The Formation of the First Stars Massimo Stiavelli STScI - PowerPoint PPT Presentation

The Formation of the First Stars Massimo Stiavelli STScI Baltimore (MD, USA)

Plan of the Lectures 1. Physical Conditions after Recombination, Cooling, Density Perturbations 2. Formation of the First Stars, Properties of the First Stars, Death of the First Stars 3. Feedback and Self-Regulation, the First Star Clusters, First Quasars 4. Photometric redshifts: methods and limitations

Why can we do it? • We do not understand star formation in our own galaxy how can we suppose to understand the formation of zero-metallicity (Population III) stars of which we have not observed even one? --> We know the initial conditions and they are simple: no metals, no magnetic fields, very low vorticity, no inhomogeneities, no feedback from previous generations of stars.

Physical conditions after recombination • Several thousands of years after the Big Bang when the radiation temperature is still many tens of thousand degrees the Universe can be described as a mixture of protons, electrons, and Helium nuclei with traces of heavier ions. • As the temperature decreases, first Helium and the Hydrogen recombine. The latter phase is known as “Recombination”. • Processes occurring during cooling are important because they set the chemical composition of the Universe after Recombination.

Physical conditions after recombination • Any object collapsing tends to heat up as it attempts to maintain hydrostatic equilibrium in a stronger gravitational field. • In order to cool any object will need a suitable cooling agent. Hydrogen is not an effective coolant as its lowest energy transition (ground state n=1 to n=2) has an energy of 10.2 eV and will be excited only at high temperatures (10 5 K with some effectiveness down to 10 4 K due to high velocity tails). • Helium is effective at even higher temperatures. • The chemical composition, i.e. the traces of other molecules, will determine whether objects able to collapse will also be able to cool and form stars.

Conditions after Recombination • Let’s start out by deriving the residual ionization fraction after Recombination. We can use Saha’s equation to characterize the equilibrium between recombinations and ionizations: • Where B 1 is 13.6 eV, n X is the number density of the species X, and T is the temperature (of radiation). Defining x = n p /(n p +n H ): • This expression is valid only as long as radiation can maintain equilibrium. This is no longer true when the number of photons more energetic than 13.6 eV is too low.

Conditions after Recombination The figure shows the results of numerical integrations: Dotted - Saha

Conditions after Recombination • When the balance between ionizations and recombinations goes out of equilibrium one needs to solve a time dependent equation. At first one could attempt to solve: • Where α B and β H are the recombination and ionization coefficients and n 2s is the number density of atoms in the state 2s (which is much easier to ionize than the ground state: 3.4 eV). The population in the ground state and the 2s state are not in thermal equilibrium as the ground state can be populated either from 2p through emission of a Lyman α photon or through two-photon decay from the 2s state. Lyman α photons so produced are absorbed by other atoms until the Hubble expansion brings them out of resonance. Peebles introduced a corrective factor to the right hand side of the equation above to capture these effects.

Conditions after Recombination The figure shows the results of numerical integrations: Dotted - Saha Long Dash - eq. no corrective factor Short dash - eq with corrective factor Solid - more complete eq. with T rad ≠ T gas X ~ 2 × 10 -4

Molecular Hydrogen • Molecular Hydrogen is the most common molecule in the early Universe and is a promising coolant because of its richness of energy level compared to atomic Hydrogen. • In the early Universe there are two major formation channels for molecular Hydrogen formation: • They rely on free electrons or protons as catalizer and therefore depend critically on the residual ionization fraction we just derived.

