Why the hole looks like that

What would you see if you looked straight at a black hole?

Light takes the straightest path there is. Near something heavy, the straightest path is curved, so a ray passing a heavy body turns a little toward it. You can measure this with the Sun. Starlight that grazes the Sun's edge bends by 1.75 arcseconds, less than a thousandth of a degree. In 1919 astronomers photographed stars next to the Sun during an eclipse, and their positions had shifted by about that much: a first confirmation that Einstein's picture of gravity was right.

Light bends much more strongly close to a black hole. A ray that passes 100 hole-radii away bends by 1.2 degrees. Pass 10 radii away and it bends by 14 degrees. Come closer and the bend grows until the ray goes all the way round and heads back the way it came. Come closer still and it never leaves.

That last fact is what you see when you look at a hole. Every ray that comes in too close is caught, and rays you trace backwards from your eye along those paths end up on the horizon, where nothing shines. So there is a shadow: a perfectly black disc, wider than the hole itself. Vesper's horizon is 30 kilometres in radius. This page works out the bending for a hole of Vesper's mass without spin, whose horizon would be 35 kilometres in radius. Its shadow is 92 kilometres in radius, 2.6 times as wide, because gravity drags in rays that would have missed.

Parallel rays pass a black hole from the left, each aimed at a different height, and every one bends toward the hole. The farther ones bend a little. One, aimed just outside the shadow edge (the larger dashed ring), winds partway round before escaping. The nearest, aimed inside it, spirals in. The smaller dashed ring is the photon sphere, where light can circle the hole.

A thin ring can appear near the shadow's edge. Rays that just barely escape have circled the hole before leaving, so a thin ring of light hugs the dark disc. That is the photon ring. And behind the hole, every star is smeared round it: gravity works like a lens, folding the sky behind the hole into arcs and a ring around the shadow.

Guess first Vesper's horizon is about 61 kilometres across. Say you look at the hole from 10,000 kilometres away. How big does its shadow look: a pinprick, about the size of a coin at arm's length, or bigger than the Moon looks from Earth?

Bigger than the Moon. For a hole of Vesper's mass without spin the shadow is 184 kilometres across, 2.6 times the width of that hole's horizon, and from that distance it spans 63 arcminutes. The Moon from Earth spans about 31. Vesper's shadow would be as wide as 2.0 Moons side by side, appearing dark inside.

The picture at the top shows what a probe in a 4,000 km orbit sees design. A hole of 12 Suns, smaller than a city, throws a shadow that covers 2.6 degrees of sky, as wide as 5.1 full Moons side by side. The shadow, the ring and the bent stars were traced ray by ray for every pixel, for a hole that does not spin. The colour and brightness of the glowing gas were set by eye styled.

How much does gravity bend light, and how big does Vesper look?

Far from a hole the bend is simple:

\text{bend} = \frac{4GM}{c^2\,b} = \frac{2\,r_s}{b}

G is Newton's gravitational constant, M the hole's mass and c the speed of light.
b is the impact parameter: how far from the hole's centre the ray would pass if gravity did not bend it.
r_s is the Schwarzschild radius, 2GM/c². It is the radius of the horizon, and a convenient ruler: writing b in units of r_s removes the mass from the formula. Whether the hole is the Sun or Vesper, a ray at 100 r_s bends by the same angle. (What is a black hole? shows where r_s comes from.)

The formula is exact only for large b. Closer in, the true bend is larger:

Ray passes at	Simple formula	Exact
1,000 r_s	0.115°	0.115°
100 r_s	1.15°	1.16°
10 r_s	11.5°	13.5°
5 r_s	22.9°	33.8°
3 r_s	38.2°	98.5°

Slide the toy below to explore how the deflection grows near the capture limit.

With JavaScript on, this is a calculator: choose a hole and how close the light passes, and it draws the ray's path and gives the bend. The table above gives the same numbers for five distances.

Where is the edge of the shadow?

