<p>A black hole with the mass of the sun is much smaller than the diameter of the sun, by a factor of about 200 000 (6 km vs 140 000 000 km).</p>
<p>The sun is, to the naked eye on earth, about the same size as the moon. We see this at every solar eclipse. If it got smaller, people would notice.</p>
<p>Even if they didn't look at the disc, people would see sharper shadows, because the black-hole sun is now basically a point source.</p>
<p>Also the spectrum and intensity of sunlight would change, that would be very obvious.</p>
<p>User GrapefruitIsAwesome tells me that the the temperature of the Hawking radiation from a black-hole of this mass would be 80 millikelvin, which is colder than the space between the stars in the sky.  The sudden lack of warm sunlight will definitely be noticed.</p>
<blockquote>
<p>distortions of the background starfield?</p>
</blockquote>
<p>Well yes, but not with the naked eye, a telescope would be needed. But the elephant in the room is that these distortions are only seen where the sun's disc would normally be. Outside the current radius of the sun the distortions are the same as they ever were (because the mass inside that sphere is the same).</p>

CC BY-SA 4.0
2022-03-04T23:23:27.937Z
Are you sure this is correct? According to [Hawking radiation](https://en.wikipedia.org/wiki/Hawking_radiation) the effective temperature of a solar mass back hole is only 60 nanokelvin.  That wouldn't be bright at all.
GrapefruitIsAwesome
CC BY-SA 4.0
2022-03-04T23:31:09.123Z
ooh, not sure at sll.I was mainly focussed on the change of diameter.  at 80nK how can it evaporate, that's colder than CBR.
Jasen
CC BY-SA 4.0
2022-03-05T01:03:13.153Z
You may enjoy playing with the [Hawking radiation calculator](https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator). The smallest stellar mass BHs are around 3 solar masses, which have a temperature around 20 nanokelvins, so they currently absorb far more heat than they radiate. It will be a *long* time before the CMB temperature drops low enough for them to start evaporating (around $10^{30}$ years, IIRC) and even longer for the SMBHs. OTOH, that's not really relevant, given how long it takes for them to evaporate.
PM 2Ring
CC BY-SA 4.0
2022-03-05T01:16:21.040Z
@Jasen It can't. For black holes which are colder than their surroundings, including the CMB, the competition between Hawking radiation and radiation absorption moves energy into the black hole, making it colder. Hawking radiation doesn't become a mass-loss mechanism until the CMB is redshifted to be colder than the black hole.
rob
CC BY-SA 4.0
2022-03-05T06:55:03.280Z
Your last point is a nice one to make, but of course you can distinguish/resolve things that have angular sizes much smaller than the disc of the Sun.
ProfRob
CC BY-SA 4.0
2022-03-04T22:26:01.390Z
2022-11-08T12:04:55.267Z
2022-11-08T12:04:55.267Z
Jasen
<p>ProfRob calculated correctly but interpreted wrongly. Any star off the BH by 0.1° would also be seen thus close (4.7&quot;) to the BH. If Mercury or Venus are in opposition and close to a node, the secondary image could be bright enough.</p>
<p>Added after ProfRob largely expanded his answer and asked for expanding mine.</p>
<p>First, let's quote and correct:</p>
<blockquote>
<p>If we argue that an angular deviation needs to be about 0.1 degree to
see with the naked eye, then β=4.7 arcseconds. i.e. The source needs
to get within 4.7 arcseconds of the black hole Sun's position for the
brighter primary image to be displaced by 0.1 degrees.</p>
</blockquote>
<p><span class="math-container">$θ_S =$</span> 4.7&quot; is small compared to <span class="math-container">$θ_E =$</span> 41&quot;, the radius of a potential Einstein ring, which would be the image of a star in line with and far behind the BH. The two images of a star 4.7&quot; off the BH would thus be off the BH by about <span class="math-container">$θ_E$</span>, slightly in- and outside, respectively.</p>
<p>The correct interpretation of the calculation yielding 4.7&quot; from 0.1° (I got 4.61&quot;) is that a star off the BH by 0.1° is seen at two positions: A primary (bright) image that is close to the true position, away from the BH, and a tiny secondary image close to the BH, away from the star, where &quot;close&quot; means off by 4.6&quot;.</p>
<p>As <span class="math-container">$u = θ_S/θ_E =$</span> 0.1°/41&quot; is only about 9, the secondary image is only two magnitudes dimmer than the star, a factor of <span class="math-container">$2/\sqrt{u^2(u^2+4)} = 0.025,$</span> after <a href="https://arxiv.org/pdf/1604.06601.pdf" rel="nofollow noreferrer">The Scales of Gravitational Lensing</a> by De Paolis <em>et al.</em> (page 8, right below Fig. 3 illustrating the two situations described above).</p>
<p>Venus in opposition has a brightness of −3.2 mag. Its secondary image would be much dimmer than the −1.2 mag calculated by the formula above, because Venus is not far behind the BH, but surely visible by the naked eye after ~11 minutes of dark adaptation, the round-trip time between the Sun and Venus.</p>
<p>Venus, however, is not likely in opposition close to a node when it happens.
Let's imagine a fusion-powered colony allowing observations for tens of thousands of years after the event, as precession of Earth's orbital plane about the <a href="https://en.wikipedia.org/wiki/Invariable_plane" rel="nofollow noreferrer">invariable plane</a> causes it to cross <a href="https://en.wikipedia.org/wiki/Regulus" rel="nofollow noreferrer">α Leo</a>.</p>

