like the Coalsack near
Crux and the Cygnus Rift, depending on how 'fussy' we wanted to be
about our final integrated magnitude figure.
Imagine a scene where one is located inside a glass cockpit of an
imaginary starship sailing the black interstellar void, some half way
between the Sun (Sol) and Alpha Centauri with a 360-degree sky view.
From this vantage point (2.2 light years away) the Sun will shine at
magnitude around -1.2 in one part of the sky, and Alpha Centauri, may
be shining a touch brighter at -1.5, in the opposite part of the sky,
with the Milky Way banding virtually all around. What a splendid view
that would make! No sunshine, just eternal night!
Bottom line question is: do the 200 odd billion stars estimated to
make up half the galaxy *this side* of the Milky Way's core,
collectively outshine say Venus (circa. -4.4 at max) or Sirius (-1.4)
in our skies? Has any studies been done in this area I wonder. If
not, does anyone like to take a guess at the fully integrated
magnitude number?!
Hope its not too much of a brain teaser.
Cheers
Abdul Ahad
http://uk.geocities.com/aa_spaceagent/spaceprojects.html
"We have lingered long enough on the shores of the cosmic ocean. We
are at last ready to set sail for the stars" - Carl Sagan.
Francis Marion
I've seen a similar question asked about the total amount of light received
from the Earth's moon over a period of one year.
As I recall, if one were to be able to concentrate "ALL" of the light
received by the moon, in all phases for the entire 365+ days, it would
amount to less than .5 second of light from the sun.
Extending this information to your question might give some idea as to the
answer you could expect.
I will be interested to see if someone has a better answer for you.
good luck,
Francis Marion
William C. Keel
> If we integrated light from the entire stretch of the Milky Way band
> across the whole night sky (what can only be done with some calculated
> "guesswork" unless I'm mistaken), I wonder how much collective light
> that would amount to?
> Even at midnight in the most favourable latitudes on Earth, the sky is
> never totally pitch black for meaningful measurements to be made which
> exclude sky background brightness. I suppose from really *far*
> orbiting satellites such measurements could be easily recorded with a
> 360-degree field, and manually edited to exclude some of the brighter
> nearby stars that happen to lie in the galactic plane. Likewise some
> adjustments could be made for obscuring areas like the Coalsack near
> Crux and the Cygnus Rift, depending on how 'fussy' we wanted to be
> about our final integrated magnitude figure.
A rough number bandied about in comparing ground- and orbitally-based
observations is that half of the sky brighness at the best ground-based
locations is from airglow and half from starlight and starlight scattered
from dust grains. Near the ecliptic plane, that includes the
local dust of the zodiacal light. The best separation of these
components has been using the photopolarimeter on Pioneer 10 while
it was beyond the asteroid belt (and hence virtually all the
zodiacal dust contribution). Here's a paper by Toller, Tanabe, and
Weinberg in which they did the tedious job of removing te light
of catalogued stars brighter than 8th magnitude and mapping what's
left over:
http://adsabs.harvard.edu/cgi-bin/nph-bib_query?bibcode=1987A%26A...188...24T&db_key=AST&high=3ea7e5a30102376
At the dimmest regions, this "diffuse" starlight amounts to about the
brightness of about 50 10th-magnitude stars per square degree
(this unit is known as S(10), here measured in the V band).
Aha! There are some numbers available in the book _The Light of the
Night Sky_ by F.E. Roach and Janet Gordon. The total visual starlight
from stars is listed at 51x brighter than Sirius, or visual magnitude
-1.58 - 2.5 log (51) = -5.85. No wonder the Moon wipes out our
view so easily.
Bill Keel
P. Edward Murray
sites, say in Zambia, Africa and other places, the light of the Milky
Way does cast shadows.
Ed
Tony Flanders
> Bottom line question is: do the 200 odd billion stars estimated to
> make up half the galaxy *this side* of the Milky Way's core,
> collectively outshine say Venus (circa. -4.4 at max) or Sirius (-1.4)
> in our skies?
That's not the right way to phrase it; most of the stars on this side
of the Milky Way's core are invisible due to obscuring dust.
In general, the total brightness of the stars below the traditional
mag 6.5 naked-eye limit -- the ones that merge to form the traditional
Milky Way -- is greater than the total brightness of the naked-eye
stars. At most times of year and locations, Venus is significantly
fainter than the total brightness of all the sub-mag-6.5 stars above
the horizon.
According to my calculations, the total brightness of *all* the stars
mag 6.5 and brighter is -4.9, just a bit brighter than Venus at its
brightest.
