Reading ionized bubbles through neutral hydrogen

Read Where the photons go when ionized bubbles meet first. We begin with two atomic energy levels, obtain a brightness temperature, and use simple waves to see what a power statistic retains. Basic trigonometry is useful later; complex numbers will package two real projections.

An atomic temperature and a background sky

The interaction between electron and proton spins splits the electronic ground state of neutral hydrogen into two nearby hyperfine levels. A transition from the upper to the lower level emits a photon with wavelength about 21 cm and rest frequency $\nu_{21}=1420.4057\,\mathrm{MHz}$. Expansion lowers its observed frequency:

At redshift eight, this is about 158 MHz. The energetic ultraviolet photons that change the ionization state and the radio photons used here to read neutral gas play different roles.

Let $n_1$ and $n_0$ be the number densities in the upper and lower hyperfine levels. Their ratio defines the spin temperature $T_{\mathrm s}$:

Here $h_{\mathrm P}$ is Planck's constant and $k_{\mathrm B}$ is Boltzmann's constant. Three counts the degenerate upper states. The splitting expressed as a temperature is only about $0.068\,\mathrm K$. Spin temperature describes the populations of these two groups; it need not equal the temperature of particle motions. Furlanetto, Oh and Briggs, 2006, §2.1

The sky also contains the microwave background at temperature $T_\gamma$. We seek a change relative to that background after radiation passes through the gas. In the Rayleigh–Jeans approximation, express the contrast as a brightness temperature:

where $\tau$ is the optical depth in the 21 cm line. A spin temperature above the background gives positive contrast; one below it gives absorption. Equal temperatures produce no contrast between gas and background.

The earlier $10^4\,\mathrm K$ ionized-phase temperature determined recombination. Neutral-atom populations require a separate spin-temperature input. We now take $T_{\mathrm s}$ far above the microwave background and assume optically thin gas, uniform density and pure Hubble expansion. These conditions allow a simple conversion from ionized volume fraction to an ideal brightness map.

Turning neutral fraction into brightness

For local coeval gas in this approximation, let $n_{\mathrm{HI}}$ be the proper neutral-hydrogen number density and $H(z)$ the Hubble rate. At spin temperatures well above the hyperfine splitting scale,

The symbols $c$ and $A_{10}$ denote the speed of light and spontaneous-transition rate, respectively; the latter has units $\mathrm{s^{-1}}$. Number density is proper, and the Hubble velocity gradient with respect to proper distance is $H(z)$. The singlet population fraction and stimulated-emission correction have already entered this expression.

In optically thin gas, $1-\mathrm e^{-\tau}\simeq\tau$. For $T_{\mathrm s}\gg T_\gamma$, the contrast factor is approximately $T_{\mathrm s}$, which cancels the inverse temperature in optical depth. With present-equivalent mean hydrogen density $\bar n_{\mathrm H,0}$, the result is

SI inputs give $T_0$ in kelvin; convert units for a millikelvin display. The cell quantity $f_j$ is still ionized-phase volume fraction. Optical thinness permits this linear average over unresolved neutral volume. Hydrogen density includes the prescribed mass fraction $X_{\mathrm H}=0.76$. A hydrogen-only reaction network does not assign all baryonic mass to hydrogen.

For the fixed inputs, $T_0\simeq25.43\,\mathrm{mK}$ at redshift eight. A cell with $f_j=0.3$ has ideal brightness about $17.8\,\mathrm{mK}$; one with $f_j=0.8$ gives about $5.09\,\mathrm{mK}$. More ionized volume leaves less neutral-hydrogen signal. This is the physical content of the familiar high-spin-temperature expression. Mesinger, Furlanetto and Cen, fixed v1, equation 1

The mean counts the mean neutral fraction

Averaging over equal-volume cells gives

Two source arrangements with the same $Q$ therefore have the same mean signal under these uniform assumptions. Bubble positions, sizes and boundaries can still differ.

