General relativity · gravitational lensing

Black Hole Light Lab

Why can one thin accretion disk appear both above and below the black-hole shadow? Spacetime bends light around the black hole. Light from the far side of the disk curves over the top and under the bottom of the hole on its way to us, so the same thin disk shows up as several images around the shadow.

Every pixel launches a ray from the observer and integrates it backward through a nonspinning Schwarzschild spacetime. A ray can fall through the horizon, cross the accretion disk, or escape toward a gravitationally distorted celestial sphere.

Spacetime modelSchwarzschild

Shadow diameter

d_{\mathrm{sh}}

$6\sqrt{3}\,r_g\approx10.39\,r_g$

Viewing inclination

i

$i=4^{\circ}$

Disk inner edge

r_{\mathrm{in}}

$r_{\mathrm{in}}=6.0\,r_g$

Approaching-side shift

g_{\max}

$g_{\max}=0.73$

Physics tutorial

Imaging a black hole: the horizon is not what we see

BackgroundA black hole gives off no light of its own. A telescope records light from distant sources and from hot surrounding matter, arriving from directions that curved spacetime has bent. The dark center is therefore not a photograph of the horizon surface; it is the black-hole shadow, cast by light paths that are captured.

Why it mattersTelling apart the shadow, the photon sphere and the multiple disk images turns a spectacular picture into a measurement of spacetime geometry.

Start with the essentials

Focus question: Why can a geometrically thin, nearly edge-on accretion disk appear above, below, and close to the center of the black-hole shadow?
One-sentence intuition: Each pixel on the screen matches one null geodesic arriving at the observer. Tracing it backward tells you whether the light came from the celestial sphere, crossed the disk or fell through the horizon.

Core mathematical model

Null geodesic of light

k^\nu\nabla_\nu k^\mu=0

The photon four-wavevector is parallel transported along itself. Integrating the spatial projection of this equation numerically gives the bent path for every pixel.

Schwarzschild black-hole shadow

d_{\mathrm{sh}}=6\sqrt{3}\,r_g\approx10.39\,r_g

Here $r_g=GM/c^2$ . The shadow is larger than the event horizon because critical rays are strongly bent and captured before they reach it.

Observed frequency-shift factor

g=\frac{\sqrt{1-2r_g/r}}{\gamma\left(1-v_{\mathrm{los}}/c\right)}

The numerator gives the gravitational redshift, and the denominator gives the relativistic Doppler shift from orbital motion. The approaching side is usually brighter and bluer.

Innermost stable circular orbit

r_{\mathrm{ISCO}}=6\,r_g

For a nonspinning Schwarzschild black hole, no stable circular orbit exists inside this radius, so the default disk begins here.

Common difficulties

The dark region is not a horizon photograph

Typical misconceptionThe rim of the central dark circle is the event horizon, so measuring it directly gives the horizon size.

Better mental modelThe dark circle is a shadow formed by critical light paths. For a Schwarzschild black hole its diameter is about ten point four gravitational radii, substantially larger than the horizon diameter.

The bright ring is not a fixed material loop

Typical misconceptionThe narrow light around the shadow is a physical tube of glowing matter at one fixed location.

Better mental modelRays close to the critical path travel around the black hole longer and can pick up light from the background or the disk more than once. The narrow brightening comes mainly from this structure of light paths, not from a ring of matter.

Run the experiment

01
Begin with a grazing view
Choose Grazing cinema view and compare the direct disk image with the higher-order images lifted above and below the shadow.
What to observe: One flat disk seems to wrap around the black hole because light from its far side bends into the observer’s line of sight.
02
Raise the viewing inclination
Increase the inclination until the disk is seen more directly.
What to observe: The upper and lower arcs merge toward the direct image, and relativistic Doppler beaming makes the approaching side brighter than the receding side.
03
Move the disk inner edge
Move $r_{\mathrm{in}}$ outward from the innermost stable circular orbit.
What to observe: The center gets dimmer and a wider dark gap opens between the disk and the critical curve, but the shadow size, which the geometry sets, stays fixed.