by raattgift | 2025-05-20 12:26:46+00:00
> accelerating observer

Note that this is proper acceleration which is measurable locally with e.g. a weighted spring, contact with a piezoelectric scale, or any other sort of accelerometer apparatus.

> a black hole reuires an accelerated observer to not be pulled in

There are an infinite number of free-fall trajectories outside a black hole, most of which never go near the black hole in the first place. There are also infinite numbers of hyperbolic orbits which "graze" a black hole and an infinity of various elliptical and (quasi-)circular orbits.

An accelerometer on one of these trajectories will report zero proper acceleration. Yet none of these trajectories cross the horizon from the outside.

There are additionally free-fall trajectories which cross the black hole horizon from the outside. Who knows what happens not long after that: flat-space physics like the Standard Model of Particle Physics curved spacetime physics like General Relativity give conflicting answers.

Finally there are also an infinity of trajectories which are somewhere properly accelerated. Most of these won't cross the horizon, but some can: one can turn on one's rocket engines and carefully steer a course that crosses the horizon.

(There are multiple types of horizon; the interesting one is the apparent horizon which can be measured locally with various types of apparatus. An event horizon -- if there is one -- can only be determined with reference to the entire global spacetime and all its contents from infinite past to infinite future. There's a variety of other horizons too. Visser catalogued some of them in <https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.12...>).

Note that trajectories in my paragraphs above correspond to everywhere non-spacelike curves of all sorts. Free-falling trajectories are geodesics (timelike or lightlike) and one can grind out the geodesic equation for the central mass in a Schwarzschild spacetime, for example.

Finally, proper acceleration is difficult to maintain for long, let along perpetually. (Although one can stand on the surface of a rocky planet and with an accelerometer measure over a very very very long term practically constant acceleration, and might for practical reasons want to assign that approximately constant acceleration to e.g. little "g"). So most curves where there is some proper acceleration also have some free-falling segments eventually.

One consequence is that a traveller who undergoes different accelerations (including none -- free-fall) along its path through spacetime will count different numbers of particles at different points along the traveller's sometimes-geodesic/sometimes-accelerated timelike curve.

A black hole doesn't really change this other than that the formation of some types of horizon induces an acceleration between freely-falling observers before the horizon forms and observers after. The later freely-falling observers see more particles than the earlier ones, and they bunch up not very far (but not preciesly on) the appropriate horizon.

The Unruh effect introduces an Unruh horizon attached to an observer undergoing proper acceleration. That observer sees more particles when it is accelerating than when it is/was freely-falling.

Some of the radiation is massless or has so little mass that it flies out to infinity. Around some types of horizon near a central mass, we'd call that Hawking radiation (as opposed to e.g. Unruh radiation, or radiation associated with e.g. a cosmic horizon in an expanding universe). And as Baez notes, over lonnnnnnnnnng times one would want to take into consideration how the central mass evolves as these particles fly away. Over much shorter durations (fuel is limited!) one would want to take into consideration how Unruh radiation and proper acceleration co-evolve.

The idea in the paper (which Baez dismantles in the linked blog entry at the top) is that no type of horizon is necessary for an apparently particle-free vacuum to look like a particle-rich patch of spacetime. That's a pretty wild claim. Why don't electrons evaporate on their own, in that case?