One saved native frame of the uniform synthetic model, at redshift about 8.64, has $Q\simeq0.458$ and $T_0\simeq26.3\,\mathrm{mK}$, giving mean brightness about $14.3\,\mathrm{mK}$. These coeval values illustrate the mean relation. The fixed-record tutorial checks the original field and finite-mode power independently. Its time, order and grid comparisons cover the declared fixed presets and selected ionized volumes; equal powers can still accompany different shapes. Same-frame values and provenance

To learn how the bubbles are arranged, examine departures from the mean. Broad light and dark patches contribute differently from fine speckles at different spatial scales.

Compare the map with waves

In a periodic cube of side $L$, write a cell-centre position as $\boldsymbol r_j$. An integer triplet $\boldsymbol n$ labels the direction and number of wave cycles. Its wavenumber magnitude is

Let $\Delta T_j=\delta T_{\mathrm b,j}-\overline{\delta T_{\mathrm b}}$ be the brightness departure from its mean. Its overlaps with cosine and sine waves can be combined into a complex coefficient:

The imaginary unit obeys $\mathrm i^2=-1$; the exponential is a cosine minus $\mathrm i$ times a sine. This packages two real projections. Coefficient amplitude describes the strength of the wave, and phase records where its peaks occur. Using actual cell-centre positions fixes the phase origin too.

A common power convention is

With brightness in mK and comoving length in Mpc, power has units $\mathrm{mK^2\,cMpc^3}$. A power calculated first from dimensionless neutral fraction instead has volume units; multiply it by $T_0^2$ to obtain brightness power.

Consider a deliberately constructed continuous stripe pattern, uniform in the other two directions:

First compare this continuous function at three positions for $\phi=0$ and $\phi=\pi$. For a discrete Fourier calculation, sample it at cell centres with at least eight cells per side.

Both maps average to 10 mK and contain waves of identical amplitudes. Changing the phase of the second wave changes how peaks and troughs combine. Each mode's power stays the same because an absolute square discards phase. Those powers cannot distinguish the two spatial maps in the table.

Analyses often group equal-length wavevectors with different directions. The first four groups used here have $|\boldsymbol n|^2$ equal to one, two, three and four, with six, twelve, eight and six signed modes. Averaging each group leaves four scale statistics and discards additional directional information. Images, value distributions and power can therefore answer different questions; semi-numerical 21 cm studies compare them separately. Mesinger et al., §2.4

Problems with explained answers

1. Given $T_0=20\,\mathrm{mK}$, find the brightness of a cell with $f=0.75$ and the mean brightness for $Q=0.4$. If the high-spin assumption is removed, does zero contrast establish full ionization?
2. Translate a single cosine wave by a quarter wavelength. What happens to its phase, power and peak location? Then explain why separately changing the second wave's phase in the example changes the combined shape.
3. A cosine wave along one axis of a $20\,\mathrm{cMpc}$ cube has amplitude $2\,\mathrm{mK}$. Find the magnitudes of its positive and negative fundamental coefficients, their individual powers and the mean power of the first six-mode group. The other four directions have zero amplitude in this example.

The first problem gives $20(1-0.75)=5\,\mathrm{mK}$ and $20(1-0.4)=12\,\mathrm{mK}$. Neutral gas with $T_{\mathrm s}=T_\gamma$ also has zero contrast. Interpreting a dark region as ionized needs the stated spin-temperature condition.

For the second, translation changes phase and moves peaks, with power unchanged. A translation of the entire two-wave pattern changes their phases in a related way. Adjusting only the second phase changes where its peaks meet the first wave's peaks and troughs, so the sum has a different shape. Power retains each wave's strength.

In the third, both coefficient magnitudes are $1\,\mathrm{mK}$. Each power is $20^3\times1^2=8000\,\mathrm{mK^2\,cMpc^3}$. Average those two nonzero modes together with the four zero modes to obtain about $2.67\times10^3\,\mathrm{mK^2\,cMpc^3}$ for the first group. Single-mode power and a group average have different values.

Real radio data also pass through foreground treatment, instrumental response and finite observing windows. We have first obtained an ideal signal from the same neutral-gas distribution, then examined how statistics compress spatial information. The later paper-reading chapter will ask which information an observed power upper limit can constrain.