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Indeed, the Cosmic Microwave Background is the light that's still left over from a time when the Universe was much hotter and filled with plasma. At a certain p

Indeed, the Cosmic Microwave Background is the light that's still left over from a time when the Universe was much hotter and filled with plasma. At a certain point, the Universe cooled down enough to become neutral, which made it mostly transparent, and allowed the photons that had been bouncing around in the plasma to travel indefinitely. > its distance is also not limited by light travel distance At cosmological scales, there is no single, correct definition of "distance." There is coordinate distance, light travel distance, angular diameter distance, luminosity distance, and other measures.[0] All these distances are the same at very small scales, but when you start looking at larger scales, General Relativity starts to matter and the geometry of the Universe is no longer Euclidean. 0. https://en.wikipedia.org/wiki/Distance_measure https://en.wikipedia.org/wiki/Distance_measure

dimator 4y ago
wow, your last paragraph filled a little empty gap in my brain, thank you!
sigmoid10 4y ago
On a global scale, the universe as a whole is actually remarkably close to euclidean [1] (which is a mystery of its own). So it's not curvature that's the problem, but expansion. To account for this, people usually quote the comoving distance [2] (which factors out expansion) when talking about very distant things like the microwave background. I.e. the distance light would have to travel today if you could freeze the evolution of the universe for the duration of the journey. [1] https://en.wikipedia.org/wiki/Shape_of_the_universe https://en.wikipedia.org/wiki/Shape_of_the_universe [2] see your own source
- DiogenesKynikos 4y ago
  Only a constant-coordinate-time slice of the Universe is Euclidean. That's a 3D subspace of the 4D spacetime volume. The 4D metric of the Universe is not close to flat (or Minkowski, which is the closest thing you'll get to Euclidean in Relativity). People quote different distances, depending on what they're using it for. Probably the most widely used distance measure, from Earth's perspective, is redshift. It's linear with distance nearby, and ask the other distances are relatively simple functions of it.
  - sigmoid10 4y ago
    >The 4D metric of the Universe is not close to flat (or Minkowski, which is the closest thing you'll get to Euclidean in Relativity) It is. See the source provided above. Technically the ΛCDM universe follows the FLRW metric, but if you look at its line element, you'll see it's usually just Minkowski in spherical coordinates times a scale factor (capturing the expansion) for spacelike coordinates: ds^2 = -dt^2 + a(t)^2 dΣ^2. Note that a scale factor that affects all spacial coordinates the same does not induce curvature. The generalised form of the FLRW metric includes a 1/(1-k) factor for the radial coordinate, but the curvature k for our actual universe is measured to be 0 within experimental precision, which, again, is a mystery. There's no good reason why a universe containing matter/energy should be globally flat (*cough* inflation). Factor out a(t) in a line integral and you'll get comoving distances over more or less euclidean space, i.e. the common format in cosmology literature. Things like luminosity distances are more interesting when you e.g. want to know the absolute magnitude of a far away object, but that doesn't really make sense for the CMB. >Probably the most widely used distance measure, from Earth's perspective, is redshift. Redshift is used to measure these distances sans any reasonable alternatives. But when you want to make e.g. a map, you can't use redshift (or proper distance for that matter), unless you want your map to be distorted.
    - DiogenesKynikos 4y ago
      I'm well aware of what the FLRW metric looks like. That a(t) factor means that the geometry of spacetime is not flat. For a full explanation, see this StackExchange answer: [0]. You're confusing the flatness of a particular 3D hypersurface (of constant coordinate time) with flatness of 4D spacetime. The k parameter describes the curvature of that hypersurface, not the curvature of spacetime. This is all to say that we don't live in a flat Minkowski spacetime. We live in a curved spacetime, which means that things like distance measures are much less straightforward. The curvature of 3D slices of spacetime is not the primary complication - the overall curvature of 4D spacetime (which involves the scale factor a(t)) is the primary complication. 0. https://physics.stackexchange.com/a/689320 https://physics.stackexchange.com/a/689320
      - 4y ago
- chrisweekly 4y ago
  "global" in your first sentence was a tough word choice
  - sigmoid10 4y ago
    It's a common term in mathematics and when talking about field theories.
    - chrisweekly 4y ago
      I know. But context matters. When talking about distance and scale and the physical universe, "global" will be read as "planetary" by many. (shrug, no big deal)

dimator 4y ago

wow, your last paragraph filled a little empty gap in my brain, thank you!

