<p>You are right that the image is not a visible light image. It was taken using very, very short wavelength microwaves -- the wavelength is <em>1.3 mm</em> which is only a bit shorter than the Far IR!</p>

<p>I think you may be reading a bit too much into the word "photo" which, as you say, has the everyday meaning of "taken with a camera using lenses in visible light."  I think most astronomers would use the term "image" rather than "photo" -- if you look at the <a href="https://www.nsf.gov/news/news_summ.jsp?cntn_id=298276&amp;org=NSF&amp;from=news" rel="noreferrer">NSF's press release</a>, for example, you'll see that it uses the term "image".  (Links to all six of the papers the EHT team has published are in the NSF press release if you want further details.)</p>

<p>(I'd bet that "image" was changed to "photo" somewhere in the journalistic process as being "more meaningful" to the average reader.)</p>

<p>The appearance to the naked eye would probably not be terribly different, since the photon orbits don't depends on frequency.</p>

<p>As far as whether it got "boosted" or not, you're right to be very skeptical -- university publicity offices are prone to exaggerate and sometimes pump relatively ordinary science into amazingly important-sounding "breakthroughs."  In this case it all seems pretty straightforward, though. And there's nothing special about a visible-light image over a microwave image -- both are useful and neither tell us everything. (It is not uncommon for images in the visible to be among the least useful -- with X-ray or radio actually telling us much more.)</p>

CC BY-SA 4.0
2019-04-11T12:53:28.800Z
Thank you, this also had me interested in the fact that the appearance to the naked eye would look similar. Would the color be close to that one though?
Maurizio Carcassona
CC BY-SA 4.0
2019-04-11T13:04:05.290Z
The color would *probably* be that of a blackbody of whatever the temperature of the light-emitting region is.  At a guess it would be quite hot -- I haven't dug through the papers to see if they make any estimates, but this is a seriously energetic system so it probably is much hotter than the surface of the Sun -- and would thus be much whiter than the colors used for the image.
Mark Olson
CC BY-SA 4.0
2019-04-11T13:09:10.330Z
That's very interesting, thank you for you very rich answers!
Maurizio Carcassona
CC BY-SA 4.0
2019-04-11T16:52:57.267Z
I suspect an optical image would be nothing like this because of obscuration by dust.
ProfRob
CC BY-SA 4.0
2019-04-18T14:41:34.223Z
Even at those temperatures, dust would play such a big role?
Maurizio Carcassona
CC BY-SA 4.0
2019-04-11T12:01:53.437Z
2019-04-11T13:04:38.210Z
2019-04-11T13:04:38.210Z
Mark Olson
<p>I'm still at my bachelor of electronic and communications engineering, so I'm not an expert of radiotelescopes; however, I surely know that, typically, radio communications are characterized by a RX that has at least one bandpass filter. </p>

<p>As a result, I am led to think that the "photo" of the black hole, recently released to the public, is a reconstruction picture taken from various radiofrequency signal samples that got mapped to the frequencies of visible light spectrum (since the Event Horizon Telescope project is based on radio telescopes), hence it doesn't depict what we would actually see (neglecting any impossibilities in doing so), but rather, it shows the radio emission from whatever is close to the event horizon of the black hole. </p>

<p>Is this correct, or is it actually a photo capturing visible light?
I mean, scientifically, this is a huge step; however, I have the feeling that the news got "boosted" a little too much, in medias, as they just talk about "photos" and "light" surrounding a black hole - not specifying anything else for a inexperienced audience.</p>

CC BY-SA 4.0
2019-04-11T13:41:43.333Z
This is a related, but different question that was just migrated from [Space Exploration SE](https://space.stackexchange.com/questions/) to here just a few minutes ago. [What part of the EM spectrum was used in the black hole image?](https://astronomy.stackexchange.com/q/30362/7982) You may find it interesting as well.
uhoh
CC BY-SA 4.0
2019-04-11T13:59:52.317Z
@uhoh thank you! It's definitely interesting as well! (It's kinda annoying I can't upvote though, ughh)
Maurizio Carcassona
CC BY-SA 4.0
2019-04-11T14:10:12.427Z
I can see you enjoy astronomy and will probably ask and answer more questions, so your reputation (points) will soon pass the threshold, which [according to this](https://astronomy.stackexchange.com/help/privileges) is only 15!
uhoh
CC BY-SA 4.0
2019-04-11T15:24:32.880Z
Oh yeah, for some reason my reputation is higher now and I finally can :)
Maurizio Carcassona
<black-hole><radio-astronomy><supermassive-black-hole><radio-telescope>
Does the "photo of a black hole" taken by EHT actually depict a visible light-based scenario?
Maurizio Carcassona
# Does the "photo of a black hole" taken by EHT actually depict a visible light-based scenario?