Molecular Hydrogen • In both channels the first reaction (H+e or H+p) is the slowest and thus sets the overall rate. • For both channels there is a critical redshift when formation is most effective. At higher redshift the products on the RHS are efficiently destroyed. At lower redshift the reaction is ineffective because the density is too low. The table shows the effective reaction rates. The fraction of H 2 produced can be estimated at z=z eff as: fH 2 ~ x R n HI t Hubble

Molecular Hydrogen • The figure shows the computed H 2 fraction from full integration of the chemical evolution equations. • Dotted - H - channel • Dashed - H 2 + channel • Solid - total • fH 2 ~ 2 × 10 -6

Cooling • It is customary to describe cooling in terms of a cooling function Λ (T). • dE/dt = Λ (T) n • In the Figure we show the cooling function for atomic Hydrogen and that for molecular Hydrogen for two H 2 fractions.

Cooling • Looking at the H 2 cooling function in a Log-Log plot it appears that over a reasonable range of temperature it could be simply approximated by a straight line.

Cooling • The following approximation of the cooling function per H 2 molecule is very good for temperature between 120 and 6400K.

Cooling with metals • One may wonder for which metallicity the cooling function begins to change significantly. • For temperatures below H 2 f=10 -4 1000 K even a metallicity of 1/1000 Z=10 -2 Z  solar can be significant. Z=10 -3 Z 

Saha’s Equation • The Equation describes the equilibrium condition for a reaction of the type: A + B = C + γ • In the case of interest for us it was A = e- B=p C=H. • The equilibrium condition is obtained from the equality of the chemical potential: µ A + µ B = µ C + µ γ • We now need to derive the chemical potentials for the species we are interested in.

Chemical potential - R1 • The chemical potential can be obtain as the derivative with respect to the number of particles of the thermodynamical free energy F(T, V, N). • Let’s derive the free energy for radiation. The energy as a function of the temperature is: E = V a T 4 • Where a = 8 π 5 k 4 /(15 h 3 c 3 ), V is the volume and T the temperature. • Note that E is not the thermodynamical energy which is a function of the entropy rather than the temperature E = E (S, V, N) but it has the same numerical value.

Chemical potential - R2 • The free energy F(T, V, N) is obtained from the thermodynamical energy by a Legendre transform replacing the variable S by the variable T (entropy and temperature are conjugate): F = E - TS • Using now the value of E we find: F = E - TS = E - T ∂ F/ ∂ T • This is now a differential equation for F(T, V, N) that has solution: F = V (a/3) T 4 + f(V, N) • The 3rd principle of thermodynamics ensures that the function f(V,N) =0.

Chemical potential - R3 • Thus the chemical potential of radiation vanishes: µ = ∂ F/ ∂ N = 0 • This was to be expected because the energy density of radiation depends only on its temperature.

Chemical potential - P1 • Let’s now derive the chemical potential for particles. We do so by writing the free energy as the log of the partition function for a multiple particle system. The global partition function is written in terms of the single particle partition function: Z N,X = (Z X ) N / N! • We can start out by writing the partition function for a single particle system with one internal energy level: • Where gX is the multiplicity of the one level.

Chemical potential - P2 • Integrating we find: • The free energy is obtained as:

Chemical potential - P3 • From the expression in terms of the single particle partition function and using Stirling’s approximation we find: • From the equality of the chemical potentials, dividing by (-kT) and taking the exponential of both sides we find:

Chemical potential - P4 • Replacing now the form of the single particle partitional function we find: • Where Δ E = E A +E B -E C is the energy difference between the two sides. • The classical Saha equation is obtained by ignoring the mass difference between protons and neutral hydrogen.

Cosmology Brief - 1 • We live in a Universe with a cosmological constant. Fortunately, at the redshifts of interest the Universe behaves like the simpler Ω =1 cosmology. We can show this by considering the expression for the Hubble constant as a function of redshift: • At z<100 the contribution of radiation to energy density can be ignored so that Ωγ ~ 0. • With Ω M = 0.26 and Ω Λ = 0.74 we have that Ω R = 0 and the equation becomes

Cosmology Brief - 2 • The critical density at a given redshift is given in terms of the Hubble constant as: • The matter density is instead given by: • Thus the Ω M as a function of redshift is given by: