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Comment by nico | Hacker News Reader

How big would the telescope/mirror/lens need to be to get a picture of something in the Alpha Centauri system, 4.37 light years away?

Also, could the image be created by “scanning” a big area and then composing the image from a bunch of smaller ones?

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It's a lot easier to reason about this using angular resolution, because that's normally what the diffraction limit formula is in reference to. If you know the angular diameter of the system (α) and the wavelength (say λ=500 nm for visible), you can use α ≈ λ/d and solve for the aperture of the telescope (d).

That puts a basic limit on the smallest thing you can resolve with a given aperture. You can use the angular diameter of the planet and the resolution you're after. For Alpha Centauri A it's 8.5 milli arc-second, so O(1 μas) for a 100px image? That's just for the star!

The Event Horizon Telescope can achieve around 20-25 μas in microwave; you need a planet-scale interferometer to do that. https://en.wikipedia.org/wiki/Event_Horizon_Telescope It's possible to do radio measurements in sync with good clocks and fast sampling/storage, much harder with visible.

I'm not super up to date on visible approaches, but there is LISA which will be a large scale interferometer in space. The technology for synchronising the satellites is similar to what you'd need for this in the optical.

https://www.edmundoptics.com/knowledge-center/application-no...

https://arxiv.org/abs/astro-ph/0303634

How far off are we still for doing this with visual light?

Let's say you build single photon detectors and ultra precise time stamping. Would that get us near? Today, maybe we don't have femtosecond time stamping and detectors yet. But that is something I can imagine being built! Timing reference distribution within fs over 100s of km? Up to now, nobody needed that I guess.

The biggest issue is the sheer separation required. EHT operates in mm wave light, visible is 4-6 orders of magnitude shorter wavelength. There are several smaller scale interferometers. They can already do quite impressive things because even a 50m baseline is better than any optical telescope that exists.

The way that timing works for EHT is each station has a GPS reference that's conditioned with a very good atomic clock - for example at SPT we use a hydrogen maser. The readout and timing system is separate from the normal telescope control system, we just make sure the dish is tracking the right spot before we need to start saving data (sampling around 64 Gbps).

I'm not sure what the timing requirements are for visible and how the clock is distributed, but syncing clocks extremely well over long distances shouldn't be insurmountable. LISA needs to solve this problem for gravitational waves and that's a million+ km baseline.

Some problems go away in space. You obviously need extremely accurate station keeping (have a look how LISA Pathfinder does it, very cool), but on Earth we also have to take continental drift into account.

Is there another limit in terms of just: how many photons from X object even hit an area of Y telescope apeture size from distance Z in like, say a year? We can't see the thing if no photons from it even intersect our telescope, right? Or maybe that limit is way way less restrictive than the other...

The number of photons themselves is not too restrictive (i think the voyager probe still emits 6ish photons per second directed at the receiving dish). And we easily build sensors that detect every photon (far above 99% levels). The tricky part will be differentiating between “source photons” and “background photons” (for Voyager we exactly know what to look for, here we wouldn’t have any baseline for distinguishing)

It's linear, so if it is 25 times closer then the telescope can be 25 times smaller. At 4.37 light-years we'd need an 18 kilometer telescope to image at Jupiter-sized planet at 100x100 pixel resolution.

If you only wanted 10x10 resolution you could get by with a 1.8 kilometer telescope.

Wikipedia has more: https://en.wikipedia.org/wiki/Angular_resolution. The Rayleigh criterion is the equation to calculate this.

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