Why the sky is dark in the night

Discussion in 'Astronomy, Exobiology, & Cosmology' started by The God, Mar 23, 2016.

  1. The God Valued Senior Member

    Messages:
    3,546
    ....Sorry, but I am still not able to get the answer how the scattered lux in the night time is 0.001....My issue is that its a fact that lux in the night is 0.001, but can it be proved with those 3 points that is finiteness of universe, limiting value of light speed and expansion of universe or these are just the qualitative statements ?

    Infact I am getting a feeling that it is being pushed as other way round.....that is the dark night is the proof of expanding universe, finite universe and finite speed of light..and of course BB (?).

    As far as Dark or Clear is concerned, they are qualitative words, again the point is intensity being of the order of 0.001 Lux in the (clear) night...
     
  2. Google AdSense Guest Advertisement



    to hide all adverts.
  3. The God Valued Senior Member

    Messages:
    3,546
    So, how did you get 0.57 Degree ?

    You messed up those posts while attemping to take daecon through this simple calculations....

    Billy T has already given simple straightforward caluclations which yield 0.53 Deg, you talk of trig function and all without understanding the geometry of the triangle, I will help you in that..

    if the angle subtended by the diametrical opposite points of the sun on an observer on the earth is A, then

    A/2 = Arctan (0.7/150) = 0.266 Deg, so A = 0.53.......So whether you use trig function or simple radian equation both give you the same answer...

    You do not understand a simple triangle geometry and bloody you had the gal to offer me to take me through some calculations !! Read some basics firsts, sonny, you are busted......If you want some help, ask the way Daecon has asked, I will educate you.
     
  4. Google AdSense Guest Advertisement



    to hide all adverts.
  5. Russ_Watters Not a Trump supporter... Valued Senior Member

    Messages:
    5,051
    I didn't make the rules and this issue is bigger than just this specific problem. It's not foolish, it is convention: positions in the sky are measured in degrees (declination and right ascension), similar to the way lat and long are measured on Earth's surface. So it makes sense to measure angular distances that way too. It is also easier to deal with (in many cases) non-decimal numbers for the angular size of things, such as 4 degrees for Andromeda and 40 arcsec for Jupiter.
     
  6. Google AdSense Guest Advertisement



    to hide all adverts.
  7. paddoboy Valued Senior Member

    Messages:
    21,225
    Admirable stuff by both of you, but don't expect anything similar from our delusional egotistical friend...he does not want to learn...he is pushing an agenda to try and invalidate cosmology ....and he will never admit to any error, rather attempt to muddy the waters and/or disguise the same errors with clownish humour and insults, which he then accuses you of.

    Please Register or Log in to view the hidden image!

     
  8. Russ_Watters Not a Trump supporter... Valued Senior Member

    Messages:
    5,051
    I'm actually not sure -- I can't reproduce it now, so it may have been a typo when I did the calc. 0.53 is the correct number. Also, I didn't use that procedure, I used the atan function on the calculator. BillyT's objection aside, it actually requires fewer button pushes to do it the way I suggested.

    So now you're interested in doing your calculations?
     
  9. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    Many small calculators do not have any trig functions, especially those giving the angles (arc tan etc.)

    My approach requires two multiplications and one division, which all calculators can do to get sun (or star's) angle.

    I. e. in arcminutes the angle is 3,437.75 x (star diameter / star distance) but even more simple if you can use radians. I. e. then just one division: In radians the angle is: (star diameter / star distance).

    As far as the final result (How much bigger the sun's angle is than the star's) any front number works, including none (one). That is one could use 3 (instead of 3,437.75) and the accuracy of the answer would still be limited by the data. That is:

    Even if you don't like radians but want to know ( the original question) how much bigger the sun's angle is than a star's angle you don't need to be concerned with what units the angle is calculated in as the answer has no units - it is just a large number, which you can get with just three divisions, same (or less)* than your method's number of "button pushes."