The other major sources of light in a "dark" sky are the zodiacal light
and airglow. Added together, a "dark" sky doesn't seem dark at all;
my first impression is always that it is dazzlingly bright. And in
an open area, especially on a reflective surface, it casts plenty
enough light to walk around confidently, to see one's telescope
easily, and so on.
- Tony Flanders
Abdul Ahad
> aa_spa...@yahoo.co.uk (Abdul Ahad) wrote in message news:<3416b228.04031...@posting.google.com>...
>
> > Bottom line question is: do the 200 odd billion stars estimated to
> > make up half the galaxy *this side* of the Milky Way's core,
> > collectively outshine say Venus (circa. -4.4 at max) or Sirius (-1.4)
> > in our skies?
>
> mag 6.5 and brighter is -4.9, just a bit brighter than Venus at its
> brightest.
>
Looking through some formulae I scribbled in an old [school] exercise
book of mine many years ago, I found the method for integrating
magnitudes of two stars as follows:-
If we let M(1) = magnitude of brighter star, M(2) = magnitude of
fainter star and D = M(2) minus M(1). Then the "brightness ratio", R =
10^[0.4 * D].
The integrated magnitude of the two stars is then:-
M(1)+M(2) = M(2) - 2.5 * Log (R+1)
So, for example, in a close telescopic binary system of two stars of
magnitudes +5.2 and +6.2, the combined magnitude of the two stars as
seen with the naked eye (appearing as one star) is +4.8.
This methodology is bound to prove cumbersome if one is intgegrating
*several thousand* stars down to say magnitude 5 or six. Also, given
that most catalogues record magnitudes accurate to just 1 decimal
place, I wonder if the accumulation of rounding errors over thousands
of calculations becomes significant?
Is there a more general algorithm available for aggregating magnitudes
of "n" stars, such that the total integrated magnitude, M(I) = M(1) +
M(2) + M(3) + ... M(n)?
Abdul Ahad
Brian Tung
> This methodology is bound to prove cumbersome if one is intgegrating
> *several thousand* stars down to say magnitude 5 or six. Also, given
> that most catalogues record magnitudes accurate to just 1 decimal
> place, I wonder if the accumulation of rounding errors over thousands
> of calculations becomes significant?
> Is there a more general algorithm available for aggregating magnitudes
> of "n" stars, such that the total integrated magnitude, M(I) = M(1) +
> M(2) + M(3) + ... M(n)?
Sure. The formula you found is based on adding the relative luminosities
of the stars. That is, suppose you have two stars of magnitudes M1 and
M2. Then their luminosities L1 and L2 are related by
L2/L1 = 10^[0.4*(M1-M2)]
The luminosity of the pair of stars is L1 + L2 = L1(1 + L2/L1), and their
combined magnitude is then
Mc = M1 - 2.5 log10 (1 + L2/L1)
We can generalize this by computing all the ratios
Li/L1 = 10^[0.4*(M1-Mi)]
for all stars i from 2 through n. The combined magnitude is then
Mc = M1 - 2.5 log10 (1 + L2/L1 + L3/L1 + ... + Ln/L1)
Actually, the Hipparcos and Tycho-2 catalogues list magnitudes to two
decimal places, which ought to be good enough for a ballpark estimate.
An error of 0.01 in the magnitude corresponds to about a 1 percent
error in luminosity.
Brian Tung <br...@isi.edu>
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.txt
Abdul Ahad
> In sci.astro.amateur Abdul Ahad <aa_spa...@yahoo.co.uk> wrote:
> > Imagine a scene where one is located inside a glass cockpit of an
> > imaginary starship sailing the black interstellar void, some half way
> > between the Sun (Sol) and Alpha Centauri with a 360-degree sky view.
> > From this vantage point (2.2 light years away) the Sun will shine at
> > magnitude around -1.2 in one part of the sky, and Alpha Centauri, may
> > be shining a touch brighter at -1.5, in the opposite part of the sky,
> > with the Milky Way banding virtually all around. What a splendid view
> > that would make! No sunshine, just eternal night!
>
> At the dimmest regions, this "diffuse" starlight amounts to about the
> brightness of about 50 10th-magnitude stars per square degree
> (this unit is known as S(10), here measured in the V band).
>
> Aha! There are some numbers available in the book _The Light of the
> Night Sky_ by F.E. Roach and Janet Gordon. The total visual starlight
> from stars is listed at 51x brighter than Sirius, or visual magnitude
> -1.58 - 2.5 log (51) = -5.85. No wonder the Moon wipes out our
> view so easily.
