I am quite impressed with the recently published image of M87* black hole. It appears that this image achieves spatial resolution well above and beyond what was available until today. But how high this resolution is, exactly?

The article "First M87 Event Horizon Telescope Results. I. The Shadow of the Supermassive Black Hole" states that Event Horizon Telescope has theoretical diffraction-limit resolution of ~25 μas.

Another wiki article states that M87* has Swartzschild radius of 5.9×10-4 parsecs. Combined with the distance of 16.4 million parsecs, this gives me angular size of 14.8 microarceconds.

Even with the shadow of the black hole being 2.5 bigger that its radius, the resolution of a widely distributed picture seems to be much higher than 25 microarcseconds stated above. Is the image actually blown up well beyond its original resolution (i.e. becoming more a less an "artistic impression"), or Event Horizon Telescope is actually capable of higher resolutions (10 microarcseconds or less)?

M87* image

  • $\begingroup$ $\lambda$/R$_{Earth}$ in radians $\endgroup$
    – eshaya
    Apr 12, 2019 at 18:50

1 Answer 1


The resolution of an interferometer at a wavelength $\lambda$ is $\theta\sim\lambda/b$, where $b$ is the longest baseline (distance between two telescopes) in the array. According to the first of this week's six papers, with observations at $\lambda=1.3\text{ mm}$, the longest baseline was $b\approx10700\text{ km}$, giving us $\theta\sim25.1\mu\text{as}$. The team expects this limit to be lowered even further with the addition of more observatories.

The picture, of course, seems much finer. The above limit can be lowered a bit by algorithms like CLEAN or regularized maximum likelihood; this brings the resolution down to about $20\mu\text{as}$. This still doesn't explain why the image seems even finer than that figure; that comes from image synthesis techniques that smooth out the data, rather than leaving it in a coarser form.


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