    *I think you need to call up the arctan twice and still do a pair of distance divisions, one each for its inputs and then one more divivision to get the anwser. That is 5 "button pushes" with neither of us counting the data entry button pushes. Last time, I checked 5>3.
     
    Last edited: Apr 6, 2016
  10. rpenner Fully Wired Staff Member

    Messages:
    4,833
    Incorrect.

    • Finite speed of light
    • Finite age of universe
    • Expanding universe
    In an infinite, uniform universe, given a certain finite average differential density of stars of radius r, \(\rho(r)\), then in a spherical shell from R to R+dR which has volume \(dV = 4 \pi R^2 dR \), we have a finite number of stars of radius between r and r+dr given by \(dN = \rho(r) dr dV = 4 \pi R^2 \rho(r) dr dR\). Each star has an angular radius of \(\frac{r}{R}\) radians or angular square measure relative to the sky as a whole of \(f(r,R) = \frac{1}{4 \pi} \times \pi \frac{r^2}{R^2} = \frac{r^2}{4 R^2}\). Thus the fraction of the sky occupied by stars at a distance between R to R+dR and of a radius between r and r+dr is \(f(r,R) dN = \pi r^2 \rho(r) dr dR\) which is independent of distance! Thus the differential fraction of the sky taken up by all stars between R to R+dR is \(F = \int \limits_0^{\infty} \pi r^2 \rho(r) dr > 0\) is some finite number larger than zero. Thus the logarithm of the fraction of the sky not taken up by any stars between R to R+dR is \(- F dR\) and therefore the logarithm of the fraction of the sky that is dark is \( \int \limits_0^{S} -F dR = -F S\) and the fraction of the sky that is dark is \(e^{-F S}\) where S is the maximum radius considered.

    But F is positive and S is positive and infinite so \(\lim \limits_{S\to\infty} e^{-F S} = 0\) and this is the paradox for the sky is mostly dark.

    Does the finite speed of light change this by itself, no. But it does make us aware that looking far away is looking back in time.

    The finite age of the universe means that S is not infinite, so the sky is allowed to be dark. Except that the Big Bang proposes the universe was very hot in the distant past which would mean the sky would be bright everywhere. //Edit: So even without taking a position on the finiteness or infiniteness of the universe, the portion which is observable is finite due to the finite time light has had to propagate. Also, changed "is dark" to "is allowed to be dark" because if the age is very old, then it might be possible that the sky is bright. A finite age turns the logical argument that the sky must be bright into a quantitative one about the average density of stars and age of the universe. And quantitative arguments are what seals the deal as we address the darkness of the all-pervasive CMBR:

    But the expanding universe redshifts very old light and rendered it not just invisible but cryogenic. Thus the sky should be (and is) dark. // Edit: Similarly, the galaxies of the Hubble Deep Field are redshifted, and their angular sizes are distorted due to similar consequences of GR models of Big Bang cosmology.
     
    Last edited: Apr 6, 2016
  11. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    While repenner's answer is 100% correct, I think his method is very slightly wrong. He calculates the fraction of the sky blocked by small spherical shell then integrates that to show it is approaches 100% blocked as the "max radius, S" (or upper limit of the integral) becomes ever larger. The error is one of "over counting." I. e. some of the blocking made by shell n will also be blocked by shell n-1 stars, (Two stars perfectly alined with earth radius) etc. His integral converges to infinity a little more slowly, if this "double blocking" were not included, counted at least twice.

    repenner's math skills are so far above mine, that I probably won't get this opportunity to correct him again, so I had to take this rare opportunity. This computational "error" is of no importance, his answer is correct, as they always are.

    For people interested I note the probability of "over blocking" by shell n-1 of shell n goes to zero as the shell thickness goes there, but the number of shells is going to infinity. I. e. there is the product of NT to consider, when N is going to infinity and T, the shell thickness, is going to zero. What that product is doing is beyond my pay grade. If it is zero, then repenner does not even have a very small error in the method.
     