>
It appears that my *hypothetical* interstellar star ship crossing the
vast 270,000 A.U. expanse between Pluto and the outer edge of the
Alpha Centauri system will indeed be deprived of natural sky light!
The inhabitants of such a journey will not be able to grow plants in
domed enclosures solely by the feeble light output of 200 odd billion
suns shining in the cosmic night sky all around the ship...
Photosynthesis and the flowering and fruiting cycles will invariably
need artificial lighting. Sun tans for lighter skinned people will be
out of the question too, but deep sky objects (for those with
astronomical inclines) will exhibit strong colours and stunning
details even with the naked eye!
And who knows what colour the Oort Cloud might appear when backlit by
the distant Sun or the diffuse glow of the Milky Way?
We dream of the far distant future.
Abdul Ahad
Tombo
> It appears that my *hypothetical* interstellar star ship crossing the
> vast 270,000 A.U. expanse between Pluto and the outer edge of the
> Alpha Centauri system will indeed be deprived of natural sky light!
>
> The inhabitants of such a journey will not be able to grow plants in
> domed enclosures solely by the feeble light output of 200 odd billion
> suns shining in the cosmic night sky all around the ship...
> Photosynthesis and the flowering and fruiting cycles will invariably
> need artificial lighting. Sun tans for lighter skinned people will be
> out of the question too, but deep sky objects (for those with
> astronomical inclines) will exhibit strong colours and stunning
> details even with the naked eye!
> And who knows what colour the Oort Cloud might appear when backlit by
> the distant Sun or the diffuse glow of the Milky Way?
>
> We dream of the far distant future.
>
> Abdul Ahad
Nice work "Abdul"..you trolled some of the best suckers here hook, line and
sinker. LOL
Abdul Ahad
> Abdul Ahad wrote:
> > This methodology is bound to prove cumbersome if one is intgegrating
> > *several thousand* stars down to say magnitude 5 or six. Also, given
> > that most catalogues record magnitudes accurate to just 1 decimal
> > place, I wonder if the accumulation of rounding errors over thousands
> > of calculations becomes significant?
> > Is there a more general algorithm available for aggregating magnitudes
> > of "n" stars, such that the total integrated magnitude, M(I) = M(1) +
> > M(2) + M(3) + ... M(n)?
>
> Sure. The formula you found is based on adding the relative luminosities
> of the stars. That is, suppose you have two stars of magnitudes M1 and
> M2. Then their luminosities L1 and L2 are related by
>
> L2/L1 = 10^[0.4*(M1-M2)]
>
> The luminosity of the pair of stars is L1 + L2 = L1(1 + L2/L1), and their
> combined magnitude is then
>
> Mc = M1 - 2.5 log10 (1 + L2/L1)
>
> We can generalize this by computing all the ratios
>
> Li/L1 = 10^[0.4*(M1-Mi)]
>
> for all stars i from 2 through n. The combined magnitude is then
>
> Mc = M1 - 2.5 log10 (1 + L2/L1 + L3/L1 + ... + Ln/L1)
>
respective magnitudes: m1 +4.5, m2 +4.8 and m3 +5.2, what figure do
you make the combined magnitude, Mc?
Ernie Wright
> So using this generalized formula for "n" stars, for 3 stars of
> respective magnitudes: m1 +4.5, m2 +4.8 and m3 +5.2, what figure do
> you make the combined magnitude, Mc?
L2/L1 = 10^(0.4 * (M1 - M2)) = 0.76
L3/L1 = 10^(0.4 * (M1 - M3)) = 0.52
Mc = M1 - 2.5 * log10 (1 + L2/L1 + L3/L1)
= M1 - 2.5 * log10 (1 + 0.76 + 0.52)
= 4.5 - 2.5 * log10 2.28
= 3.6
But there's another (equivalent) way to do this that might be easier to
remember.
The reason you can't just add up the magnitudes is that they're
logarithmic, and adding logarithms is the same as *multiplying* the base
numbers, which isn't what you want. You need to convert the logarithms
to a linear scale before adding, then convert back to a logarithm
afterward.
The above formulas do this in a funny way, by converting the differences
between magnitudes. But you can skip the difference step and convert
the magnitudes directly.
Lc = 0.4^M1 + 0.4^M2 + ... + 0.4^Mn
Mc = -2.5 * log10 Lc
Easy peasy. For the example you gave,
Mc = -2.5 * log10 (0.4^4.5 + 0.4^4.8 + 0.4^5.2) = 3.6
The disadvantage of this is that you can't easily use base-10 logarithm
tables or sliderules to calculate it, but hopefully most people by now
have access to a calculator that can raise 0.4 to a fractional power.
