The first-ever image of a black hole released this week was taken with radio telescopes, suggesting that it is a false-color image and doesn't represent what a person would see in reality, or visible light. This seems to be confirmed by this Wired article, but I am perplexed as to why this is not widely mentioned, including the announcement on the telescope's official website and Wikipedia (image file, article on black holes). Perhaps not mentioning it makes for better headlines, but it feels wrong.

What's going on here?


Yes, it's a false-color image. Radio telescopes don't detect visible light.

In the first of the six technical papers (which seem to be open access! they are listed in this summary article), the color scale is assigned a physical interpretation:

famous photo plus labels

The interpretation is at the bottom: it's some kind of an effective temperature, with dark being "cold" and the brightest being six billion kelvin. There's a huge amount of detail in the technical papers.

I decline to speculate on why news organizations do or don't say things. But it sure is less confusing to look at false-color images that use modern perceptually-uniform color mappings.

  • 2
    $\begingroup$ Note that brightness temperature has little to do with physical temperature, or at least not always. It's more closely related to flux density, i.e. the amount of radio emission. science.nrao.edu/facilities/vla/proposing/TBconv $\endgroup$ – Kyle Oman Apr 12 at 22:21
  • $\begingroup$ @KyleOman I appreciate the clarification; writing the intentionally vague "some kind of" was the best I could manage in the time that I had. $\endgroup$ – rob Apr 12 at 23:35
  • $\begingroup$ Thank you! False color image it is then. I personally think such images should always go together with a color scale legend like the one here. I would think most people don't know how a radio telescope works, and in this case really think that this is what they'd see if they observe the black hole with their own eyes. $\endgroup$ – Artyom Apr 13 at 21:24

suggesting that it is a false-color image and doesn't represent what a person would see in reality, or visible light

Pretty much every image you see in astronomy is not what you'd see in person.

Sometimes it is a composite image that adds colors (from non-visible wavelengths) or enriches existing colors; other times it's a false-color image because just how massive the electromagnetic spectrum is compared to the visible light band, as demonstrated in the opacity of the sky diagram:

enter image description here

There are a bunch of processes that give rise to optical emissions (H$\alpha$, O I, N II, He I, etc) and they sometimes can give great pictures, such as the Crab nebula:

enter image description here
(source--note that this is still a composite image, so it's still not what you'd see in person)

Other times you don't get much in optical, such as the case for Cassiopeia A:

enter image description here

But things can get pretty interesting when you look at it in other wavelengths (still Cass A),

enter image description here

What's happening is that other processes (synchrotron, metal transitions, inverse Compton, pion-decay (see my question here), bremsstrahlung, etc) occur at frequencies well outside the optical bands (i.e., are non-visible).

Since we're still getting data (e.g., fluxes) from the source, we still can present an image of what it is that the observatory has observed, even though it's not something someone would observe themselves if they were there (because we see in optical, not in radio/infrared/X-ray/$\gamma$-ray).

So at this point, the only choice is to represent the data (intensities, fluxes, etc.) along some color scale and display that.1 Hence, pretty much every astronomical image you see is "faked",2 at least in part.

1. If/when you do make images for public display from your data, please, as rob points out, don't use the 'rainbow heat map', take a look at Color Brewer instead.
2. In the sense that it's not actually what you'd see in person.

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    $\begingroup$ Can I suggest putting faked in scare quotes (that is "faked") to emphasize that you are going to address exactly what that word means. The point that these images are processed data is well worth making, but the word carries implications that the fact of processing is intrinsically problematic. $\endgroup$ – dmckee Apr 13 at 18:55
  • $\begingroup$ @dmckee i am begrudgingly obliging your request. This site is expressly for active researchers, academics and students of physics and astronomy, so I don't think it's a problem to use language that researchers, academics & students would understand but laypeople may not (an opinion, mind you, that does not give a lot of credence to the intelligence of laypeople and their reading comprehension). $\endgroup$ – Kyle Kanos Apr 13 at 19:22
  • $\begingroup$ Thanks for providing a fantastic background on astronomy images! I hope more "laypeople" would now be aware of this after seeing your answer :) $\endgroup$ – Artyom Apr 13 at 21:42

VLBI (like various synthetic aperture radar techniques and medical ultrasound) is a coherent imaging process, so many words used in standard photography are simple not applicable.

The article "VLBI From Ground and Space", E. Pruess, In ESA, Space Science and Fundamental Physics p 105-116, says:

"All radioastronomical interferometers work according to the same basic principles. They all measure primarily the coherence of a radio-wavefield by correlating complex wave amplitudes measured at two space-time points by antenna signals prior to quadratic detection (i.e. voltages and not intensity!)"

So color photographs reveal light intensity ($\propto ||E^2||$) at 3 frequencies, averaged over some time, usually matching the human eye's response. Anything else is "false color".

Black and white photos look just at intensity. (note: it's "false not color").

With VLBI you're looking at 2 signals that can be a thousand times smaller than the noise, and then filtering out their correlation to get a number at each pixel, which then assigns a relative value for an image. So yes, it is very "false".


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