What am I looking at?

Details

• Black hole is a Schwarzschild, unit radius BH.
• Observer distance oscillates between ~5 and 10 Schwarzschild radii.
• Redshift is ignored.
• No relative velocity effects (focusing of celestial sphere/aberration). The algorithm assumes the observer is hovering (accelerating, of course) at fixed Schwarzschild \((r,\theta,\phi) \).
• Full aberration of light from orbital velocity. Observer is following a circular orbit around the BH. (The time delta is not to scale)

Controls

The simulation is interactive. This is what you can do:

• Click/Tap on simulation to set distance from black hole. Top of the canvas: photon sphere (1.5 Schwarzschild radii), bottom: 10 radii.
• Select whether to look directly at the black hole or to your right (ahead in your orbit). If you are wondering why the BH is not centered while orbiting, read on: it's a legitimate physical effect.
• Select a texture for the sky.
• Choose resolution or activate Fullscreen. Note that fullscreen mode uses the resolution you set, not that of your screen. Higher resolutions mean higher load on the GPU, of course.
• Select whether to orbit the black hole (inertial) or to push on your rockets to stay stationary at a fixed Schwarzschild \(r,\theta,\phi\) point.

Mathematical background

Our black hole is the simplest of its kind, a Schwarzschild BH, with zero charge and angular momentum:

I'm using, of course, units where \( c = 1 \) and \(r_S = 1 \). From here, one could derive the Christoffel symbols, write down the geodesic equation, and integrate it for every pixel. Easy, right? Of course not. This is a massive computation I have no intention of ever performing in my life. Other people have done full on geodesic raytracing (this guy is pretty awesome, but it's orthogonal to real-time; this one was off to a fantastic start, doing real-time 4D raytracing on the GPU, but got the Schwarzschild metric wrong when converting to cartesian coordinates! And this thing here is fun to play with, but seems very unaccurate, especially after taking a look at the code), but this seems very prohibitive, and very wasteful for the very symmetric Schwarzschild spacetime! Isometries of spacetime will correspond to constants of motion for free-fall geodesics. This means that the geodesic equation will take on a much simpler form.

In particular, we center our reference frame on the (temporarily massive) particle's orbital plane (\( \theta = \pi/2 \). Notice that through this we fixed one of the constants of motion, namely the direction of the pseudo-angular-momentum. We have two additional constants, which we can derive directly from the metric: \[ \left( 1 - \frac{1}{r} \right) \frac{dt}{d\tau} = e \] \[ r^2 \frac{d\phi}{d\tau} = l \] These are recognized as the pseudo-energy and pseudo-angular momentum associated to the time-translation and phi-rotation Killing vectors. Plugging back these conservation laws in the metric yields the equation for the trajectory: \[ \left(\frac{dr}{d\phi}\right)^2 = \frac{r^4}{b^2} - \left( 1 - \frac{1}{r} \right) \left( \frac{r^4}{a^2} + r^2 \right) \] where \(b = l/e\) and \(a = l\); however, inspired by Newtonian tradition, we perform the very useful change of coordinates \(r \rightarrow u = \frac{1}{r} \): \[\left(\frac{du}{d\phi}\right)^2 = \frac{1}{b^2} - (1-u)\left(a^{-2} + u^2\right) \] Finally, we send \(a \rightarrow \infty \) to obtain the trajectory for a massless particle such as a photon. We encounter no difficulties with this limit, since we have factored out proper time from our equation.

And here, I recognized that the equation above was of the form of an energy conservation equation for a standard, Newtonian mechanical system, with \(\frac{du}{d\phi}\) in the role of velocity. LHS is kinetic energy, RHS is minus potential. The corresponding Newton's equation is: \[ u''(\phi) = - u \left( 1 - \frac{3}{2} u^2 \right) \] Which is so satisfyingly simple, and took a relatively minuscule amount of work. This second order equation is preferrable to the above first order equation because the latter is a numerical nightmare rich with square roots, inflection points, and ambiguities. The \(u''\) equation must be understood as the real equation of motion for \(u(\phi)\), while the \(u'^2 \) one is akin to a Hamiltonian conservation law, the implicit equation of orbits in the \(u,\pi \) plane - \(\pi\) being the conjugate momentum to \(u\).

Strong emphasis must be put on the fact that the equation's simplicity is dependant on the spacetime's simmetries.

Circular Orbit

Circular orbits of massive objects are possible at any radius above the photon sphere \(\left(\frac{3}{2} r_S \right)\) - though they're only stable when \(r>3 r_S\), but we'll allow our observer to also force himself on these unstable orbits. I define the orbital speed as: \[ \beta = r \frac{d\phi}{dt} = \frac{1}{\sqrt{2}} (r-1)^{-\frac{1}{2}} \] Where the latter result's derivation is sketched, for example, here. This velocity actually is the relative velocity between a fixed-Schwarzschild hovering observer and a circularly orbiting observer when they pass through the same event.

This ultimately implies that we can obtain the view of the orbiting observer just by correcting for the local Lorentz transformation the data we already acquired for the Schwarzschild observer. Namely, we only need to apply aberration, or relativistic beaming (redshift is ignored for clarity). Incoming photons in our eyes have their 4-momenta transformed as (in local inertial cartesian coordinates): \[ k^\mu = (k^0,\vec k) \rightarrow k'^\mu = \Lambda^\mu_\nu k^\nu = \] \[ = \left(\gamma k^0 + \beta k_3, \left(\beta k^0 + \gamma k^1\right), k^2, k^3\right) \] (I assumed the left-right direction, that "tangent" to the orbit, was the x direction). The change in in energy is the redshift/blueshift that we ignore. The change in momentum instead is aberration and affects the direction from which we see the light ray come. To compute the final momentum, we need the original energy, but since these are photons, we know that \[ k^\mu k_\mu = 0 \Rightarrow k^0 = \left| \vec k \right| \] (We can use \( g_{\mu\nu} = \eta_{\mu\nu} \) since we are expressing \(k\) and \(k'\) in local inertial cartesian coordinates). This allows us to to correct for aberration.

Implementation

I'll be very schematic:

Photon orbits starting at the photon sphere (r=1.5) with varying directions. Note therefore that the circular orbit at the photon sphere is unstable.

Scattering of incoming photons from infinity with various impact parameters. Note how the "cross section" for photon absorption is much larger than the event horizon. Consequently, the image of the horizon will be easily more than twice as big than the horizon itself, and all of its surface is simultaneously visible.

The deflection angle map. Colour bands are due to the coding algorithm (1 colour channel is not precise enough)

F.A.Q.