<p>Aside from trying to phase-sync optical scopes across such large distances, the problem is one of spatial frequency filtering.  If you look thru any decent book on optics (I highly recommend Smith's Engineering Optics), you'll see that, due to the &quot;magic&quot; of Fourier Transforms,  a small lens (or primary mirror) handles low spatial frequencies well but not high spatial frequencies.  A large lens will handle higher and higher spatial frequencies as well.   But if you use a &quot;doughnut&quot; lens, i.e. with large central obscuration, you lose all the lower spatial frequencies.</p>
<p>So if you scatter a bunch of phased-up optical scopes, you will at best only get some of the higher spatial frequencies in your final image.</p>

CC BY-SA 4.0
2022-06-11T02:27:39.207Z
Certainly [donuts do not good telescopes make](https://astronomy.stackexchange.com/a/34247/7982), but array telescopes are never donuts with receivers only at the edges. In fact they're the opposite. Look at any arrangement of ALMA or VLA and density is almost always highest in the center. This isn't possible with the EHT but those exposures exploit the rotation of the Earth to allow many receivers to pass through the center of the "aperture".
uhoh
CC BY-SA 4.0
2022-06-11T16:07:10.223Z
@uhoh  I completely agree.  Just sort of warning that you can't  just randomly scatter 'scopes around and expect a "large aperture equivalent" performance.
Carl Witthoft
CC BY-SA 4.0
2022-06-10T12:51:10.600Z
2022-06-10T12:51:10.600Z
Carl Witthoft
<p>In regards of resolution a virtual telescope acts identical to a normal telescope: size matters:</p>
<p>More accurately, the optic's diameter <span class="math-container">$D$</span> and the wavelength <span class="math-container">$\lambda$</span> determine the resolution or optical resolving power: <span class="math-container">$\theta = 1.22\cdot\frac{\lambda}{D}$</span>. This is called the <a href="https://en.wikipedia.org/wiki/Angular_resolution#The_Rayleigh_criterion" rel="nofollow noreferrer">Rayleigh criterion</a> which states that at this fraction two adjacent dots can be distinguished as then their airy disks of the diffraction patterns of the two adjacent point sources are sufficiently separate to indentify them separately in the image. Note that the area of your telescope does not play a role in the resolution, but only the diameter of the optics.</p>
<p>Now, why use a large disk? The larger your collecting area is, the more of the signal you gather, the better is your signal to noise ratio, the brighter the source compared to the background (like you see with binoculars fainter objects than with your naked eye, and with a larger telescope fainter ones than with a binocular). Thus the sensitivity simply scales with the actual collecting area of your instrument. Thus more or larger disks do help to detect fainter sources.</p>
<p>The <a href="https://www.atnf.csiro.au/outreach/education/senior/astrophysics/resolution_sensitivity.html" rel="nofollow noreferrer">Australian Telescope National Facility</a> has a nice illustration of these two phenomena.</p>
<p>So in summary: a virtual earth-sized telescope is as good as one big telescope when it comes to see the spatial detail. But more and larger telescope as part of this virtual telescope help increase the contrast, thus allow to identify spatially-resolved details which only vary slightly in brightness.</p>

CC BY-SA 4.0
2022-06-10T11:31:03.543Z
This isn't actually quite true. EHT goes through heroic efforts to try to hide the diffraction due to its woefully-under-sampled aperture. And large ratio telescopes depend on long exposures such that the Earth's rotation moves the array around in UV space in order to try to get cleaner images, but it still has to be post-processed. [Math behind a uv plot in interferometry?](https://astronomy.stackexchange.com/q/26413/7982)
uhoh
CC BY-SA 4.0
2022-06-10T06:11:54.347Z
2022-06-10T06:11:54.347Z
planetmaker
<p>The Event Horizon Telescope emulates an Earth-sized telescope by syncing a bunch of radio telescopes across the planet to do take pictures with a small enough angular resolution to take pictures of a black hole. Would the angular resolution, or perhaps just general performance, of an Earth-sized telescope be appreciably higher than this?</p>

CC BY-SA 4.0
2022-06-10T16:01:05.870Z
Obviously, the received light doesn't emulate such an Earth-sized telescope.
Felicia
CC BY-SA 4.0
2022-06-11T02:28:30.143Z
@Felicia that's such an important point that I think it deserves an answer of its own. Resolution based on aperture assumes a long enough exposure that it's not limited by signal to noise ratio (SNR). With low SNR you can have uncertainties in size and position much larger than what a simple point-spread function (PSF) might suggest. *Consider adding an answer?*
uhoh
<observational-astronomy><radio-astronomy><radio-telescope><interferometry><event-horizon-telescope>
In terms of results, how similar is a bunch of telescopes across the globe to an Earth-sized telescope?
zucculent
# In terms of results, how similar is a bunch of telescopes across the globe to an Earth-sized telescope?

