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An amusing paper about the size of the Universe 
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#1
Nov1012, 02:42 PM

P: 28

I wrote this while I was bored at work earlier.
Measurable Theoretical Breakdown of the Universe: A Paper for Amusement by Pete Lerza 11/10/2012 I’ve done all mathematics with fairly loose approximations. The purpose of this is not for scientific accuracy, but for a general understanding of a theoretical size of the Universe. The Names of the Large Numbers are in the American Standard form [then the European Standard Form] but not SI prefix or Greekbased forms. I decided to determine what percentage of the theoretical volume of the Universe that Earth consumes, then what percentage of the theoretical volume of the Universe that you consume. To write a number in scientific notation: Put the decimal after the first digit and drop the zeroes. In the number 256,000,000,000 The coefficient will be 2.56 To find the exponent count the number of places from the decimal to the end of the number. In 256,000,000,000 there are 11 places. Therefore we write 256,000,000,000 as: 2.56 x 1011 To begin, we assume that the Universe is 13.7 billion years old & that it takes light 3 x 109 s to travel one meter. We then convert the age of the Universe into seconds, divide the age of the Universe in seconds by the amount of seconds it takes for light to travel one meter & have an approximate radius of the Universe in meters. (13,700,000,000 years)(365 days)(24 hours)(60 minutes)(60 seconds) = 4.32 x 1017 s {432,000,000,000,000,000 s} {432 quadrillion [billiard] seconds} 4.32 x 1017 s is the age of the Universe in seconds (4.32 x 1017 s) / (3 x 109 s) = 1.44 x 1026 m or 1.44 x 1023 km 1.44 x 1023 km is the Theoretical Radius of the Universe A Diameter is determined by multiplying the Radius by 2 The Theoretical Diameter is 2.88 x 1023 km We know the Universe is not a Circle or Sphere, but we do not have enough information to calculate the dimensions of an ellipse or elliptical sphere, so we will practice under the theory of which the Universe is a Circle, which should give us approximately the same percentages at the end. Circumference of a Circle = π x Diameter π (2.88 x 1023 km) = 9.05 x 1023 km {905,000,000,000,000,000,000,000 km} {905 sextillion [trilliard] kilometers} 9.05 x 1023 km is our Theoretical Circumference of the Universe Area of a Circle = πr^2 π(1.44 x 1023 km)2 = 6.51 x 1046 km2 {6,510,000,000,000,000,000,000,000,000,000,000,000,000,000 km2} {65.1 quattuordecillion [septilliard] square kilometers} 6.51 x 1046 km2 is the theoretical Area of the Universe Volume of a Sphere = (4/3)r3 (4/3)(1.44 x 1023 km)3 = 3.98 x 1069 km3 {398,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00 0,000,000,0000,000,000 km3} {3.98 duovigintillion [undecilliard] cubic kilometers} 3.98 x 1069 km3 is the Theoretical Volume of the Universe Measurements in Miles: 1 km = 0.62 Miles Radius: 8.9 x 1022 Miles Diameter: 1.79 x 1023 Miles Circumference: 5.62 x 1023 Miles Area: 4.05 x 1046 Square Miles Volume: 2.47 x 1069 Cubic Miles The given Radius of Earth is 6,378.1 km. Earth is a Sphere. Volume of a Sphere = (4/3)r3 (4/3)(6,378.1)3 = 3.46 x 1011 km3 {346,000,000,000 km3} {346 billion [milliard] cubic kilometers} The Volume of Earth is 3.46 x 1011 km3 To determine what percentage of the Universe is Earth, we divide the Volume of Earth by the Volume of the Universe. To divide numbers in Scientific Notation, we divide the coefficients, then subtract the exponents. (3.46 x 1011 km3) / (3.98 x 1069 km3) = (3.46 / 3.98) x 10 (11 69) = 8.7 x 1059 % {0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 087 %} {8.7 octodecillionths [nonilliardths] of a percent} The Earth consumes 8.7 x 1059 % of the Theoretical Volume of the Universe The average given Volume of a Human Being is 6.93 x 102 m3 We know the Volume of the Earth & the Theoretical Volume of the Universe in kilometers, so to convert 6.93 x 102 m3 to km, we divide by 1,000 = 6.93 x 105 km3 To determine what percentage of the Earth is a Human Being (You), we divide the Volume of You by the Volume of the Earth. (6.93 x 105 km3) / (3.46 x 1011 km3) = 2.00 x 1016 % {0.000 000 000 000 000 2 %} {2 quadrillionths [billiardths] of a percent} You are 2.00 x 1016 % of the Earth’s Volume To determine what percentage of the Universe is a Human Being (You), we divide the Volume of You by the Theoretical Volume of the Universe. (6.93 x 105 km3) / (3.98 x 1069 km3) = 1.74 x 1074 % {0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 01 74 %} {174 trevigintillionths [duodecillionths] of a percent} You consume 1.74 x 1074 % of the Universe’s Theoretical Volume Let me know your thoughts & feel free to contact me about anything that I mentioned, was vague about, or messed up on at <personal email deleted> 