CC BY-SA 4.0
2022-03-04T21:10:17.273Z
This is presumably because you can trace the light rays back the other way? I must admit I'm struggling to visualise what's going on.
ProfRob
CC BY-SA 4.0
2022-03-04T21:17:16.770Z
Yes, think of the BH and its close vicinity as a raindrop scattering light.
Rainald62
CC BY-SA 4.0
2022-03-04T21:25:34.350Z
Yes, but direct light from the star to the observer would hardly be deviated. SO I think what you are saying is that there would be a unresolved (to the naked eye) point of light associated with the black hole which would be scattering light in our direction from all the point sources in the sky. Isn't this the photon ring I referred to in my answer?
ProfRob
CC BY-SA 4.0
2022-03-04T22:06:01.157Z
The photon ring is exactly that, scattering from all over the sky. The small-angle scattering happens outside the ring. The smaller the scattering angle, the farther away and the larger is the image (at constant surface brightness). Note: The average surface brightness of the ring is the average brightness of the sky.
Rainald62
CC BY-SA 4.0
2022-03-05T02:25:01.257Z
My answer [here](https://physics.stackexchange.com/a/680961/123208) has code running on the SageMathCell server that plots photon trajectories around a Schwarzschild black hole. But it's intended for close-up plots, not viewing the BH from 50,000,000 times the Schwarzschild radius.
PM 2Ring
CC BY-SA 4.0
2022-03-05T07:46:03.977Z
I now understand. You are talking about a secondary image inside the Einstein radius. Have you worked out what the very large demagnification factor is for a source with a true position at 15 times the Einstein radius?
ProfRob
CC BY-SA 4.0
2022-03-06T08:35:08.997Z
If the sun is not radiating visible light (to my understanding, Hawking radiation is generally not visible), can you still see Mercury and Venus with the naked eye?
Kevin
CC BY-SA 4.0
2022-03-07T05:34:40.807Z
@Kevin Hawking radiation is essentially a black body radiation at a very low temperature.  So all frequencies are in principle radiated, but the low temperature means that the vast majority starts well into the infrared, and then gets further redshifted from getting out of the gravity well to a distant observer.
zibadawa timmy
CC BY-SA 4.0
2022-03-10T13:40:20.617Z
There is a good additional answer here waiting to be epanded upon. It is true that there is a secondary image inside the Einstein ring as suggested. What is not discussed is that this secondary image tends to be much fainter - it is demagnified by a factor that increases the further away the primary image is from the Einstein ring at ~40 arcsec radius. Thus it may only be applicable to very bright stars that cross within say 0.1 degrees of the Sun (are there any?)
ProfRob
CC BY-SA 4.0
2022-03-10T17:30:22.253Z
As correctly pointed out in comments, if the Sun is a black hole then Venus and Mercury are effectively invisible and so cannot act as "stars" for the purposes of this image distortion.
ProfRob
CC BY-SA 4.0
2022-03-04T20:27:14.687Z
2022-03-15T03:13:35.387Z
2022-03-15T03:13:35.387Z
Rainald62
<p>On May 29, 1919, Frank Watson Dyson and Arthur Stanley Eddington <a href="https://en.wikipedia.org/wiki/Eddington_experiment" rel="nofollow noreferrer">measured</a> the gravitational lensing of the Sun. They had to do it during an eclipse because normally the sunlight would drown out the starlight. If the Sun had no radiation other than Hawking radiation, scientists could perform the experiment at any time. And given that the radius of the Sun would be much smaller, there would be rays that have even more pronounced of an effect.</p>

