<p>The picture isn't a &quot;colour&quot; picture - it is monochrome. i.e. It is obtained at a <em>single</em> microwave wavelength of 1.3 mm, and so not at any wavelength you could see (<a href="https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7" rel="noreferrer">Akiyama et al. 2019</a>).
There isn't therefore any spectral information that would reveal the expected Doppler effect. Any difference of colour in the &quot;false-colour picture&quot; is purely related to the <em>intensity</em> of the emission, not its wavelength.</p>
<p>If one were able to obtain coverage at multiple wavelengths then you might expect a &quot;bluer&quot; (i.e. shorter average wavelength) colour to be associated with the brightest regions.</p>

CC BY-SA 4.0
2020-12-16T12:45:39.247Z
Thank you! This makes sense. I always thought that, because a orangeish color was chosen, different wavelengths were still projected to visible color.
jng224
CC BY-SA 4.0
2020-12-17T12:19:52.757Z
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](https://en.wikipedia.org/wiki/Astronomical_interferometer) There is is no photography involved whatsoever.
Peter - Reinstate Monica
CC BY-SA 4.0
2020-12-16T12:41:01.860Z
2020-12-17T12:49:46.213Z
2020-12-17T12:49:46.213Z
ProfRob
<p>Last year, the first photo of a black hole in Messier 87 was published:</p>
<p><a href="https://i.sstatic.net/LLZRS.jpg" rel="noreferrer"><img src="https://i.sstatic.net/LLZRS.jpg" alt="enter image description here" /></a><br />
(Source: <a href="https://eventhorizontelescope.org/press-release-april-10-2019-astronomers-capture-first-image-black-hole" rel="noreferrer">EHT</a>)</p>
<p>It is quite obvious that about the lower half of the accretion disk is brighter.</p>
<p><a href="https://astronomy.stackexchange.com/questions/30343/why-are-there-uneven-bright-areas-in-this-photo-of-black-hole">This question</a> (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 &quot;Interstellar&quot; looked like:</p>
<p><a href="https://i.sstatic.net/nITWg.jpg" rel="noreferrer"><img src="https://i.sstatic.net/nITWg.jpg" alt="enter image description here" /></a><br />
(<a href="https://www.newscientist.com/article/dn26966-interstellars-true-black-hole-too-confusing/" rel="noreferrer">Source</a>)</p>
<p>So why is this not the case for the photo of the M87 black hole?</p>

CC BY-SA 4.0
2020-12-16T22:42:11.380Z
different but related and possibly of interest to future readers: [Do the actual false colours in the M87 black hole picture convey information?](https://astronomy.stackexchange.com/q/30350/7982) and [How does the black hole at the center of M87 look like in the visible spectrum?](https://astronomy.stackexchange.com/q/30476/7982) and [If a black hole does not emit light, how can one take a picture of the black hole itself?](https://astronomy.stackexchange.com/q/30424/7982)
uhoh
CC BY-SA 4.0
2020-12-17T06:34:40.530Z
The applicability of the term "photo" to astronomy varies from "somewhat" to "not at all", with most things (the Milky Way galaxy as a whole being one of the rare exceptions) outside of our solar system being  on the "not at all" side. This is a visual depiction of electromagnetic radiation, but that's about the extent to which the term "photo" is applicable.
Acccumulation
CC BY-SA 4.0
2020-12-17T14:21:41.807Z
*Interstellar* is a movie.  It is fiction.  It represents a guess at what a black hole *might* look like at much, much closer range than anything we're likely to see an image of in our lifetimes.  Possibly an educated guess, and maybe more accurate than previous depictions of black holes, but it's still just a guess, and quite likely exaggerated in order to make for more impressive visuals.  I wouldn't treat it as anything more than pure fantasy.
Darrel Hoffman
<black-hole><photography><supermassive-black-hole><doppler-effect><m87>
Why is there no color shift on the photo of the M87 black hole?
jng224
# Why is there no color shift on the photo of the M87 black hole?

(Asked by: username_0 on 2020-12-16)

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

[![enter image description here](https://i.sstatic.net/LLZRS.jpg)](https://i.sstatic.net/LLZRS.jpg)  
(Source: [EHT](https://eventhorizontelescope.org/press-release-april-10-2019-astronomers-capture-first-image-black-hole))

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

[This question](https://astronomy.stackexchange.com/questions/30343/why-are-there-uneven-bright-areas-in-this-photo-of-black-hole) (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](https://i.sstatic.net/nITWg.jpg)](https://i.sstatic.net/nITWg.jpg)  
([Source](https://www.newscientist.com/article/dn26966-interstellars-true-black-hole-too-confusing/))

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

**Tags:** black-hole, photography, supermassive-black-hole, doppler-effect, m87

> **Comments:**
> **username_1**: different but related and possibly of interest to future readers: [Do the actual false colours in the M87 black hole picture convey information?](https://astronomy.stackexchange.com/q/30350/7982) and [How does the black hole at the center of M87 look like in the visible spectrum?](https://astronomy.stackexchange.com/q/30476/7982) and [If a black hole does not emit light, how can one take a picture of the black hole itself?](https://astronomy.stackexchange.com/q/30424/7982)  
> **username_2**: The applicability of the term "photo" to astronomy varies from "somewhat" to "not at all", with most things (the Milky Way galaxy as a whole being one of the rare exceptions) outside of our solar system being  on the "not at all" side. This is a visual depiction of electromagnetic radiation, but that's about the extent to which the term "photo" is applicable.  
> **username_3**: *Interstellar* is a movie.  It is fiction.  It represents a guess at what a black hole *might* look like at much, much closer range than anything we're likely to see an image of in our lifetimes.  Possibly an educated guess, and maybe more accurate than previous depictions of black holes, but it's still just a guess, and quite likely exaggerated in order to make for more impressive visuals.  I wouldn't treat it as anything more than pure fantasy.  

---

## 1 answers

### Answer by username_4 on 2020-12-16. Score: 25 | Accepted answer

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](https://iopscience.iop.org/article/10.3847/2041-8213/ab0ec7)).
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.

> **Comments:**
> **username_0**: Thank you! This makes sense. I always thought that, because a orangeish color was chosen, different wavelengths were still projected to visible color.  
> **username_5**: 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](https://en.wikipedia.org/wiki/Astronomical_interferometer) There is is no photography involved whatsoever.  

---