#2
Nov1012, 03:51 PM

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P: 11,839

From wiki:
The comoving distance from Earth to the edge of the observable universe is about 14 gigaparsecs (46 billion light years or 4.3×10^{26} meters) in any direction. The observable universe is thus a sphere with a diameter of about 29 gigaparsecs[14] (93 Gly or 8.8×10^{26} m).[15] Assuming that space is roughly flat, this size corresponds to a comoving volume of about 1.3×10^{4} Gpc^{3} (4.1×10^{5} Gly^{3} or 3.5×10^{80} m^{3}). 


#3
Nov1012, 04:03 PM

P: 28

Addendum
The observable Universe has a radius of about 46.5 billion light years. The age of the universe does not equate to the radius, due to expansion (Thanks to SpeedFreek). As well, the Universe may be a sphere. One light year is about 9.46 x 1012 km. (4.65 x 1010 light years)(9.46 x 1012 km/light year) = 4.40 x 1023 km The Radius of the Universe is 4.40 x 1023 km The Diameter of the Universe is 8.80 x 1023 km Circumference of a Circle = π x Diameter π(8.80 x 1023 km) = 2.76 x 1024 km {2,760,000,000,000,000,000,000,000 km} {2.76 septillion [quadrillion] kilometers} 2.76 x 1024 km is the Circumference of the Universe Area of a Circle = πr^2 π(4.40 x 1023 km)2 = 6.08 x 1047 km2 {608,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000 km2} {608 quattuordecillion [septilliard] square kilometers} The Area of the Universe is 6.08 x 1047 km2 Volume of a Sphere = (4/3)r3 (4/3)( 4.40 x 1023 km)3 = 1.14 x 1071 km3 {114,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,000,00 0,000,000,000,000,000,000 km3} {114 duovigintillion [undecilliard] cubic kilometers} The Volume of the Universe is 1.14 x 1071 km3 Margin of Error: By assuming the size of the Universe by its age, the calculated volume was 3.98 x 1069 km3. With the known Radius of the Universe, the calculated volume is 1.14 x 1071 km3. 1.14 x 1071 km3  3.98 x 1069 km3 = 1.10 x 1071 km3 The given Radius of Earth is 6,378.1 km. Earth is a Sphere. Volume of a Sphere = (4/3)r3 (4/3)(6,378.1)3 = 3.46 x 1011 km3 {346,000,000,000 km3} {346 billion [milliard] cubic kilometers} The Volume of Earth is 3.46 x 1011 km3 To determine what percentage of the Universe is Earth, we divide the Volume of Earth by the Volume of the Universe. To divide numbers in Scientific Notation, we divide the coefficients, then subtract the exponents. (3.46 x 1011 km3) / (1.14 x 1071 km3) = (3.46 / 1.14) x 10 (11 71) = 3.04 x 1060 % {0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 003 04 %} {3.04 novemdecillionth [decillionth] of a percent} The Earth consumes 3.04 x 1060 % of the Universe The average given Volume of a Human Being is 6.93 x 102 m3 We know the Volume of the Earth & the Theoretical Volume of the Universe in kilometers, so to convert 6.93 x 102 m3 to km, we divide by 1,000 = 6.93 x 105 km3 To determine what percentage of the Earth is a Human Being (You), we divide the Volume of You by the Volume of the Earth. (6.93 x 105 km3) / (3.46 x 1011 km3) = 2.00 x 1016 % {0.000 000 000 000 000 2 %} {2 quadrillionths [billiardths] of a percent} You are 2.00 x 1016 % of the Earth’s Volume To determine what percentage of the Universe is a Human Being (You), we divide the Volume of You by the Volume of the Universe. (6.93 x 105 km3) / (1.14 x 1071 km3) = 6.08 x 1076 % {0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 608 %} {608 trevigintillionths [duodecillionths] of a percent} You consume 6.08 x 1076 % of the Universe’s Volume 


#4
Nov1012, 04:05 PM

P: 28

An amusing paper about the size of the Universe
& I didn't have to quote wikpedia ^.^



#5
Nov1012, 04:25 PM

Astronomy
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It's really misleading for you to have referred several times to the observable portion of the universe as "the Universe".
If you don't mind and can still edit, why don't you correct that. All your "universe" size numbers are about the observable portion. As far as we know this is not a fixed region, it keeps growing as time goes on and more light has time to reach us, and it is only a modest portion of the whole universe. BTW conventional mainstream cosmology models the whole universe and there are lowerbound estimates of it's present volume. But the professionals do not consider the U to be spatially a spherical volumei.e. a ball of a certain radius. So if you want to estimate the spatial volume you would not use the volume formula one learns in highschool namely (4/3)πR^{3}. You might for instance use the formula for the volume of a 3D hypersphere as a function of its radius of curvature (for which there is a lowerbound estimate). It could be quite a bit more amusing for you, to do it that way. 


