If the Sun was replaced with a sun-mass black hole, would it be visually detectable?

Question

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

Answer 1

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, 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...

Answer 2

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 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.

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).

Answer 3

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.

Answer 4

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.