Brian Tung
> The above formulas do this in a funny way, by converting the differences
> between magnitudes. But you can skip the difference step and convert
> the magnitudes directly.
>
> Lc = 0.4^M1 + 0.4^M2 + ... + 0.4^Mn
> Mc = -2.5 * log10 Lc
>
> Easy peasy.
Duh. What can I say, I had a Slow Moment. Although we should really
use 10^(-0.4) instead of 0.4, and 10^(0.4) instead of 2.5. Of course,
available precision probably means the two are mostly indistinguishable.
BTW, on the other thread--the one about Google--I figured that it was
something about reformatting, but I couldn't figure out their algorithm
for doing so. I couldn't figure out why some lines were slid over and
others weren't. Do you understand it in that detail and can you explain
it?
Ernie Wright
>> Easy peasy.
>
> Duh. What can I say, I had a Slow Moment.
Fortunately, that's only accurate for unusual values of "slow," and in
any event this occurs infrequently. That way I can relax and learn
something while mostly lurking. :)
> Although we should really use 10^(-0.4) instead of 0.4, and 10^(0.4)
> instead of 2.5. Of course, available precision probably means the
> two are mostly indistinguishable.
I agree, except for a small thing that will take me a while to explain,
after which you'll probably want to kill me. But I've waded in this
far...
We should also say *why* we're using those numbers: Magnitudes are
logarithmic, and 10^(-0.4) is the base.
A difference of five magnitudes corresponds to a difference of 100 in
luminosity, so that the log base of the magnitude scale is the value of
B in the equation
B^5 = 1/100
which is 100^(-1/5) or 10^(-2/5). It's the inverse of 100 rather than
simply 100 because larger magnitudes correspond to smaller brightness.
Knowing that B = 10^(-2/5), we can convert magnitudes to a linear scale
just by raising B to the power M (the magnitude). For example,
M = -1.0, B^M = 2.512
M = 0.0, B^M = 1.0
M = 1.0, B^M = 0.398 = B
M = 2.0, B^M = 0.158
M = 3.0, B^M = 0.0631
M = 4.0, B^M = 0.02512
M = 5.0, B^M = 0.01
So magnitude 0.0 corresponds to a "brightness" of 1.0, and mag 5.0 is
1/100 on our linear scale, like we'd expect.
We use the other number, -2.5, to convert between bases when we go in
the other direction, from linear brightness to magnitude. Computers and
calculators don't have built-in functions for dealing with logarithms of
base B, but they do have functions for log base 10. To convert from
base 10 to base B (the magnitude), we just have to multiply by the
constant
1 / log10 B
which is
1 / (log10 10^(-0.4)) = 1/-0.4 = -2.5
Note that this is -2.5 exactly.
By a perverse coincidence, it's very close to another number that comes
up in explanations of the magnitude scale. Very often it is claimed
that the base of the magnitude log scale is the solution of a slightly
different equation,
B^5 = 100
which is 100^(1/5) or 10^(2/5), or about 2.5119. But because of the
inverse relationship of magnitude and brightness, this is *incorrect*,
and causes precisely the confusion we have before us now.
In the second line of the formula in the previous post,
Mc = -2.5 * log10 Lc
I multiplied by -2.5 because I'm converting from the log base 10 to
the log base B. You do this by multiplying by 1 / log10 B, *not* by
multiplying by B (or 1/B).
To summarize, yes, replace 0.4 with 10^(-0.4), but don't replace 2.5.
It's good just the way it is.
> BTW, on the other thread--the one about Google--I figured that it was
> something about reformatting, but I couldn't figure out their algorithm
> for doing so. I couldn't figure out why some lines were slid over and
> others weren't. Do you understand it in that detail and can you explain
> it?
I don't have personal knowledge, so I'm just reverse-engineering, but in
the case of the diagram that led you to ask about it,
http://groups.google.com/groups?selm=c3aoo0%2498b%241%40zot.isi.edu
http://groups.google.com/groups?selm=c3au1m%249i1%241%40zot.isi.edu
it's fairly clear what happened.
Google decided that your ASCII diagram was a block of quoted text, and
it came to this decision after seeing a sequence of lines beginning with
(ignoring spaces) the vertical bar character, "|", as you know a common
quote delimiter (although not as common as ">").
It then decided that the few lines in this block that did not begin with
this character were actually the dangling fragments of the previous
lines, and it therefore "repaired" the broken lines by moving each
fragment to the end of the preceeding line.
