<p>If you look at an object with a pair of telescopes separated by <span class="math-container">$D$</span> and the object is &quot;off-axis&quot; by an angle <span class="math-container">$\theta$</span>, then as @uhoh's answer shows, there is a phase difference between the signal received by the two telescopes of <span class="math-container">$2\pi D \sin \theta/\lambda$</span>.</p>
<p>Now suppose we arrange it so that this phase difference is a multiple of <span class="math-container">$2\pi$</span> so that the signals are &quot;in phase&quot;. What we can then ask is &quot;by how much could we change <span class="math-container">$\theta$</span> for the signals to go &quot;out of phase&quot; - i.e. for the phase difference to be <span class="math-container">$\pi$</span>. Because, if we can tell the difference between a source at an angle <span class="math-container">$\theta$</span> and at an angle <span class="math-container">$\theta + \Delta \theta$</span> by observing the interference pattern from the two telescopes go from being in phase to out of phase, we can claim to have &quot;resolved&quot; objects separated by that angle <span class="math-container">$\Delta \theta$</span>. (NB: This definition of the limiting angular separation is a bit handwavy and more exact definitions are usually used).</p>
<p>The limiting value of <span class="math-container">$\theta$</span> is therefore given by
<span class="math-container">$$ \frac{2\pi}{\lambda} D \sin (\theta + \Delta\theta) - \frac{2\pi}{\lambda} D\sin\theta = \pi\ .$$</span>
Using standard trigonometric identities and using the fact that <span class="math-container">$\Delta \theta \ll 1$</span> radian
<span class="math-container">$$ \frac{2\pi}{\lambda}D\sin(\Delta \theta) \cos\theta \simeq \frac{2\pi}{\lambda} D\cos\theta\  \Delta\theta = \pi\ ,$$</span>
<span class="math-container">$$\Delta \theta \simeq  \frac{\lambda}{2D\cos\theta}\ .$$</span>
Thus the larger <span class="math-container">$D$</span> is, the smaller the separation <span class="math-container">$\Delta \theta$</span> is between two close sources that can be resolved.</p>
<p>The problem with this is that if we make <span class="math-container">$\Delta \theta$</span> a bit larger, then the signals start to come into phase again. However, if we had information from a pair of telescopes with a <em>smaller</em> separation, we would see the signals from those telescopes go out of phase. Combining the information from the two pairs of telescopes we might come to an understanding of exactly what the separation of the sources is (so long as it is <span class="math-container">$\geq$</span> the limiting <span class="math-container">$\Delta \theta$</span>).</p>
<p>The above illustrates how we can say that a pair of telescopes separated by <span class="math-container">$D$</span> behaves like a single telescope of diameter <span class="math-container">$D$</span> (if the source is overhead and <span class="math-container">$\cos \theta \sim 1$</span>) in terms of the smallest angular separation between two sources that can be resolved. However, to provide unambiguous information about the separation between the sources we could also do with information from telescopes at smaller separations. This of course is naturally provided by a telescope with &quot;filled&quot; dish, where all baselines from zero to the diameter of the dish are sampled and leads to the formation of an image.</p>
<p>With interferometers one tries to &quot;fill the dish&quot; in two ways (i) as the Earth rotates <span class="math-container">$\cos \theta$</span> can change and provide a range of baselines (though this wouldn't be helpful if the source were variable!) (ii) we can use an array of telescopes with different baselines to fill in that information.</p>