CC BY-SA 4.0
2022-03-07T03:56:36.210Z
2022-03-07T03:56:36.210Z
Acccumulation
<p>Yes, easily with a telescope, but not with the naked eye.</p>
<p>It is a matter of routine to detect the 1.7 arcsecond shifts caused to stellar positions when seen near to the limb of the Sun.</p>
<p>The removal of a photosphere while keeping the mass constant would mean that starlight could travel past the Sun with impact parameters all the way down to the minimum possible at 2.6 times the Schwarzschild radius. This produces deviation angles that can in principle be of any size - light can loop around and come back again or complete several orbits before escaping.</p>
<p>However, the big distortions happen within a few Schwarzschild radii of the black hole. The photon ring of the Sun, where light can undergo an unstable circular orbit, would only be 15 km in diameter. This subtends an angle of <span class="math-container">$0.02$</span> arcseconds at the Earth, several orders of magnitude below what is resolvable with the naked eye.</p>
<p>Edit:</p>
<p>Having said that, it is always possible that by careful observation, one might see a strange change in position of a naked eye star if the black-hole Sun passed close enough to it whilst it travelled along the ecliptic.</p>
<p>The lensing effect of a point mass can be characterised in terms of the <a href="https://en.wikipedia.org/wiki/Einstein_radius" rel="noreferrer">Einstein radius</a>, which for the scenario of a very distant star being viewed from the Earth via a black hole Sun, could be written as
<span class="math-container">$$\theta_E = \left(\frac{4GM_\odot}{c^2 \times 1 {\rm au}}\right)^{1/2} = 2\times 10^{-4}\ \ {\rm radians}$$</span></p>
<p>When a distant star is seen beyond a lens there are in general <em>two</em> images, one inside the Einstein ring and one outside. The angular deviation caused by the lens is <span class="math-container">$\alpha = \theta - \beta$</span>, where <span class="math-container">$\theta$</span> is the observed angular separation of the lensing object and the star and <span class="math-container">$\beta$</span> is the angular separation if there were no lensing effect. The two solutions for <span class="math-container">$\theta$</span> are given by
<span class="math-container">$$ \theta_{\pm} = \frac{\beta}{2} \pm \left(\frac{\beta^2}{4} + \theta_E^2\right)^{1/2}\ .$$</span>
If <span class="math-container">$\beta \gg \theta_E$</span> then we can approximate
<span class="math-container">$$ \theta_{\pm} = \beta + \frac{\theta_E^2}{\beta}\ \ \ {\rm or} \ \ \ -\frac{\theta_E^2}{\beta}$$</span>
The first solution indicates that the first (and brightest) image will get increasingly close to the unlensed position
<span class="math-container">$$ \alpha \simeq \frac{\theta_E^2}{\beta} = \frac{4\times 10^{-8}}{\beta}\ ,$$</span>
where <span class="math-container">$\alpha$</span> and <span class="math-container">$\beta$</span> are in radians. If we argue that an angular deviation needs to be about 0.1 degree to see with the naked eye, then <span class="math-container">$\beta = 4.7$</span> arcseconds. i.e. The source needs to get within 4.7 arcseconds of the black hole Sun's position for the brighter primary image to be displaced by 0.1 degrees.</p>
<p>This can be thought of a in a different way.
Let's say you could detect shifts of about 0.1 degree with the naked eye (that's about a fifth of the diameter of the full moon). For small angle deflections
<span class="math-container">$$\Delta \theta \simeq 4 \frac{GM}{bc^2}\ ,$$</span>
where <span class="math-container">$b$</span> is the impact parameter (roughly speaking, how close the light gets to the black hole) and <span class="math-container">$\theta$</span> is in radians.</p>
<p>If we let <span class="math-container">$\theta = 0.1 \times \pi/180$</span> radians, then <span class="math-container">$b=3400$</span> km. Thus star positions would be significantly deviated if they crossed within a region of angular radius <span class="math-container">$\sim 3400/1.5\times 10^{8}\times 180/\pi = 0.0013$</span> degrees or 4.7 arcseconds from the black hole Sun. i.e. The same result.</p>
<p>Thus we might expect to see major and perhaps visible deviations in the position of a star if the black hole Sun got within 5-10 arcseconds of it on the sky.</p>
<p>It would then be an exercise to see whether any sufficiently bright stars do pass within 5-10 arcseconds of the Sun's path along the ecliptic...</p>

