Start here
A radio source sends noise: a hiss, like the static between stations. Two antennas some
way apart hear the same hiss, one of them a moment later. Multiply what they hear and
average, and that small delay survives as a phase. This one number is what an
interferometer measures, and enough of them make a picture of the sky.
The page tells it in four steps, the numbered buttons at the top. 1 · One source :
the noise itself, and what multiplying and averaging do to it. 2 · Two sources, two
antennas : drag antenna B along the ground and watch the dot in the complex plane wind
round. 3 · The uv plane : the correlation a pair would measure at every separation, in
colour; drag the ● and the live dot settles on the colour under it. 4 ·
Many baselines : a whole array, the turning Earth, the image that comes out, and three ways of
making it.
Look for this: one antenna alone sees no pattern at all, but the average of the product
of two antennas makes fringes as their separation grows; the fringes of different sources
add; and the map of every separation is the Fourier transform of the sky.
1 · One source
1. Band-limited noise: lumps that last about 1/Δν
One source of noise, through a band 5% as wide as the carrier ν₀ (Δν = 45 MHz at
915 MHz). The top plot is the voltage it sends: a sine at $\nu_0$ whose size and phase
wander, and the grey curve is its envelope. The lumps last about $1/\Delta\nu$ = 20 cycles
(22 ns): that is the coherence time . The spectrum below is flat across the band
and empty outside it. Drag the band's edge on the spectrum, or the slider, to 1%: the lumps
stretch to 100 cycles. The receiver's filter is what makes the hiss "nearly one frequency"
(quasi-monochromatic) for that long.
2. Multiply and average: the autocorrelation, one lag at a time
The dashed curve is the same signal a lag $\tau$ later. The middle plot is their
product $x(t)\,x(t+\tau)$, blue above zero and red below, and its average is
$R(\tau)$, the autocorrelation , the dot on the autocorrelation plot. At $\tau=0$ the
product is $x^2$, never negative, and its average is the power. Here $\tau$ is half a
cycle: the copy is upside down, the product is mostly negative, and $R=-0.99$ (the
running average wanders a few hundredths either side of it: it is an average of noise,
and its scatter is the radiometer's, about $1/\sqrt{N}$ for $N$ independent stretches
of $1/\Delta\nu$; raise the averaging to steady it). Drag
$\tau$ along: $R$ oscillates at $\nu_0$, a cycle per cycle of lag, under an envelope that
first reaches zero at $\tau=1/\Delta\nu$, 20 cycles, where the lumps no longer line up.
That envelope is the Fourier transform of the band (the Wiener–Khinchin theorem): a flat
band of width $\Delta\nu$ gives $\mathrm{sinc}(\Delta\nu\,\tau)$.
3. White noise: no carrier, no lumps, a spike
The band as wide as it goes, 0 to $2\nu_0$: white noise. Its spectrum is flat everywhere
and its autocorrelation is a spike half a cycle wide: each moment is independent of the
next. Narrow the band to 20%: the carrier reappears, in lumps 5 cycles long. A radio
source's noise is white over a wide range; the receiver's band-pass filter is what lets two
antennas see it as the same wave for a while.
4. Negative frequencies and the complex envelope
The spectrum of a real signal has two halves: the band at $+\nu_0$ and its mirror image
at $-\nu_0$, because $\cos 2\pi\nu_0t=\tfrac12(e^{2\pi i\nu_0t}+e^{-2\pi i\nu_0t})$. The
negative half carries nothing new: it is the complex conjugate of the positive half.
Complex envelope is on: the receiver mixes the signal down, keeping one side only
and shifting it to zero. What is left is the envelope $s(t)$, two slow curves called I and
Q (solid and dashed), with $x(t)=\sqrt2\,\mathrm{Re}[s(t)\,e^{2\pi i\nu_0t}]$. Its spectrum
is one band at 0, and its autocorrelation is the envelope without the oscillation. The
correlator works with $s$: a delay $\tau$ shifts $s$ by $\tau$ and turns it by the phase
$e^{-2\pi i\nu_0\tau}$.
5. Two independent noises: the average of their product is zero
Now the dashed curve is a second source, with its own noise. Their product swings
positive and negative with no preference, and however you drag $\tau$, its average dies
away (the autocorrelation plot's curve flattens towards zero as the averaging goes on). This
is the key to everything after: two sources that have nothing to do with each other add
nothing to a correlation but noise that averages away.
