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In this question I am not considering the sun, but only other stars.

Stars are big, but they are also far away. https://en.wikipedia.org/wiki/Star

The nearest star to the Earth, apart from the Sun, is Proxima Centauri, which is 39.9 trillion kilometres, or 4.2 light-years away.

In pictures like this:

Stars

I believe that the size of the spots is due to overflow of charge in the CCD, and that the four rays is due to optical artifacts in the telescope. (Please correct me, if I am wrong)

What is the largest space angle of any star? What is the best angular resolution attainable by any current telescope? Is there any theoretical limit to angular resolution attainable by telescopes?

So is it possible to take a picture of a star, where the star is not just a single point or pixel?

Update: The spot size is not due to charge overflow - see answer by Chris White.

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Infrared interferometry reconstructed image of Betelgeuse - with indication of sunspots. –  Mark Rovetta Jun 11 '13 at 16:55
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Related: "Apart from the Sun, the star with the largest angular diameter from Earth is R Doradus, a red supergiant with a diameter of $0.05$ arcsecond. Because of the effects of atmospheric seeing, ground-based telescopes will smear the image of a star to an angular diameter of about $0.5$ arcsecond; in poor seeing conditions this increases to $1.5$ arcseconds or even more." Source –  Glen The Udderboat Jun 11 '13 at 17:02
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The rays are indeed diffraction spikes. To my untrained eye they appear to be the result of support struts in the optical path (such as the case with Hubble), rather than polygonal aperturing. See physics.stackexchange.com/q/35935/10851 –  Chris White Jun 11 '13 at 17:25

3 Answers 3

up vote 10 down vote accepted

In most cases you are right, stars are pointlike. They are spread over multiple pixels not because of charge overflow (this can be overcome with shorter exposures and/or better equipment; there really should never be overflow in the image) but because the point spread function (PSF) of the telescope is larger than a single pixel (which is a good thing; you want to oversample the image).

An ideal optical setup will take an infinitesimal point source and produce an image that is not a point but rather an Airy pattern, assuming the aperture is circular, as this is the Fourier transform of a disk. Based on the size of the central peak in the pattern, we often say the diffraction limit for a telescope is $$ \theta_\mathrm{DL} = \frac{1.22\lambda}{D}, $$ where $\lambda$ is the wavelength of the light being considered and $D$ is the diameter of the telescope. $\theta_\mathrm{DL}$ is the smallest angular size you expect to resolve in your image.

For visible light, $\lambda$ ranges from $4\times10^{-7}$ to $7\times10^{-7}$ meters. If a star like the Sun (diameter $1.4\times10^{9}\ \mathrm{m}$) were placed somewhere between $4$ and $1000$ light years ($3.8\times10^{16}$ to $9.5\times10^{18}$ meters) away, its angular size would range from $1.5\times10^{-10}$ to $3.7\times10^{-8}$ radians. The diameter of a telescope needed to see resolve this would range from $13\ \mathrm{m}$ (just at the limit of what we have built so far, for seeing a very close-by star in violet) to $5.8\ \mathrm{km}$ (for resolving a typical star in red light at a distance of about $1\%$ the diameter of the galaxy).

All that being said, you can do better. Interferometry is the technique of using separated telescopes to beat the diffraction limit in a sense (it only cares about the furthest-separated points in the collecting area, not about how much of that area is actually filled in). Moreover, it might help to consider other wavelengths.

Also, stars can be very, very large, about the size of the inner Solar System in some extreme cases. As a result of all these considerations, we actually have resolved a small handful of stars. In fact, Wikipedia has a list of such objects (though I suspect it is not presently complete).

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Yes, though the only star that appears as a disk rather than a point is Betelgeuse. Even though it's 520 light years away, about 100 times farther than the closest star, it can just be resolved because it's a red supergiant and absolutely enormous!

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R Doradus seems to be ranked higher in terms of angular diameter. Why does that then not appear as a disk if Betelgeuse does? Is it because Betelgeuse is brighter? –  Glen The Udderboat Jun 11 '13 at 17:13
    
@Gugg: to be honest I don't really know. Possibly it's simply because R Doradus is in the southern hemisphere and traditionally all the big stellar interferometers have historically (I don't know I this is still the case) been in the northern hemisphere. Betelgeuse is a lot brighter than R Doradus, but I don't know how big a difference this makes for imaging studies. –  John Rennie Jun 11 '13 at 17:41
    
@Gugg They are almost the same brightness in IR. And the image here looks resolved...ish... Certainly we can directly measure the angular diameter of R Doradus, as described in this article. –  Chris White Jun 11 '13 at 18:00
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We took pictures of Betelgeuse using a 4m telescope in the early 90s with about 4-6 pixels across the star –  Martin Beckett Jun 12 '13 at 23:45

I think no.

The diameter of Riegel Kentaurus A has an average diameter of $1.4 \cdot 10^6\,\mathrm{km}$. Say it is only four lightyears away. Than the viewing angle would be $3.6995 \cdot 10^{-8}\,\mathrm{rad}$. If you want a picture with some 100 Pixels of that star, you will need a resolution of $10^{-11}\,\mathrm{rad}$. I think $\mathrm{\mu rad}$ is feasable, but not that kind of angular resolution.

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