CC BY-SA 4.0
2022-03-04T19:17:55.917Z
"not with the naked eye [unless] any sufficiently bright stars do pass within 5-10 arcseconds of the Sun's path along the ecliptic." (which just begs the question)
Mazura
CC BY-SA 4.0
2022-03-04T19:54:53.367Z
Quickly did a glance through in Stellarium over the morning; I certainly didn't see any eye-bright stars that close to 0° ecliptic latitude.
notovny
CC BY-SA 4.0
2022-03-05T02:13:33.737Z
The Sun's ecliptic latitude isn't exactly 0°, it can get 8 or so arcsecs away. Here's a dump for Greenwich (with no atmospheric refraction). https://ssd.jpl.nasa.gov/api/horizons.api?format=text&MAKE_EPHEM=YES&COMMAND=10&EPHEM_TYPE=OBSERVER&CENTER=000%40399&QUANTITIES=31&REF_SYSTEM=ICRF&CAL_FORMAT=BOTH&APPARENT=AIRLESS&CSV_FORMAT=YES&OBJ_DATA=NO&START_TIME=2021-Jan-1&STOP_TIME=2022-Jan-1&STEP_SIZE=7D
PM 2Ring
CC BY-SA 4.0
2022-11-07T20:57:35.020Z
Could we detect changes in *brightness* of a star (compared to nearby stars) as it gets magnified by the lens? That would help break up the boredom of walking to Hawaii.
Kevin Kostlan
CC BY-SA 4.0
2022-11-09T16:08:49.823Z
@PM 2Ring  In most years, the sun's greatest *geocentric* ecliptic latitude w.r.t. ecliptic of date is less than +/-1" and only in a minority of years reaches +/- ~1.05".  It fluctuates +(N) and -(S) roughly every month (see daily values in Astronomical Almanac w.r.t. ecliptic of date). I suspect that the figure of about 8" that you quoted comes about largely through the addition of parallax effects, because the file dump that you linked is not geocentric, but *topocentric* for Greenwich. In your data the intra-monthly fluctuations are seen mainly in the 4th d.p. in degrees.
terry-s
CC BY-SA 4.0
2022-11-09T16:43:44.657Z
@terry I figured that it was appropriate to use topocentric data, since the OP is asking about observable effects. It's a bit difficult to make observations from the geocentre. ,)
PM 2Ring
CC BY-SA 4.0
2022-03-04T07:50:40.627Z
2022-03-05T08:37:15.483Z
2022-03-05T08:37:15.483Z
ProfRob
<p>Assuming it had no accretion disk, could we still detect e.g. distortions of the background star field?</p>

