Homework #4 Returned: Friday, March 25, 2016 1. Growth of structure with neutrinos. [30 points; 5 points each] In most of our studies of the matter-dominated era in class, we have neglected the neutrinos. However, if the neutrinos are massive, then when their momentum becomes less than their mass they act as non-relativistic particles and contribute to the density of the universe, even in the matter-dominated era. (a) What is the number density n(νe + ν¯e ) of relic electron neutrinos and anti-neutrinos from the Big Bang? Express your result in units of particles per cm3 . What about the µ and τ neutrinos? The number density of neutrinos in thermal equilibrium at high temperature (where the masses are negligible) can be expressed by integrating over the blackbody spectrum: Z ∞ 4πp2 1 n(νe + ν¯e ) = 2 dp 3 ecp/kT + 1 h 0 Z 8π(kT )3 ∞ x2 = dx (hc)3 ex + 1 0 12πζ(3)(kT )3 = , (1) (hc)3 where we substituted x = cp/kT , and used the Riemann ζ-function in evaluating the integral. Since – again at high redshfit where the neutrinos are effectively massless – the neutrino temperature is Tν = 1.95(1 + z) K, we substitute into this equation and find n(νe + ν¯e ) = 113(1 + z)3 cm−3 . Once neutrinos stop interacting, their number density scales as the inverse volume of the Universe, i.e. ∝ (1 + z)3 , regardless of the blackbody curve. We then see that the number density today is 113 cm−3 . The µ and τ neutrinos have the same number density, since they were at the same initial temperature and the above calculation applies to them as well. (b) Suppose that the three species of neutrinos have masses {mi }3i=1 . Show that today the neutrinos contribute to the density in accordance with P3 mi Ων h2 = i=1 . (2) 94 eV The mass density of neutrinos today is the sum of the density over the three species: ρ=

3 X

mi n(νi + ν¯i ) =

i=1

3 X

mi × 113 cm−3 .

(3)

i=1

To find Ων , we must compare this to the critical density: 3H02 3(3.24 × 10−18 h s−1 )2 = = 1.88 × 10−29 h2 g/cm3 = 1.69 × 10−8 h2 erg/cm3 = 1.05 × 104 h2 eV/cm3 . 8πG 8π × 6.672 × 10−8 cm3 /g/s2 (4) We conclude that the neutrino density parameter is P3 P3 −3 mi i=1 mi × 113 cm = i=1 . (5) Ων = 4 2 3 2 1.05 × 10 h eV/cm 93h eV ρcrit =

(c) What is the comoving distance that a neutrino can travel in a Hubble time today, if it has mass mi ? (Express your answer in units of h−1 Mpc, and evaluate it for m = 1 eV, 0.1 eV, and 0.01 eV.) Let us first recall that the momentum of a typical neutrino is p∼

3kT = 2.7 × 10−26 (1 + z) g cm/s = 5.0 × 10−4 (1 + z) eV/c, c

(6)

where we used Tν = 1.95(1 + z) K again. Since momenta redshift as 1 + z (i.e. as the inverse of a de Broglie wavelength), this formula remains valid today.

2 Then the distance travelled in a Hubble time is p p 5.0 × 10−4 eV/c 1.5h−1 Mpc tH = = = . −1 mν mν H0 mν × c/(2998h Mpc) mν /(eV/c2 )

(7)

Thus the distance a neutrino travels per Hubble time today would be 1.5h−1 Mpc (1 eV mass neutrino), 15h−1 Mpc (0.1 eV), or 150h−1 Mpc (0.01 eV). (d) Explain why, on scales smaller than the answer to (c), you expect the neutrinos to be uniformly distributed. On scales larger than the neutrino free-streaming length (Eq. 7), the neutrinos are effectively motionless and there is no difference between their behavior and that of cold dark matter. On smaller scales, the growth of structure takes ∼a Hubble time, and in that time the neutrinos propagate through many perturbation wavelengths. Thus they have passed through many crests and troughs of the perturbations and have not been consistently pulled in one direction throughout the history of the Universe (as the dark matter has). (e) Suppose that the neutrinos are massive and uniformly distributed, and that a fraction fν of the mass in the universe is neutrinos (and hence 1 − fν is baryonic or dark matter). Return to the growth of structure equation and show that √

G(a) ∝ a(−1+

25−24fν )/4

≈ a1−3fν /5 ,

(8)

where the last approximation holds if fν  1. Conclude that neutrinos slow the growth of structure on small scales. The growth equation is δ¨m + 2H δ˙m − 4πG¯ ρm δm = 0.

(9)

In a universe with mostly matter (baryonic+dark) and neutrinos, we use the “m” subscript to denote the matter. If the neutrinos are non-relativistic, then from the perspective of cosmic expansion they act like matter. Thus it is still true that a ∝ t2/3 and H = 2/(3t), and that the total cosmic density is ρ¯ is the critical density. However the (clustering) matter density is ρ¯m = (1 − fν )¯ ρ = (1 − fν )ρcrit = (1 − fν )

1 − fν 3H 2 = . 8πG 6πGt2

(10)

Substituting into Eq. (9) gives 4 2(1 − fν ) δ¨m + δ˙m − δm = 0. 3t 3t2

(11)

This is a dimensionally homogeneous equation with solutions δm ∝ tα , where the characteristic equation is 4 2(1 − fν ) α(α − 1) + α − = 0. 3 3 The solution to this equation is   s √ 2  1 1 1 −2(1 − fν )  −1 ± 25 − 24fν α= − ± − − 4(1) = .  2 3 3 3 6

(12)

(13)

The growing mode is thus √

G(t) ∝ t(−1+

25−24fν )/6

,

(14)

25−24fν )/4

(15)

and using t ∝ a3/2 we find G(a) ∝ a(−1+

√

.

Taylor expansion of the square root then yields Eq. (8). (f) If all 3 species of neutrino had a mass of ≈ 0.1 eV, what is fν ? By how much do you expect them to reduce the amplitude of density perturbations today?

3 In this case, we would have Ων h2 =

3 × 0.1 eV = 0.0032, 94 eV

(16)

whereas for current cosmological parameters Ωm h2 = 0.12. Thus we have fν =

0.0032 = 0.027. 0.12

(17)

To understand the impact on the growth of structure, it helps to re-write G(a) using G(a)|with neutrinos a1−3fν /5 ∝ = a−3fν /5 = e−(3fν /5) ln a . G(a)|no neutrinos a

(18)

Thus if 53 fν = 0.016, then each e-fold of expansion leaves G(a) 1.6% smaller than it would have been without the massive neutrinos. To complete the problem, we must determine for how many e-folds of expansion the above result applies. From (c), the typical neutrino momentum is 5.0 × 10−4 (1 + z) eV/c, so a 0.1 eV neutrino would have become nonrelativistic at 1 + z = 200. Therefore there are ln 200 = 5.3 e-folds of expansion during the time when the neutrinos are effectively massive (before 1 + z = 200 there is no difference relative to massless neutrinos). We therefore expect that the amplitude of small-scale density perturbations today is reduced in the massive neutrino case relative to the massless neutrino case by a factor of e−0.016×5.3 = 0.92.