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The balloon-surface model works if the universe is actually positively curved. It bends around in on itself and comes back. So you would be able to see yourself

The balloon-surface model works if the universe is actually positively curved. It bends around in on itself and comes back. So you would be able to see yourself (eventually) after light made its way around the surface and back to you. As far as anyone can tell, though, the universe is flat. So photons traveling outwards from you will never come back around- they'll just keep going. I'm partial to the raisin bread model of the universe: https://www.loc.gov/rr/scitech/mysteries/universe.html https://www.loc.gov/rr/scitech/mysteries/universe.html It's a tad misleading: the raisin bread is, probably, infinite in extent. As a practical matter, we can only see a finite portion of the raisin bread. As we gaze more deeply into the sky, we're looking further into the raisin bread. Because light only travels at a finite speed, the further something is away, the older the image we see of it is. At a certain point, all we can see is the goopy bread batter that the raisin bread used to be: that's (sort of, kinda) the Cosmic Microwave Background: it's the oldest light in the universe, and we can't see anything older than it. So the universe looks like a bunch of galaxies, with us at the center, and a sphere of microwave radiation at the edge. But every single observer sees a similar sphere around their exact location, so it's a kind of illusion.

eridius 8y ago
Does the expansion of the universe mean that galaxies that are too far away from us are seemingly moving faster than the speed of light, and this is why we can't see any galaxies past a given point (because that's where they "accelerate" faster than light)?
- roywiggins 8y ago
  A lot of the visible universe actually appears to be receding faster than the speed of light: https://phys.org/news/2015-10-galaxies-faster.html https://phys.org/news/2015-10-galaxies-faster.html It's because the space inbetween is getting bigger. But light keeps moving towards us from those galaxies, so we see those photons eventually anyway. You can still hear things that are going above the speed of sound, as long as they're headed away from you- if they're headed toward you, you'll not hear them (probably?). But the waves they throw off will expand behind them just fine.
  - acqq 8y ago
    > if they're headed toward you, you'll not hear them (probably?) Actually, you would not have much time to hear, but what you would hear (among the other effects) would be the sound “in reverse”(!) https://physics.stackexchange.com/questions/14880/hearing-a-sound-backwards-because-of-doppler-effect https://physics.stackexchange.com/questions/14880/hearing-a-...
    - rcxdude 8y ago
      For a practical (if dangerous!) example, see supersonic rounds: if one is shot towards you, you hear the bullet whizzing by before you hear the gunshot.
      - LgWoodenBadger 8y ago
        Well, a supersonic round wouldn't whizz until it drops down to subsonic speeds. You'd hear the crack of the round then the crack from the original shot.
simonh 8y ago
If the universe originated with the big bang and then expanded, then unless it expanded infinitely fast it must be bounded. But if it's bounded, how can it be flat?
- roywiggins 8y ago
  The math indicates that the singularity was infinitely dense and infinitely hot... if you're already allowing that something infinitely dense could be real, then infinitely fast expansion doesn't seem less crazy. Going from an infinite temperature to a finite temperature is an infinite amount of cooling, which also seems impossible; likewise, an infinite density to a finite one. You'd need infinite expansion to do that- if it was merely finite, you'd still end up with an infinite density. Expanding an infinitely dense point to a bowling ball still gets you an infinite density; you need to expand it out "infinitely far" (whatever that means) to spread stuff out enough to get a merely finite density. https://en.wikipedia.org/wiki/Initial_singularity https://en.wikipedia.org/wiki/Initial_singularity It seems more likely that the singularity at t=0 in general relativity isn't physical- what does an infinite density even mean?
  - raattgift 8y ago
    > what does an infinite density even mean? What does t=0 even mean? If we switch to some coordinates where t_0 \def t_now = 0 and t_past \gt 0, as with the scale factor in cosmology, we don't have a special time coordinate at t_past ~ 13.8 Gyr, so we can still think about the stress-energy (and the mechanisms that generate it) and the curvature it sources locally. We can also think about things under time-reversal: A comoving test observer in vacuum just free falls without end. Our test observer is a point with no charges (including "active" gravitational charge) and no internal structure, so it feels no tidal effects. If we have two such observers starting at arbitrarily large space-like separation, they eventually they come close to each other and ultimately end up freely falling on the same trajectory, forever. If we complicate the test observers from this no-repulsive-interaction picture, we develop pressure (represented in the stress-energy tensor, T^11 T^22 T^33). The local increase in pressure generates local curvature in response, and if the repulsive interaction is finite then eventually the two interacting test observers end up freely falling on the same trajectory, forever. The eventuality is because the background has infinite curvature, and that and the contribution from the self-gravitation of pressure wins out over any finite repulsion. We can complicate test object interactions further by introducing: repulsive and attractive interactions; extended observers that are composites of these test observers; and even have these objects obey the statistics for bosons and fermions. Same-charged fermions in effect resist being pushed onto the same trajectory, but the resistance is almost certainly finite. Eventually some daughter product free-falls forever. (We can see some of this in sufficiently massive objects that repulsion or exclusion is overcome, as in the cores of stars or in neutron stars, where internal pressure tends to dominate the local stress-energy.) What the daughter products of squashed-together Standard Model particles might be is a matter of research, both in terrestrial laboratories and in astrophysical processes (including black holes). Conversely, mirroring the time-reversal thinking above, the Standard Model particles have to freeze out of something hotter and denser that exists at greater lookback time. (cf. electroweak symmetry breaking). Worse, at a(t_0) ("now") there is a lot of dark matter, and we don't know the details of how that interacts with the Standard Model. Those details will almost certainly matter in the time-reversal picture above. Consequently we don't know that repulsion/exclusion is finite. Perhaps some crystalline structure develops in the time-reverse picture (or in black holes), and things just accumulate in that. Or perhaps there is a sudden drop in pressure where daughter products fly away from some ultrastrong squashing-together (cf. pair-instability supernovae). There are quite a few options. And of course the Robertson-Walker background is already just an approximation; the real metric is likely to be very different at high lookback times. Finally, Carroll has a nice deck (and it's filled with references) at https://www.slideshare.net/seanmcarroll/what-we-dont-know-about-the-beginning-of-the-universe https://www.slideshare.net/seanmcarroll/what-we-dont-know-ab...