CC BY-SA 4.0
2022-03-04T15:52:39.703Z
Well, for the first 8 minutes, we wouldn't even notice.  After that, we'd probably have bigger concerns, e.g. the impending end of all life on Earth.  Wouldn't be immediate, but wouldn't take very long either.  (I believe that question has already been asked a few times...)
Darrel Hoffman
CC BY-SA 4.0
2022-03-04T19:34:05.203Z
yes :)  maybe is the title perhaps "can we observe a black hole with the naked eye at the distance of 1 AU?"
Mike M
CC BY-SA 4.0
2022-03-04T23:48:24.367Z
Like, apart from the fact that the bright fiery disc was replaced with total darkness?
Vilx-
CC BY-SA 4.0
2022-03-05T06:59:20.610Z
@Vilx it would not be total darkness. There would be an unresolved photon ring which would mean the black hole Sun would look like a star unless you were much closer to it. I am not sure how bright it would be.
ProfRob
CC BY-SA 4.0
2022-03-05T12:20:14.663Z
@Vilx- Indeed. Seeing Sirius in the middle of the "day" in summer might give a hint that something's changed.
Eric Duminil
CC BY-SA 4.0
2022-03-05T12:42:37.457Z
If the question is "Is it visually detectable that the sun has been replaced by a black hole"? Then the answer is an obvious "yes" because a black hole emits a lot less light than the sun!
Stef
CC BY-SA 4.0
2022-03-05T16:03:13.257Z
@ProfRob - I tried to read about the "Photon sphere" but did not understand it. Where would this photon ring come from? A black hole doesn't produce light, right? It can only bend the paths of light rays going past it. Which means that whatever brightness it looks like, it cannot be brighter than the total brightness of the sky behind it. Which isn't much.
Vilx-
CC BY-SA 4.0
2022-03-05T16:04:47.380Z
If you took all the light from all the stars that you can see in the sky at night, it would still be way, way less than what the sun is sending our way - as is evidenced by the fact that it's dark at night.
Vilx-
CC BY-SA 4.0
2022-03-05T16:07:23.230Z
@Vlix it would not be "total darkness".
ProfRob
CC BY-SA 4.0
2022-03-07T03:55:45.740Z
Not sure its possible for such a BH to form naturally? The smallest Black Hole ever found is a little under 4x the Sun's mass.
T.E.D.
CC BY-SA 4.0
2022-03-24T19:10:32.707Z
IIRC, theory requires a mass of at least 1.4M☉ for a star to be capable of singularity collapse.  I don't know the specifics, such as the type of star nor at what point in its lifecycle it needs that much mass, but our Sol isn't a candidate.  FWIW.
RoUS
<black-hole><gravitational-lensing><hypothetical><detectors>
If the Sun was replaced with a sun-mass black hole, would it be visually detectable?
2080
# If the Sun was replaced with a sun-mass black hole, would it be visually detectable?

(Asked by: username_0 on 2022-03-04)

Assuming it had no accretion disk, could we still detect e.g. distortions of the background star field?

**Tags:** black-hole, gravitational-lensing, hypothetical, detectors

> **Comments:**
> **username_1**: Well, for the first 8 minutes, we wouldn't even notice.  After that, we'd probably have bigger concerns, e.g. the impending end of all life on Earth.  Wouldn't be immediate, but wouldn't take very long either.  (I believe that question has already been asked a few times...)  
> **username_2**: yes :)  maybe is the title perhaps "can we observe a black hole with the naked eye at the distance of 1 AU?"  
> **username_3**: Like, apart from the fact that the bright fiery disc was replaced with total darkness?  
> **username_4**: @username_3 it would not be total darkness. There would be an unresolved photon ring which would mean the black hole Sun would look like a star unless you were much closer to it. I am not sure how bright it would be.  
> **username_5**: @username_3- Indeed. Seeing Sirius in the middle of the "day" in summer might give a hint that something's changed.  
> **username_6**: If the question is "Is it visually detectable that the sun has been replaced by a black hole"? Then the answer is an obvious "yes" because a black hole emits a lot less light than the sun!  
> **username_3**: @username_4 - I tried to read about the "Photon sphere" but did not understand it. Where would this photon ring come from? A black hole doesn't produce light, right? It can only bend the paths of light rays going past it. Which means that whatever brightness it looks like, it cannot be brighter than the total brightness of the sky behind it. Which isn't much.  
> **username_3**: If you took all the light from all the stars that you can see in the sky at night, it would still be way, way less than what the sun is sending our way - as is evidenced by the fact that it's dark at night.  
> **username_4**: @Vlix it would not be "total darkness".  
> **username_7**: Not sure its possible for such a BH to form naturally? The smallest Black Hole ever found is a little under 4x the Sun's mass.  
> **username_8**: IIRC, theory requires a mass of at least 1.4M☉ for a star to be capable of singularity collapse.  I don't know the specifics, such as the type of star nor at what point in its lifecycle it needs that much mass, but our Sol isn't a candidate.  FWIW.  

---

## 4 answers

### Answer by username_4 on 2022-03-04. Score: 32 | Accepted answer

Yes, easily with a telescope, but not with the naked eye.

