# Download Astronomy 275 Lecture Notes, Spring 2009 by Edward L. Wright, Phd, UCLA PDF

By Edward L. Wright, Phd, UCLA

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8788 Ω◦ h2 × 10−29 gm/cc. 000000000000000000000000000000000000000000000000000000000002 (99) which is really stretching the bounds of coincidence. This is known as the flatness-oldness problem in cosmology. 1 at a time 10−43 seconds after the Big Bang, it would recollapse in 10−42 seconds. Thus in order to have a Universe as old as ours with Ω◦ close to 1 requires Ω very very close to 1 at early times. 1. The Flatness-Oldness Figure The calculations involved in the flatness-oldness figure are as follows: Ωm /a + Ωv a2 + Ωr /a2 + Ωk a˙ = H◦ = (100 km s−1 Mpc−1 ) ωm /a + ωv a2 + ωr /a2 + ωk (100) with ωm = Ωm h2 etc.

We can express the redshift in terms of the flux by defining ζ = S1 /S where S1 = L/(4π(c/H◦ )2 ) is the flux the source would have in a Euclidean Universe with distance d = c/H◦ . Think of √ ζ (“zeta”) as a Euclidean distance in redshift units. Solving ζ = z(1 + z/2) gives z = −1 + 1 + 2ζ so for Ω = 0 the N(S) law is n◦ (L/4π)3/2 1 n◦ (L/4π)3/2 dN = = dS 2S 5/2 (1 + 2ζ)2 2S 5/2 1 1+2 S1 /S For an Ω = 2 matter dominated Universe we get cz dL = H◦ c d(dL ) = dz H◦ cdt c 1 √ = dz H◦ (1 + z)2 1 + 2z n◦ (L/4π)3/2 1 dN √ = 5/2 3 dS 2S (1 + z) 1 + 2z 2 (163) (164) For large z we have S ∝ z −2 in this model, so the source counts flatten to dN/dS ∝ S −3/4 .

160) dN n◦ (L/4π)3/2 1 + O(z 2 ) = dS 2S 5/2 (1 + z)4 (161) The q◦ dependence cancels out in the first two terms so We see that the correction term decreases the source counts below the Euclidean expectation. This flattening of dN/dS avoids the divergence implied by Olber’s paradox. 4. 50 We can easily work out the exact form of the relativistic correction for a few simple cases. For the empty Universe we get cz (1 + z/2) dL = H◦ c d(dL ) = (1 + z) dz H◦ cdt c 1 = dz H◦ (1 + z)2 n◦ (L/4π)3/2 1 dN = (162) dS 2S 5/2 (1 + z)4 For large z we have dL ∝ z 2 in this case so S ∝ z −4 .