#6
Nov1012, 04:30 PM

P: 28

I welcome the criticism. I am not a physics major, not even a college student actually. Just doing this to amuse myself so all of the comments are welcomed because I love to learn. That being said, "The purpose of this is not for scientific accuracy, but for a general understanding of a theoretical size of the Universe."



#7
Nov1012, 04:38 PM

P: 28

But, in fairness, I should insert "observable universe" since it is rather long & that first paragraph is well out of our short term memory by the time we get to the end lol



#8
Nov1012, 04:39 PM

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PF Gold
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#9
Nov1012, 04:42 PM

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#10
Nov1012, 05:07 PM

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and the 3D volume of the SURFACE of that ball is 2π^{2}R^{3} (this is called the "3sphere of radius R") talking about the standard model of the U as a whole, the lowest estimate volume I know of is based on a recent report from a radio telescope at the south pole called SPT (south pole telescope). They essentially gave a 95% confidence interval for the radius R. If we pick the shortest R in that range, which I think is 140 billion LY, then plug that R into 2π^{2}R^{3} then its very unlikely that the volume of the whole U could be less than that. It could be much larger, could even be infinite, but at least we'd have a lowerbound estimate"at least suchandsuch". Are you up for that? http://physicsforums.com/showthread....80#post4152380 


#11
Nov1012, 05:26 PM

P: 28

I definitely will edit it so I dont have to trace back in the future. So would the 2pisquared R cubed function plus the pi squared over two function with 140 LY R be the most accurate volume?



#12
Nov1012, 05:30 PM

P: 28

It might save you time if you know of a page that will show me what you are talking about with the 3D hypersphere function. I'd love to get these numbers as accurate as I can, it's a lot more fun when the numbers aren't. 96.7% wrong like I had them the first time around lol



#13
Nov1012, 06:08 PM

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Science is based on doubt, skepticism, admitting we don't know a lot of stuff. That's why it is exciting. There's always more to ask about. All we have, that I know about, is a 95% confidence LOWER BOUND on the volume, and that is based on the standard model universe that astronomers fit their data to and think is the best model so far. You do the best you can and keep testing. What they have is merely the best fit model SO FAR. And it comes in a range of sizes including infinite. But the range is gradually being narrowed down. In the conventional standard model, space has no boundary, no edge. And is roughly uniformly spread with matter..galaxies etc. If space is finite and no boundary then probably the most common picture is a hollow hypersphere with no inside or outside. Analogous to the surface of a balloona 2D world with all existence concentrated on the surfaceno inside or outsidebut the 3D version of that. That is the socalled 3sphere picture. It has a certain volume measured in cubic lightyears, or whatever (cubic centimeters if you like those better than cubic lightyears). According to the SPT report that just came out the most compact smallest volume that 3sphere could have is (well 2 π^{2} ≈20) so 20 (140 billion)^{3} cubic lightyears 


#14
Nov1012, 06:48 PM

P: 28

Awesome! I'm revamping my numbers lol thanks so much! This stuff is awesome & you are definitely right about how not knowing makes it so exciting. If you want to explain other related aspects of this send me an email. Thanks again!



#15
Nov1012, 07:08 PM

P: 28

By the way, with that SPT volume, a person is 1.53 quinvigintillionth of a percent of the observable universe. 1.53x 10^78 %



#16
Nov1112, 09:10 AM

P: 28

So why exactly do we have to take the volume of the surface of the 4 D sphere?



#17
Nov1112, 09:16 AM

P: 28

Also, are you sure about that 140 LY figure? I can't find it online, only coming up with the 46.5 billion that I've already used.



#18
Nov1112, 05:19 PM

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There is a standard way to get the radius of curvature, given the socalled HUBBLE RADIUS and the Ω_{k} they give H_{0} = 69.xx km/s per Mpc (do you see that?) that means the Hubble radius R_{hub} is 14.0 billion LY they give the upper limit on Ω_{k} = 0.0099 ≈ 0.01 (do you get that?) and the square root of that is 0.1 R_{curv} = R_{hub}/sqrt(Ω_{k}) = 14/0.1 billion LY = 140 billion LY. The upper limit on curvature leads to the lower limit on the radius of curvature (a car that turns the tightest corners has the shortest turning radius) It just a straightforward putting together of the standard definitions. the SMALLEST they think it could be corresponds to a curvature radius of 140 billion the LARGEST they think it could be corresponds to a curvature radius of 2010 billion (LY). You get the 2010 number the same way, using the other end of their Ω_{k} range. Basically they are saying that with 95% probability the curvature radius has to be something in between those two limits. So one can calculate the SPATIAL VOLUME from that and get that the spatial volume has to be something between the corresponding smallest and largest volumes (or is 95% likely to be). 


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