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

  • 19
    $\begingroup$ 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...) $\endgroup$ Mar 4 at 15:52
  • 5
    $\begingroup$ yes :) maybe is the title perhaps "can we observe a black hole with the naked eye at the distance of 1 AU?" $\endgroup$
    – Mike M
    Mar 4 at 19:34
  • 21
    $\begingroup$ Like, apart from the fact that the bright fiery disc was replaced with total darkness? $\endgroup$
    – Vilx-
    Mar 4 at 23:48
  • 7
    $\begingroup$ If the question is "Is it visually detectable that the sun has been replaced by a black hole"?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! $\endgroup$
    – Stef
    Mar 5 at 12:42
  • 4
    $\begingroup$ 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. $\endgroup$
    – Vilx-
    Mar 5 at 16:04

4 Answers 4


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.


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, 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...

  • 1
    $\begingroup$ "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) $\endgroup$
    – Mazura
    Mar 4 at 19:17
  • 2
    $\begingroup$ 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. $\endgroup$
    – notovny
    Mar 4 at 19:54
  • 2
    $\begingroup$ 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). ssd.jpl.nasa.gov/api/… $\endgroup$
    – PM 2Ring
    Mar 5 at 2:13

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. (6Km vs 140 000 000km)

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 GrapefruitIsAwesome tells me that the the temperature of the hawking radiation from a black-hole of this mass wuld be 80 millikelvin, which is colder than the space between the stars in the sky. the sudden lack of warm sunlight will definately 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)

  • 3
    $\begingroup$ Are you sure this is correct? According to Hawking radiation the effective temperature of a solar mass back hole is only 60 nanokelvin. That wouldn't be bright at all. $\endgroup$ Mar 4 at 23:23
  • $\begingroup$ 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. $\endgroup$
    – Jasen
    Mar 4 at 23:31
  • 1
    $\begingroup$ You may enjoy playing with the 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. $\endgroup$
    – PM 2Ring
    Mar 5 at 1:03
  • 3
    $\begingroup$ @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. $\endgroup$
    – rob
    Mar 5 at 1:16
  • 1
    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Mar 5 at 6:55

ProfRob 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 ProfRob 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 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 causes it to cross α Leo.

  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Mar 4 at 21:10
  • $\begingroup$ Yes, think of the BH and its close vicinity as a raindrop scattering light. $\endgroup$
    – Rainald62
    Mar 4 at 21:17
  • 1
    $\begingroup$ My answer here 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. $\endgroup$
    – PM 2Ring
    Mar 5 at 2:25
  • 2
    $\begingroup$ 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? $\endgroup$
    – Kevin
    Mar 6 at 8:35
  • 1
    $\begingroup$ 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. $\endgroup$
    – ProfRob
    Mar 10 at 17:30

On May 29, 1919, Frank Watson Dyson and Arthur Stanley Eddington measured 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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .