21joanna12 said:
I am trying to wrap my brain around the evidence for accelerating expansion of the universe from type 1a supernovae. From what I understand, it was first realized that the universe was expanding at an increasing rate from discrepancies between the calculated distances to type 1a supernovae using the method of their apparent magnitude and another method of considering their red shift and thus their recession velocity, and using this in Hubble's law, v=Hd. What I can't quite understand is which one of these would give a greater value of distance to imply that the expansion is accelerating...
21joanna12 said:
Thank you for all of the brilliant replies, although I still can't quite wrap my head around this statement...
IMHO this is a good sign. I don't think it is intuitive.
If you have time, have a look at some underlying integral calculus involving "change of variable". From somewhere I've gotten the impression that you have some college calculus. If not, or if too busy, just ignore this.
Define s = z+1, the factor by which wavelengths and distances have been enlarged while the light was traveling to us. It is just a convenience, we could write z+1 all over the place, but this is cleaner.
s=1 denotes the present.
s(t) decreases as time increases. So when we do 'change of variable' there will be a minus sign or else we swap the upper and lower limits of integration.
What about the distance the light has covered (aided by expansion)---the distance NOW to the source where it was emitted? I denote the two times "em" for emitted and "now" for received. $$D(S) = \int_{em}^{now} s(t) cdt$$
in each little bit of time the light travels cdt, and then the stretch factor s(t) enlarges that. I'll assume c = 1 and omit it.
Here's where you need calculus: you know that H = a'/a the fractional growth of scale factor. and you know s = 1/a.
Show that H is also equal to s'/s except for a minus sign. In fact s'/s = -H
Now s'dt = ds so we can do a change of variable and put in ds/s' for dt
$$D(S) = \int_{em}^{now} s(t) cdt = \int_S^1\frac{s}{s'}ds = -\int_1^S\frac{s}{s'}ds = \int_1^S\frac{ds}{H(s)}$$
Now here is the nice thing.
Friedmann equation tells how H(s) depends on s, because as you go back in time (in matter era) the energy density* grows as s3. A particularly simple form of Friedmann equation(assuming spatial flatness) that applies any time after the matter era kicks in is
$$H^2 - \frac{\Lambda}{3} = [const]\ \rho = [const]\ \rho_{now}s^3 $$
$$H^2 - H_\infty^2 = [const]\ \rho $$
If we go back to s = 3, distances are then a third the size and energy density 27 times that of today.
As energy density thins out to zero, H
2 must approach Λ/3 so we give Λ/3 the special name of H
∞2
*I don't include in energy density anything about the cosmological constant---Lambda is simply a curvature constant as in the classic (1917) Einstein GR equation.
The density is that of matter and radiation, principally dark and ordinary matter.
$$H_{now}^2 - H_\infty^2 = [const]\ \rho_{now} $$
$$H(s)^2 - H_\infty^2 = (H_{now}^2 - H_\infty^2) s^3 $$
This gives us a way to evaluate the distance integral I wrote earlier depending only on the value we know for H
now and the values we TRY for H
∞ by comparing with supernova data points consisting of (s, D) a stretch s = z+1, and a standard candle distance D. The H(s) term in the denominator of that integral is
$$H(s) = \sqrt{(H_{now}^2 - H_\infty^2) s^3 + H_\infty^2}$$
The neatest way to treat that is probably to think rates per billion years, e.g. H
now as 1/14.4 and H
∞ as for example 1/17.3 (a current estimate, to try alternatives against). Then divide both sides by H
∞ or in effect multiply through by 17.3 billion years.
$$\frac{H(s)}{H_\infty} = \sqrt{((\frac{17.3}{14.4})^2 - 1) s^3 + 1}$$
So here's the punchline. Suppose you know that today's Hubble rate is 1/14.4 per billion years and you want to compare two possible values for H
∞: 1/17.3 per billion years, and 1/20 per billion years, to see which fits the set of (s, D) data better. So for each S value you compute the D in two different ways. Each will give the answer in billions of lightyears.
$$D(S) = 17.3\int_1^S\frac{ds}{\sqrt{((\frac{17.3}{14.4})^2 - 1) s^3 + 1}}$$
$$D(S) = 20\int_1^S\frac{ds}{\sqrt{((\frac{20}{14.4})^2 - 1) s^3 + 1}}$$
Which integral would you imagine gives the larger distance? (There is a term in the denominator of the integrand which is being amplified by the cube of s = z+1, essentially by the cube of the redshift-plus-one if you are used to thinking in terms of redshift. If that term is large it will tend to make the denominator larger and the integrand smaller. This will overcome the merely linear effect of a larger coefficient out front.)