2 · Two sources, two antennas
6. One source, two antennas: the fringe is in the separation
A source 10° from straight up, antennas A and B 20 cm apart, at 915 MHz ($\lambda$ =
32.8 cm). The waves of noise come down at 10° and reach B first: B is nearer the source by
$D\sin\theta$ (the thick bar), 0.106 wavelengths, so B hears everything 0.106 cycles before
A. What they hear (Antenna A hears and Antenna B hears , under the hint) is the same
noise, shifted. The correlator multiplies
A by the conjugate of B and averages: the dot in the complex plane settles at the source's
brightness, 1, at a phase of $-2\pi D\sin\theta/\lambda$ = −38°. Drag B to the right: the
magnitude stays and the phase winds round, once for every $\lambda/\sin\theta$ = 1.89 m.
The real part, $\langle x_Ax_B\rangle$, is the handout's $I\cos(2\pi\theta D/\lambda)$:
the heavy curve under the live dot on the fringe plot. That is the interference fringe,
not on a screen but in the separation of the antennas. The orange line is the power B
receives: the same wherever B is. One antenna alone sees no fringe; only the product of
two does. Drag the source: the fringes get finer as $\theta$ grows.
7. Two sources: the correlations add, the cross terms average away
A second source, half as bright, at −25°. Each antenna hears the sum of two independent
noises, and the product of A and B has four parts: source 1 at A times source 1 at B, the
same for source 2, and two cross terms, source 1 at one antenna times source 2 at the
other. The complex plane draws them: each source's own term as an arrow in its colour,
laid end to end, and the cross terms as the grey diamond, which wanders and dies, as in
experiment 5. What is left is the sum of the two fringes, and the heavy curve beats as
you drag B (the slides' "superposition of both sources"). Because the sources are
incoherent, the correlation of the sky is a sum of one fringe per source,
$V=\sum_k I_k\,e^{-2\pi i\,u\,l_k}$.
8. Coherent sources: fringes at one antenna, and no longer a sum
The same two sources, but one transmitter feeds both: they send the same noise. Now the
power at one antenna has fringes (the orange curve, from 0.4 to 3.2, a fringe every
$\lambda/(\sin\theta_1-\sin\theta_2)$ = 55 cm): this is Young's double slit, the two
waves adding in step at some places and out of step at others, a pattern fixed on the
ground. And the correlation is no longer the sum of two fringes: the cross terms stay.
Drag A as well: with incoherent sources only the separation matters; here the correlation
changes when the pair moves together. Turn Coherent sources off: the orange line
goes flat again.
9. A wide band and a long baseline: why the correlator delays one antenna
The source 30° from straight up, a band of 40% (lumps 2.5 cycles long), the antennas 3 m
apart, and delay tracking off. B hears the noise $D\sin\theta/\lambda$ = 4.6 cycles before
A: the lumps no longer overlap (look at the two streams) and the correlation is almost
gone (0.09), not because anything is incoherent but because A and B are multiplying
different stretches of the same noise. The cross-correlation against lag shows where it
went: its peak is at the delay. The spectrum shows it too: the phase of A·B* winds across
the band, so its parts cancel. Press Delay tracking : the correlator delays B by
the geometric delay towards the source (the reference direction), the lumps line up, and
the dot comes back to 1. Real arrays have delays of microseconds, millions of cycles: the
electronic delay is what sets the direction the array looks at.
3 · The uv plane
10. One baseline measures one set of stripes on the sky
One source, off to one side. The pair's stripes are drawn over the sky: the pattern
$e^{-2\pi i(ul+vm)}$, crests a fringe spacing $1/|b|$ apart and across the baseline,
coloured by phase. The source sits on one colour, and the live dot settles on that
colour: what the pair measures is the colour of the stripe the source sits on. Drag the ●
on the uv plane: the stripes turn with the baseline and close up as it grows. For one
source $|V|$ is the same everywhere and only the phase winds: the uv plane is a rainbow
of straight stripes, $1/l$ apart. Choose real part in the menu: the handout's
$\cos 2\pi(ul+vm)$, red where it is positive and cyan where it is negative.
