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Sandbox Physics
Interferometric imaging from wave delay to EHT data

AC05 · L1

From wave delay to complex visibility

Sandbox Physics ·

A telescope pair measures one Fourier component of the brightness. The real part compares the sky with a cosine pattern; the imaginary part compares it with a shifted sine pattern. Deriving that statement from the wave delay tells us what a baseline measures and how to keep the image’s direction and flux correct.

Choose the sky and baseline directions

Place the tracked direction at the origin. Let ll increase eastward and mm northward; both are small angles in radians. Let the directed physical baseline be b=r2−r1\mathbf b=\mathbf r_2-\mathbf r_1. Project it onto those two sky directions and divide by wavelength to obtain uu and vv. A value of one billion means a separation of one billion wavelengths, not one billion metres.

The angular frequency channel has wavelength λ\lambda. In the browser it is 1.3 mm1.3\,\mathrm{mm}. A 3900 km3900\,\mathrm{km} projected baseline is then 3.00×1093.00\times10^9 wavelengths. Use the projected separation: a long ground baseline can appear short when viewed nearly along its length.

A radio image often places east on the left and north at the top. The equations still use east-positive coordinates. Reversing the display columns produces the astronomical view; changing the Fourier sign would also reverse the phase information and is a different operation.

Reading: NRAO Essential Radio Astronomy, Chapter 3, aperture synthesis

A small angular displacement produces a phase

After the correlator removes the propagation delay toward the field center, the remaining path difference from a nearby direction is bll+bmmb_l l+b_m m. Dividing this length by the wavelength gives the remaining number of cycles. Multiplying by 2π2\pi gives a phase in radians.

Choose the correlation convention in which the tracked first signal multiplies the complex conjugate of the second. A point with spectral flux density FF then contributes the phasor below. This convention fixes the negative forward exponent. Reversing the baseline reverses the phase and conjugates the visibility.

V(u,v)=Fe−2πi(ul+vm)=Fcos⁡ϕ−iFsin⁡ϕ,ϕ=2π(ul+vm)V(u,v)=F e^{-2\pi i(ul+vm)}=F\cos\phi-iF\sin\phi,\qquad \phi=2\pi(ul+vm)
The phasor length is the point’s flux density. Its direction records position relative to the tracked origin. A central point has zero phase on every baseline.

Reading: NRAO Essential Radio Astronomy, Chapter 3, aperture synthesis

Why brightness adds after averaging

Different distant emitters fluctuate independently. When their electric fields are correlated and averaged, cross terms between unrelated emitters average away. Each small patch contributes its own brightness times a phase factor. Adding those patches gives an integral over the sky.

Write I(l,m)I(l,m) for spectral brightness in jansky per steradian. In a small field the solid-angle element is approximately dl dmdl\,dm. The visibility has units of jansky because the integration adds brightness over angle. One jansky is 10−26 W m−2 Hz−110^{-26}\,\mathrm{W\,m^{-2}\,Hz^{-1}}.

V(u,v)=∬I(l,m)e−2πi(ul+vm) dl dm,V(0,0)=FtotalV(u,v)=\iint I(l,m)e^{-2\pi i(ul+vm)}\,dl\,dm,\qquad V(0,0)=F_{\mathrm{total}}
The zero baseline adds every patch without a phase shift, so it measures total flux. The relation assumes a constant primary-beam response across the field and neglects the noncoplanar wide-field term.

Reading: NRAO Essential Radio Astronomy, Chapter 3, aperture synthesis · EHT Collaboration, First M87 Results IV, arXiv:1906.11241v1

A point tells us the sign; two points reveal a size

Take a 0.60 Jy0.60\,\mathrm{Jy} point 10 μas10\,\mu\mathrm{as} east of the origin and a purely east–west 3.00 Gλ3.00\,\mathrm{G}\lambda baseline. One microarcsecond is 4.8481368×10−12 rad4.8481368\times10^{-12}\,\mathrm{rad}. The phase is about −0.914 rad-0.914\,\mathrm{rad}, or −52.4∘-52.4^{\circ}. Its amplitude remains 0.60 Jy0.60\,\mathrm{Jy}. A position shift changes phase even when no amplitude changes.

Now place equal points at east coordinates +θ+\theta and −θ-\theta, with total flux FF. Adding their conjugate phasors cancels the imaginary parts. The visibility oscillates as a cosine, with its first zero at u=1/(4θ)u=1/(4\theta). For θ=15 μas\theta=15\,\mu\mathrm{as}, that is 3.44 Gλ3.44\,\mathrm{G}\lambda, about 4470 km4470\,\mathrm{km} at the browser wavelength.