eridius 8y ago

Does the expansion of the universe mean that galaxies that are too far away from us are seemingly moving faster than the speed of light, and this is why we can't see any galaxies past a given point (because that's where they "accelerate" faster than light)?

roywiggins 8y ago
A lot of the visible universe actually appears to be receding faster than the speed of light: https://phys.org/news/2015-10-galaxies-faster.html https://phys.org/news/2015-10-galaxies-faster.html It's because the space inbetween is getting bigger. But light keeps moving towards us from those galaxies, so we see those photons eventually anyway. You can still hear things that are going above the speed of sound, as long as they're headed away from you- if they're headed toward you, you'll not hear them (probably?). But the waves they throw off will expand behind them just fine.
- acqq 8y ago
  > if they're headed toward you, you'll not hear them (probably?) Actually, you would not have much time to hear, but what you would hear (among the other effects) would be the sound “in reverse”(!) https://physics.stackexchange.com/questions/14880/hearing-a-sound-backwards-because-of-doppler-effect https://physics.stackexchange.com/questions/14880/hearing-a-...
  - rcxdude 8y ago
    For a practical (if dangerous!) example, see supersonic rounds: if one is shot towards you, you hear the bullet whizzing by before you hear the gunshot.
    - LgWoodenBadger 8y ago
      Well, a supersonic round wouldn't whizz until it drops down to subsonic speeds. You'd hear the crack of the round then the crack from the original shot.

simonh 8y ago

If the universe originated with the big bang and then expanded, then unless it expanded infinitely fast it must be bounded. But if it's bounded, how can it be flat?

roywiggins 8y ago
The math indicates that the singularity was infinitely dense and infinitely hot... if you're already allowing that something infinitely dense could be real, then infinitely fast expansion doesn't seem less crazy. Going from an infinite temperature to a finite temperature is an infinite amount of cooling, which also seems impossible; likewise, an infinite density to a finite one. You'd need infinite expansion to do that- if it was merely finite, you'd still end up with an infinite density. Expanding an infinitely dense point to a bowling ball still gets you an infinite density; you need to expand it out "infinitely far" (whatever that means) to spread stuff out enough to get a merely finite density. https://en.wikipedia.org/wiki/Initial_singularity https://en.wikipedia.org/wiki/Initial_singularity It seems more likely that the singularity at t=0 in general relativity isn't physical- what does an infinite density even mean?
- raattgift 8y ago
  > what does an infinite density even mean? What does t=0 even mean? If we switch to some coordinates where t_0 \def t_now = 0 and t_past \gt 0, as with the scale factor in cosmology, we don't have a special time coordinate at t_past ~ 13.8 Gyr, so we can still think about the stress-energy (and the mechanisms that generate it) and the curvature it sources locally. We can also think about things under time-reversal: A comoving test observer in vacuum just free falls without end. Our test observer is a point with no charges (including "active" gravitational charge) and no internal structure, so it feels no tidal effects. If we have two such observers starting at arbitrarily large space-like separation, they eventually they come close to each other and ultimately end up freely falling on the same trajectory, forever. If we complicate the test observers from this no-repulsive-interaction picture, we develop pressure (represented in the stress-energy tensor, T^11 T^22 T^33). The local increase in pressure generates local curvature in response, and if the repulsive interaction is finite then eventually the two interacting test observers end up freely falling on the same trajectory, forever. The eventuality is because the background has infinite curvature, and that and the contribution from the self-gravitation of pressure wins out over any finite repulsion. We can complicate test object interactions further by introducing: repulsive and attractive interactions; extended observers that are composites of these test observers; and even have these objects obey the statistics for bosons and fermions. Same-charged fermions in effect resist being pushed onto the same trajectory, but the resistance is almost certainly finite. Eventually some daughter product free-falls forever. (We can see some of this in sufficiently massive objects that repulsion or exclusion is overcome, as in the cores of stars or in neutron stars, where internal pressure tends to dominate the local stress-energy.) What the daughter products of squashed-together Standard Model particles might be is a matter of research, both in terrestrial laboratories and in astrophysical processes (including black holes). Conversely, mirroring the time-reversal thinking above, the Standard Model particles have to freeze out of something hotter and denser that exists at greater lookback time. (cf. electroweak symmetry breaking). Worse, at a(t_0) ("now") there is a lot of dark matter, and we don't know the details of how that interacts with the Standard Model. Those details will almost certainly matter in the time-reversal picture above. Consequently we don't know that repulsion/exclusion is finite. Perhaps some crystalline structure develops in the time-reverse picture (or in black holes), and things just accumulate in that. Or perhaps there is a sudden drop in pressure where daughter products fly away from some ultrastrong squashing-together (cf. pair-instability supernovae). There are quite a few options. And of course the Robertson-Walker background is already just an approximation; the real metric is likely to be very different at high lookback times. Finally, Carroll has a nice deck (and it's filled with references) at https://www.slideshare.net/seanmcarroll/what-we-dont-know-about-the-beginning-of-the-universe https://www.slideshare.net/seanmcarroll/what-we-dont-know-ab...