(Asked by: username_0 on 2019-04-11)

I'm still at my bachelor of electronic and communications engineering, so I'm not an expert of radiotelescopes; however, I surely know that, typically, radio communications are characterized by a RX that has at least one bandpass filter.

As a result, I am led to think that the "photo" of the black hole, recently released to the public, is a reconstruction picture taken from various radiofrequency signal samples that got mapped to the frequencies of visible light spectrum (since the Event Horizon Telescope project is based on radio telescopes), hence it doesn't depict what we would actually see (neglecting any impossibilities in doing so), but rather, it shows the radio emission from whatever is close to the event horizon of the black hole.

Is this correct, or is it actually a photo capturing visible light?
I mean, scientifically, this is a huge step; however, I have the feeling that the news got "boosted" a little too much, in medias, as they just talk about "photos" and "light" surrounding a black hole - not specifying anything else for a inexperienced audience.

**Tags:** black-hole, radio-astronomy, supermassive-black-hole, radio-telescope
> **Comments:**
> **username_1**: This is a related, but different question that was just migrated from [Space Exploration SE](https://space.stackexchange.com/questions/) to here just a few minutes ago. [What part of the EM spectrum was used in the black hole image?](https://astronomy.stackexchange.com/q/30362/7982) You may find it interesting as well.  
> **username_0**: @uhoh thank you! It's definitely interesting as well! (It's kinda annoying I can't upvote though, ughh)  
> **username_1**: I can see you enjoy astronomy and will probably ask and answer more questions, so your reputation (points) will soon pass the threshold, which [according to this](https://astronomy.stackexchange.com/help/privileges) is only 15!  
> **username_0**: Oh yeah, for some reason my reputation is higher now and I finally can :)  

---

## 1 answers

### Answer by username_2 on 2019-04-11. Score: 7 | Accepted answer

You are right that the image is not a visible light image. It was taken using very, very short wavelength microwaves -- the wavelength is *1.3 mm* which is only a bit shorter than the Far IR!

I think you may be reading a bit too much into the word "photo" which, as you say, has the everyday meaning of "taken with a camera using lenses in visible light." I think most astronomers would use the term "image" rather than "photo" -- if you look at the [NSF's press release](https://www.nsf.gov/news/news_summ.jsp?cntn_id=298276&org=NSF&from=news), for example, you'll see that it uses the term "image". (Links to all six of the papers the EHT team has published are in the NSF press release if you want further details.)

(I'd bet that "image" was changed to "photo" somewhere in the journalistic process as being "more meaningful" to the average reader.)

The appearance to the naked eye would probably not be terribly different, since the photon orbits don't depends on frequency.

As far as whether it got "boosted" or not, you're right to be very skeptical -- university publicity offices are prone to exaggerate and sometimes pump relatively ordinary science into amazingly important-sounding "breakthroughs." In this case it all seems pretty straightforward, though. And there's nothing special about a visible-light image over a microwave image -- both are useful and neither tell us everything. (It is not uncommon for images in the visible to be among the least useful -- with X-ray or radio actually telling us much more.)

> **Comments:**
> **username_0**: Thank you, this also had me interested in the fact that the appearance to the naked eye would look similar. Would the color be close to that one though?  
> **username_2**: The color would *probably* be that of a blackbody of whatever the temperature of the light-emitting region is.  At a guess it would be quite hot -- I haven't dug through the papers to see if they make any estimates, but this is a seriously energetic system so it probably is much hotter than the surface of the Sun -- and would thus be much whiter than the colors used for the image.  
> **username_0**: That's very interesting, thank you for you very rich answers!  
> **username_3**: I suspect an optical image would be nothing like this because of obscuration by dust.  
> **username_0**: Even at those temperatures, dust would play such a big role?  

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