11. The lab's pentagon: five sets of stripes add up to a Fourier transform
Five sources on a pentagon, as in the lab notebook. Each makes its own stripes in the
uv plane, and they add: the uv plane is
$V(u,v)=\sum_kI_k\,e^{-2\pi i(ul_k+vm_k)}$, the two-dimensional Fourier transform of the
sky. The thin arrows in the complex plane are the five sources' terms, each in the colour
of its stripe, laid end to end; the live dot settles where they end, on the colour under
the ●. Its axes: the separation in metres along the bottom and the left, and $u=\Delta
x/\lambda$, $v=\Delta y/\lambda$ in wavelengths along the top. The ○ is the mirror point:
swapping A and B conjugates the correlation, so one pair gives two points. Drag a source:
its stripes turn and change spacing, and the whole pattern changes.
12. A disk: the Airy pattern, and where its first dark ring is
A uniform disk in place of the points. Its transform is the Airy pattern: a bright
centre and rings, the first dark ring where the baseline is $1.22/D$ wavelengths for a disk
$D$ across. Here $D$ = 9.2°, so that is at 7.6 wavelengths, 2.5 m. Drag the ● out along a
radius: the dot's size falls to zero there, then comes back with its phase turned half a
turn (the colour jumps to the opposite hue). Michelson measured the diameter of a star
this way in 1920: by finding where the fringes vanish. For a picture the live simulation
uses 200 points drawn at random from its brightness, so the dot settles on the × (what
those points give), a little off the ring (the picture's exact transform). Try the ring,
the galaxy and Fourier's portrait, or your own picture.
13. The Fourier Etch-a-Sketch
Keep what you measure is on: every measurement is painted into the sampled uv
plane where the pair is, and at its mirror point, and the image from what you kept is the
inverse transform of everything painted so far. Drag one line through the middle of the uv plane:
the image is the sky smeared out across your line. Draw lines at three angles, then fill
in a disk: the three sources come apart. Press Autopilot to let a spiral fill the
plane. This is the optics lab's Fourier Etch-a-Sketch, done with the phases kept: there,
two collectors in pseudo-thermal light correlate intensities and only $|V|^2$ comes out
(experiments 16 and 17).
14. The lab bench's near field: where the pair sits matters
The sources 5 m above the antennas and 3 m from the axis, as on the lab bench and in the
notebook: the exact distances now, not plane waves. The plane-wave picture needs sources
much farther away than $D^2/\lambda$, 12 m for this 2 m baseline, and these are not: their
wavefronts are still curved across the pair. So the correlation is no longer the far-field
transform under the ● (4.13): the × in the complex plane, what the exact distances give,
is at 2.72. And Slide the pair , the correlation as the pair moves sideways with
its separation kept, is far from flat: its real part swings between −2.3 and 2.7 as the
pair moves a metre either way. This is the notebook's plot of the correlation against ΔX
and the mean X. Drag on it to move the pair. Radio sources in the sky are so far away
that only the separation matters.
15. Why this needs incoherent sources
The pentagon, but every source sends the same noise, with a quarter turn of phase
between neighbours, as if one transmitter fed five antennas. The cross terms between
sources no longer average away, and the dot leaves the colour under the ●: what the pair
measures is not the Fourier transform of the brightness any more. Worse, look at
Slide the pair : the correlation now changes as the pair moves with its
separation fixed, so there is no uv plane to map at all. Turn Coherent sources
off (under More): the curve goes flat and the dot returns to its colour. The van
Cittert–Zernike theorem, that the correlation of the fields far from an incoherent source is
the Fourier transform of its brightness, is exactly the statement that the cross terms
vanish. ("Spatially incoherent sources make coherent fields far away.")
16. Correlate intensities instead: the phase problem
HBT is on: the correlator multiplies the intensities $|v_A|^2$ and $|v_B|^2$ (their
swings about the mean) instead of the voltages, as Hanbury Brown and Twiss did in 1956 and
as the optics lab's Etch-a-Sketch does with photodiodes. For noise like this
$\langle\Delta I_A\,\Delta I_B\rangle=|V|^2$: the dot stays on the real axis, the uv picture
shows $|V|^2$ in grey, and the phase is gone. A triangle and the same triangle turned by
180° give the same map (press Triangle, turned ): $|V|^2$ cannot tell a sky from its
mirror image through the centre. That is the phase problem of intensity interferometry and
of crystallography; radio arrays avoid it by correlating voltages. The optics lab meets it
in Part 5
of the Fourier Etch-a-Sketch , with this triangle and a letter F; experiment 17 paints the
map and looks at the image it makes.