A ray escapes only if it is aimed outside a critical impact parameter:

b_c = \frac{3\sqrt{3}}{2}\,r_s = \frac{\sqrt{27}}{2}\,r_s \approx 2.6\,r_s

For a hole of Vesper's mass without spin that is 92.1 kilometres. Inside it lies the photon sphere, at 3/2 r_s or 53.2 kilometres from the centre, where light itself can circle the hole. The orbit is unstable: a ray there falls in or flies off at the slightest nudge. Where ships can orbit safely is the subject of Where you can live around a hole and a star.

The shadow doubles as a target. Anything that wants to dump heat into the hole must aim at this disc, which for Vesper has an area of 2.7 × 10¹⁰ square metres, 27/4 times the horizon's own cross-section. That small window is why a hole makes a poor freezer; the full story is on The black hole is not your freezer.

How big is Vesper in the sky?

For a viewer far away, the shadow's angular size is its width over the distance. From Vesper's partner Ember, one orbital separation away, the shadow is 0.17 arcseconds across: a speck, and even its Einstein ring there is only 120 arcseconds in radius. Ember, seen from Vesper, is a disc 14 arcminutes across, about half the Moon's width from Earth. From 100,000 km, a distance picked for the example, it spans 6.3 arcminutes. At 4,000 km, the probe's orbit, it is 2.6 degrees.

The rest of the sky is bent too. A star exactly behind the hole turns into a full ring, the Einstein ring, whose angular radius for a source far behind a hole at distance d is

\theta_E = \sqrt{\frac{2\,r_s}{d}}

From 100,000 km it is 1.5 degrees in radius, a ring 3.1 degrees across, far bigger than the shadow it surrounds. At the probe's 4,000 km it reaches about 7.6 degrees, so only arcs of it fit in the picture. Stars a little off the line become two images, one outside the ring and a fainter one inside. Near the hole, every star in the sky is copied.

Does this match real holes?

The 2019 image of the black hole in the galaxy M87 gives a check. For its mass, 6.5 billion Suns, at its distance of 16.8 million parsecs, the formula predicts a shadow 39.7 microarcseconds across. The ring of glowing gas the Event Horizon Telescope measured was 42 ± 3 microarcseconds across, a little larger, as gas shining just outside the shadow should be. Because the team obtained the mass by matching the ring to simulations, the agreement shows only that the ring has the size the shadow geometry gives for that mass. These numbers are for a hole that does not spin. Vesper spins (0.7 of the most a hole can), which shifts its shadow a little to one side and flattens it slightly.

Open everything

Why does light curve that way, and where does the shadow come from?Null paths in Schwarzschild, a perturbation, and an unstable orbit

A ray's path as a one-line equation

In the spacetime around a non-spinning hole, a ray of light moves along a path where the spacetime interval is zero. It stays in one plane through the hole's centre, so we can use the radius $r$ and the angle $\varphi$ round the hole. Two quantities are conserved along the path: the energy $E$ and the angular momentum $L$ of the light. Their ratio $b = L/E$ is the impact parameter, because far from the hole it is exactly the distance by which the ray would miss the centre.

Putting the zero interval together with those two conserved quantities gives the radius as a function of angle:

$\left(\frac{dr}{d\varphi}\right)^2 = \frac{r^4}{b^2} - r^2\left(1 - \frac{r_s}{r}\right)$

Change variable to $u = r_s / r$ and measure $b$ in units of $r_s$ . The equation becomes

$\left(\frac{du}{d\varphi}\right)^2 = \frac{1}{b^2} - u^2 + u^3$

Differentiating once more gives an equation in the form of an oscillator with a push:

$\frac{d^2u}{d\varphi^2} + u = \tfrac{3}{2}\,u^2$

Without the $u^2$ term the solution is a straight line, $u = \sin\varphi / b$ . The $u^2$ term is gravity.

