Lyman-alpha forest
By Bruce A. Peterson (The Australian National University). May, 2013.
Edit by Yuan-Sen Ting (Harvard University).
The hot early-Universe cools as it expands, and after about half a million years, the temperature has dropped enough for protons and electrons to combine to form hydrogen atoms. At this point the Universe is transparent to radiation with wavelengths longer than the Lyman-limit at 912 Å, except at wavelengths that correspond to ground state hydrogen absorption lines.
Radiation with wavelengths shorter than the Lyman-limit will ionize an atom of neutral hydrogen and be absorbed. Radiation shorter than the Lyman-alpha line at 1215 Å travels outward from a quasar and is seen redshifted along its path. It will eventually be redshifted to the wavelength of the Lyman-alpha line, where the radiation will excite the electron in a neutral hydrogen atom from the ground state to the first level, and be re-emitted by the atom in a random direction as it decays back to the ground state, effectively scattering the radiation.
Consider a photon emitted by a quasar with wavelength \(\lambda_e < \lambda_\alpha\), where \(\lambda_\alpha\) is the wavelength of the Lyman-alpha line. Along the line of sight, suppose there is a hydrogen atom at some redshift \(z\), where the photon is redshifted to the wavelength of the Lyman-alpha line according to \(\lambda_\alpha = \lambda_e (1+z_Q)/(1+z)\), where \(z_Q\) is the redshift of the quasar. This atom will scatter the photon out of the line of sight. Thus, Lyman-alpha line scattering by neutral hydrogen atoms along the line of sight will remove light from a quasar spectrum with wavelengths \(\lambda_o = (1+z) \lambda_\alpha\), where \(0 \leq z \leq z_Q\), producing an absorption trough in a quasar spectrum from \((1+z_Q)\lambda_\alpha\) down to \(\lambda_\alpha\). Of course, if the distribution of neutral hydrogen is not smooth, but clumped, then the optical depth along the line of sight will vary, being deeper where there are clumps.
The optical depth is given by \(p\), the probability of scattering of a photon in a proper length interval \(\mathrm{d}l = c\, \mathrm{d}t\) at cosmic time \(t\), which is \begin{equation} \mathrm{d}p = n(t) \sigma(\nu_s) c \, \mathrm{d} t, \end{equation} where \(c\) is the speed of light in vacuum, \(n(t)\) is the number density of neutral hydrogen atoms at time \(t\), and \(\sigma(\nu)\) is the radiative excitation cross-section for the Lyman-alpha transition. This has the form \begin{equation} \sigma(\nu) = \frac{\pi e^2}{m_e c} f \phi (\nu - \nu_\alpha), \end{equation} where \(e\) is the electric charge, \(m_e\) is the electron mass, \(f\) is the oscillator strength (here \(f =\) 0.416) and \(\phi\) is the line profile function, which is strongly peaked at \(\nu = \nu_\alpha\), the Lyman-alpha frequency (\(\nu_\alpha =\) 2.46 \(\times\) 1015 Hz), and has the integral \begin{equation} \int^\infty_{-\infty} \phi(x) \, \mathrm{d}x = 1. \end{equation}
Let the redshift of the quasar observed be \(z_Q\), and suppose the redshift of the scattering atoms as seen from here is \(z\), where \(z< z_Q\). If the observed frequency is \(\nu\), then the frequency seen by an observer stationary with respect to the scattering atoms is \(\nu_s = \nu (1+z)\). Thus the total optical depth at \(\nu\) is \begin{equation} p = \int \mathrm{d} p = \int^{z_Q}_0 n[t(z)] \sigma [\nu (1+z) ] c \frac{dt}{dz} dz. \end{equation} where (the details can be found in any cosmology textbook) \begin{equation} \frac{dt}{dz} = \frac{-1}{(1+z)H}. \end{equation} \(H\) is known as the Hubble parameter.