(Asked by: username_0 on 2022-06-10)

The Event Horizon Telescope emulates an Earth-sized telescope by syncing a bunch of radio telescopes across the planet to do take pictures with a small enough angular resolution to take pictures of a black hole. Would the angular resolution, or perhaps just general performance, of an Earth-sized telescope be appreciably higher than this?

**Tags:** observational-astronomy, radio-astronomy, radio-telescope, interferometry, event-horizon-telescope
> **Comments:**
> **username_1**: Obviously, the received light doesn't emulate such an Earth-sized telescope.  
> **username_2**: @Felicia that's such an important point that I think it deserves an answer of its own. Resolution based on aperture assumes a long enough exposure that it's not limited by signal to noise ratio (SNR). With low SNR you can have uncertainties in size and position much larger than what a simple point-spread function (PSF) might suggest. *Consider adding an answer?*  

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## 2 answers

### Answer by username_3 on 2022-06-10. Score: 2

Aside from trying to phase-sync optical scopes across such large distances, the problem is one of spatial frequency filtering. If you look thru any decent book on optics (I highly recommend Smith's Engineering Optics), you'll see that, due to the "magic" of Fourier Transforms, a small lens (or primary mirror) handles low spatial frequencies well but not high spatial frequencies. A large lens will handle higher and higher spatial frequencies as well. But if you use a "doughnut" lens, i.e. with large central obscuration, you lose all the lower spatial frequencies.

So if you scatter a bunch of phased-up optical scopes, you will at best only get some of the higher spatial frequencies in your final image.

> **Comments:**
> **username_2**: Certainly [donuts do not good telescopes make](https://astronomy.stackexchange.com/a/34247/7982), but array telescopes are never donuts with receivers only at the edges. In fact they're the opposite. Look at any arrangement of ALMA or VLA and density is almost always highest in the center. This isn't possible with the EHT but those exposures exploit the rotation of the Earth to allow many receivers to pass through the center of the "aperture".  
> **username_3**: @uhoh  I completely agree.  Just sort of warning that you can't  just randomly scatter 'scopes around and expect a "large aperture equivalent" performance.  

---

### Answer by username_4 on 2022-06-10. Score: 2

In regards of resolution a virtual telescope acts identical to a normal telescope: size matters:

More accurately, the optic's diameter $D$ and the wavelength $\lambda$ determine the resolution or optical resolving power: $\theta = 1.22\cdot\frac{\lambda}{D}$. This is called the [Rayleigh criterion](https://en.wikipedia.org/wiki/Angular_resolution#The_Rayleigh_criterion) which states that at this fraction two adjacent dots can be distinguished as then their airy disks of the diffraction patterns of the two adjacent point sources are sufficiently separate to indentify them separately in the image. Note that the area of your telescope does not play a role in the resolution, but only the diameter of the optics.

Now, why use a large disk? The larger your collecting area is, the more of the signal you gather, the better is your signal to noise ratio, the brighter the source compared to the background (like you see with binoculars fainter objects than with your naked eye, and with a larger telescope fainter ones than with a binocular). Thus the sensitivity simply scales with the actual collecting area of your instrument. Thus more or larger disks do help to detect fainter sources.

The [Australian Telescope National Facility](https://www.atnf.csiro.au/outreach/education/senior/astrophysics/resolution_sensitivity.html) has a nice illustration of these two phenomena.

So in summary: a virtual earth-sized telescope is as good as one big telescope when it comes to see the spatial detail. But more and larger telescope as part of this virtual telescope help increase the contrast, thus allow to identify spatially-resolved details which only vary slightly in brightness.

> **Comments:**
> **username_2**: This isn't actually quite true. EHT goes through heroic efforts to try to hide the diffraction due to its woefully-under-sampled aperture. And large ratio telescopes depend on long exposures such that the Earth's rotation moves the array around in UV space in order to try to get cleaner images, but it still has to be post-processed. [Math behind a uv plot in interferometry?](https://astronomy.stackexchange.com/q/26413/7982)  

---