It is a matter of routine to detect the 1.7 arcsecond shifts caused to stellar positions when seen near to the limb of the Sun.

The removal of a photosphere while keeping the mass constant would mean that starlight could travel past the Sun with impact parameters all the way down to the minimum possible at 2.6 times the Schwarzschild radius. This produces deviation angles that can in principle be of any size - light can loop around and come back again or complete several orbits before escaping.

However, the big distortions happen within a few Schwarzschild radii of the black hole. The photon ring of the Sun, where light can undergo an unstable circular orbit, would only be 15 km in diameter. This subtends an angle of $0.02$ arcseconds at the Earth, several orders of magnitude below what is resolvable with the naked eye.

Edit:

Having said that, it is always possible that by careful observation, one might see a strange change in position of a naked eye star if the black-hole Sun passed close enough to it whilst it travelled along the ecliptic.

The lensing effect of a point mass can be characterised in terms of the [Einstein radius](https://en.wikipedia.org/wiki/Einstein_radius), which for the scenario of a very distant star being viewed from the Earth via a black hole Sun, could be written as
$$\theta\_E = \left(\frac{4GM\_\odot}{c^2 \times 1 {\rm au}}\right)^{1/2} = 2\times 10^{-4}\ \ {\rm radians}$$

When a distant star is seen beyond a lens there are in general *two* images, one inside the Einstein ring and one outside. The angular deviation caused by the lens is $\alpha = \theta - \beta$, where $\theta$ is the observed angular separation of the lensing object and the star and $\beta$ is the angular separation if there were no lensing effect. The two solutions for $\theta$ are given by
$$ \theta\_{\pm} = \frac{\beta}{2} \pm \left(\frac{\beta^2}{4} + \theta\_E^2\right)^{1/2}\ .$$
If $\beta \gg \theta\_E$ then we can approximate
$$ \theta\_{\pm} = \beta + \frac{\theta\_E^2}{\beta}\ \ \ {\rm or} \ \ \ -\frac{\theta\_E^2}{\beta}$$
The first solution indicates that the first (and brightest) image will get increasingly close to the unlensed position
$$ \alpha \simeq \frac{\theta\_E^2}{\beta} = \frac{4\times 10^{-8}}{\beta}\ ,$$
where $\alpha$ and $\beta$ are in radians. If we argue that an angular deviation needs to be about 0.1 degree to see with the naked eye, then $\beta = 4.7$ arcseconds. i.e. The source needs to get within 4.7 arcseconds of the black hole Sun's position for the brighter primary image to be displaced by 0.1 degrees.

This can be thought of a in a different way.
Let's say you could detect shifts of about 0.1 degree with the naked eye (that's about a fifth of the diameter of the full moon). For small angle deflections
$$\Delta \theta \simeq 4 \frac{GM}{bc^2}\ ,$$
where $b$ is the impact parameter (roughly speaking, how close the light gets to the black hole) and $\theta$ is in radians.

If we let $\theta = 0.1 \times \pi/180$ radians, then $b=3400$ km. Thus star positions would be significantly deviated if they crossed within a region of angular radius $\sim 3400/1.5\times 10^{8}\times 180/\pi = 0.0013$ degrees or 4.7 arcseconds from the black hole Sun. i.e. The same result.

Thus we might expect to see major and perhaps visible deviations in the position of a star if the black hole Sun got within 5-10 arcseconds of it on the sky.

It would then be an exercise to see whether any sufficiently bright stars do pass within 5-10 arcseconds of the Sun's path along the ecliptic...

