Last year, the first photo of a black hole in Messier 87 was published:

enter image description here
(Source: EHT)

It is quite obvious that about the lower half of the accretion disk is brighter.

This question (or rather, the asnwers) explain that it is caused by Doppler beaming. Since Doppler beaming partly constitutes of Doppler effect, I would expect to see a blueshift on the brighter areas and a redshift on the fainter areas; similarly to what the original rendering of the black hole from the movie "Interstellar" looked like:

enter image description here

So why is this not the case for the photo of the M87 black hole?


1 Answer 1


The picture isn't a "colour" picture - it is monochrome. i.e. It is obtained at a single microwave wavelength of 1.3 mm, and so not at any wavelength you could see (Akiyama et al. 2019). There isn't therefore any spectral information that would reveal the expected Doppler effect. Any difference of colour in the "false-colour picture" is purely related to the intensity of the emission, not its wavelength.

If one were able to obtain coverage at multiple wavelengths then you might expect a "bluer" (i.e. shorter average wavelength) colour to be associated with the brightest regions.

  • $\begingroup$ Thank you! This makes sense. I always thought that, because a orangeish color was chosen, different wavelengths were still projected to visible color. $\endgroup$
    – jng224
    Dec 16, 2020 at 12:45
  • 1
    $\begingroup$ You could make it crystal clear to the casual reader that this wavelength is not visible light but radiation on a radio frequency (actually close to infrared, but considered radio in the publication). The image was created with radio interferometry: Signals from several telescopes are digitally processed and combined There is is no photography involved whatsoever. $\endgroup$ Dec 17, 2020 at 12:19

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