Look-Back Time, Observable Universe, and Cosmic Horizon

Light takes time to travel through space, so we are always observing the past when we look at distant objects.
The key concept is look-back time: the actual time that light took to reach us from an object.
Distances stated in this textbook (in light-years) are based on look-back times; i.e., a galaxy at distance D light-years implies we are seeing it as it was D years ago.
Relationship to time: if an object is D light-years away, the light we see left the object D years ago. In other words, look-back time tlb equals the distance D when D is expressed in light-years: t{lb} = D \text{(if } D ext{ is in light-years)}.
In an expanding universe, distance is ambiguous (distance now vs distance when the light left). The transcript emphasizes look-back time as the basis for these distances.

Alpha Centauri (nearest star system): distance ≈ 4 ext{ ly}, so we see it as it was about 4 ext{ years ago}.
Orion Nebula: distance ≈ 1{,}350 ext{ ly}, so we see it as it looked about 1{,}350 ext{ years ago}.
Andromeda Galaxy: distance ≈ 2.5 imes 10^{6} ext{ ly}, so we see it as it looked about 2.5 imes 10^{6} ext{ years ago}.
General principle: the farther away we look, the further back in time we see.

A galaxy labeled as 7 imes 10^{9} ext{ ly} away means its light took 7 imes 10^{9} ext{ years} to reach us, so we see it as it was 7 imes 10^{9} ext{ years ago}.
If the light left when the universe was younger, we can infer the universe’s age at that time:
- Current age: T_0 \,\approx\, 14 \times 10^{9} \text{ years}.
- Light from the galaxy at 7 \times 10^{9} ext{ years} ago means the universe’s age then was T_0 - 7 \times 10^{9} = 7 \times 10^{9} \text{ years} (about half its current age).
A galaxy at 1.2 \times 10^{10} ext{ ly} away (i.e., 12 \times 10^{9} ext{ ly}) implies we see it as it was 12 \times 10^{9} ext{ years ago}; the universe’s age then would be T_0 - 12 \times 10^{9} = 2 \times 10^{9} ext{ years}.
Therefore, looking far away lets us see the universe when it was very young (e.g., only a few billion years old).

If we observe a galaxy at a distance with a light-travel time equal to the age of the universe, i.e., about 14 imes 10^{9} ext{ years}, we are seeing light from the Big Bang era.
The distance corresponding to this look-back time is about 14 imes 10^{9} ext{ light-years}, and defines the boundary of the observable universe.
This boundary is a boundary in time, not in space: light from anything beyond this horizon has not had time to reach us in a universe that is only about 14 imes 10^{9} ext{ years} old.
Note: this does not imply the entire universe is only 14 \times 10^{9} ext{ light-years} across; the full universe could be far larger than our observable patch.

In an expanding universe, distant galaxies are receding as light travels toward us, which creates ambiguity about what distance we mean (distance now, distance when the light left, or some intermediate value).
The distances given in the text (7, 12, 14 billion light-years) are based on look-back time, i.e., the time it took the light to reach us.
The convention in this textbook: distances stated in light-years are based on look-back times.

Light from near the Big Bang era has been detected as the cosmic microwave background (CMB).
The CMB represents the oldest light we can observe, coming from when the universe was only a small fraction of its current age.
The idea of a horizon imposes a fundamental limit on the portion of the universe we can observe at any given time.

Suppose a supernova is located in a galaxy at distance = 10 imes 10^{9} ext{ ly} (i.e., D = 10^{10} ext{ ly}).
This means the supernova occurred t_{lb} = 10^{10} ext{ years} ago.
The universe’s current age is T_0 \,\approx\, 14 \times 10^{9} ext{ years}.
The age of the universe at the time the supernova occurred would be:
T{ ext{then}} = T0 - t_{lb} = 14 \times 10^{9} - 10 \times 10^{9} = 4 \times 10^{9} ext{ years}.
Therefore: the supernova occurred when the universe was about 4 \times 10^{9} ext{ years old}.
Answer to the multiple-choice question: 4 \times 10^{9} ext{ years} (option b).

The farther we look, the further back in time we see, approaching the era of the Big Bang as distance increases toward the horizon.
The observable universe is bounded by a time-based horizon corresponding to a look-back time of about 14 imes 10^{9} ext{ years}.
The cosmic microwave background provides a glimpse of the universe shortly after the Big Bang, setting a practical observational limit.
Distances described as light-years in this context are shorthand for look-back times, not current spatial separations, due to cosmic expansion.

Alpha Centauri: 4 \text{ ly} \Rightarrow t_{lb} = 4 \text{ years}
Orion Nebula: 1{,}350 \text{ ly} \Rightarrow t_{lb} = 1{,}350 \text{ years}
Andromeda: 2.5 \times 10^{6} \text{ ly} \Rightarrow t_{lb} = 2.5 \times 10^{6} \text{ years}
A far galaxy: 7 \times 10^{9} \text{ ly} \Rightarrow t_{lb} = 7 \times 10^{9} \text{ years}
A farther galaxy: 12 \times 10^{9} \text{ ly} \Rightarrow t_{lb} = 12 \times 10^{9} \text{ years}
Universe current age: T_0 \approx 14 \times 10^{9} \text{ years}
Horizon/look-back boundary: D \,=\, 14 \times 10^{9} \text{ light-years} (look-back time = age of universe)
Supernova in a galaxy at D = 10^{10} \text{ ly}: occurred when the universe was T_0 - D = 4 \times 10^{9} \text{ years old}.