3 ms·

A black hole containing all matter in the universe wouldn't be much denser than the current average density of the universe. Spacetime expanding sorta helps kee

by zaarn 6y ago

A black hole containing all matter in the universe wouldn't be much denser than the current average density of the universe. Spacetime expanding sorta helps keep things from blowing apart. It's an interesting factoid that blackhole mass isn't related to their volume but their surface area, so by adding mass they grow a lot faster than you'd expect and density goes down fast. Larger black holes are barely denser than water.

CGamesPlay 6y ago
I was surprised by this and I don't think your sentence is entirely clear, but the ending figure seems to check out. I think it's more clear to say "the radius of a black hole increases sub-linearly with its mass" (specifically, the square root of mass). Sagittarius A*, the closest black hole, is 4.1 million solar masses, so its Schwarzchild radius is 12.1 million km, so its density is 0.25 kg/m^3, which is about 1/5 as dense as air at sea level--much less than water. A black hole made from Jupiter would have a Schwarzchild radius of 2.8 meters, so its density would be... denser than a neutron star.
- 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 ago
    Depends 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)
- marcosdumay 6y ago
  The relevant figure is that the radius grows faster than mass^(1/3). What it does, because it grows with mass^(1/2). If it grew more than linearly with mass, we wold see the same reducing density, just faster.

by zaarn 6y ago

CGamesPlay 6y ago

I was surprised by this and I don't think your sentence is entirely clear, but the ending figure seems to check out. I think it's more clear to say "the radius of a black hole increases sub-linearly with its mass" (specifically, the square root of mass). Sagittarius A*, the closest black hole, is 4.1 million solar masses, so its Schwarzchild radius is 12.1 million km, so its density is 0.25 kg/m^3, which is about 1/5 as dense as air at sea level--much less than water. A black hole made from Jupiter would have a Schwarzchild radius of 2.8 meters, so its density would be... denser than a neutron star.

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 ago
  Depends 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)
marcosdumay 6y ago
The relevant figure is that the radius grows faster than mass^(1/3). What it does, because it grows with mass^(1/2). If it grew more than linearly with mass, we wold see the same reducing density, just faster.