Ki_Ryn Posted March 30, 2009 Report Posted March 30, 2009 I'm looking for a number of opinions on the question reproduced below as I'm currently in a disagreement regarding the results. The observed density of matter (both luminous and dark) is roughly 34% of the critical density needed for a flat universe. That part is not under dispute. What is under dispute is which answer (a or :umno: follows from this fact. 1. How does the observed density of matter (both luminous in stars and galaxies, and "dark") in the Universe compare with the critical density? a) The density of luminous and dark matter is currently far smaller than the critical density. :umno: The density of luminous and dark matter is currently slightly smaller than the critical density. Quote
freeztar Posted March 30, 2009 Report Posted March 30, 2009 "far" and "slightly" are both subjective terms. But, I'm wondering where you get 34% from? the best known values for H and G give a value for the critical density of about 1x10-29 grams per cubic centimeter. This seems very small, but we have to remember that most of the universe is practically empty. Recent measurements indicate that the actual density of our universe is very close to the critical density.How was the critical density of the universe calculated? Quote
modest Posted March 31, 2009 Report Posted March 31, 2009 I'm looking for a number of opinions on the question reproduced below as I'm currently in a disagreement regarding the results. The observed density of matter (both luminous and dark) is roughly 34% of the critical density needed for a flat universe. That part is not under dispute. Perhaps it should be. The concordance model has [math]\Omega_M = 0.266[/math]. How did you get 34%? What is under dispute is which answer (a or b) follows from this fact. 1. How does the observed density of matter (both luminous in stars and galaxies, and "dark") in the Universe compare with the critical density? a) The density of luminous and dark matter is currently far smaller than the critical density. b) The density of luminous and dark matter is currently slightly smaller than the critical density. Your question appears to compare two things directly which are not necessarily directly comparable. The matter density is significantly smaller than the critical density. We can easily figure the critical density:[math]\rho_c=\frac{3H^2}{8 \pi G} = \frac{(3)(2.3 \times 10^{-18})^2}{(8)(3.14)(6.67 \times 10^{-11})} \approx 10^{-26} \ Kg/m^3[/math]which is about 5 atoms per cubic meter. The density of baryonic matter and dark matter together is suspected of being 26.6% of that which would be about [math]2.66 \times 10^{-27}[/math]. As to whether 2.66E-27 is far smaller or slightly smaller than 1E-26... well... it's exactly 73.4% smaller. But, your question is a bit misleading comparing the matter density to the: "critical density needed for a flat universe". That would be true if you added the following: "if the cosmological constant were zero". Otherwise, you cannot directly compare the matter density to curvature. If the cosmological constant (dark energy) is currently 73.2% of the energy density of the universe then the "density of luminous and dark matter is currently" exactly what is needed for a flat universe. If you plot [math]\Omega_M[/math] (matter density) on one axis and [math]\Omega_{\Lambda}[/math] (dark energy density) on another axis then you get this pretty little graph: The blue line is a flat universe and the axis on the left is the ratio of matter density to critical density ([math]\rho_m \div \rho_c[/math]). The matter density doesn't have to equal the critical density for a flat universe. At [math]\Omega_M = .26[/math] the universe will be flat if [math]\Omega_\Lambda = 0.74[/math]. This is predicted to be the case, so our universe is expected to be cosmically flat. ~modest freeztar 1 Quote
freeztar Posted March 31, 2009 Report Posted March 31, 2009 I'm curious as to the discrepancy between our numbers, Modest. I've read several estimates, but most seem to hover around [math]10^{-30}[/math]. :confused: Quote
