4 ms·

The crazy thing is the amount of stuff we are able to observe is constantly shrinking. As the universe expands faster and faster, eventually the stuff at the fa

The crazy thing is the amount of stuff we are able to observe is constantly shrinking. As the universe expands faster and faster, eventually the stuff at the far edges is moving away faster than the speed of light so we will never get there or even know it exists without records.

vbezhenar 6y ago
Sounds like we're in a black hole.
- maneesh 6y ago
  Or in a video game where the background is an unreachable layer.
- TeMPOraL 6y ago
  Ironically, I realized yesterday that if the universe was destined for the Big Crunch (until quite recently, this was believed to be a possibility), then we'd be literally living inside the event horizon of an universe-sized black hole!
- 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.
- lrem 6y ago
  For extra mind blowing, do compare the Schwarzschild mass of the observable universe and its radius. Spoiler: gurl ner abg whfg fvzvyne, gurer ner gurbergvpny erfhygf gung gurl unir gb or rdhny. Hasbeghangryl gur cncre vgfrys nffhzrq jnl zber culfvpf xabjyrqtr guna V unir. Ohg, guvf znxrf bhe havirefr pbafvfgrag jvgu gur fvzcyr qrsvavgvba bs oynpx ubyr ("n znff ragveryl jvguva vgf Fpujnemfpuvyq enqvhf").
  - grandchild 6y ago
    So I quickly did a back-of-envelope calculation using [0] for source of mass and diameter of the universe. Then I used the formula given in [1]: r = (2 G M) / (c^2). You have: (2 * G * 1.5 * 10^56 grams ) / (c^2) You want: lightyears * 2.3548373e+10 So about 23,548,373,000 ly of SR, while the radius of the observable universe is given as ~46,500,000,000 ly. So the ordinary-matter ratio is about half the density of a black hole. Not quite there, but still _much_ closer than I would have thought! If we account for not just ordinary matter but dark matter (a ratio of 4.9% to 26.8%) the picture is different: You have: (2 * G * (1.5 * 10^56 grams + (1.5/4.9*26.8) * 10^56 grams) ) / (c^2) You want: lightyears * 1.5234355e+11 So the SR then would be about 5 times larger. But I have no idea whether dark matter would count towards the black hole mass and SR, and I don't know if anyone does. [0] https://en.wikipedia.org/wiki/Observable_universe https://en.wikipedia.org/wiki/Observable_universe [1] https://en.wikipedia.org/wiki/Schwarzschild_radius https://en.wikipedia.org/wiki/Schwarzschild_radius (The scientific calculator I used, which is awesome every day is "units" btw: https://linux.die.net/man/1/units https://linux.die.net/man/1/units)
    - gnramires 6y ago
      Yes, dark matter has mass, this is its defining characteristic: it seems to be a detectable mass, just doesn't interact electromagnetically. In any case, it's important to note the Schwarzchild model is a spherically symmetrical ball. Mass in our Universe seems more or less uniformly dispersed (with some anisotropy). Without a center for mass to collapse, I suspect the black hole dynamics don't apply.
  - layoutIfNeeded 6y ago
- fsflover 6y ago
  https://www.nationalgeographic.com/news/2014/2/140218-black-hole-blast-explains-big-bang/ https://www.nationalgeographic.com/news/2014/2/140218-black-...

vbezhenar 6y ago

Sounds like we're in a black hole.

maneesh 6y ago
Or in a video game where the background is an unreachable layer.
TeMPOraL 6y ago
Ironically, I realized yesterday that if the universe was destined for the Big Crunch (until quite recently, this was believed to be a possibility), then we'd be literally living inside the event horizon of an universe-sized black hole!
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.
lrem 6y ago
For extra mind blowing, do compare the Schwarzschild mass of the observable universe and its radius. Spoiler: gurl ner abg whfg fvzvyne, gurer ner gurbergvpny erfhygf gung gurl unir gb or rdhny. Hasbeghangryl gur cncre vgfrys nffhzrq jnl zber culfvpf xabjyrqtr guna V unir. Ohg, guvf znxrf bhe havirefr pbafvfgrag jvgu gur fvzcyr qrsvavgvba bs oynpx ubyr ("n znff ragveryl jvguva vgf Fpujnemfpuvyq enqvhf").
- grandchild 6y ago
  So I quickly did a back-of-envelope calculation using [0] for source of mass and diameter of the universe. Then I used the formula given in [1]: r = (2 G M) / (c^2). You have: (2 * G * 1.5 * 10^56 grams ) / (c^2) You want: lightyears * 2.3548373e+10 So about 23,548,373,000 ly of SR, while the radius of the observable universe is given as ~46,500,000,000 ly. So the ordinary-matter ratio is about half the density of a black hole. Not quite there, but still _much_ closer than I would have thought! If we account for not just ordinary matter but dark matter (a ratio of 4.9% to 26.8%) the picture is different: You have: (2 * G * (1.5 * 10^56 grams + (1.5/4.9*26.8) * 10^56 grams) ) / (c^2) You want: lightyears * 1.5234355e+11 So the SR then would be about 5 times larger. But I have no idea whether dark matter would count towards the black hole mass and SR, and I don't know if anyone does. [0] https://en.wikipedia.org/wiki/Observable_universe https://en.wikipedia.org/wiki/Observable_universe [1] https://en.wikipedia.org/wiki/Schwarzschild_radius https://en.wikipedia.org/wiki/Schwarzschild_radius (The scientific calculator I used, which is awesome every day is "units" btw: https://linux.die.net/man/1/units https://linux.die.net/man/1/units)
  - gnramires 6y ago
    Yes, dark matter has mass, this is its defining characteristic: it seems to be a detectable mass, just doesn't interact electromagnetically. In any case, it's important to note the Schwarzchild model is a spherically symmetrical ball. Mass in our Universe seems more or less uniformly dispersed (with some anisotropy). Without a center for mass to collapse, I suspect the black hole dynamics don't apply.
- layoutIfNeeded 6y ago
fsflover 6y ago
https://www.nationalgeographic.com/news/2014/2/140218-black-hole-blast-explains-big-bang/ https://www.nationalgeographic.com/news/2014/2/140218-black-...