3 ms·
Hang on, it’s been a long time since I studied physics at uni in the mid 80s, but isn’t the Schwarzchild radius just the boundary which marks the point within
by phs318u 6y ago
Hang on, it’s been a long time since I studied physics at uni in the mid 80s, but isn’t the Schwarzchild radius just the boundary which marks the point within which even light cannot escape gravity? A black hole is a singularity, a point (if non-rotating) of infinite density. I didn’t think the mass of the back hole was evenly distributed within the Schwarzchild boundary (which is a mathematical boundary, not a physical boundary such as the surface of a body). Or have I missed something?
- zaarn 6y agoDepends on how you look at it. If you look into holographic theory (which I prefer personally) then the entire mass of a black hole is represented as information on the surface of the black hole, explaining the relation between mass and surface not mass and volume. This solves some problems with the information theorem (no information can be lost) by presenting all information contained within a black hole on it's surface where it can be read, atleast theoretically. (Here information and mass would be considered equivalent) To some extend, it still holds true if you just consider the mass of a black hole to be it's singularity and it's volume the volume that fills the event horizon. edit: Also consider; if you placed an equivalent amount of mass in space that occupies less volume than the event horizon would, it would automatically be a black hole (and note; the schwarzschild radius is not necessarily the event horizon, the event horizon can be tighter if the black hole rotates)