Nucleosynthesis and the early Universe

Transcription

1 Nucleosynthesis and the early Universe First a caveat. Liddle does a good job of covering this material but makes some bad mistakes in deriving the time temperature or time energy relationships in chapter 11 of his book (as he notes himself on his errata page on his website). These should be corrected in the revised 2008 edition, but are present in all earlier versions. He uses these wrong values in chapter 12 on nucleosynthesis. You should not derive those relationships as he did, you should follow the solution given in the second tutorial sheet, which I ve reproduced here for reference. You need to be able to make calculations such as these yourself. Friedmann is (ȧ a ) 2 = 8πG 3 ρ kc2 a 2 but since this is the early universe you can consider k = 0. Radiation scales as ρ radiation = ρ radiation,0 a 4 with ρ radiation,0 = ǫ radiation,0 c 2 = / = kg/m 3 Integrate the Friedmann equation ( ) 32πG 1/2 t = ρ radn,0 a 2 3 Since T = T 0 (1 + z) with T 0 = 2.7K ( ) 32πG 1/2 ( ) 2 T0 t = ρ radn,0 3 T Plug in the actual numbers for T 0 and ρ radiation,0, to get ( with T in Kelvin): t = T 2 To convert instead to energies simply equate an energy to a temperature E = k B T

2 For example for 1Mev, the energy is J, so T is T(1MeV) = / = K and hence t(1mev) = /( ) 2 = 1.3s Nucleosynthesis In stars fusion occurs at 10 7 K but density in stellar cores is greater than in early universe. Need higher temperatures still (> 10 9 K) to boost the collision rate between particles to get the fusion reactions to occur. The nucleosynthesis calculation is a detailed numerical one balancing all the competing reactions etc. This version only gets decent as opposed to accurate estimates. Basic assumptions are that by the time fusion occurs the main elements that make up atoms, namely protons and neutrons, are entirely non-relativistic. This is actually pretty reasonable since the simplest atom other than hydrogen is deuterium and it has a binding energy of only 2.2MeV. The neutron has a mass of 939.6MeV and the proton 938.3MeV (in particle physicists units), or a difference of only 1.3MeV. The net result is that the statistics of the populations for both protons and neutrons is that given by Maxwell-Boltzmann statistics. The appropriate energy to use in these statistics is the combination of both kinetic and rest-mass energy. However since the thermal energy at this point is much less than the rest-mass energy, only the latter actually matters. Note this implies these populations are still in thermal equilibrium. This gives the equation for number densities as: n m 3/2 exp ( ) mc 2 k B T The missing constants here are unimportant since we only want the ratio of the proton and neutron number densities. n n n p = ( mn m p ) 3/2 exp ( (mn m p )c 2 ) k B T At high enough T (ie k B T m n c 2 ) this ratio is simply 1 since the numbers are kept in balance through the weak interaction chains: n + ν e p + e

3 n + e + p + ν e These chains obviously terminate near the difference in mass between proton and neutron, ie k B T 1MeV. After this the heavier particle, the neutron, decays unless it is locked into an atom only free neutrons decay. This is the point at which weak interactions freeze-out ie they become one way only since the thermal bath the universe provides is no longer hot enough to allow the two way processes to continue. This freeze-out process happens a lot in cosmology and particle physics, and is at the basis of how things like grand unified theories for forces work. After freeze-out the Maxwell-Boltzmann statistics no longer apply. In practice the point at which the freeze-out takes full effect is about k B T = 0.8MeV. At that energy the ratio gives 0.2 neutrons for every proton. Unfortunately, just like with the CMB, although deuterium can form at temperatures below 2.2MeV there are just too many energetic photons. The actual temperature where the high energy tail dies out is about 0.06MeV (note Liddle says 0.1 but this was to make the numbers agree with his wrong values for the age!). This difference is important since the universe is only 2s old at 0.8MeV, but 360s old at 0.06MeV. And neutrons have a half-life of 614s. The abundance changes due to decay according to the standard formula: n(t) = n 0 exp( λt) λ = ln2 t 1/2 So at 0.06MeV the number of neutrons n(t) = 0.66n(0.8MeV). Reaction Chains There are many possible reactions. The main things to consider though are (i) what elements are formed and (ii) why any others aren t. p + n 2 D 2 D + p 3 He 2 D + 2 D 4 He 2 D + 2 D 3 H + p 3 H + 2 D 4 He + n + γ