    Last edited: Apr 6, 2016
  12. The God Valued Senior Member

    Messages:
    3,546
    Thanks, I will get back, if any issues remain on this.....

    I find BillyT response meaningful in a sense, can we attempt some maths by which if the front obstructing layers can result into non infinite flux at our end even if universe is infinite ?
     
  13. rpenner Fully Wired Staff Member

    Messages:
    4,833
    I believe if you would check again, I have already addressed that, because I addressed the fraction of the sky which is DARK.

    Obviously, you do get an overcounting problem if you try and add fractions of the sky covered by each shell because \(\lim \limits_{S\to\infty} FS > 1\). How can you have more than 1 full sky? But, as with other probability arguments, focusing on the opposite outcome -- the fraction of the shell not blocked by stars -- is the way to go.

    Thus each dR shell has a fraction of sky with stars, F dR, and a portion of sky without stars, 1 - F dR.

    So we want to multiply the "dark" part of all the shells. If \( dR = \frac{S}{n} \) then we want \(\lim \limits_{n\to\infty} \prod \limits_{k=1}^{n} ( 1 - F \frac{S}{n} )\) which is almost like an integral: \(\int \limits_{0}^{S} g(R) dR = \lim \limits_{n\to\infty} \sum \limits_{k=1}^{n} g \left( \frac{k - \frac{1}{2}}{n} S \right) \times \frac{S}{n} \). But we can made this analogy precise by replacing \(\prod g \left( \frac{k - \frac{1}{2}}{n} S \right) \) with \(\sum \ln g \left( \frac{k - \frac{1}{2}}{n} S \right) \). So what is \(\ln \left( 1 - F dR \right) \) ?

    The natural logarithm of this latter quantity is, -F dR. This is because \(\lim \limits_{x \to 0} \ln (1 + x) = x + O(x^2)\).

    Thus the logarithm of free sky of k shells is -k F dR and for all shells out to distance S is -F S.

    Say \(F = 10^{-7} \, \textrm{pc}^{-1}\) (ridiculously high, but useful for illustration). Compare the two models for fraction of sky covered by stars:

    \( \begin{array}{l|ll} S & FS & 1 - e^{-FS} \\ \hline \\ 100,000 \textrm{pc} & 0.01 & 0.009950166 \\ 200,000 \textrm{pc} & 0.02 & 0.01980133 \\ 500,000 \textrm{pc} & 0.05 & 0.04877058 \\ 1 \textrm{Mpc} & 0.1 & 0.09516258 \\ 2 \textrm{Mpc} & 0.2 & 0.1812692 \\ 5 \textrm{Mpc} & 0.5 & 0.3934693 \\ 10 \textrm{Mpc} & 1 & 0.6321206 \\ 20 \textrm{Mpc} & 2 & 0.8646647 \\ 50 \textrm{Mpc} & 5 & 0.9932621 \\ 100 \textrm{Mpc} & 10 & 0.9999546 \end{array}\)

    // Edited to change f -> g for the general function in illustration of integration because my previous post used f in a problem-specific manner
     
    Last edited: Apr 6, 2016
  14. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    Thanks. I won't "check you" I have learned that is not needed, but I'm sorry I caused you the trouble of another post.
     
  15. rpenner Fully Wired Staff Member

    Messages:
    4,833
    Russ_Watters likes this.
  16. The God Valued Senior Member

    Messages:
    3,546
    Rpenner,



    I like your salvaging this guy....I do understand that 1.5 can be typo to 1.4. But the point is, he was not following this calculation, infact it was opposed by him, when first Billy T and then I put on record.....so that does not make your typo explanation obvious......It is a straw man attempt on his part to latch on to your relief..

    But, you missed the more significant problem with his maths...thats perpetuating mistake...You know very well (its a simple isosceles triangle trigonometry for you) that the angle subtended in this case is 2arctan(radius of sun/distance) = 2arctan(0.7/150)......it is not arctan(1.4 or 1.5/150)....Basic mistake and the guy has the guts to take me on and you gave inadvertant support to his math ? Please correct the situation as you found out typo on his behalf, not your call. [You also would like to correct his and paddoboy gaffe (or it is lack of understanding?) as asked by Expleticve Deleted in that other thread.]
     