sigmoid10 4y ago

On a global scale, the universe as a whole is actually remarkably close to euclidean [1] (which is a mystery of its own). So it's not curvature that's the problem, but expansion. To account for this, people usually quote the comoving distance [2] (which factors out expansion) when talking about very distant things like the microwave background. I.e. the distance light would have to travel today if you could freeze the evolution of the universe for the duration of the journey. [1] https://en.wikipedia.org/wiki/Shape_of_the_universe https://en.wikipedia.org/wiki/Shape_of_the_universe [2] see your own source

DiogenesKynikos 4y ago
Only a constant-coordinate-time slice of the Universe is Euclidean. That's a 3D subspace of the 4D spacetime volume. The 4D metric of the Universe is not close to flat (or Minkowski, which is the closest thing you'll get to Euclidean in Relativity). People quote different distances, depending on what they're using it for. Probably the most widely used distance measure, from Earth's perspective, is redshift. It's linear with distance nearby, and ask the other distances are relatively simple functions of it.
- sigmoid10 4y ago
  >The 4D metric of the Universe is not close to flat (or Minkowski, which is the closest thing you'll get to Euclidean in Relativity) It is. See the source provided above. Technically the ΛCDM universe follows the FLRW metric, but if you look at its line element, you'll see it's usually just Minkowski in spherical coordinates times a scale factor (capturing the expansion) for spacelike coordinates: ds^2 = -dt^2 + a(t)^2 dΣ^2. Note that a scale factor that affects all spacial coordinates the same does not induce curvature. The generalised form of the FLRW metric includes a 1/(1-k) factor for the radial coordinate, but the curvature k for our actual universe is measured to be 0 within experimental precision, which, again, is a mystery. There's no good reason why a universe containing matter/energy should be globally flat (*cough* inflation). Factor out a(t) in a line integral and you'll get comoving distances over more or less euclidean space, i.e. the common format in cosmology literature. Things like luminosity distances are more interesting when you e.g. want to know the absolute magnitude of a far away object, but that doesn't really make sense for the CMB. >Probably the most widely used distance measure, from Earth's perspective, is redshift. Redshift is used to measure these distances sans any reasonable alternatives. But when you want to make e.g. a map, you can't use redshift (or proper distance for that matter), unless you want your map to be distorted.
  - DiogenesKynikos 4y ago
    I'm well aware of what the FLRW metric looks like. That a(t) factor means that the geometry of spacetime is not flat. For a full explanation, see this StackExchange answer: [0]. You're confusing the flatness of a particular 3D hypersurface (of constant coordinate time) with flatness of 4D spacetime. The k parameter describes the curvature of that hypersurface, not the curvature of spacetime. This is all to say that we don't live in a flat Minkowski spacetime. We live in a curved spacetime, which means that things like distance measures are much less straightforward. The curvature of 3D slices of spacetime is not the primary complication - the overall curvature of 4D spacetime (which involves the scale factor a(t)) is the primary complication. 0. https://physics.stackexchange.com/a/689320 https://physics.stackexchange.com/a/689320
    - 4y ago
chrisweekly 4y ago
"global" in your first sentence was a tough word choice
- sigmoid10 4y ago
  It's a common term in mathematics and when talking about field theories.
  - chrisweekly 4y ago
    I know. But context matters. When talking about distance and scale and the physical universe, "global" will be read as "planetary" by many. (shrug, no big deal)