- #1
Kekkuli
- 9
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I find it interesting that the more massive the black hole, the weaker the fall acceleration at the distance of the Schwarzschild radius - that's why you wouldn't necessarily notice anything special in the event horizon.
##GM / R^2## is not "the fall acceleration" except in a very technical sense: it is the "redshifted" proper acceleration of an observer "hovering" at ##R##. So, for example, if an observer at infinity were holding up an object at ##R## using a very long rope, ##GM / R^2## is the force per unit mass that the observer at infinity would have to exert on the rope. But the object at ##R## would not experience that acceleration; the object's proper acceleration would be ##GM / (R^2 \sqrt{1 - 2GM / (c^2 R)})##.Kekkuli said:the fall acceleration at the distance of the Schwarzschild radius
No, it isn't. You wouldn't notice anything special falling through the horizon because spacetime is locally Lorentzian there just like it is everywhere else. It has nothing to do with "fall acceleration".Kekkuli said:that's why you wouldn't necessarily notice anything special in the event horizon
Saying "notice" you seem to think of tidal force or spaghettification. Yes, the larger the black hole the less you feel it, as already said in #4. The reason is tidal force goes with 1/M².Kekkuli said:I find it interesting that the more massive the black hole, the weaker the fall acceleration at the distance of the Schwarzschild radius - that's why you wouldn't necessarily notice anything special in the event horizon.