The Journal of Things I Don't Understand
Mostly questions that bug me about whats going on and doesn't seem to fit with reality.
Some values of interest:
p will be used as short hand for photons.
pc will be used as short hand for parsec.
sh will be used as shorthand for shell.
ly will be used as shorthand for light year.
aoo will be used as shorthand for Bohr radius
s will be used for seconds
E = hf :Energy = Plancks constant * frequency = 3.5780772384e-19Joules/photon @ 555nm
1photon/s = 3.5780772384e-19 Watt
1/683W = 1 lm (lumin)
1/683J/s = (1/3.5780772384e-19J/photon)* (1/683W/nit) = 4.0919e+15 photons/nit
pc = 4.30856775807e+17m
ly = 9.454e+15m
pc = 3.2616334ly
aoo = 0.529177249(24)e−10m
Age of universe is approximately. 1.5e+10yr
The earth is approximately 1.5e+11m. from the sun.
Table 1: Approximate luminance's of various sources 
* Assuming 555nm Photons only
** 555nm Photons corrected using 2.27...e-4
Since the suns radiation is transmitted equally in all directions, the photons per m2 on the earth would be the total photons emitted divided by the surface area of a sphere with a radius of 1.5e+11m.
(1) 4.2e+44p/(4 * pi * r2) = 4.2e+44/(4 * pi * (1.5e+11)2m) = 1.485e+21p/m2·s
Note – This differs from that observed value of 6.547e+24 in Table 1 column Luminance (nits) because the values in this column are for all observable frequencies, not just the 555nm frequency. Values in the 555 Photons/s are adjusted by 1/4.4e+3 for the final comparison.
Assuming 1 star per cubic parsec as was done in this explanation , we get:
(2) 4 * pi = 12.566 stars at 1 parsec from the sun
(3) p/m2·star = 4.2e+44p/s·star/ (4 * pi * (4.30856775807e+17m/pc)2) =
For the sphere of 12.566 stars you get a total of
(4) 1.8004e+8p/m2·s·star·pc2 * 12.566star/pc/2 = 1.1312e+9 p/m2·s·sh
(You have to divide the number of stars by 2 since the night is only half the sky and the calculation is for a sphere.)
Which is the number of photons we expect for each parsec spaced shell.
The required size of the universe to be as bright as the sun would be the photons per second the earth receives from the sun divided by the photons per second received per 1 parsec section. This computes to:
(5) 1.485e+21p/m2·s/1.1312e+9 p/m2·s·sh = 1.313e+12sh.
shells to be as bright as the sun.
The required size in light years to be as bright as the sun would be the number of shells times the parsecs per shell times then number of light years per parsec.
(6) 3.262 ly/pc*1pc/sh*1.313e+12sh = 4.283e+12 ly
A value at least 150 times the age given for the universe. This would indicate that relying on the big bang theory would say that the night sky should be within 150% of the brightness of the sun, instead of almost a million million times less.
So, how bright should the night be using the values we have?
It should be:
(7) ((1.5e+10ly/(3.262ly/pc*1pc/sh))*1.1312e+9 p/m2·s·sh ) = 5.202e+18 p/m2·s
Or about 1e13 times brighter than it actually is. This must mean that our initial assumption that being within 150 times the estimated age of the universe is not accurate enough and we are not considering all the factors.
The 1 star per cubic parsec works for within a galaxy, but the space between galaxies and galaxy clusters is great and reduces this value considerably. Lets look at it from a different perspective to see if that will get us a more reasonable value.
There are an estimated 7e+22 stars in the universe.
The universe is 1.5e+10ly (4.598922736e+9pc) in radius.
Volume of Sphere :
(8) 4/3 * pi * r3 = (4 * pi * (4.6e+9)3)/3 = 4.077e+29pc3
in the Universe.
Therefor there are:
(9) 7e+22star/4.077e+29pc3 = 1.717e-7stars/pc3
not 1 star per cubic parsec.
Therefor the average intensity per cubic region is only:
(10) 1/ 1.717e-7stars/pc3 = 5.824285e+6pc3 /star
This means that each star is within a 1.8e+2 sided cube, or approximately 1.8e+2 pc from any other star in the universe.
Recalculating the effect of each sphere we get:
(11) p/m2·s = 4.2e+44 p/s·star / (4 * pi * (1.799e+2 pc/star*4.30856775807e+17m/pc)2) =
The first section contains 12.566 stars, for a total intensity of
(12) (5.557e+3 p/m2·s * 12.566)/2 = 3.491e+4 p/m2·s·sh .