17. The Etch-a-Sketch with intensities: the sky’s autocorrelation
HBT and Keep what you measure are both on, and the autopilot is painting a
spiral, as the optics lab's Fourier
Etch-a-Sketch does. What gets painted is $|V|^2$, without its phase, and its inverse
transform is not the sky: it is the sky's autocorrelation , the sky slid over itself
and multiplied, $A(\vec r)=\sum_{j,k}I_jI_k\,\delta(\vec r-(\vec r_j-\vec r_k))$. Every pair
of sources makes a point at their separation, and another at minus it; every source adds to
a point at the centre.
a binary
its autocorrelation
a triangle
its autocorrelation
The binary's two sources (1 and 0.6, 8.6° apart) give three points in a line: the centre,
$1^2+0.6^2=1.36$, and a point 8.6° to either side, $1\times0.6=0.6$, so 0.44 of the centre's
height (0.41 in the picture: the brush that paints the uv plane is a few cells wide, and that
dims the image away from its middle). Press Triangle , Clear and
Autopilot : three sources make six points on a hexagon, one for each ordered pair,
round a centre three times as bright. Then Triangle, turned : the same hexagon. A
sky and the same sky turned by 180° have the same autocorrelation, which is experiment 16's
phase problem, seen in the image. The autocorrelation still tells sizes and separations,
and from it a clever guess can sometimes find the sky (phase retrieval, the lab's last
question).
4 · Many baselines
18. Few antennas, many baselines
The slides' eight antennas on a line, with gaps of 8, 10, 1, 3, 2, 7 and 8 units: their
28 pairs measure every spacing from 1 to 24 units. The second row appears: under the ideal
uv plane, the sampled uv plane ($V$ where a pair measures it, black elsewhere);
under the sky, the image the samples give, their inverse transform (the dirty
image ), with the arrow for the flow. The small panel is the dirty beam , the
image one point source would give. Pairs on a line measure only one direction of the uv
plane, east–west, so the image is smeared out north–south, across the line.
19. A Y of 27 dishes: 351 pairs at once
A Y like the Very Large Array: 27 dishes make $27\cdot26/2$ = 351 pairs, and so 702 points
of the uv plane at once, a six-pointed star of them. The dirty beam has six spokes, and
the dirty image is the sky blurred by it. Drag a dish on the ground: its 26 baselines move
together.
Why the image goes blue (negative) in places: no pair measures $u=v=0$, the total flux of
the sky (that would need two dishes in the same place); in a real array the spacings shorter
than a dish is wide are missing too. So the dirty image has no average: it is the sky with its broadest,
smoothest part taken away, and wherever the sky is fainter than its surroundings the image
dips below zero. A bright point on a dark sky barely shows this; a picture with a bright
background, like Fourier's portrait, shows it everywhere. Real arrays add a single dish's
own map to fill the hole in the middle.
20. The Earth turns, and the tracks fill the plane
The same Y, 8 hours of a source at declination 40°. As the Earth turns, each pair's point
in the uv plane sweeps an ellipse of axis ratio $\sin\delta$ (the white tracks); the sampled
plane fills in, the beam's spokes fade and the image sharpens. Press Turn the
Earth to watch it fill. This is Earth-rotation synthesis (Martin Ryle, Nobel prize
1974). Set the declination to 85°: the tracks are nearly circles; to 5°: nearly straight
lines, and the beam is a streak that can hardly tell north from south.
21. Long baselines see fine detail, short ones the broad shapes
Two point sources 5″ apart at 5 GHz ($\lambda$ = 6 cm), and a small Y whose longest
baseline is 1.8 km: the finest fringe, $\lambda/B$ = 6.8″, is coarser than their
separation, and the image shows one blob with a shoulder. Spread the array out: at a
spread of 0.1 (2.8 km, $\lambda/B$ = 4.4″) a dip opens between them, and at 0.2 (5.6 km,
2.2″) they stand apart. The longest baseline sets the resolution, as the
diameter of a mirror does: an interferometer is a huge telescope with most of its mirror
missing. The shortest baseline sets the other limit: anything larger than
$\lambda/B_\mathrm{min}$ lives at separations no pair measures.
