How one baseline separates smooth foregrounds from hydrogen

You need the previous chapter's result that the hydrogen signal is a few to a few hundred millikelvin, and the idea of a Fourier transform as a sum of oscillating terms. Delays are in nanoseconds, baselines in metres and wavenumbers in $h\,\mathrm{Mpc^{-1}}$.

What one baseline measures

Two antennas separated by a vector $\mathbf b$ multiply their voltages and average. The result, the visibility, is a weighted sum over the sky:

Here $I$ is the sky brightness, $A$ is the antenna's power pattern, or beam, and the exponential is the extra path a wavefront travels to reach the second dish. Parsons et al. 2012, §2.1 For an east-west baseline of length $b$ only the east-west direction cosine $l$ enters, and the phase is $2\pi\nu b l/c$.

Now read the exponent along frequency. A source at fixed $l$ adds a phase that grows linearly with $\nu$, at a rate set by its geometric delay $\tau_g=bl/c$. The visibility is a sum of such linear phase ramps, one per direction.

The delay transform

Multiply the spectrum by a taper $w(\nu)$ that goes smoothly to zero at the band edges, and transform:

A single source with a flat spectrum turns into a peak at $\tau=\tau_g$. Since $|l|\le1$ for anything above the horizon,

This is the horizon delay. Parsons et al. 2012, §2.1 It does not depend on frequency or on where the beam points.

A real source is not flat. Its spectrum $S(\nu)$, times the beam and the taper, multiplies the phase ramp, so in delay the peak is convolved with the transform of that product. A smooth spectrum has a narrow transform and the peak stays near $\tau_g$. Hydrogen fluctuates by order unity from one megahertz to the next, because each channel sees a different slab of the universe, so its transform is broad.

The finite band also sets the delay resolution. A band of width $B$ gives delay samples spaced $1/B$ apart. The Lab uses $128$ channels of $125\,\mathrm{kHz}$, so $B=16\,\mathrm{MHz}$ and the spacing is $62.5\,\mathrm{ns}$.

Worked example: horizon delays for HERA's spacings

HERA dishes are $14\,\mathrm m$ across and packed hexagonally, and its shortest baseline is $14.6\,\mathrm m$. HERA Collaboration 2022, §1 and Table 1

1. Shortest baseline: $\tau_h=14.6/(2.998\times10^8)\,\mathrm s=48.7\,\mathrm{ns}$, less than one delay sample.
2. Three spacings, $43.8\,\mathrm m$: $\tau_h=146.1\,\mathrm{ns}$.
3. Eight spacings, $116.8\,\mathrm m$: $\tau_h=389.6\,\mathrm{ns}$.

Longer baselines push foregrounds to larger delays. That is why the cleanest region is reached first with short baselines.

From delay to cosmological wavenumber

Frequency maps to distance along the line of sight, and angle maps to distance across it. Near band centre write these conversions as $Y$, in $h^{-1}\mathrm{Mpc}$ per megahertz, and $X$, in $h^{-1}\mathrm{Mpc}$ per radian. A delay is a Fourier partner of frequency, and a baseline in wavelengths is a Fourier partner of angle, so

For the Lab's Planck 2018 cosmology at $159\,\mathrm{MHz}$, $z=7.93$, the numbers are $X=6149\,h^{-1}\mathrm{Mpc}$ per radian and $Y=11.25\,h^{-1}\mathrm{Mpc}$ per megahertz. Planck Collaboration 2018, Table 2

Check against HERA: a $200\,\mathrm{ns}$ delay gives $k_\parallel=2\pi\times0.200/11.25=0.112\,h\,\mathrm{Mpc^{-1}}$. HERA quotes $0.11\,h\,\mathrm{Mpc^{-1}}$ for its $200\,\mathrm{ns}$ buffer at $z=7.9$. HERA Collaboration 2022, §4.3

The wedge

Put the horizon delay into both conversions. Both $k_\parallel$ at the horizon and $k_\perp$ are proportional to $b$, so their ratio does not depend on the baseline:

On a plot of $k_\parallel$ against $k_\perp$ the horizon is a straight line through the origin. Smooth-spectrum foregrounds live below it, and the region they fill is called the foreground wedge. Parsons et al. 2012, Fig. 9 At $z=7.93$ the slope is $6149/(11.25\times159)=3.44$; at $z=14.4$ it is $5.14$. The wedge is steeper at higher redshift, so foregrounds take up more of the plane exactly where the signal is oldest.

The region above the line is often called the window. Inside it the hydrogen is measured directly, without modelling the foregrounds at all. The price is that every mode below the line, including most of the large-scale structure along the line of sight, is thrown away.