Each shell would then contribute 3.491e+4 p/m2·s per shell, considerably less than the number for 1 star per cubic parsec.
There would be a total of:
(13) 4.598922736e+9pc/1.799e+2pc/sh = 2.556e+7 sh.
And since each shell contributes 3.491e+4 p/m2·s, the night sky should shine with a brightness of:
(14) 3.491e+4 p/m2·s·sh * 2.556e+7 sh = 8.923e+11 p/m2·s
This still leaves the night sky one million times brighter than it actually is.
Since astronomy accepts that over 95% of the mass of the universe is “dark mass” and not observable, it may be that this mass is absorbing the electromagnetic radiation on its way to the earth. Let us examine this possibility.
We know the mass of the sun is 1.8143695e+30 kg
If this is only 5% of the mass of the suns cube, then the total mass would be:
(15) 1.8143695e+30 kg/0.05 = 3.628739e+31 kg
Knowing that the majority of the sun is hydrogen, and using the worst case scenario:
Mass of Hydrogen atom = 1.7*10-27kg
(16) 3.628739e+31 kg/1.7*10-27kg = 2.135e+58 Atoms.
Assuming a 1.799e+2pc square face on each cube, we get:
(17) 4.30856775807e+17m/pc*1.799e+2pc/sh =7.75111e+19m/sh
So the area is
(18) (7.75111e+19m/sh)2 = 6.008e+39m2/sh .
Since a Hydrogen atoms area = pi*r2
(19) pi * aoo2 = pi*(0.529177249e−10m)2 = 8.797e-21 m2 /Atom
and there are 2.135e+58 Atoms/sh, they can obstruct
(20) 8.797e-21 m2 /Atom* 2.135e+58 Atom/sh = 1.878e+38 m2/sh.
This means that of our 6.008e+39m2/sh of area, 1.878e+38 m2/sh or about 3.125% of the area of any one shell is blocked by the hydrogen atoms. But the light must pass through more than 1 cube to get to us. On the average, it must pass through 2.556e+7 sh/2 = 1.278e+7 sh. In this case, the hydrogen between the stars and the night sky would cover an effective area of
(21) 1.878e+38 m2/sh* 1.278e+7sh = 2.4e+45 m2.
Assuming the condition of 6.008e+39m2 by 1 atom thick planes stacked one on top of the other, we would have the equivalent to a liquid hydrogen cube:
(22) (2.4e+45 m2/6.008e+39m2/sh) * (0.529177249e−10m/atom) * 2 =
(23) 4.23e-5m thick by 7.75111e+19m wide by 7.75111e+19m high.
or about a 0.04mm thick liquid hydrogen barrier, this is not thick enough to eliminate much of the light.
Also, considering that hydrogen obstructs only a few lines in the electromagnetic spectrum we need to determine if there is anything else that would block the light. What does obstruct light is interstellar dust, which makes up about 1% of the mass of interstellar material. Knowing this, we know that the dust “thickness” is 1% of what we calculated for hydrogen, or:
(24) 4.23e-5m * 0.01 = 4.23e-7m = 423nm
thick solid dust layer. Since this would be approximately equivalent to the coating applied to some sunglasses with a 75% transmission (500nm of coating). This would indicate that the dust only attenuates the light by about 25%, to:
(25) 8.923e+11 p/m2·s * 0.75 = 6.692e+11 p/m2·s
or, about 1 million times as brighter than it actually is.
So, there is no intervening material which seems to block the light, so why is the night so dark?
Any way you calculate it, if we accept that Olbers Paradox says that the universe cannot be infinite, then we must also accept that it says that the universe was not born of a big bang. Or we must accept the fact that we don't know enough about light and or the universe in general to say that Olbers paradox has anything whatsoever to do with the age or origin of the universe.
Personally I prefer the latter thought.Citations
<1>National Physical Laboratory, UK; http://www.npl.co.uk/publications/light/spectrum.htm, 04/09/2008
<2>JAMES NOBORU IMAMURA, http://zebu.uoregon.edu/~imamura/123/lecture-5/olbers.html, 06/21/2008
<3>Robert Roy Britt, http://www.space.com/scienceastronomy/star_count_030722.htm, 06/21/2008