The bend, to first order

Far from the hole $u$ is small, so treat the push as a correction. Write $u = \sin\varphi / b + \delta$ , with $\delta$ small. Then $\delta'' + \delta = \tfrac{3}{2}\sin^2\varphi / b^2 = \tfrac{3}{4b^2}(1 - \cos 2\varphi)$ . A solution is

$\delta = \frac{3}{4b^2} + \frac{1}{4b^2}\cos 2\varphi$

The ray reaches infinity where $u = 0$ . Near $\varphi = \pi$ , put $\varphi = \pi + \epsilon$ : then $\sin\varphi \approx -\epsilon$ , so $0 = -\epsilon/b + \delta(\pi) = -\epsilon/b + 1/b^2$ , which gives $\epsilon = 1/b$ . The ray leaves at angle $\pi + 1/b$ and, by symmetry, entered at $-1/b$ . The total bend is

$\alpha = \frac{2}{b} \quad\Longrightarrow\quad \alpha = \frac{2 r_s}{b} = \frac{4GM}{c^2 b}$

Carrying the correction to the next order gives

$\alpha \approx \frac{2}{b} + \frac{15\pi}{16}\,\frac{1}{b^2}, \qquad \frac{15\pi}{16} = 2.95$

with $b$ in units of $r_s$ . Books that measure lengths in $GM/c^2$ , half of $r_s$ , write the same term as $15\pi/4$ . Integrating the exact equation numerically reproduces it: at $b = 1000\,r_s$ , exact minus first-order, times $b^2$ , matches the coefficient to within 0.2 per cent, and that leftover is the size of the next term, $16/(3b^3)$ .

The exact bend and the shadow edge

The exact total sweep is $2\int_0^{u_0} du \big/ \sqrt{1/b^2 - u^2 + u^3}$ , where $u_0$ is the turning point, the smaller root of $u^2 - u^3 = 1/b^2$ . The bend is that sweep minus $\pi$ .

A turning point exists only if $1/b^2$ is no more than the largest value of $u^2 - u^3$ . That function peaks where its slope $2u - 3u^2$ vanishes, at $u = 2/3$ , that is $r = 3r_s/2$ , and the peak height is $4/27$ . So

$\frac{1}{b^2} \le \frac{4}{27} \quad\Longleftrightarrow\quad b \ge \frac{\sqrt{27}}{2}\,r_s$

If $b$ is smaller than $\sqrt{27}/2$ , the right-hand side of the path equation never reaches zero, the ray never turns round, and $u$ climbs to 1 at the horizon. That is the shadow's edge, and it is why the shadow's radius is 2.6 horizon radii. Charles Darwin, the physicist grandson of the naturalist, gave this capture limit in 1959. The radius $r = 3r_s/2$ itself is the photon sphere: a ray there has $du/d\varphi = 0$ and $d^2u/d\varphi^2 = 0$ at once, which is a circular path.

Why the photon ring has a stack of copies

Look at what happens to a ray skimming the circular path at $u_c = 2/3$ . Write $u = u_c + \delta u$ in $u'' + u = \tfrac{3}{2}u^2$ : the right-hand side expands to $\tfrac{3}{2}u_c^2 + 3u_c\,\delta u$ , the constant terms cancel, and what is left is

$\delta u'' = (3u_c - 1)\,\delta u = \delta u$

So a small error grows as $e^{\varphi}$ : the circular path is unstable, and the growth is $e^{\pi}$ for each half-lap. A ray that goes half a lap further needs a starting aim closer to the critical one by that factor, so successive images of the same far-away source are squeezed toward the critical curve by

$e^{-\pi} = 0.0432$

each, about 23 times narrower, and about that much fainter, than the one before. That is the photon subring sequence: one bright ring on top of a stack that shrinks exponentially. Darwin saw it too: every star in the field shows, besides its direct image, a series of faint ghosts lying just outside the shadow's edge. The factor depends on spin for a spinning hole.

