1. Dec 13, 2015

astrobird

- 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, shouldnt 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 travelling 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:)

2. Dec 13, 2015

Chalnoth

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.

3. Dec 13, 2015

astrobird

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?

4. Dec 14, 2015

Chalnoth

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).

5. Dec 14, 2015

Thanks!