AY127 Problem Set 3, Winter 2018

Transcription

1 California Institute of Technology AY27 Problem Set 3, Winter 28 Instructor: Sterl Phinney & Chuck Steidel TA: Guochao (Jason) Sun February 7, 28 Problem Detailed Solution : (a) We work in natural units ( = c = k B = ) and convert to more conventional units at the end of the problem. For Fermi-Dirac/Bose-Einstein gases in the ultra-relativistic limit (temperature is high enough such that M-B distribution is no longer a valid approximation), the distribution function (a.k.a. occupation number) is given by f(p) = e E/T ±. () Then the energy density can be written as an integral of E times the density of states d 2 N /dv x dv p = g/h 3 = g/(2π) 3 (recall that = h/2π = ) times the distribution function over the momentum space, namely ρ(t ) = g Ep 2 dp e E/T ±, (2) where we have assumed distribution function has no angular dependence and g describes the degeneracy of the species of interest. For bosons, the denominator takes negative sign and we can write E = p. Then gsun@astro.caltech.edu Courtesy of Daniel Grin ρ(t ) = g p 3 dp e p/t x 3 dx e x e x dx e x x3 (5) dx x 3 e x(n+) (6) dxx 3 e x(n+) (7) Γ(4) (n + ) 4 = 3gT 4 π 2 ζ(4) = π2 gt 4 3 (3) (4) (8)

2 For fermions, we can use the fact that e x + = e x 2 e 2x as well as the similar trick used for bosons to show that each fermionic integral is equal to 7/8 times the corresponding bosonic integral. For the degeneracy, we know that g = 2 for photons (two polarization states), each family of neutrinos (one spin state for neutrinos and anti-neutrinos, respectively) and electrons/positrons (two spin states). Therefore, we have (all in natural units) (9) ρ γ = π2 T 4 5 ρ ν = ρ ν = 7 π 2 T ρ e = ρ e + = 7 π 2 T or (after converting into conventional SI or cgs units) () () (2) ρ γ = at 4 (3) ρ ν = ρ ν = 7 6 at 4 (4) ρ e = ρ e + = 7 8 at 4 (5) where a = π 2 k 4 B /[5( c)3 ]. (b) For a relativistic species, P = ρ/3, so given the entropy density s = (ρ + P )/T (c.f. MvdBW), in a physical volume V we have S = sv = 4ρV 3T Eq of (6) (c) Since entropy is conserved, (at ) 3 g,s (c.f. Eq of MvdBW). When particle species become non-relativistic (and thus are no longer legitimately considered as radiation), or annihilate, they dump their entropy into the remaining particle species, so T decreases less rapidly than it would otherwise. At today s low temperatures, the only relativistic degrees of freedom are photons and neutrinos (assuming neutrinos have no or very small masses), so g,s (today) = (T ν/t γ ) 3. The reaction e + +e γ +γ freezes out at about T m e.5 MeV, slowing the cooling of the photons as electrons-positrons annihilate. On the other hand, neutrinos are able to maintain a thermal equilibrium, although being decoupled from photons, by weak reactions of the form ν + ν e + + e. In the Fermi effective theory for the weak interaction, the cross sections for these processes are calculated, to lowest order, by evaluating single vertex Feynman diagrams, so M G F, and σ = G 2 F T 2 (T 2 comes from demanding that the cross section have the right units). For relativistic particles, v = in natural units and n T 3, so the reaction rate Γ nσv G 2 F T 5. Equating this to the Hubble parameter allows one to solve for the freeze-out temperature for these weak reactions to be T f MeV. Thus, neutrinos decouple before electron-positron annihilation and their temperature redshifts as /a from there on. For those particles still in equilibrium with photons at electron-positron annihilation, entropy is conserved. Before annihilation, for the photon fluid we have photons, and electron-positron pairs, so g,s = (2 + 2) = /2, and after annihilation, g,s = 2. Conservation of entropy before and after annihilation then yields, for photon fluid and decoupled neutrinos respectively, (at γ ) 3 (a T γ, ) 3 = (7) 4 2

3 and (at ν ) 3 =. (8) (a T γ, ) 3 Thus, we have T ν = (4/) /3 T γ. A small neutrino mass will not effect this result much, as at electronpositron annihilation these (light) neutrinos are still relativistic and we can continue to use the same values for g,s that we did before, as these neutrinos are still relativistic degrees of freedom. (d) Although neutrinos fall out of equilibrium around T MeV, we can still use their equilibrium abundance, as their effective temperature shifts as T /a if they are massless. A phase space integral similar to that we had in part (a) gives us n ν = 3 4π ζ(3)gt cm 3, where g = 6 for the 3 flavors of neutrinos 2 and their anti-neutrinos. These neutrinos move roughly at the speed of light, and I estimate my surface area as about 3 cm 2, so the number of neutrinos passing through me per unit time is roughly given by Γ = n ν σv 3 6 s. For the case of massive neutrinos, g, T or n ν will not change, due to the relatively high temperature of electron positron freeze-out. Even if they are massive, neutrinos will be relativistic at T =.5 MeV. Thus the flux of neutrinos through my body will change only due to the resulting smaller velocity of neutrinos. Recall that in special relativity, the energy of a particle is given by E = γβmc 2 β, so if E T, then = β 2 T/m and we can find Γ m=. ev 6 5 s (9) Γ m=. ev 7 3 s (2) (e) Now let s calculate how many neutrinos interact with atoms in our body, assuming a realistic weak interaction cross section, σ ν G 2 F T 2 ν GeV 2 63 cm 2. This is for a single nucleon. The number density of nucleons in our body is about 24 cm 3. So the mean free path between neutrino collisions with nucleons in our body is l /nσ 4 38 cm. Assuming a path length through our body of roughly cm, we can calculate the number of collisions experienced by a single human over their lifetime to be N = l/l Γ T, where T 25 years. Thus so a single person, N 2. Today, there are of order people, so there is a % chance that any human has interacted with a neutrino. 3