In the Friedman-Lemaître Universe, the derivative of the scale factor, \(a(t)\), is determined by \begin{equation} \frac{\dot{a}^2}{a^2} = \frac{8\pi G \rho}{3} - \frac{kc^2}{a^2} + \frac{\Lambda c^2}{3}. \end{equation} where \(\rho\) is the mass density of the Universe, \(G\) is the gravitational constant, \(\Lambda\) is the cosmological constant, \(k\) is a constant given a particular solution of the Friedman-Lemaître equation. We define the dimensionless cosmological parameters \begin{equation} \Omega = \frac{8\pi G \rho}{3H^2}, \qquad \lambda = \frac{\Lambda c^2}{3 H^2}, \qquad \kappa = \frac{kc^2}{a^2 H^2}, \end{equation} These parameters are subject to the constraint \begin{equation} \Omega_0 + \lambda_0 = \kappa_0 + 1. \end{equation} The subscript 0 designates the present epoch, \(z =\) 0. With \(\rho = (1+z)^3 \rho_0\), we have \begin{equation} H(z) = H_0 (1+z) \sqrt{\Omega_0 (1+z) - \kappa_0 + \lambda_0/(1+z)^2} \end{equation} where \(H_0\) is also known as the Hubble constant. Finally, \begin{equation} p = \frac{\pi e^2}{m_e c} f \frac{c}{H_0} \int^1_{1+z_Q} \Bigg[ \frac{n(z)}{\nu(1+z)^2 \sqrt{\Omega_0 (1+z) - \kappa_0 + \lambda_0/(1+z)^2}} ~ \Bigg] \phi [ \nu (1+z) - \nu_\alpha ] \nu~ \mathrm{d} ( 1+z)~. \end{equation} The function \(\phi\) is strongly peaked at zero. Thus we can take the factor in the brace out of the integral, evaluated at \((1+z)=\nu_\alpha/\nu\). The integral of \(\phi\) that remains is unity, and we obtain \begin{align} p &= \frac{\pi e^2}{m_e \nu_\alpha} f \frac{1}{H_0} \frac{n_s}{(1+z)\sqrt{\Omega_0 (1+z) - \kappa_0 + \lambda_0/(1+z)^2}} \\ &= \frac{4.14 \times 10^{10} n_s}{h_0(1+z) \sqrt{\Omega_0 (1+z) - \kappa_0 + \lambda_0/(1+z)^2}} \end{align} where \(h_0 = H_0/\)100 km s-1 Mpc-1 and \(n_s\) is the number density of neutral hydrogen atoms in the scattering region. The flux is reduced by the factor \(e^{-p}\).
With \(\Omega_0 =\) 0.3, \(\lambda_0 =\) 0.7 and \(h_0 =\) 0.7, the mass density of the Universe is given by \begin{equation} \rho = \frac{3 H_0^2 \Omega_0}{8 \pi G} (1+z)^3 = 2.76 \times 10^{-30} (1+z)^3 \; [\mathrm{g/cm^3}]. \end{equation} Most of this is dark matter. The baryonic density is only about 10 per cent of the total matter density. About 1/4 of the baryons are in the primordial 4He and about 80 per cent are in galaxy clusters. Thus the expected number density of intergalactic hydrogen is \begin{equation} n_{\small \mathrm{H}} = 0.1 \times 0.75 \times 0.2 \times \rho/m_{\small \mathrm{H}} = 2.5 \times 10^{-8} (1+z)^3. \end{equation} Setting \(p =\) 1 in equation (12) gives the value for \(n_s\) that produces an optical depth of one. WIth equation (14), one finds \(n_{\small{\mathrm{H}}}/n_s >\) 104 over the range \(z=\) 2\(-\)6. In other words, over the redshift range 2 \(< z_Q <\) 6, less than 1 in 104 hydrogen atoms in the neutral ground state will produce an optical depth greater than one and therefore a significant absorption trough.
The ground based spectra of quasars with redshifts in the range 2 \(< z_Q <\) 6 have large numbers of absorption lines at wavelengths \(\lambda_o\) shorter than \((1+z_Q)\lambda_\alpha\). The number density of absorption lines in a quasar spectrum increases rapidly with the redshift of the quasar. At a redshift of \(z_Q \sim\) 6, the absorption lines have merged into a continuous absorption trough. This phenomenon provides clear evidence for the reionization of the clumpy Universe, starting at a redshift somewhat beyond six.