Are there inherent limits to Interferometer size?

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biffus22
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The recent high angular resolution images of M87's inner black hole taken by radio telescopes around the globe all linked together in a computer to simulate a giant interferometer, suggest that to achieve still better angular resolution we would need to supplement the terrestrial receiving antennas with similar, but much more distant, telescopes in space. The question thus arises, are there some inherent limits on how far from one another the various components of such a massive interferometer can be before getting the various inputs to interfere becomes impossible? If there IS such a limit, how can it be found? If there is NOT such a limit, how can one explain how the photons received at one far outlying antenna could still be made to interfere with other photons received by the other space-born antennas each of which might well be, say, a few billion miles away?
 

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Baluncore
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The question thus arises, are there some inherent limits on how far from one another the various components of such a massive interferometer can be before getting the various inputs to interfere becomes impossible?
To maintain a higher resolution in the u and v axes of the image, you will need to expand the array in two orthogonal directions.

If your interferometer array is wide and sparse, then there may be many possible time delays that will correlate. That may introduce artefacts into the image. It may take a longer time to establish the correlation between some sites. How much correlator time can you afford?

You will need clocks with lower phase noise for more widely spaced arrays. That is a square law problem, better resolution requires better clocks, and those clocks must be distributed over a wider space for the bigger array.

You might do better by increasing the frequency, which reduces the wavelength, thereby increasing the angular resolution for the same array. You will then need proportionally better clocks.

I see no brick wall limit, but there is an economic budget with rapidly diminishing returns.

If there is NOT such a limit, how can one explain how the photons received at one far outlying antenna could still be made to interfere with other photons received by the other space-born antennas each of which might well be, say, a few billion miles away?
You correlate waves, not photons.
 
  • #3
tech99
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Another interesting technique is the Intensity Interferometer, invented by Hanbury-Brown (https://en.wikipedia.org/wiki/Robert_Hanbury_Brown) This uses a correlator to compare the noise envelope from the source when received on spaced antennas, and so does not require phase locking.
 
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Baluncore
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The intensity interferometer was used to measure the diameter of sources. It cannot be used for imaging.
 
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sophiecentaur
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how can one explain how the photons received at one far outlying antenna could still be made to interfere with other photons received by the other space-born
Yet again, there's a notion that using the word "photon" automatically raises the Physics to another level of sophistication very often it doesn't. This particular topic is based on waves. Photons are nowhere in particular and have' no time' in particular so would it be easy to base interference using them? In a situation like interferometry, no doubt there is a calculation, involving photons, that will explain how interferometry works. But after messing about with wave functions etc etc, the exact same equations will fall out and predict the exact same thing about the formation of an image.

Photons come into their own in many other situations - the photoelectric effects best described using them. It's horses for courses.
 
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Andy Resnick
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[...] The question thus arises, are there some inherent limits on how far from one another the various components of such a massive interferometer can be before getting the various inputs to interfere becomes impossible? If there IS such a limit, how can it be found? [...]
Size limits for interferometers can be estimated from the coherence of the source. For Michaelson-type interferometers, the maximum path length difference ΔL is given by:

ΔL = λ2/Δλ, where λ is the average wavelength and Δλ the bandwidth.

For Young-type interferometers, the maximum aperture separation ΔL is given by the size of the source

ΔL = √(λ2/ΔΩ), where ΔΩ is the solid angle subtended by the source.

Does that help?
 
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