I went to https://eventhorizontelescope.org/array and read about the ten sites listed as part of the EHT, I have a mashed-up screen shot of them below.

I made a little script also shown below with approximate coordinates, and saw that Sgr A* is at a declination of about -29.5 degrees (South). I spun it around the earth, calculated the dot product between each of the sites' position vectors and all normals pointing to Sgr A* as it rotates around the Earth.

I then plotted the elevation angle above the horizon for all ten sites. I was quite surprised!

  1. Sgr A* is never above the horizon for the Greenland telescope!
  2. Sgr A* is never simultaneously visible even to all of the remaining nine telescopes

Have I made a mistake? If not, how do these ten sites combine their data to image Sgr A*?

Event Horizon Telescope visibility of Sgr A*

Event Horizon Telescope

class Site(object):
    def __init__(self, name, lat, lon, alt):
        self.name = str(name)
        self.lat  = rads * float(lat)
        self.lon  = rads * float(lon)
        self.alt  = km   * float(alt)
        (clat, slat), (clon, slon) = [[f(x) for f in (np.cos, np.sin)] for x in (self.lat, self.lon)]
        self.norm = np.array([clon*clat, slon*clat, slat])

data = (('Northern Extended Millimeter Array', (44.634, 5.908, 2550.)),
        ('IRAM 30 meter telescope', (37.066, -3.393, 2850.)),
        ('The Greenland Telescope now near Thule Air Base', (76.531, -68.703, 10.)),
        ('Combined Array for Research in Millimeter-wave Astronomy CARMA', (37.280, -118.142, 2196.)),
        ('Kitt Peak National Observatory 12 meter Submillimeter Telescope (SMT)', (1.9583, -111.5967,  2096.)),
        ('Mt. Graham International Observatory 12 meter ALMA prototye', (32.701, -109.892, 3191.)),
        ('The Large Millimeter Telescope Alfonso Serrano', (18.985, -97.315, 4600.)),
        ('ALMA', (-22.971, -67.703, 4800.)),
        ('Caltech Submillimeter Observatory', (19.823, -155.476, 4140.)),
        ('South Pole Telescope', (-90.0, 0.0, 2800.)))

# https://eventhorizontelescope.org/array
# https://astronomy.stackexchange.com/questions/26413/math-behind-a-uv-plot-in-interferometry

datadict = dict(data)

import numpy as np
import matplotlib.pyplot as plt
from skyfield.api import Topos, Loader, EarthSatellite
from mpl_toolkits.mplot3d import Axes3D

halfpi, pi, twopi = [f*np.pi for f in (0.5, 1, 2)]
degs, rads = 180/pi, pi/180
km         = 0.001

sites = [Site(a, *b) for a, b in datadict.items()]

Dec = rads * -29.5  # SgrA_star
cDec, sDec = [f(Dec) for f in (np.cos, np.sin)]
th         = twopi * np.linspace(0, 1, 100001)[:-1]
cth, sth   = [f(th) for f in (np.cos, np.sin)]
zth, oth   = np.zeros_like(th), np.ones_like(th)

SgrA_star = np.stack([cDec*cth, cDec*sth, sDec*oth], axis=1)

for site in sites:
    site.elev = np.arcsin(np.dot(SgrA_star, site.norm))

if True:
    for site in sites:
    plt.plot(zth, '-k', linewidth=2)
    plt.ylim(-90, 90)
    plt.ylabel('elevation (deg)', fontsize=16)
  • 1
    $\begingroup$ Better change your script for M87! $\endgroup$
    – ProfRob
    Apr 10 '19 at 13:30
  • $\begingroup$ @RobJeffries uhoh! :-0 $\endgroup$
    – uhoh
    Apr 10 '19 at 16:03

I don't think it is necessary for all the telescopes to view the target simultaneously if you can make the assumption that the source you are observing is only varying on timescales longer than it takes the source to be viewable by all the telescopes. Sure, the maximum baseline you can get will be defined by the most distant pair of simultaneously observing telescopes, but that doesn't stop you adding together images taken at other times.