How the dark disc looks from different places

A viewer who hovers at radius $r$ sees the shadow's edge at an angle $\theta$ from the direction of the hole, where $\sin\theta = (b_c / r)\sqrt{1 - r_s/r}$ and $b_c = (\sqrt{27}/2)\, r_s$ , a result J. L. Synge worked out in 1966. Far away this is $b_c / r$ , the small-angle size used above. At $r = 3r_s/2$ the right-hand side is exactly 1, so $\theta = 90^\circ$ : at the photon sphere the shadow fills half the sky. At the probe's 113 $r_s$ , the square-root factor moves the edge by only 0.4 per cent. The real probe orbits rather than hovers, and its speed tilts and squeezes its view a little; the traced picture includes that.

Lensing as a mirror: the Einstein ring

Take the weak bend $\hat\alpha = 2r_s/b$ and let $b = D_l\,\theta$ , where $D_l$ is the distance to the hole and $\theta$ the angle at which you see an image. Put a source at distance $D_s$ beyond it, and let $\beta$ be the angle at which you would see the source if the hole were absent. For small angles, geometry gives $\theta D_s = \beta D_s + \hat\alpha\,D_{ls}$ , with $D_{ls}$ the hole-to-source distance, so

$\theta - \beta = \frac{\theta_E^{\,2}}{\theta}, \qquad \theta_E^{\,2} = \frac{2 r_s D_{ls}}{D_l D_s}$

When the source is exactly behind the hole, $\beta = 0$ and every point on the circle $\theta = \theta_E$ is a solution: a ring. For a source far behind ( $D_{ls} \approx D_s$ ) this is $\theta_E = \sqrt{2 r_s / d}$ , the formula in the numbers stretch. For $\beta \neq 0$ the equation has two roots, $\theta_\pm = \tfrac{1}{2}\left(\beta \pm \sqrt{\beta^2 + 4\theta_E^{\,2}}\right)$ , one outside the ring and one negative: an image on the far side of the hole, inside the ring.

How a computer draws it

A picture of a black hole is made backwards. For each pixel the computer starts a ray at the camera, pointing the way that pixel looks, and follows it through the curved space. The tracer behind the picture at the top, built for M87 Descent, does not step the one-line equation itself but its twin in ordinary $x, y, z$ coordinates: the ray moves like a particle pulled toward the hole with acceleration $\tfrac{3}{2} r_s b^2 / r^4$ , $b$ being the ray's impact parameter, which gives exactly $u'' + u = \tfrac{3}{2}u^2$ . It advances with a fourth-order Runge-Kutta step, shorter near the hole, and watches for three endings. If the ray crosses the horizon, the pixel is black. If it escapes to a great distance, the pixel takes the colour of the star field in the direction it left. Each time it crosses the disc of hot gas, the pixel picks up some of that gas's light, so rays that circle and cross again collect it several times. Each pixel is sampled many times over, so the thin arcs near the shadow come out smooth. The shadow in the finished picture matches the analytic size to within 0.5 per cent.

Far from a hole, tracing every pixel is wasted work. The simulation is designed to switch there to a table, a lookup from impact parameter to total bend, loops near the photon sphere included, applied in a single pass over the star field; where Vesper is far away, its small bend and Ember's even smaller one simply add.

Who has seen it, and what is still argued?The first computed image, Interstellar, the Event Horizon Telescope and dark binaries

The first images

In 1979 Jean-Pierre Luminet computed what a non-spinning black hole with a thin glowing disc looks like to a distant observer slightly above the disc, and drew the result by hand from the computed points. The disc arches over the shadow, a thinner second image of it lies underneath, and one side is far brighter because that side moves toward the viewer. In 2015 Oliver James, Eugénie von Tunzelmann, Paul Franklin and Kip Thorne published how they drew a spinning hole for the film Interstellar. Their code followed bundles of rays instead of single rays, so stars and the disc stayed smooth, and it handled a camera close to a spinning hole, a case most earlier studies had skipped.