22. An Earth-sized telescope sees a black hole's ring
The nine sites of the Event Horizon Telescope (2018), at 1.3 mm, looking at a ring 42 µas
across like the one around the black hole in M87, for 12 hours. Baselines nearly as long
as the Earth is wide (12,700 km) give $\lambda/B$ = 21 µas, enough to see the ring; the
slides' round numbers are 20 µas for 1 mm and 10,000 km. With so few stations the
sampled plane is sparse and the dirty beam has strong sidelobes, yet the ring shows. A
track stops where the source sets for one of its stations.
4 · Many baselines: making the image
23. The dirty image: every spacing no pair measured counts as zero
The galaxy, watched by the Y for 8 hours. Under Making the image ,
Dirty image is chosen: the inverse transform of the sampled uv plane, with a zero
wherever no pair measured. The strip at the bottom shows the same thing the other way
round: the sky convolved with the dirty beam (⊛), each point of the sky replaced by a copy
of the beam, sidelobes and all. Look for what is missing: the galaxy's faint outer disk,
and round it a blue bowl. The broadest, smoothest parts of a sky live at the shortest
spacings, and the shortest of all, the zero spacing (the total flux), no pair measures (the
black middle of the strip's uv plane): add up the dirty image and you get 0.000. The
number under the image is how far it is from the true sky seen through the clean beam
(root-mean-square, as a fraction of its peak): 4.4%, the number to beat.
24. CLEAN takes the dirty beam out, point by point
Högbom's CLEAN (1974) supposes the sky is a few points. The strip's first row is one
step: it finds the brightest pixel of the residual (at first the dirty image), the ✚, takes
a tenth of it (the loop gain) out of the residual as a copy of the dirty beam centred there,
and puts a point of that much into its list: 0.0997 at the first step, a tenth of a source
of brightness 1. Two sources, two hours of a small Y: press Step a few times and
watch the ✚ hop between the sources and the list grow, then Watch . After 84 steps
the residual's peak is below 1% of the dirty image's and CLEAN stops, with 2 points. The
second row restores the image: the points smoothed by the clean beam , a Gaussian
fitted to the dirty beam's main lobe, plus what is left; it is 0.03% off the true sky,
against 2.5% for the dirty image. The points' uv plane (top right) is CLEAN's guess at every
spacing, measured or not: what two points would give. Raise the loop gain to 1 and press
Watch : 2 steps, a whole source each. That works for points standing alone; on
anything else a gain of a tenth or so is the custom, so that a step taken in the wrong place
costs little.
25. Maximum entropy: the smoothest sky that fits
The galaxy and the Y again, with Maximum entropy (Gull and Daniell, 1978). It
looks for the positive sky that agrees with the measurements and is otherwise as smooth and
flat as it can be: it starts from a flat guess $M$ holding the total flux (which one dish
alone can measure) and minimises the misfit to the data less $\alpha$ times the entropy.
The answer obeys one rule at every pixel, $I=M\,e^{R/\alpha}$, $R$ the residual there: a
pixel rises above the flat guess only where the measurements still want more, and it can
never go negative. Watch the strip's first row: the image starts flat, the residual is the
dirty image, and each iteration moves every pixel towards $M\,e^{R/\alpha}$; in 40
iterations the two agree, and the residual has faded. The outer disk the dirty image lost
is back, and the image is 0.61% off the true sky. Its uv plane shows how: in the gaps, and
in the hole in the middle, it has made smooth guesses. Raise $\alpha$ (the entropy's
weight) to 0.1: done in 10 iterations, but smoother and 2.1% off; lower it to 0.003: 77
iterations, 0.57% off, closer to the data.
26. Points or a smooth sky: CLEAN against maximum entropy
The binary and the Y for 8 hours, with CLEAN allowed 1000 steps. It finds the two points
in 83 steps and its image is 0.01% off the true sky: points are what it looks for. Choose
Maximum entropy : 32 iterations, 0.11% off, the two points sitting in faint rings of
the dirty beam, what the entropy held back. Now press Galaxy : maximum entropy takes
40 iterations and is 0.61% off. Choose CLEAN : its 1000 steps have put down 402
points, a speckle along the arms (the strip's second row), and its image is 2.9% off, the
outer disk still missing. CLEAN builds a smooth sky out of thousands of points, slowly;
maximum entropy builds points out of a smooth sky, a little softened. Real arrays use both,
and newer guesses too (multiscale CLEAN, compressed sensing): each is a different guess at
what no pair measured.
