Comment by raattgift
1 day ago
The relevant quantities are the curvature scalars near the horizon, and for a sizable black hole they are small there. As an example, consider the Kretschmann scalar (KS). The KS is the sum of the squares of all components of a tensor. In Schwarzschild spacetime KS looks like R_{\mu\nu\lambda\rho}R^{\mu\nu\lambda\rho} = (48 G^2M^2)/(c^4r^6), where R is the Riemann curvature tensor, and we can safely set G=1 and c=1 so (48 M^2)/r^6. In this setting, KS is proportional to the spacetime curvature. At r = 2M, the Schwarzschild radius, the number becomes very small as we increase M, the black hole's mass. However, for any M at r = 0, the Kretschmann scalar diverges.
For a large-M black hole, there is "no drama" for a free-faller crossing the event horizon, as the KS gradient is tiny.
Since the crosser is in "no drama" free-fall he can raise his hands, toss a ball between his hands, throw things upwards above his head, and so forth. The important thing though is that all these motions are most easily thought of in his own local self-centred freely-falling frame of reference, and not against the global Schwarzschild coordinates. His local frame of coordinates is inexorably falling inwards. Objects moving outwards in his local frame are still moving inwards against the Schwarzschild coordinates.
You might compare with a non-freely-falling frame of reference. Your local East-North-Up (ENU) coordinates let you throw things upwards or eastwards, but in less-local coordinates your ENU frame of reference is on a spinning planet in free-fall through the solar system (and the solar system is in free-fall through the Milky Way, and the galaxy is in free-fall through the local group). That your local ENU is not a freely-falling set of coordinates does not change that the planet is in free-fall, and your local patch of coordinates is along for the ride.
A comparison here would be a long-running rocket engine imparting a ~ 10 m s^-1 acceleration to a plate you stand on. In space far from the black hole, you and the rocket engine would tend to move away from the black hole, but you'd be able to do things like juggle or jump up and down, and it'd feel like doing it on Earth's surface. This is a manifestation of the equivalence principle. Inside the horizon the rocket would still be accelerating the plate and you at ~ 10 m s^-1, but you, the plate, and the rocket would all be falling inwards.
No comments yet
Contribute on Hacker News ↗