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Unformatted text preview: Ph 444 Solutions for Problem Set 6 1. (Ryden 9.3) This problem examines the recombination of helium in the early universe. For simplicity, it considers a universe containing only He and assumes that the amount of doubly ionized helium is negligible. The latter assumption is reasonable for temperatures close to that at which He becomes neutral. Then the relevant reaction is He + He + + e . The ionization energy for He is Q He = 24 . 6 eV. Applying the MaxwellBoltzmann equation for the number densities of He, He + , and e yields, in analogy with Ryden equation (9.22) n He n He + n e = g He g He + g e parenleftbigg m He m He + m e parenrightbigg 3 / 2 parenleftBigg kT 2 h 2 parenrightBigg 3 / 2 exp parenleftBigg [ m He + + m e m He ] c 2 kT parenrightBigg . (1) Now m He m He + , the ratio of statistical weights is 1 / 4, and the difference in the rest masses is Q He . Thus, n He n He + n e = 1 4 parenleftBigg m e kT 2 h 2 parenrightBigg 3 / 2 exp parenleftbigg Q He kT parenrightbigg . (2) This is the equivalent of Ryden equation (9.23) for He. The fractional ionization of He is defined as X n He + n He + + n He = 4 n He + n baryon = 4 n e n baryon . (3) The above equation uses the recommended n baryon = 4( n He + + n He ). Equation 3 implies n He = 1 X X n He + = 1 X X n e (4) and so equation 2 becomes 1 X X = n He + 4 parenleftBigg m e kT 2 h 2 parenrightBigg 3 / 2 exp parenleftbigg Q He kT parenrightbigg . (5) Now equation 3 also implies that photontobaryon ratio can be written as n baryon n = 4 n He + Xn = 4 n He + . 243 X parenleftBigg hc kT parenrightBigg 3 . (6) The last step in the above equation uses the formula for the number density of black body photons at temperature T (Ryden equation 9.27). Using equation 6 to eliminate n He + from equation 5 produces the final equation for X in terms of and T : 1 X X = . 243 X 4 4 parenleftBigg kT hc parenrightBigg 3 parenleftBigg m e kT 2 h 2 parenrightBigg 3 / 2 exp parenleftbigg Q He kT parenrightbigg (7) 1 X X 2 = 0 . 239 parenleftBigg kT m e c 2 parenrightBigg 3 / 2 exp parenleftbigg Q He kT parenrightbigg . (8) 1 This equation must be solved numerically for the T that corresponds to a given X . It is useful to introduce the dimensionless variable y Q He / ( kT ) and take the logarithm of both sides of equation 8: ln parenleftbigg 1 X X 2 parenrightbigg = ln(0 . 239 ) + 3 2 ln parenleftbigg Q He m e c 2 parenrightbigg 3 2 ln( y ) + y (9) ln parenleftbigg 1 X X 2 parenrightbigg = 16 . 34 + ln( ) 3 2 ln( y ) + y (10) For X = 1 / 2 and = 5 . 5 10 10 , this becomes 38 . 36 = y 3 2 ln( y ) . (11)...
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 Fall '08
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