CC BY-SA 4.0
2023-01-30T13:45:44.327Z
2023-01-30T15:57:23.850Z
2023-01-30T15:57:23.850Z
ProfRob
<p><a href="https://astronomy.stackexchange.com/a/51738/7982">@JamesK's answer</a> explains in prose, but we can also address the OP's <a href="https://astronomy.stackexchange.com/questions/51737/how-does-very-long-baseline-interferometry-work/51738#comment116217_51738">comment</a></p>
<blockquote>
<p>On what grounds do you say that a larger diameter leads to larger phase difference?</p>
</blockquote>
<p>more rigorously.</p>
<p>The diagram below shows two dish antennas on the ground separated by a distance <span class="math-container">$D$</span>. Each dish is <em>tilted</em> so that it is pointed directly at the source, so that the phase of the received signal is constant and independent of where the signal hits the dish.</p>
<p>However, The signal is coming in at an angle with respect to the surface of the Earth, so the phase will be different between one dish and the next.</p>
<p>If we draw a big right triangle, with the hypotenuse being the baseline between the two, and define <span class="math-container">$\theta$</span> as the angle of incidence with respect to the zenith, the triangle construction will show that length of the side representing the phase difference will be <span class="math-container">$D \sin \theta$</span>. The extra phase <span class="math-container">$\Delta \phi$</span> accumulated along that length for the signal collected by the further  antenna with respect to the closer antenna will then be given by</p>
<p><span class="math-container">$$\Delta \phi = 2 \pi \frac{D}{\lambda} \sin \theta$$</span></p>
<hr />
<p><a href="https://i.sstatic.net/C7cMx.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/C7cMx.png" alt="enter image description here" /></a>
<a href="https://commons.wikimedia.org/wiki/File:Emmaalexander_inter_dishes.png" rel="nofollow noreferrer">Source</a></p>

CC BY-SA 4.0
2023-01-30T10:41:19.777Z
Extend to explain why this behaves like one big dish of diameter $D$ (and also in what respects that it doesn't).
ProfRob
CC BY-SA 4.0
2023-01-30T12:19:16.793Z
@ProfRob Because I love that bit so much, it is really challenging for me to do it in a reasonably short space suitable for an SE answer post. I'll give some thought to how to do so. My intention here was only to address that one comment under the other answer, not to write the chapter on radio interferometric imaging that your suggestion deserves.
uhoh
CC BY-SA 4.0
2023-01-30T20:39:44.680Z
@ProfRob [nicely done](https://astronomy.stackexchange.com/a/51753/7982)!
uhoh
CC BY-SA 4.0
2023-01-28T23:16:24.853Z
2023-01-28T23:16:24.853Z
uhoh
<p>Suppose, instead of thinking about interferometry, you first consider an ordinary mirror. If you are using this mirror to create an image of a point source of light, you would hope that the mirror would focus the point of light to a single point in the image.</p>
<p>This is what &quot;geometric optics&quot; (in which you treat light as moving in straight lines) would predict.</p>
<p>But there is a problem, the light reaches the mirror in phase and so when it is reflected, there is constructive and destructive interference around the point, which blurs the point of light.  This problem is greater with small mirrors, since with a small mirror, the phase difference across the diameter of the mirror is small.  The larger the mirror, the greater is the difference in phase and the smaller is the blurring.</p>
<p>You can even achieve less blurring by having two mirrors that are separated but acting as one  (in practice, getting two mirrors aligned accurately enough is very difficult, you need nano-metre precision to bring the light together to form a single image)</p>
<p>The formation of an image depends on the process of bringing light together.  Now if you are observing in radio waves, not light, you can actual measure the peaks and troughs of the radio wave, and not just the intensity of the wave. (by contrast, when you take a photograph, you only record the intensity).</p>
<p>This means you can record the pattern of the waves, and bring the waves together computationally.  You are essentially using a computer to do what a mirror or lens does: bring the waves together to form an image.</p>
<p>You are limited by exactly the same blurring problem that occurs with real mirrors, but the separation of the radio telescopes can be much greater.  Mirrors and lenses depend on nanometer accuracy to achieve a good image.  But if you can combine the radiowaves in a computer, you can compensate for that, so it is possible to have a telescope with an effective diameter of 10000km.</p>
<p>So the process of VLB interferometry is doing with a computer what a mirror naturally achieves: bringing together radiowave/light from different places to create an image.</p>

