How redshift/brightness lead to acceleration/deceleration?

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SUMMARY

The discussion centers on the relationship between redshift and brightness in type Ia supernovae, particularly how these observations relate to the universe's expansion dynamics. When observing type Ia supernovae at medium distances, they appear fainter than expected due to accelerated expansion, while at large distances, they appear brighter due to decelerated expansion. The integral D = c ∫ {dz / H(z)} is used to calculate distances, revealing that accelerated expansion results in a nearly constant H(z), leading to higher observed luminosity compared to decelerated expansion.

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  • Understanding of type Ia supernovae as standard candles
  • Familiarity with cosmological redshift and its implications
  • Knowledge of the Hubble expansion rate (H(z))
  • Basic grasp of integrals in cosmology, specifically D = c ∫ {dz / H(z)}
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  • Explore the implications of redshift in cosmology
  • Study the relationship between luminosity and distance in type Ia supernovae
  • Investigate the concept of the cosmological constant and its effects on H(z)
  • Learn about the evidence for the accelerating universe from observational data
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Astronomers, astrophysicists, and students of cosmology seeking to understand the dynamics of cosmic expansion and its effects on supernova observations.

astrobird
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I have read several articles about this topic, on Wikipedia but for example also this article http://supernova.lbl.gov/PDFs/PhysicsTodayArticle.pdf and also several books that discuss the subject. In very brief my understanding is that:
- 1. When observing type 1a supernovae (standard candle) at medium-range distance, they appear fainter than you would expect based on their redshift value, because the expansion of the universe has been accelerating since the light we are observing left the supernovae.
- 2. When observing type 1a supernovae at very large distance, they appear brighter than you would expect based on their redshift value, because the expansion of the universe was decelerating still when the light we are observing left the supernovae.

My question now is how such conclusions can be made from the redshift and brightness observations. As is explained in the various texts the redshifts arises from the fact that the photons are stretched while on their way from the supernovae to earth. I understand how this works, but if the photons are stretched because of the expansion it also means that (because of the expansion) the distance between the photon and Earth becomes longer, is that right?
When then looking at point 1 above. The fact that the supernovae appear fainter than expected implies that they are more distant than expected (based on their redshift), however, if they are more distant than expected because the expansion accelerated, shouldn't this acceleration of expansion have affected the redshift in exactly the same way so that in the end distance from the object(s) to Earth as well as the redshift of the photons traveling from the object(s) to Earth would be equally affected?

I'm sure I'm missing something obvious as in most texts I read the point 1 and 2 above are presented as very logical conclusions based on the observations, I just don't see how, please enlighten me:)
 
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It's probably better to think of the redshift as the amount that the universe has expanded since the photon was emitted. A redshift of z=1, for example, means that the wavelength of that photon has doubled. This also means that distances within the universe have, on average, doubled since that photon was emitted.

Imagine, now, that we're going to compare two hypothetical supernovae, each at z=1. For one supernova, let's imagine that the there was an accelerated expansion. For the other, there was deceleration. Which supernova will appear brighter? It turns out that the way this is calculated is by using the following integral:

D = c \int {dz \over H(z)}

Now, an accelerated expansion has an H(z) that is nearly constant, so that the distance from z=0 to [itez]z=1[/itex] is just D = c/H_0, where H_0 is the current Hubble expansion rate.

The decelerating expansion, on the other hand, has an H(z) that was much bigger in the past. Because as H(z) gets bigger, 1/H(z) gets smaller, the faster expansion in the past lowers the value of this integral, which lowers the distance.

So at the same redshift, the observed luminosity will be higher for a decelerated expansion.
 
Thanks, that helps a lot! It's indeed easier to understand when thinking of redshift as how much the universe expanded during the photon's travel.
I Googled a bit to fully understand the formulas you gave and read the bit under "evidence for acceleration" at https://en.m.wikipedia.org/wiki/Accelerating_universe which also helped.

One thing I don't get though is why an accelerating universe would have an H(z) that is nearly constant, wouldn't it change just like in a decelerating universe?
 
astrobird said:
Thanks, that helps a lot! It's indeed easier to understand when thinking of redshift as how much the universe expanded during the photon's travel.
I Googled a bit to fully understand the formulas you gave and read the bit under "evidence for acceleration" at https://en.m.wikipedia.org/wiki/Accelerating_universe which also helped.

One thing I don't get though is why an accelerating universe would have an H(z) that is nearly constant, wouldn't it change just like in a decelerating universe?
The simplest accelerating universe is a universe with no matter or radiation, just a cosmological constant. That universe has a constant H(z). The recession velocity between any two test particles in an expanding universe is v = Hd. So if H is a constant, as the distance between any two test particles increases, so does their recession velocity. Thus things accelerate away from one another.

In our current universe, H(z) is of course changing, but it is changing slowly enough that there is accelerated expansion at present (and for the last few billion years).
 
Thanks!
 

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