V(u,0)=F2e−2πiuθ+F2e2πiuθ=Fcos⁡(2πuθ)V(u,0)=\frac F2e^{-2\pi iu\theta}+\frac F2e^{2\pi iu\theta}=F\cos(2\pi u\theta)
A zero occurs when the two point contributions cancel. Beyond it the real visibility changes sign; the amplitude is its absolute value. The source’s total flux stays fixed.

Write a finite image without losing an area factor

For a pixel image, define xpx_p as flux already integrated over pixel pp, in jansky per pixel. The forward model is a sum of those fluxes times their phasors. No extra pixel area belongs in this sum. If the stored values were brightness per steradian, each term would need a solid-angle factor instead.

The saved browser default uses 24×2424\times24 pixels over a 120 μas120\,\mu\mathrm{as} square. Pixel centers are separated by 5 μas5\,\mu\mathrm{as}. Each synthetic source is evaluated on this grid and scaled to a total of 0.600 Jy0.600\,\mathrm{Jy}. Thus the synthetic disk is a finite-pixel disk; its exact visibility need not equal the continuum circular-disk formula.

Vk=∑pxpexp⁡[−2πi(uklp+vkmp)],∑pxp=FtotalV_k=\sum_p x_p\exp[-2\pi i(u_kl_p+v_km_p)],\qquad \sum_p x_p=F_{\mathrm{total}}
The grid stores flux, not surface brightness. This same discrete operator generates the synthetic record and predicts the signals of every candidate reconstruction.

Three quick checks before making an image

First, put all flux in a central point: every predicted visibility should equal the total flux with zero imaginary part. Next, move that point east: a positive eastward baseline should acquire the negative phase calculated above. Finally, reverse every baseline: the predictions should be complex conjugates.

The conjugacy follows because sky brightness is real. Pairing (u,v)(u,v) with (−u,−v)(-u,-v) can make an image real, but it supplies no independent observation. The next chapter uses the actually distinct projected baselines to explain what Earth’s rotation adds.

Reading: NRAO Essential Radio Astronomy, Chapter 3, aperture synthesis

Saved synthetic default and method

Four spherical-Earth stations, nine hour angles over a three-hour track, 1.3 mm wavelength. An exact discrete Fourier sum generates a noiseless finite-pixel source. A separate ideal point calibrator and total-power measurement enter both ninety-step constrained inversions.

Image grid and field24×24,120 μas24\times24,\quad120\,\mu\mathrm{as}
Central Greenwich hour angleH=105∘H=105^{\circ}
Total fluxF=0.600 JyF=0.600\,\mathrm{Jy}
Recorded complex samples54
First directed closure phase0.287631circ0.287631^{circ}
Smooth-prior signal RMS7.0408,mathrmmJy7.0408,mathrm{mJy}
Compact-prior signal RMS5.9340,mathrmmJy5.9340,mathrm{mJy}

The full station coordinates, sampled signals and reconstructions are saved in default-synthetic-record.json; the six-preset summary is model-benchmarks.json. Image pixels store jansky per pixel, dirty images use jansky per unit-peak beam.

Predict how a finite source resolves

  1. Choose Resolve a stellar disk and note which spatial frequencies the array samples.
  2. Drag a telescope outward from the others. Compare the changed dirty beam with the reconstructed disk.
  3. Explain why the smallest beam feature and the amount of collected flux answer different questions.

Exercises with explanations

  1. A source shifts north by ten microarcseconds. Which pure east–west visibility changes?

    Read the explanation

    None in this narrow-field model. A pure east–west baseline has zero northward spatial frequency, so its phase contains no north coordinate. A baseline with a northward component is needed to detect this displacement.

  2. A program multiplies flux-per-pixel values by the pixel solid angle again. What happens when the grid is refined?

    Read the explanation

    For the same physical source, each pixel already contains its share of flux. Multiplying by area a second time makes the total visibility shrink as the pixels become smaller. Checking the zero-baseline sum exposes the mistake immediately.

  3. For the equal pair at ±15 μas\pm15\,\mu\mathrm{as}, what happens at twice the first-null baseline?

    Read the explanation

    The cosine argument grows from π/2\pi/2 to π\pi, so the visibility is −F-F. Its amplitude is again FF, with a phase of π\pi. A visibility amplitude alone hides this sign.

Sources and calculation scope

  1. NRAO Essential Radio Astronomy, Chapter 3, aperture synthesis

    The narrow-field Fourier relation and Earth-rotation synthesis, especially equations 3.180–3.200. The browser uses a spherical Earth, a single wavelength and a finite pixel sky.

  2. EHT Collaboration, First M87 Results IV, arXiv:1906.11241v1

    Fixed version of the 2019 imaging paper, DOI 10.3847/2041-8213/ab0e85. Read Figures 1, 3, 5 and 7, the station-gain equation 7, regularized imaging equation 14, and Appendix A. Figures remain with the publisher and authors; this guide links to them.

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