The Event Horizon Telescope

In 2019 the Event Horizon Telescope, a network of radio dishes working as one Earth-sized instrument, resolved a ring of glowing gas 42 ± 3 microarcseconds across around the hole at the centre of M87, with a dark depression in the middle. By matching it to simulations the team found an angular gravitational radius of 3.8 ± 0.4 microarcseconds and a mass of 6.5 ± 0.7 billion Suns, consistent with a spinning (Kerr) hole. In 2022 the same approach gave a ring 51.8 ± 2.3 microarcseconds across at the centre of the Milky Way, consistent with the hole of about 4 million Suns that stellar orbits there imply. The two holes differ in mass by more than a factor of a thousand, and their rings agree with general relativity at both ends.

Is the dark patch the shadow?

The observed dark patch depends on the emitting gas as well as the capture geometry. Samuel Gralla, Daniel Holz and Robert Wald pointed out in 2019 that the textbook shadow is what a hole looks like lit from every direction. For gas glowing near the hole, the size of the dark patch is set mostly by where the gas shines and how much its light is reddened climbing out. They also showed that the photon ring, from light that orbits near the hole before escaping and arrives near a ring-shaped critical curve on the image, brightens the image only slightly, too little to matter in the images of 2019. A wider "lensing ring", a small image of the back of the disc from light that makes only part of an orbit, can stand out more, but what dominates the picture is the gas itself.

Photon subrings

Michael Johnson and colleagues showed in 2020 that the subrings, though each is narrower and fainter than the last, leave strong and universal signatures on very long baselines, each subring needing a longer baseline than the one before. The images of 2019 and 2022 are too coarse to separate them. Measuring them would give precise values of a hole's mass and spin and a test of general relativity.

Dark holes seen through their effect on light and stars

Most black holes are dark, and there are two ways to find the quiet ones. One does not use lensing at all: watch a star wobble. Gaia, a space telescope that measures star positions to a tiny fraction of an arcsecond, caught a Sun-like star circling a dark companion of 9.62 ± 0.18 Suns, 480 parsecs away, every 185.6 days (Gaia BH1; Kareem El-Badry and colleagues, 2023). Gaia BH2 is a red giant with a dark companion of 8.9 ± 0.3 Suns on a 1,277-day orbit, 1.16 kiloparsecs away. Gaia BH3, announced by the Gaia Collaboration in 2024, is a hole of 32.70 ± 0.82 Suns on an 11.6-year orbit, 590 parsecs away, more massive than any other stellar black hole then known in the Milky Way. Vesper was found the same way, by Ember's wobble (A pair too wide to hear).

The other way is lensing. A lens of 7.1 Suns at 1.58 kiloparsecs, the event below, has an Einstein ring at most 6.0 milliarcseconds in radius, far too small to see as a ring. What can be measured is a background star brightening and its apparent position shifting as the dark lens passes in front. In 2022 Kailash Sahu and colleagues followed one such event with the Hubble Space Telescope and reported the first unambiguous detection of an isolated stellar-mass black hole: 7.1 ± 1.3 Suns at 1.58 ± 0.18 kiloparsecs. Casey Lam and colleagues, analysing the same event, saw a shift of more than 1 milliarcsecond but found a lens of 1.6 to 4.4 Suns, most likely a neutron star or a low-mass black hole, with significant tension between models fitted to the brightness and to the position. Przemek Mróz, Andrzej Udalski and Andrew Gould traced the disagreement to systematic errors, including faint light from companions of the background star, and found a black hole of 7.88 ± 0.82 Suns. In 2023 Lam and Jessica Lu reanalysed the event with new Hubble data, found their earlier positions biased by a bright star nearby, and got a black hole of about 6 Suns (1.0 below to 1.2 above). The three analyses after the first all find a black hole and differ in its mass.

Open edges

Spin shifts and flattens the shadow and changes the subring factor. The images of 2019 and 2022 only partly constrain how closely M87 and Sgr A* follow the Kerr model. Resolved subrings would provide a more precise test.