Falling Into a Black Hole

Fall straight at a black hole from rest far away and three things happen to the view at once. Lensing bends the light of stars behind the hole around it, so a dark disc with a thin bright rim grows ahead of you. Your speed, which reaches the speed of light at the horizon, squeezes the whole sky towards the direction you are travelling, the same aberration that tilts rain towards a running observer. And every photon arrives with its frequency changed: the light you are rushing into is blueshifted, the light catching you up from behind is redshifted, and gravity blueshifts everything on top.^[1]

Loading the sky and tracing geodesics…

20.00 r / r_s

22% of c

×1.29 ahead

×0.82 behind

Distance from the centre — 20.00 r_sExposure — +0.0 stops

colour shift

Drag to look around. You are falling straight at the black hole from rest far away, and the stars are the real naked-eye sky with the hole placed towards the Galactic centre. Every star is drawn as a blackbody at the temperature its colour implies; what you see is that blackbody after the combined gravitational and Doppler shift for that direction, so its colour genuinely moves — red stars turn white, then blue, and brighten — rather than just getting brighter. Ahead, aberration crowds the whole sky into a shrinking patch and the shift runs away; behind, the universe reddens towards a factor of ½ at the horizon. The black disc is the set of directions whose light would have had to come out of the horizon. Untick colour shiftto see the same view with only the brightening, which is what most renderings show.

Most renderings of this journey show the first two effects and print a number for the third. Here the colours really move. Each star is treated as a blackbody at the temperature its colour index implies, and a blackbody whose light is shifted by a factor $g$ is exactly a blackbody at temperature $g\,T$ with the same brightness scale — so an orange star seen with $g = 2$ looks like a white one, and with $g = 4$ like a blue one, with no further modelling. Looking back, the shift settles towards $\tfrac12$ at the horizon: the universe behind you turns red and dim but never vanishes. Looking ahead it has no limit, which is why the forward view saturates and why the exposure slider exists.^[2] The stars are the real naked-eye sky, with the hole placed in the direction of the Galactic centre.^[3]

Deep dive · The shift factor, and how the picture is made

Work in units where the Schwarzschild radius $r_s = 1$ and $c = 1$ . An observer who fell from rest at infinity passes radius $r$ at speed

\beta = \sqrt{\frac{1}{r}}, \qquad \gamma = \frac{1}{\sqrt{1 - 1/r}},

relative to a static observer hovering there. A static observer already sees distant light blueshifted by $\gamma$ (the gravitational shift). The faller, moving at $\beta$ through that frame, adds a Doppler factor $\gamma\,(1 + \beta\cos\psi)$ , where $\psi$ is the angle between the photon's direction of travel and the outward radial. Together,

g \;=\; \frac{\nu_{\text{seen}}}{\nu_{\text{emitted}}} \;=\; \gamma^{2}\,(1 + \beta\cos\psi).

Light overtaking you from behind travels inward, $\cos\psi = -1$ , and $g = \gamma^2(1-\beta) = 1/(1+\beta) \to \tfrac12$ at the horizon. Light reaching you from ahead has been bent around the hole and is travelling outward, $\cos\psi = +1$ , and $g = 1/(1-\beta) \to \infty$ .^[4]

Where each pixel's light comes from. A pixel is a direction in the faller's frame. Special relativity's aberration formula turns it into a photon direction in the static frame. From there the photon's path is a Schwarzschild null geodesic, fixed by its impact parameter $b = r\sin\psi / \sqrt{1 - 1/r}$ . Traced backwards, it either came in from infinity (possibly looping the photon sphere first, which is what paints the thin rim) or it would have had to come up out of the horizon, which no light does — those directions are black. The deflection $\Delta\varphi = \int du\big/\sqrt{1/b^2 - u^2 + u^3}$ (with $u = 1/r$ ) is tabulated once for a grid of $r$ and $\psi$ , then looked up per pixel in a shader.

Why a blackbody stays a blackbody. The Planck spectrum is $B_\nu(T) \propto \nu^3 / (e^{h\nu/kT} - 1)$ , and the invariant along a light ray is $I_\nu/\nu^3$ . A photon emitted at $\nu$ arrives at $g\nu$ , so the observed spectrum is $g^3 B_{\nu/g}(T) = B_\nu(gT)$ : the same curve at temperature $gT$ . Brightness therefore scales as $g^4$ in total, while the colour follows the Planckian locus. The lookup table here integrates the Planck curve against the CIE colour matching functions for temperatures from 500 K to two million kelvin, so a 3000 K red giant at $g = 40$ is drawn as the 120 000 K object it would appear to be.

What is left out. The faller's own time runs normally, but this picture is a snapshot at each $r$ , not an account of how long each stage takes. Stars fainter than the naked-eye limit, the diffuse Milky Way, and interstellar dust are omitted, as is any accretion disc: this is a bare black hole in a clear sky. The hole is non-rotating.