The three lines it did this to were
x---s | | p = exit pupil
F---------------x | f = eyepiece focal length
x-----------------------D
In the Google version of the post, you'll find these at the ends of the
lines that came before them.
The subtlety of the algorithm is in how it decides which blocks of text
are quoted. Not all of the lines in the block need to start with the
presumed quote delimiter (in fact, if they did, there'd be no need for
Google to repair the line breaks). But I don't know anything about how
it decides that.
As an experiment, I'm going to copy your diagram, but put a period at
the beginning of each line. This isn't a common quote character, so
the hope is that your diagram won't be treated as quoted text. It's
also fairly unobtrusive, so it won't be seen as part of the diagram by
readers.
. T-------x-------M-------x------ p
. | | | |
. | A | | T = true field of view
. | | | M = magnification
. x---s | | p = exit pupil
. | | | A = apparent field of view
. | | | s = eyepiece field stop
. | | | F = objective focal length
. F---------------x | f = eyepiece focal length
. | | | D = aperture
. | f | r = focal ratio
. | |
. x-----------------------D
. |
. |
. |
. r
Assuming this works, it'll be one way you can keep these from being
"repaired" into junk in the Google archives.
Brian Tung
> In the second line of the formula in the previous post,
>
> Mc = -2.5 * log10 Lc
You're right, that's exact. My mistake again.
> As an experiment, I'm going to copy your diagram, but put a period at
> the beginning of each line. This isn't a common quote character, so
> the hope is that your diagram won't be treated as quoted text. It's
> also fairly unobtrusive, so it won't be seen as part of the diagram by
> readers.
No need, actually. Any additional level of quoting that doesn't use
the same character seems to do the trick. Note in Google's expansion
of that very thread, the post in which I quoted my own diagram (with
initial character ">") turned out just fine.
Actually, what surprised me was not that it turned out wrong in the
usual presentation. What surprised me was that even when I asked for
the "original format," I got some line scrambling. As I recall, it
wasn't even the same scrambling, but I haven't gone back to check.
Ernie Wright
>> As an experiment
>
> No need, actually. Any additional level of quoting that doesn't use
> the same character seems to do the trick.
I know, but presumably you don't want to use ">" or whatever to protect
your diagrams, since Google, and possibly your readers, will interpret
your diagrams as quoted text. I was hoping to preserve not only the
formatting but also the fact that it's yours, by using a character that
Google won't interpret as a quote delimiter.
> Actually, what surprised me was not that it turned out wrong in the
> usual presentation. What surprised me was that even when I asked for
> the "original format," I got some line scrambling. As I recall, it
> wasn't even the same scrambling, but I haven't gone back to check.
Looks like it's trying to undo its repairs. The line breaks are put
back, but in the process, a few leading spaces are lost.
Abdul Ahad
> Abdul Ahad wrote:
>
> > So using this generalized formula for "n" stars, for 3 stars of
> > respective magnitudes: m1 +4.5, m2 +4.8 and m3 +5.2, what figure do
> > you make the combined magnitude, Mc?
> Lc = 0.4^M1 + 0.4^M2 + ... + 0.4^Mn
> Mc = -2.5 * log10 Lc
>
> Easy peasy. For the example you gave,
>
> Mc = -2.5 * log10 (0.4^4.5 + 0.4^4.8 + 0.4^5.2) = 3.6
>
Oh yes, of course the beauty of Algebra and rule of indices is if you
don't like an expression one way, just flip it round to better suit
your *style*! We are in agreement, as I made it 3.603552 (3.6 to 1
d.p.).
Even better, importing a text file of 1,476 selected stars (mainly
down to mag. 5.5) from a mixture of the "Yale Bright Star Catalouge"
and the "FK5" catalogue on the US Naval Observatory site:-
http://asa.usno.navy.mil/SecH/2004/brightstar_2004.txt
I did the sums in a spreadsheet on the given visual magnitudes (V) to
get an integrated total magnitude of -4.43 for the sample!
This result is in ballpark agreement with William Keel's earlier quote
on this thread (from a book) of -5.85 and Tony Flanders' own
calculation of -4.9.
Well done folks!
Abdul Ahad
>
> - Ernie http://mywebpages.comcast.net/erniew
Ernie Wright
> As an experiment, I'm going to copy your diagram, but put a period at
> the beginning of each line.
For what it's worth, this works. It maintains the integrity of the
diagram, isn't parsed or displayed as quoted text by Google, and doesn't
look like quoted text to human readers.
http://groups.google.com/groups?selm=n6SdnWOfk8uAcvzdRVn-uw%40comcast.com