> **Comments:**
> **username_17**: "not with the naked eye [unless] any sufficiently bright stars do pass within 5-10 arcseconds of the Sun's path along the ecliptic." (which just begs the question)  
> **username_18**: Quickly did a glance through in Stellarium over the morning; I certainly didn't see any eye-bright stars that close to 0° ecliptic latitude.  
> **username_11**: The Sun's ecliptic latitude isn't exactly 0°, it can get 8 or so arcsecs away. Here's a dump for Greenwich (with no atmospheric refraction). https://ssd.jpl.nasa.gov/api/horizons.api?format=text&MAKE_EPHEM=YES&COMMAND=10&EPHEM_TYPE=OBSERVER&CENTER=000%40399&QUANTITIES=31&REF_SYSTEM=ICRF&CAL_FORMAT=BOTH&APPARENT=AIRLESS&CSV_FORMAT=YES&OBJ_DATA=NO&START_TIME=2021-Jan-1&STOP_TIME=2022-Jan-1&STEP_SIZE=7D  
> **username_19**: Could we detect changes in *brightness* of a star (compared to nearby stars) as it gets magnified by the lens? That would help break up the boredom of walking to Hawaii.  
> **username_20**: @username_11  In most years, the sun's greatest *geocentric* ecliptic latitude w.r.t. ecliptic of date is less than +/-1" and only in a minority of years reaches +/- ~1.05".  It fluctuates +(N) and -(S) roughly every month (see daily values in Astronomical Almanac w.r.t. ecliptic of date). I suspect that the figure of about 8" that you quoted comes about largely through the addition of parallax effects, because the file dump that you linked is not geocentric, but *topocentric* for Greenwich. In your data the intra-monthly fluctuations are seen mainly in the 4th d.p. in degrees.  
> **username_11**: @terry I figured that it was appropriate to use topocentric data, since the OP is asking about observable effects. It's a bit difficult to make observations from the geocentre. ,)  

---

### Answer by username_9 on 2022-03-04. Score: 4

A black hole with the mass of the sun is much smaller than the diameter of the sun, by a factor of about 200 000 (6 km vs 140 000 000 km).

The sun is, to the naked eye on earth, about the same size as the moon. We see this at every solar eclipse. If it got smaller, people would notice.

Even if they didn't look at the disc, people would see sharper shadows, because the black-hole sun is now basically a point source.

Also the spectrum and intensity of sunlight would change, that would be very obvious.

User username_10 tells me that the the temperature of the Hawking radiation from a black-hole of this mass would be 80 millikelvin, which is colder than the space between the stars in the sky. The sudden lack of warm sunlight will definitely be noticed.

> distortions of the background starfield?

Well yes, but not with the naked eye, a telescope would be needed. But the elephant in the room is that these distortions are only seen where the sun's disc would normally be. Outside the current radius of the sun the distortions are the same as they ever were (because the mass inside that sphere is the same).

> **Comments:**
> **username_10**: Are you sure this is correct? According to [Hawking radiation](https://en.wikipedia.org/wiki/Hawking_radiation) the effective temperature of a solar mass back hole is only 60 nanokelvin.  That wouldn't be bright at all.  
> **username_9**: ooh, not sure at sll.I was mainly focussed on the change of diameter.  at 80nK how can it evaporate, that's colder than CBR.  
> **username_11**: You may enjoy playing with the [Hawking radiation calculator](https://www.vttoth.com/CMS/physics-notes/311-hawking-radiation-calculator). The smallest stellar mass BHs are around 3 solar masses, which have a temperature around 20 nanokelvins, so they currently absorb far more heat than they radiate. It will be a *long* time before the CMB temperature drops low enough for them to start evaporating (around $10^{30}$ years, IIRC) and even longer for the SMBHs. OTOH, that's not really relevant, given how long it takes for them to evaporate.  
> **username_12**: @username_9 It can't. For black holes which are colder than their surroundings, including the CMB, the competition between Hawking radiation and radiation absorption moves energy into the black hole, making it colder. Hawking radiation doesn't become a mass-loss mechanism until the CMB is redshifted to be colder than the black hole.  
> **username_4**: Your last point is a nice one to make, but of course you can distinguish/resolve things that have angular sizes much smaller than the disc of the Sun.  

---

### Answer by username_13 on 2022-03-04. Score: 3

username_4 calculated correctly but interpreted wrongly. Any star off the BH by 0.1° would also be seen thus close (4.7") to the BH. If Mercury or Venus are in opposition and close to a node, the secondary image could be bright enough.

Added after username_4 largely expanded his answer and asked for expanding mine.

First, let's quote and correct:

> If we argue that an angular deviation needs to be about 0.1 degree to
> see with the naked eye, then β=4.7 arcseconds. i.e. The source needs
> to get within 4.7 arcseconds of the black hole Sun's position for the
> brighter primary image to be displaced by 0.1 degrees.