How it works
Every source $k$ emits its own complex Gaussian noise $s_k(t)$, independent of every
other source (incoherent), filtered to a flat band of width $\Delta\nu$ round the carrier
$\nu_0$. The page works with the complex envelope: the real voltage is
$x(t)=\sqrt2\,\mathrm{Re}[s(t)\,e^{2\pi i\nu_0t}]$, so that $\langle x^2\rangle=\langle|s|^2\rangle$
is the source's brightness $I$ (the handout's $I=V^2/2$ for a sine of amplitude $V$), and a
delay $\tau$ shifts the envelope by $\tau$ and turns it by $e^{-2\pi i\nu_0\tau}$. A
receiver at $\vec r_j$ hears a source in the direction $\hat s_k$ (direction cosines $l_k$,
$m_k$) at the time $\tau_{jk}=-\hat s_k\cdot\vec r_j/c$, earlier if it is nearer the source:
$$v_j(t)=\sum_k s_k(t-\tau_{jk})+n_j(t),$$
with its own noise $n_j$. On the lab bench's near field the exact distances $R_{jk}$ and
amplitudes $z/R_{jk}$ are used instead, as in the lab notebook. The correlator forms, for
each chunk of 1024 samples (512 cycles of the carrier), $c=\langle v_A\,v_B^*\rangle$, and a
running average $\bar c\leftarrow\bar c+\alpha(c-\bar c)$, as the GNU Radio flowgraph's
single-pole filter does. Its real part is $\langle x_Ax_B\rangle$, the average of the product
of the two real voltages; its imaginary part is the same with B a quarter cycle later.
With the baseline $(u,v)=(\vec r_B-\vec r_A)/\lambda$ and incoherent sources, the cross
terms $\langle s_k s_{k'}^*\rangle$ ($k\ne k'$) average to zero, and
$$\bar c\to V(u,v)=\sum_k I_k\,e^{-2\pi i(ul_k+vm_k)}\to\int I(l,m)\,e^{-2\pi i(ul+vm)}\,dl\,dm,$$
the van Cittert–Zernike theorem. (The notebook conjugates A instead and gets
$e^{+2\pi i(\dots)}$; nothing physical changes.) Each source's term is one set of stripes
in the uv plane, $1/l_k$ apart; the sum of them is the sky's Fourier transform. The same
sum read the other way: one baseline $(u,v)$ weighs the sky with one set of stripes on it,
$e^{-2\pi i(ul+vm)}$, crests $1/|b|$ apart.
With a band of width $\Delta\nu$ the envelopes at A and B overlap only if their relative
delay is within the coherence time $1/\Delta\nu$, so each source's term carries the band's
coherence $\gamma(\Delta\tau)\approx\mathrm{sinc}(\Delta\nu\,\Delta\tau)$, the Fourier
transform of the band (Wiener–Khinchin). The correlator's electronic delay removes the
delay towards the reference direction; what is left smears the edges of a wide field
(bandwidth smearing). In the complex plane the ring is the ideal $V$ and the × what this
band and these delays give. The scatter of one chunk's correlation is the radiometer
value $\sigma=\sqrt{P_AP_B/2N}$ with $N=\Delta\nu\,T$ independent samples in the chunk. The
simulation makes everything in the frequency domain: a white Gaussian process has white
Gaussian Fourier coefficients, so each source is a set of Gaussian bins inside its band, a
delay is a phase ramp across them, and the spectra and correlations come from the same
bins. The time streams on the screen are a slowed-down recording of the same kind of
noise, composed with the antennas' delays as they are now; the correlator averages 512
cycles each frame. For a picture the simulation uses 200 points drawn at random from its
brightness; the ideal $V$ is the picture's exact transform.