CC BY-SA 4.0
2023-01-28T12:55:28.963Z
On what grounds do you say that a larger diameter leads to larger phase difference?
Ambica Govind
CC BY-SA 4.0
2023-01-29T10:28:52.537Z
"The larger the mirror, the greater is the difference in phase and the smaller is the blurring." Why does a greater phase difference lead to lesser blurring?
Ambica Govind
CC BY-SA 4.0
2023-01-28T11:46:19.930Z
2023-01-28T11:46:19.930Z
James K
<p>According to internet sources, this technique increases the effective diameter of the telescope, thus improving the angular resolution of the telescope. However I don't understand how this actually works. The way I understand it, I know that signals from different locations in the boundary of the black hole may constructively or destructively interfere, and from the information about the relative separation between the telescope can tell us about the shape of the black hole. <br>
However how does any of this have anything to do with the 'effective diameter' of the telescope?</p>

<black-hole><radio-astronomy><supermassive-black-hole><radio-telescope><interferometry>
How does Very Long Baseline Interferometry work?
Ambica Govind
# How does Very Long Baseline Interferometry work?

(Asked by: username_0 on 2023-01-28)

According to internet sources, this technique increases the effective diameter of the telescope, thus improving the angular resolution of the telescope. However I don't understand how this actually works. The way I understand it, I know that signals from different locations in the boundary of the black hole may constructively or destructively interfere, and from the information about the relative separation between the telescope can tell us about the shape of the black hole.   
However how does any of this have anything to do with the 'effective diameter' of the telescope?

**Tags:** black-hole, radio-astronomy, supermassive-black-hole, radio-telescope, interferometry

---

## 3 answers

### Answer by username_1 on 2023-01-30. Score: 2 | Accepted answer

If you look at an object with a pair of telescopes separated by $D$ and the object is "off-axis" by an angle $\theta$, then as @uhoh's answer shows, there is a phase difference between the signal received by the two telescopes of $2\pi D \sin \theta/\lambda$.

Now suppose we arrange it so that this phase difference is a multiple of $2\pi$ so that the signals are "in phase". What we can then ask is "by how much could we change $\theta$ for the signals to go "out of phase" - i.e. for the phase difference to be $\pi$. Because, if we can tell the difference between a source at an angle $\theta$ and at an angle $\theta + \Delta \theta$ by observing the interference pattern from the two telescopes go from being in phase to out of phase, we can claim to have "resolved" objects separated by that angle $\Delta \theta$. (NB: This definition of the limiting angular separation is a bit handwavy and more exact definitions are usually used).

The limiting value of $\theta$ is therefore given by
$$ \frac{2\pi}{\lambda} D \sin (\theta + \Delta\theta) - \frac{2\pi}{\lambda} D\sin\theta = \pi\ .$$
Using standard trigonometric identities and using the fact that $\Delta \theta \ll 1$ radian
$$ \frac{2\pi}{\lambda}D\sin(\Delta \theta) \cos\theta \simeq \frac{2\pi}{\lambda} D\cos\theta\ \Delta\theta = \pi\ ,$$
$$\Delta \theta \simeq \frac{\lambda}{2D\cos\theta}\ .$$
Thus the larger $D$ is, the smaller the separation $\Delta \theta$ is between two close sources that can be resolved.

The problem with this is that if we make $\Delta \theta$ a bit larger, then the signals start to come into phase again. However, if we had information from a pair of telescopes with a *smaller* separation, we would see the signals from those telescopes go out of phase. Combining the information from the two pairs of telescopes we might come to an understanding of exactly what the separation of the sources is (so long as it is $\geq$ the limiting $\Delta \theta$).

The above illustrates how we can say that a pair of telescopes separated by $D$ behaves like a single telescope of diameter $D$ (if the source is overhead and $\cos \theta \sim 1$) in terms of the smallest angular separation between two sources that can be resolved. However, to provide unambiguous information about the separation between the sources we could also do with information from telescopes at smaller separations. This of course is naturally provided by a telescope with "filled" dish, where all baselines from zero to the diameter of the dish are sampled and leads to the formation of an image.