4 3 He + n 4 He So it is easy to make deuterium (but also to destroy it) and to make both He-3 and He-4. What about the heavier elements? 3 H + 4 He 7 Li 3 He + 4 He 7 Be The lack of a stable element with atomic weight of 8 makes it difficult (ie slow) to get beyond this. Before any of the slow reactions can occur the universe has cooled too much, and all the nuclear fusion stops. (The lack of the element with atomic weight of 8 is also why carbon-12 in stars is formed by a three-body process). The following plot summarises the actual evolution with time. It s from a slightly hidden bit of Ned Wright s webpage at wright/bbns Actually the whole page is worth a look for background. How much helium? Effectively all the neutrons get locked up in helium-4 at the end. The other light elements exist as traces only. This means that for every neutron one proton is locked into helium. Therefore the mass fraction for helium, usually written as Y 4 is given by 1 minus the hydrogen mass fraction Y 4 = 1 n ( p n n = n ) 1 p n n + n p n n

5 Therefore at 0.06MeV, since n n n p = = 0.13 Y 4 = 0.23 ie 23% of the mass in the universe was originally helium. If the neutron half-life was less there would be more helium, very much longer there would be none. Y 4 is dependent on the baryon density, since fewer baryons means fewer reactions. Equally, higher densities lead to less deuterium since it ends up in helium instead. The following figure is from Liddle s textbook. There s a similar plot on Ned Wright s page which may be more up-to-date. There is also a constraint on the number of species of massless neutrino here too, since the time temperature relation depends on the radiation density. ( ) 32πG 1/2 ( ) 2 T0 t = ρ radn,0 3 T

6 Add in another species of neutrino and you increase the radiation density. The value normally given is split 1:0.68 for photons to neutrinos, so adding another massless neutrino makes it 14% larger. This means you multiply the right hand side of the equation by The universe expands more rapidly early on if there is more radiation (more radiation means eventually more braking due to gravity, but this takes effect gradually remember to get to the same point seen now a higher density universe expands more rapidly early on since it must decelerate more by now). Therefore for a fixed temperature such as 0.06MeV the universe is younger, so fewer neutrons decay and more helium is formed. Observations The basic idea is simple: deuterium is destroyed in all stars and never made. Lithium is destroyed in stellar cores and only rarely made. Helium is formed in stars so we cannot rely on what we see being primordial. All these can be corrected for however. The one only rarely used is helium-3 since its signature is very weak compared to helium-4 and therefore its abundance poorly constrained. For helium-4 the idea is to look at galaxies which have had few previous generations of star formation. This should mean they have more primordial gas left in them. As successive generations of star formation occur more and more metals are formed. A good indicator of metal content is actually oxygen. Therefore the trick is to determine the correlation between oxygen and helium in metal-poor galaxies. The figure shows Y 4 as a function of the oxygen abundance from a paper by Olive & Skillman (ApJ, 2004, 617, 29). Extrapolate back to zero oxygen and you have an estimate of the primordial helium abundance: the point shown at O/H=0 is actually that extrapolated value, and is Y = ± We see lithium as an absorption line from the atmospheres of warm-ish stars and very young stars. The abundance is usually measured from the former. Stars with effective surface temperatures > 6000K do not

7 have convective envelopes so there is no mixing from the surface down to the core and back. Therefore the lithium in the surface layers is never destroyed. At least that s the simple principle. Unfortunately the observed values come out a factor of 2 3 lower than the other elements would lead one to expect and most people now accept that other processes affect the lithium abundance in these stars. If you believe we can accurately determine the corrections made for these processes the lithium abundance is still a good probe. The magnitude of these corrections is still debated however. The strongest constraint actually comes from the most easily destroyed element, deuterium. There are two reliable means to determine its abundance. Recently groups have used background quasars to look for absorption due to deuterium to measure a genuinely primordial value (since the gas doing the absorbing has never been near a star at z = 1). The more common method, though one that caused problems initially, is to look for absorption along the line of sight to stars in our galaxy, This seemed to give low values but it has now been realised that deuterium can be locked up in dust grains without destroying it. Therefore we should always take the maximum value seen. Here s what the abundance as a function of distance from us looks like (from d2h solved.html): Note the most distant of these is still this side of the Galactic centre. The low values near us caused people to believe the universe was denser until the more recent higher results were obtained. Remember deuterium is never created in stars so the highest value seen is the correct one to use. The results obtained from the quasars agree with this (though with larger errors).

8 Conclusions Deuterium gives the strongest constraints on nucleosynthesis, helium-4 the most accurate values. Between them the likely range of allowed baryon density is Ω baryon,0 h 2 = 0.02 ± Therefore there is a serious deficit between the baryon density and the observed mass density in the universe. This must be nonbaryonic (the only exclusion would be micro-black-holes formed before nucleosynthesis ever happened). It can t be relativistic since that would change the helium fraction as noted before. It can t be heavy neutrinos as neutrinos are too common, and that would make the universe matter dominated too early. It has to be either: low mass but non-relativistic (there are particles called axions that are proposed as part of unified particle physics models that would be a good candidate here) high mass but very rare Of course no one has ever detected a real dark matter particle so this is an open issue. The main conclusion to take away from this however is that cosmological nucleosynthesis really does provide us with direct evidence for what was happening in the period between 1 and 1000 seconds in the early universe.