modest Posted March 31, 2009 Report Posted March 31, 2009 I'm curious as to the discrepancy between our numbers, Modest. I've read several estimates, but most seem to hover around [math]10^{-30}[/math]. :confused: Centimeters cubed and meters cubed :) ~modest ********** EDIT *********** 1 * (10^(-29)) * (g / (cm^3)) = 1.0 × 10-26 Kg / (m^3) ********** EDIT *********** Quote
freeztar Posted March 31, 2009 Report Posted March 31, 2009 Centimeters cubed and meters cubed :umno: ~modest ********** EDIT *********** 1 * (10^(-29)) * (g / (cm^3)) = 1.0 × 10-26 Kg / (m^3) ********** EDIT *********** Ahh...It's usually the simple stuff huh? :umno: Quote
Ki_Ryn Posted March 31, 2009 Author Report Posted March 31, 2009 My data is from "Perspectives on Astronomy" by Seeds & Backman, page 238: "The amounts of deuterium and lithium-7 show that normal baryonic matter can make up only about 4 percent of the critical density. Dark matter must be nonbaryonic and makes up less than 30 percent of the critical density." As I mentioned though, that part isn't under dispute. Quote
Pyrotex Posted March 31, 2009 Report Posted March 31, 2009 [Ahem] I happen to be an authority on Far-Significantly-Slightly Less (FSSL) comparisons.(pronounced, "fizzle") In matters of chemistry, for example, we haveFar Less means > 36% less, Sig Less means > 17% less, and Sli Less means > 3% less.Or, in shorthand notation:Chemistry: [36, 17, 3] We also have,Physics: [58, 25, 7]Astrophysics: [86.3, 44, 19.1]Cosmology: [99.999, 99, 90]Kitchen Recipes: [50, 16.7, 8.1]Carpentry: [19, 10, 2.25]Plumbing: [19, 8, 1.75] {currently under re-negotiation}Sociology: [37, 29, 15]Epidemiology: [22.2, 8.8, 1.5]UFOlogy: [1.03, 1.02, 1.01] Quote
modest Posted March 31, 2009 Report Posted March 31, 2009 [Ahem] I happen to be an authority on Far-Significantly-Slightly Less (FSSL) comparisons.(pronounced, "fizzle") In matters of chemistry, for example, we haveFar Less means > 36% less, Sig Less means > 17% less, and Sli Less means > 3% less.Or, in shorthand notation:Chemistry: [36, 17, 3] We also have,Physics: [58, 25, 7]Astrophysics: [86.3, 44, 19.1]Cosmology: [99.999, 99, 90]Kitchen Recipes: [50, 16.7, 8.1]Carpentry: [19, 10, 2.25]Plumbing: [19, 8, 1.75] {currently under re-negotiation}Sociology: [37, 29, 15]Epidemiology: [22.2, 8.8, 1.5]UFOlogy: [1.03, 1.02, 1.01] :confused: :naughty: :shrug: :confused: "Plumbing... currently under re-negotiation" :naughty: My data is from "Perspectives on Astronomy" by Seeds & Backman, page 238: "The amounts of deuterium and lithium-7 show that normal baryonic matter can make up only about 4 percent of the critical density. Dark matter must be nonbaryonic and makes up less than 30 percent of the critical density." As I mentioned though, that part isn't under dispute. Yes, I understand you're not disputing the 34% part. I'm just letting you know that the best current figure is 26%. Where you read "critical density. Dark matter must be nonbaryonic and makes up less than 30 percent of the critical density", it's currently believed to be 22% which is sure enough less than 30. Perhaps you can tell us the context of this question... why does it matter if it's considered "far" or "slightly" less than the critical density? What's the implication? ~modest Quote
freeztar Posted April 1, 2009 Report Posted April 1, 2009 How is Omega-m reconciled with critical density? If Omega-m is 0.266, how does this relate to 10.5 for critical density? Quote
modest Posted April 1, 2009 Report Posted April 1, 2009 How is Omega-m reconciled with critical density? If Omega-m is 0.266, how does this relate to 10.5 for critical density? I cannot figure out why the link from my last post has 10.5 kg/m^3. It must be a typo because it makes no sense and it's very much wrong. All the sources I can find have around 1 x 10^-26 kg/m^3. Wolfram scienceworld gives: 1.9 x 10^-29 g/cm^3 and wiki's Lambda-CDM page gives 0.94 x 10^-26 kg/m^3. Those numbers makes sense, not 10.5... So, I wouldn't put too much thought into trying to figure out what that 10.5 means... I don't think it means anything at all. How Omega-m relates to critical density... Critical density and matter density are both absolute values... density.. just like water has a density of 1g/ml or gold has a density of 18g/ml, the universe's matter density is in units of mass per volume. Omega on the other hand is not a density. It is, rather, a ratio of densities which makes it