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
• wn galaxy how can we suppose to

understand the formation of zero-metallicity (Population III) stars of which we have not

• bserved 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 (105 K with some effectiveness down to 104 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 B1 is 13.6 eV, nX is the number density of the species X, and T

is the temperature (of radiation). Defining x = np/(np+nH):

• 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.

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

Conditions after Recombination

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 αB and βH are the recombination and ionization coefficients and

n2s 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

• r 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.

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 Trad ≠ Tgas

X ~ 2 × 10-4

Conditions after Recombination

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 H2 produced can be estimated at z=zeff as: fH2 ~ x R nHI tHubble

Molecular Hydrogen

• The figure shows the

computed H2 fraction from full integration of the chemical evolution equations.

• Dotted - H- channel
• Dashed - H2

+ channel

• Solid - total
• fH2 ~ 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 H2 fractions.

Cooling

• Looking at the H2

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 H2 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

1000 K even a metallicity of 1/1000 solar can be significant.

H2 f=10-4

Z=10-2 Z 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:

µA + µB = µC + µγ

• 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 T4

• Where a = 8 π5 k4/(15 h3 c3), 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) T4 + 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
• f 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 = (ZX)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 ΔE = EA+EB-EC 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 ΩM = 0.26 and ΩΛ = 0.74 we have that ΩR = 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 ΩM as a function of redshift is given by:

Cosmology Brief - 3

• It is easy to verify that ΩM ≈ 1 to better than 10% at z>3 and to better

than 1% at z>6. Similarly ΩΛ ≈ 0 at the same redshift.

• Thus our Universe at 6 < z < 100 behaves as a flat Einstein-de Sitter

ΩM = 1 Universe. This simplification is valid only for local quantities and doesn’t apply to quantities that are computing by integration to very high or very low redshift (like the Hubble time).

• It is possible to find a simplification for the Hubble time:

Perturbations

• One can compute the mass distribution of halos from linear theory. It is

convenient to introduce the fluctuation σ :

• Where P(k) is the power spectrum, b(z) is the linear growth of

perturbations normalized to 1 for z=0, and W is a filter function (e.g. Gaussian, top hat in configuration space, etc).

• σ can be used as a mass scale (for a monotonic spectrum).
• The number density of halos of a given mass is given by:
• Where f is a dimensionless mass function.

Perturbations

• The standard Press-Schechter theory is obtained with :
• Where δc is ~ 1.67. A better approximation to the results of N-

body simulations is given by the Sheth-Tormen mass function;

Perturbations

• The figure gives the mass

function of dark halos in the Sheth-Tormen approximation at different redshifts.

Perturbations normalization

• In a true ΩM=1 Universe perturbations scale as (1+z)-1 at all
• redshifts. This is true only locally in our Universe. When applying

the normalization a z=0 of the amplitude of perturbations we must face the difficulty that below z=6 our Universe is not well described by ΩM=1.

• Luckily a simple formula can be found providing the effective

normalization at z=0:

• This formula is valid for 1+z > Ωm
• 1/3.

Virialized Halos

Let‘s focus on the non-linear evolution of an overdensity. size time

Universe scalelength

The overdensity behaves like a Universe with Ω>1 and recollapses. t1 t2

t1 is the turnaround time, the collapse time t2 = 2 t1 At turnaround the perturbation has a density 9π2/16 times higher than the Universe. When the perturbation virializes at collapse time its density increases by another factor of 8 but the density of the Universe will have decreased by another factor 4 so that the perturbation has now a density ~178 times higher than the Universe.

Virialized Halos

We have seen that the virialized density of a halo is: With ξ≈178. From the mass and density we can derive a radial scale (sometimes known as the virial radius) and we can define Rh as the radius that would contain half the mass if the system was

• homogeneous. The virial theorem tells us that 2K+W = 0 and we

can estimate the potential energy per particle W as GMmp/Rh. Expressing the kinetic energy per particle K as 3/2 kB Tvir we find the expression for the virial temperature:

Bibliography

• Stiavelli, First Stars and Reionization, Wiley
• Peebles, Principles of physical cosmology,

Princeton

• Peacock, Cosmological Physics, Cambridge
• MANY papers: Galli&Palla, Abel, Oshea&Bryan.