hw1-sol - Ph 444 Solutions for Problem Set 1 1(Ryden 2.2 To...

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Unformatted text preview: Ph 444 Solutions for Problem Set 1 1. (Ryden 2.2) To decide how far one can see on average in a universe filled with spherical objects of radius R , it is simplest to think of a long cylinder along the line of sight. If an object is closer than R to the line of sight, then the line of sight intersects its surface. For a distance ` , the cylindrical volume which would contain such objects is πR 2 ` . If the density of objects is n , then the volume that will on average contain one object is defined by πR 2 `n = 1 and the average distance to which we see before our vision is blocked is ` = 1 πR 2 n . (1) You have probably seen a similar formula before as the equation for the mean-free- path between collisions of, for example, atoms in a gas. Another way to derive Equation (1) is to calculate the probability of each mean-free-path when the proba- bility of collision in each d` along the line of sight is d`πR 2 . Calculating the average value of ` using the probability distribution yields Equation (1). For a density of stars equal to 10 9 Mpc- 3 and a stellar radius of R = (7 × 10 8 m) / (3 . 086 × 10 22 m Mpc- 1 ) = 2 . 3 × 10- 14 Mpc , the average distance is ` = 1 π (2 . 3 × 10- 14 Mpc) 2 (10 9 Mpc- 3 ) = 6 . 2 × 10 17 Mpc . This distance is much larger than the distance light has traveled since the Big Bang, about 6000 Mpc, which is a large part of the reason why the night sky is dark. If galaxies have an average density of 1 Mpc- 3 and an average radius of R = 2 . × 10 3 pc = 2 . × 10- 3 Mpc, then the distance to a galaxy along the typical line of sight is ` = 1 π (2 . × 10- 3 Mpc) 2 (1 Mpc- 3 ) = 8 . × 10 4 Mpc . Thus, we might expect to see a galaxy everywhere on the sky; however, this distance is large enough that the expansion and geometry of the universe must be taken into account. 2. (Ryden 2.4) Observations of one type of neutrino ”oscillating” into another type yield two constraints on the masses of the electron, muon, and tau neutrinos ( m ( ν e ), m ( ν μ ), and m ( ν τ ), respectively): ( m ( ν μ ) 2- m ( ν e ) 2 ) c 4 = 5 × 10- 5 eV 2 (2) ( m ( ν τ ) 2- m ( ν μ ) 2 ) c 4 = 3 × 10- 3 eV 2 . (3) Equations (2) and (3) obviously require that...
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This note was uploaded on 02/06/2012 for the course PHYSICS 444 taught by Professor Staff during the Fall '08 term at Rutgers.

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hw1-sol - Ph 444 Solutions for Problem Set 1 1(Ryden 2.2 To...

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