4 Problem 2 Detailed Solution: Let s first write down the differential form of the Thomson scattering optical depth due to free electrons dτ = αds = n e σ T cdt = n e(z)σ T c ( + z)h(z) dz = n e (z)σ T c dz (2) ( + z)h E(z) For the reference cosmological model with Ω m =.3, Ω b =.45 and Ω Λ =.7, the evolution function E(z) = Ω m ( + z) 3 + Ω Λ. (22) Since the first ionization energy of helium (24.6 ev) is about double of that of hydrogen, we ignore the contribution towards n e due to helium reionization and write n e (z) = n H (z)x i (z) (23) where, for instantaneous reionization, x i (z) is simply a step function with x i = at z z r and x i = elsewhere. Meanwhile, n H (z) = ρ b, (+z) 3 ( Y )/m p = Ω b, ρ crit, (+z) 3 ( Y )/m p. So our optical depth integral effectively goes from to z r and can be written as τ = = zr zr n H (z)σ T c( + z) 2 H Ωm ( + z) 3 + Ω Λ dz (24) 3σ T ch ( Y )Ω b, ( + z) 2 8πGm p Ωm ( + z) 3 + Ω Λ dz (25) zr =.7 3 ( + z) 2 dz (26) Ωm ( + z) 3 + Ω Λ = [ ] Ωm ( + z) 3Ω 3 + Ω Λ m (27) For z r Ω m, we can drop the radiation component and we have τ = [ ] 3 ( + z) 3/2 Ω m (28) Solving for τ = in the reference model, we get z 6. 4

5 Problem 3 Detailed Solution: (a) At very high redshift, the Universe is radiation-dominated and in this case we have a(t) t /2, which implies H = ȧ/a = /2t. Then the Friedmann equation (neglecting k and Λ terms) suggests where which implies ( ) /2 3 t =, (29) 32πGρ γ ρ rad.68 ρ γ =.68 π2 k 4 B 5 3 c 5 T 4 (3) ( c 5 ) /2 ( t = 32π 3 kb 4 G T 2 9 ) 2 K 7 s. (3) T Given that the half-life of neutron decay is only 9 s, if we wait for much longer than 7 s, most of the neutron would have decayed. (b) Solving σv n b t for σv = 5 2 cm 3 /s and t = 7 s, we have n b the required baryon density appropriate for neutron capture..2 7 cm 3 as (c) We note that n b (t)a 3 (t) remains constant across cosmic time. The baryon number density today is n b, Ω b ρ crit /m p = /(.7 24 ) = cm 3. So the scale factor at T = 9 K is a(t) = (n b, /n b ) /3 =.3 8. (d) From (c), z /a(t) 7.6 7, which implies T = 9 /(7.6 7 ) 3 K. 5

6 Problem 4 Detailed Solution: (a) The abundance of a cold (non-relativistic) species χ in thermal equilibrium is described by the Maxwell- Boltzmann statistics (treating them as classical, distinguishable particles) ( ) 3/2 mχ T n χ (T f ) = g e mχ/t (32) 2π Since we are to decide the point at which the species falls out of equilibrium due to cosmic expansion, it s helpful to write down the Hubble equation 8πGρ 8πG H(T f ) = = 3 3 g π 2 3 T 4, (33) where we have assumed at freeze-out ρ is mostly attributed to some general form of radiation specified by g. Freezing out occurs when H = Γ = nσv. After a bit of algebra, this yields e mχ/t f ( ) /2 = A /2 Tf, (34) m χ where A = 64π 6 g G/(9g 2 σ 2 av 2 m 2 χ), or if we define y = T f /m χ, then y = ln A ln y (35) 2 Since one would expect that T f m χ and thus y, to the first order y / ln A is a slowly varying function of A and one can approximately write T f m χ. The number density of species χ today can be related to that at freeze-out by n χ, = n χ,f a 3 f have n χ,f H(T f )/σ a T 2 f σ a, so n χ, T f n χ,f T 3 f. Meanwhile, from the freeze-out condition, we σa m χ σa. (b) By definition, the density parameter of a non-relativistic species χ is Ω χ, = m χ n χ, /ρ crit. From the result of part (a), we know that n χ, m χ σa. Therefore, the mass dependence cancels out and the contribution of species χ to Ω only depends on σ a, not its mass m χ! For a more exact derivation, first of all we have n χ, = n χ,f (T γ, /T f ) 3 (36) = H(T ( ) 3 f ) Tγ, σ a v a T f (37) = 8π 3 k 4 g G Tγ, 3 9( c) 3, T f σ a v a (38) where we have converted from natural units to the conventional cgs units. Consequently, the density parameter becomes 8π3 k Ω χ, = 6 g G Tγ, 3 m χ c c 9 (39) ρ crit kt f σ a v a 7 27 g/cm 3 σv ( mχ c 2 ) where the last term in parentheses is a slowly varying logarithm as we have demonstrated previously, and the coefficient is evaluated for T γ, 3 K, ρ crit 29 g/cm 3 for H = 7 km/s/mpc and g 6 which is kt f (4) 6

7 the number of relativistic degrees of freedom predicted by the Standard Model for m χ GeV. (c) Given a cross-section σ a G 2 F m2 χ, we have ( σ a v 27 m ) 2 χ cm 3 /s (4) GeV From the expression of Ω χ we obtained in part (b), it s clear that it makes a significant contribution to Ω if m χ GeV. 7