unitless. All of the Omega's are ratios (Omega-M, Omega-Lambda, Omega-Total). They are ratios of the measured energy density to the critical density. It's explained (albeit, not very well) in Wiki here. So, Omega-m is the mass density divided by the critical density or 2.66 x 10^-27 Kg/m^3 divided by 1 x 10^-26 Kg/m^3 which equals 0.266 or 26.6%. This means the "measured" mass density of the universe is 26% of the critical density. I think Ki_Ryn is most-likely wondering why it is only 26% of the critical density when we often hear that our universe is almost exactly equal to the critical density. In other words: shouldn't it be only "slightly" smaller than 100%? This is because mass density isn't the only energy density contributing to the total density. There is also dark energy, radiation density, and a spatial curvature density. There are 4 energy densities all together:[math]\Omega_M[/math] - Mass (visible matter and dark matter)[math]\Omega_{\Lambda}[/math] - Dark energy (cosmological constant)[math]\Omega_R[/math] - Radiation [math]\Omega_k[/math] - Spatial CurvatureRadiation density is very negligible in the universe today and spatial curvature density is expected to be zero (our universe is expected to be flat). These values should total to equal 1 in our universe to give Omega-Total = 100%. When the concordance model values are used, they do. The concordance model gives:[math]\Omega_M \approx 0.26[/math] [math]\Omega_{\Lambda} \approx 0.74[/math] [math]\Omega_R \approx 0[/math] [math]\Omega_k \approx 0[/math] [math]\Omega_{Total} \approx 1[/math] You can also, with these omegas, express the Friedmann Equation by adding the scale factor (a), the Hubble constant (Ho), and the Hubble parameter (H) giving: [math]\frac{H^2}{H_0^2} = \Omega_R a^{-4} + \Omega_M a^{-3} + \Omega_k a^{-2} + \Omega_{\Lambda}[/math] And, that equation is what most of modern cosmology is based on... The Friedmann equation :) ~Jay Quote
sanctus Posted April 1, 2009 Report Posted April 1, 2009 Since Modest gave such a good explanation there is only one detail (modest knows of) to add which he forgot to write: the values of the [math]\Omega_x[/math] he gave are the values today. For example, early on the universe was radiation domianated and then [math]\Omega_r[/math] is not close to zero.But then also, it depends a bit on literature [math]\Omega_x[/math] refers to densitiy ratios today implicitly or not. I write all this, because I lost already a few hours on a calculation until I figured out that the ratios were time dependent... Quote
modest Posted April 1, 2009 Report Posted April 1, 2009 Good call, Sanctus. I would agree, that's a pretty essential thing to point out. I think it's also described really well here: Conventionally we use the density parameters at the present time: [math]\Omega_m = \rho_m / \rho_c, \; \Omega_{\Lambda} = \rho_{\Lambda} / \rho_c, \; \Omega = \Omega_m + \Omega_{\Lambda}[/math]This choice has the nice feature that we don't have to worry about k explicitly, as it is given by the sign of Omega-1. On the other hand, there is a drawback: there is nothing special about the present time, so we are inserting an extra scale (the Hubble time tH = 1 / H0 , i.e. the inverse of the present-day Hubble parameter) into the equations, which complicates matters a bit. Notice that because H changes with time, so does the critical density, and so the density parameters (Omegas) change with time in a moderately complicated way. The Friedman Equation On that website right under that description there's a really nice Java program. If you select the checkboxes for "Plot... Trajectories" and "Enable Animation" then you can click anywhere on the graphic to choose the current values of Omegas and it will show you the evolution of the universe by plotting the past and future omegas and the evolution of the scale factor on the image to the right. If a person wanted to get an intuitive sense of how these Omegas affected the past and future evolution of the universe and how they make different 'Friedmann universes' then I think that Java applet would really help accomplish that—I know it did for me. I spent... probably far too long playing with it :computerkeys: :hihi: ~modest Quote
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