  17. rpenner Fully Wired Staff Member

    Messages:
    4,833
    For arctan which gives answers in degrees:
    \(2 \tan^{-1} \frac{x}{2} = \frac{180}{\pi} x \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{(2k +1)4^k}= \frac{180}{\pi} \left( x- \frac{1}{12} x^3 + O(x^5) \right) \\ \tan^{-1} x = \frac{180}{\pi} x \sum_{k=0}^{\infty} \frac{(-1)^k x^{2k}}{2k +1} = \frac{180}{\pi} \left( x- \frac{1}{3} x^3 + O(x^5) \right)\)

    Thus the relative error of approximating the first by \(\frac{180}{\pi} x\) is less than \(\frac{x^2}{12}\) and the relative error of approximating the first by the second is less than \(\frac{x^2}{4}\).

    So you not only want to quibble about an error in methodology which results in a relative error of less than 0.00218% in results made with figures of only 2 significant digits (0.35% error over using precise values and up to 1.7% error over ignoring the Earth's orbit is not a circle), when a much larger error (over 7%!) was created by a keystroke error, but carry that as an off-topic grudge from thread to thread? Your pedantic concerns are misplaced and completely out of place when this figure was only being calculated with an eye to comparison with angular measures of far more distant stars where the distances are necessarily inexact.

    Take the well-known star Betelgeuse, for example. Depending on method, its distance is estimated at somewhere between 131 and 250 parsecs.

    http://iopscience.iop.org/article/10.1088/0004-6256/135/4/1430/pdf
     
    Last edited: Apr 10, 2016
    paddoboy likes this.
  18. Russ_Watters Not a Trump supporter... Valued Senior Member

    Messages:
    5,051
    Last edited: Apr 9, 2016
  19. paddoboy Valued Senior Member

    Messages:
    21,225
    Like you, expletive deleted has made many assumptions, and ignored replies in return. Like your own misunderstandings and confusions re physics and cosmology.
    With regards to this thread, many good answers have been forthcoming. In essence Olber's paradox, like the twin paradox, is not really a paradox at all.

    Please Register or Log in to view the hidden image!

     
  20. Daecon Kiwi fruit Valued Senior Member

    Messages:
    3,048
    Wow. I'm starting to wish I never asked...
     
    Russ_Watters likes this.
  21. The God Valued Senior Member

    Messages:
    3,546
    Whats happening to you ?? You are out of your depth ?

    An observer 'O' forms an isosceles triangle on Sun's dia AB with OA = OB, if the OC is the perpendicular from O to AB, then the angle subtended at O is

    = 2arctan(AB/2OC)...........this is exact maths (irrespective of values) for isosceles triangle.

    Now if you are trying to tell that this can be approximated as arctan(AB/OC)...then please stop this continued perpetuation of silliness as started by Russ watters......this expression can be rightly approximated as 360/2pi * AB/OC, which was told to him long back by both Billy T and me, but objected by him.......so what if due to small angles both the trig functions are giving same value ?

    and secondly when the guy is saying he did not use the formula as shown by you, so the suggested typo could not have happened.
     
  22. danshawen Valued Senior Member

    Messages:
    3,942
    Betelgeuse? Don't make me LOL. Think of black existing hole mergers in terms of the scale and masses of 1000+ medium sized galaxies.

    Last year we saw this:

    http://www.space.com/28664-monster-black-hole-largest-brightest-ever.html

    and this is news from just a few days ago:

    https://www.washingtonpost.com/news...e-black-holes-might-actually-be-super-common/

    Tends to indicate, black hole mergers are nearly as common as the incidence of galaxies. These monstrous bodies must have been the result of billions of mergers. And these are just the largest ones, the ones easiest for us to see.

    No wonder the sky is so relatively dark.
     

Share This Page