Coherent sources share one noise: then $\langle s_ks_{k'}^*\rangle\ne0$, the power at
one antenna has Young's fringes, and the correlation depends on both positions, not only on
their difference. HBT : for circular Gaussian noise
$\langle|v_A|^2|v_B|^2\rangle-\langle|v_A|^2\rangle\langle|v_B|^2\rangle=|\langle v_Av_B^*\rangle|^2$.
Many baselines. An array of $n$ dishes gives $n(n-1)/2$ pairs. As the Earth turns, a
pair's baseline seen from a source at hour angle $H$ and declination $\delta$ sweeps
$u=\sin H\,X+\cos H\,Y$, $v=-\sin\delta\cos H\,X+\sin\delta\sin H\,Y+\cos\delta\,Z$: an ellipse of
axis ratio $\sin\delta$ (Thompson, Moran and Swenson). The samples go onto the grid of the
picture's transform (natural weighting counts each sample, uniform each cell once); the
dirty beam is the transform of the sampling alone, the dirty image the transform of the
sampled $V$, and the dirty image is the sky convolved with the beam.
Making the image. With weights $S$ on the sampled cells (zero elsewhere), the dirty
image is $I_D=\mathcal F^{-1}[S\,V]$ and the dirty beam $B=\mathcal F^{-1}[S]$, so
$I_D=B*I$: every method has to guess $V$ where $S=0$. CLEAN (Högbom, 1974) guesses points:
it finds the residual's peak $R_p$ (inside the field), adds $gR_p$ to its model there,
subtracts $gR_p$ times the beam centred there, and repeats until the peak is under 1% of the
dirty image's; the image is then the model convolved with the clean beam (a Gaussian fitted
to the beam's main lobe above 0.35) plus the residual. Maximum entropy (Gull and Daniell
1978, Cornwell and Evans 1985) minimises
$$J(I)=\tfrac12\langle I,B*I\rangle-\langle I,I_D\rangle+\alpha\sum_i\Big[I_i\ln\frac{I_i}{M_i}-I_i+M_i\Big]$$
over positive images in the field: the first two terms are, but for a constant,
$\tfrac12\sum_kS_k|\tilde I_k-V_k|^2$, the misfit to the measured visibilities, and the last
is minus the entropy relative to the flat default $M$ holding the total flux. The gradient
is $\alpha\ln(I/M)-R$ with $R=I_D-B*I$ the residual, so the answer is
$I=M\,e^{R/\alpha}$ at every pixel, and it is unique, since $B*$ is positive semi-definite
(its eigenvalues are the weights $S\ge0$) and the entropy strictly convex. The page finds
it by conjugate gradients, preconditioned by the Hessian's diagonal $1+\alpha/I$, with an
exact line search (along $I+t\,d$, $J$ needs one convolution, $B*d$). With every spacing
measured ($B=\delta$) each pixel solves $\alpha\ln(I/M)=I_D-I$ alone: $I=\alpha\,
W\!\big(\tfrac M\alpha e^{I_D/\alpha}\big)$, with Lambert's $W$, which the unit tests check.
The error under the image is the root-mean-square difference, over the field, from the true
sky convolved with the clean beam, as a fraction of that image's peak; CLEAN's and maximum
entropy's images are restored the same way, their model through the clean beam plus the
residual.
Intensities and the autocorrelation. $|V|^2=VV^*$ is the transform of the sky
correlated with itself (the convolution theorem):
$\mathcal F^{-1}[|V|^2](\vec r)=\int I(\vec r\,')\,I(\vec r\,'+\vec r)\,d^2r'$, for points
$\sum_{j,k}I_jI_k\,\delta(\vec r-(\vec r_j-\vec r_k))$. It is symmetric through the centre
whatever the sky, which is why the sky and the sky turned by 180° cannot be told apart. The
Etch-a-Sketch paints each chunk's value into a 3 × 3 brush of uv cells; the blur this makes
in the uv plane multiplies the image by the brush's transform, which falls a little away from
the middle.
Conventions: $l$ east (to the right), $m$ north (up), and $u$, $v$ the same way; on the lab
bench sky angles are $\sin\theta$, in degrees, elsewhere small angles. The units presets
change the labels (and, for the near field, the geometry). Astronomers usually draw east on
the left, as the sky looks from below.
New in 2026 for BrainFlux, after Jason Gallicchio's radio interferometry lab and his Fourier Etch-a-Sketch. Source:
github.com/gallicchio/brainflux , MIT licence.