With interferometers one tries to "fill the dish" in two ways (i) as the Earth rotates $\cos \theta$ can change and provide a range of baselines (though this wouldn't be helpful if the source were variable!) (ii) we can use an array of telescopes with different baselines to fill in that information.


---

### Answer by username_2 on 2023-01-28. Score: 3

[@JamesK's answer](https://astronomy.stackexchange.com/a/51738/7982) explains in prose, but we can also address the OP's [comment](https://astronomy.stackexchange.com/questions/51737/how-does-very-long-baseline-interferometry-work/51738#comment116217_51738)

> On what grounds do you say that a larger diameter leads to larger phase difference?

more rigorously.

The diagram below shows two dish antennas on the ground separated by a distance $D$. Each dish is *tilted* so that it is pointed directly at the source, so that the phase of the received signal is constant and independent of where the signal hits the dish.

However, The signal is coming in at an angle with respect to the surface of the Earth, so the phase will be different between one dish and the next.

If we draw a big right triangle, with the hypotenuse being the baseline between the two, and define $\theta$ as the angle of incidence with respect to the zenith, the triangle construction will show that length of the side representing the phase difference will be $D \sin \theta$. The extra phase $\Delta \phi$ accumulated along that length for the signal collected by the further antenna with respect to the closer antenna will then be given by

$$\Delta \phi = 2 \pi \frac{D}{\lambda} \sin \theta$$

---

[![enter image description here](https://i.sstatic.net/C7cMx.png)](https://i.sstatic.net/C7cMx.png)
[Source](https://commons.wikimedia.org/wiki/File:Emmaalexander_inter_dishes.png)

> **Comments:**
> **username_1**: Extend to explain why this behaves like one big dish of diameter $D$ (and also in what respects that it doesn't).  
> **username_2**: @ProfRob Because I love that bit so much, it is really challenging for me to do it in a reasonably short space suitable for an SE answer post. I'll give some thought to how to do so. My intention here was only to address that one comment under the other answer, not to write the chapter on radio interferometric imaging that your suggestion deserves.  
> **username_2**: @ProfRob [nicely done](https://astronomy.stackexchange.com/a/51753/7982)!  

---

### Answer by username_3 on 2023-01-28. Score: 2

Suppose, instead of thinking about interferometry, you first consider an ordinary mirror. If you are using this mirror to create an image of a point source of light, you would hope that the mirror would focus the point of light to a single point in the image.

This is what "geometric optics" (in which you treat light as moving in straight lines) would predict.

But there is a problem, the light reaches the mirror in phase and so when it is reflected, there is constructive and destructive interference around the point, which blurs the point of light. This problem is greater with small mirrors, since with a small mirror, the phase difference across the diameter of the mirror is small. The larger the mirror, the greater is the difference in phase and the smaller is the blurring.

You can even achieve less blurring by having two mirrors that are separated but acting as one (in practice, getting two mirrors aligned accurately enough is very difficult, you need nano-metre precision to bring the light together to form a single image)

The formation of an image depends on the process of bringing light together. Now if you are observing in radio waves, not light, you can actual measure the peaks and troughs of the radio wave, and not just the intensity of the wave. (by contrast, when you take a photograph, you only record the intensity).

This means you can record the pattern of the waves, and bring the waves together computationally. You are essentially using a computer to do what a mirror or lens does: bring the waves together to form an image.

You are limited by exactly the same blurring problem that occurs with real mirrors, but the separation of the radio telescopes can be much greater. Mirrors and lenses depend on nanometer accuracy to achieve a good image. But if you can combine the radiowaves in a computer, you can compensate for that, so it is possible to have a telescope with an effective diameter of 10000km.

So the process of VLB interferometry is doing with a computer what a mirror naturally achieves: bringing together radiowave/light from different places to create an image.

> **Comments:**
> **username_0**: On what grounds do you say that a larger diameter leads to larger phase difference?  
> **username_0**: "The larger the mirror, the greater is the difference in phase and the smaller is the blurring." Why does a greater phase difference lead to lesser blurring?  

---