Why the edge is not sharp

Three things push foreground power above the horizon line.

The taper. A finite band, even a carefully tapered one, turns a perfectly smooth source into the taper's own delay kernel. HERA applies a Blackman-Harris taper across each band. HERA Collaboration 2022, §4.3 The four-term version used in the Lab has a main lobe that reaches four delay samples on each side. In the Lab, at the shortest baseline and $z=7.93$, the foreground falls from $4.6\times10^{7}$ to $1.4\times10^{2}\,\mathrm{mK^2}$ between $k_\parallel=0.035$ and $0.175\,h\,\mathrm{Mpc^{-1}}$, then to about $2\,\mathrm{mK^2}$ at $0.21$, even though the horizon sits at $0.027$. HERA notes the same taper footprint at low $k_\perp$. HERA Collaboration 2022, §4.3

The beam. A dish's beam narrows as frequency rises, roughly as $1/\nu$. A source away from the beam centre therefore sees its gain change across the band, which adds spectral structure the source did not have. Parsons et al. point out that any intrinsic or instrumental frequency structure widens the delay kernel. Parsons et al. 2012, §2.1 The effect grows in the sidelobes, where the response changes fastest with frequency. Furlanetto, Oh and Briggs 2006, §9.3

The instrument. Cable reflections and coupling between neighbouring antennas create delayed copies of the bright sky. HERA's systematics are discussed in the paper reading guide of this series; the Lab leaves them out.

For these reasons analyses keep only delays beyond the horizon plus a buffer. HERA set the weight to zero for $k_\parallel$ below the horizon plus $\Delta k_\parallel$, with a $200\,\mathrm{ns}$ buffer. HERA Collaboration 2022, §4.4 The Lab defaults to $300\,\mathrm{ns}$ because at the shortest baseline its taper lobe still puts more foreground than hydrogen into the delay sample at $312\,\mathrm{ns}$.

Worked example: the first clean mode

Take the Lab default: $b=14.6\,\mathrm m$, buffer $300\,\mathrm{ns}$, $z=7.93$.

1. Edge of the used region: $\tau=48.7+300=348.7\,\mathrm{ns}$.
2. Line-of-sight wavenumber: $k_\parallel=2\pi\times0.3487/11.25=0.195\,h\,\mathrm{Mpc^{-1}}$.
3. Across the sky: $k_\perp=2\pi\times14.6\times1.59\times10^8/(2.998\times10^8\times6149)=0.0079\,h\,\mathrm{Mpc^{-1}}$.
4. The total $k=\sqrt{k_\parallel^2+k_\perp^2}$ is $0.195$ to three figures. On short baselines the measured modes point almost entirely along the line of sight.

This matches the first line-of-sight mode the Lab reports. HERA's smallest reported $k$ in this band is $0.192\,h\,\mathrm{Mpc^{-1}}$, set by its $200\,\mathrm{ns}$ buffer and its own band and grid. HERA Collaboration 2022, §4.4

Lab tasks

Open the 21 cm cosmic dawn Lab and look at the delay spectrum and wedge charts.

1. In the delay spectrum, the solid line marks the horizon and the dashed line the buffer edge. Drag the right dish outward from $14.6$ to about $44\,\mathrm m$ and watch the solid line move from $0.027$ to about $0.082\,h\,\mathrm{Mpc^{-1}}$. Check that the dashed line moves by the same amount.
2. In the wedge chart, find the foreground power for each of the eight baselines. Does the bright region follow the horizon line, and does it stop at the line or a little above it?
3. Choose the chromatic-beam preset. Then, under the advanced settings, freeze the beam. Compare the foreground just beyond the dashed line in the two cases. In the Lab record at $k_\parallel=0.21$ the leaked foreground is about $1\times10^{5}\,\mathrm{mK^2}$ with the changing beam and about $2\,\mathrm{mK^2}$ with the frozen beam.

Exercises

1. Find the horizon delay of a $29.2\,\mathrm m$ baseline, and the corresponding $k_\parallel$ at $z=7.93$.
2. Show from the two conversions that a baseline twice as long moves its horizon point along the same straight line.
3. If the band were cut from $16$ to $8\,\mathrm{MHz}$, what would the delay spacing and the $k_\parallel$ spacing become at $z=7.93$?
4. A foreground source has spectrum $S\propto\nu^{-2.6}$ over $151$ to $167\,\mathrm{MHz}$. Estimate how much its brightness changes across the band, and explain why its delay transform is still narrow.
5. A colleague proposes a $0\,\mathrm{ns}$ buffer to gain more modes. Using the taper example above, what would go wrong at the shortest baseline?