$θ\_S =$ 4.7" is small compared to $θ\_E =$ 41", the radius of a potential Einstein ring, which would be the image of a star in line with and far behind the BH. The two images of a star 4.7" off the BH would thus be off the BH by about $θ\_E$, slightly in- and outside, respectively.

The correct interpretation of the calculation yielding 4.7" from 0.1° (I got 4.61") is that a star off the BH by 0.1° is seen at two positions: A primary (bright) image that is close to the true position, away from the BH, and a tiny secondary image close to the BH, away from the star, where "close" means off by 4.6".

As $u = θ\_S/θ\_E =$ 0.1°/41" is only about 9, the secondary image is only two magnitudes dimmer than the star, a factor of $2/\sqrt{u^2(u^2+4)} = 0.025,$ after [The Scales of Gravitational Lensing](https://arxiv.org/pdf/1604.06601.pdf) by De Paolis *et al.* (page 8, right below Fig. 3 illustrating the two situations described above).

Venus in opposition has a brightness of −3.2 mag. Its secondary image would be much dimmer than the −1.2 mag calculated by the formula above, because Venus is not far behind the BH, but surely visible by the naked eye after ~11 minutes of dark adaptation, the round-trip time between the Sun and Venus.

Venus, however, is not likely in opposition close to a node when it happens.
Let's imagine a fusion-powered colony allowing observations for tens of thousands of years after the event, as precession of Earth's orbital plane about the [invariable plane](https://en.wikipedia.org/wiki/Invariable_plane) causes it to cross [α Leo](https://en.wikipedia.org/wiki/Regulus).

> **Comments:**
> **username_4**: This is presumably because you can trace the light rays back the other way? I must admit I'm struggling to visualise what's going on.  
> **username_13**: Yes, think of the BH and its close vicinity as a raindrop scattering light.  
> **username_4**: Yes, but direct light from the star to the observer would hardly be deviated. SO I think what you are saying is that there would be a unresolved (to the naked eye) point of light associated with the black hole which would be scattering light in our direction from all the point sources in the sky. Isn't this the photon ring I referred to in my answer?  
> **username_13**: The photon ring is exactly that, scattering from all over the sky. The small-angle scattering happens outside the ring. The smaller the scattering angle, the farther away and the larger is the image (at constant surface brightness). Note: The average surface brightness of the ring is the average brightness of the sky.  
> **username_11**: My answer [here](https://physics.stackexchange.com/a/680961/123208) has code running on the SageMathCell server that plots photon trajectories around a Schwarzschild black hole. But it's intended for close-up plots, not viewing the BH from 50,000,000 times the Schwarzschild radius.  
> **username_4**: I now understand. You are talking about a secondary image inside the Einstein radius. Have you worked out what the very large demagnification factor is for a source with a true position at 15 times the Einstein radius?  
> **username_14**: If the sun is not radiating visible light (to my understanding, Hawking radiation is generally not visible), can you still see Mercury and Venus with the naked eye?  
> **username_15**: @username_14 Hawking radiation is essentially a black body radiation at a very low temperature.  So all frequencies are in principle radiated, but the low temperature means that the vast majority starts well into the infrared, and then gets further redshifted from getting out of the gravity well to a distant observer.  
> **username_4**: There is a good additional answer here waiting to be epanded upon. It is true that there is a secondary image inside the Einstein ring as suggested. What is not discussed is that this secondary image tends to be much fainter - it is demagnified by a factor that increases the further away the primary image is from the Einstein ring at ~40 arcsec radius. Thus it may only be applicable to very bright stars that cross within say 0.1 degrees of the Sun (are there any?)  
> **username_4**: As correctly pointed out in comments, if the Sun is a black hole then Venus and Mercury are effectively invisible and so cannot act as "stars" for the purposes of this image distortion.  

---

### Answer by username_16 on 2022-03-07. Score: 1

On May 29, 1919, Frank Watson Dyson and Arthur Stanley Eddington [measured](https://en.wikipedia.org/wiki/Eddington_experiment) the gravitational lensing of the Sun. They had to do it during an eclipse because normally the sunlight would drown out the starlight. If the Sun had no radiation other than Hawking radiation, scientists could perform the experiment at any time. And given that the radius of the Sun would be much smaller, there would be rays that have even more pronounced of an effect.


---