This is very probably the reason why M87 is in the fact the first image presented by the EHT. The timescale for significant variability around this (big) black hole will be of order a few time $2GM/c^2$, which is a few days for the M87 black hole. Imaging Sgr A* is going to be more difficult (or at least more blurry) because the variability timescale is minutes.

It looks like (from the data reduction paper) that the first steps involve treating every baseline in a pair-wise way (as I suggested above) and then combining them across the network, with whatever baselines were observing at the time, to give data at the surprisingly (to me) high time-resolution of 10s.

  • $\begingroup$ I think you need to observe simultaneously, You actually want to compare the phase of the RF signal at each telescope and the source is not phase correlated over time. $\endgroup$ Apr 10 '19 at 14:43
  • 2
    $\begingroup$ @SteveLinton Small edit - that is true in so far as the maximum baseline achievable will be determined by simultaneously observing telescopes. $\endgroup$
    – ProfRob
    Apr 10 '19 at 14:51
  • $\begingroup$ Thanks, I will have a look at the paper. I'd never thought about timescales, that will give me something to think about. $\endgroup$
    – uhoh
    Apr 10 '19 at 16:03
  • $\begingroup$ Yes, this timescale is a key issue. Thanks! $\endgroup$
    – uhoh
    Jul 25 '19 at 2:25

The recently announced M87 observations only used 8 of the 10 sites. This paper includes this diagram.Even the South Pole instrument seems to have been used only for calibratiom (the dashed lines).

enter image description here

  • $\begingroup$ Thanks for your answer, I'll have a look. $\endgroup$
    – uhoh
    Apr 10 '19 at 16:01
  • $\begingroup$ By the way, I'm going to leave my question as-is and wait for an answer about imaging Sgr A*. My question does not make any reference to anything imminent or upcoming; I'm really interested in how they will image Sgr A*! However, I will be asking a new question about M87 and I'll link back here. Thanks for the graphic! $\endgroup$
    – uhoh
    Apr 11 '19 at 3:32
  1. Sgr A* is never above the horizon for the Greenland telescope!

Well, Sgr A* is not the only focus of EHT; the website mentions it also studies other objects like the M87 galaxy in Virgo which, at +12° declination, is visible from Greenland.

  1. Sgr A* is never simultaneously visible even to all of the remaining nine telescopes

Very Long Baseline Interferometry just combines the results of the telescopes which can (and have been) pointed at a specific target at a specific time (plus or minus a few nanoseconds, because some telescopes might be further away from the target than others, and light takes slightly longer or shorter to reach them). So it will never use the full potential of all the telescopes in the array simultaneously.

At the location of the correlator the data are played back. The timing of the playback is adjusted according to the atomic clock signals on the (tapes/disk drives/fibre optic signal), and the estimated times of arrival of the radio signal at each of the telescopes. A range of playback timings over a range of nanoseconds are usually tested until the correct timing is found.

  • $\begingroup$ 1) "...just combines the results of the telescopes..." how!? 2) It's milliseconds because the telescopes are tens of thousands of kilometers away, not nanoseconds! 3) the block quote has to do with local time recovery with respect to each local atomic clock, not the interferometry. See How does the Event Horizon Telescope implement the interferometry? $\endgroup$
    – uhoh
    Apr 10 '19 at 11:15
  • $\begingroup$ That's explained in brief in the Wikipedia article. $\endgroup$
    – Glorfindel
    Apr 10 '19 at 11:17
  • $\begingroup$ I don't have to remind you that "go read Wikipedia" is not a good way to answer a Stack Exchange question. $\endgroup$
    – uhoh
    Apr 10 '19 at 11:19
  • 3
    $\begingroup$ I'm just focusing on (the part of) your question which asks about the geographic configuration of the telescopes, because that's what it focuses on (see e.g. the title). If you want to know in detail how VLBI works, you should probably ask a new question. $\endgroup$
    – Glorfindel
    Apr 10 '19 at 11:21
  • 1
    $\begingroup$ Ah, I missed that you edited the comment. $\endgroup$
    – Glorfindel
    Apr 10 '19 at 16:11

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