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Dark energy and type 1a supernovae?

  1. Apr 9, 2015 #1
    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 realised 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...

    Any help would be much appreciated!
     
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  3. Apr 9, 2015 #2

    wabbit

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    If my understanding is correct it's really one method, not two. The key is the relation between distance (estimated from luminosity) and velocity (derived from redshift). The data is noisy but it is spread around a certain curve.

    Hubble's law v=Hd describe the curve for "small" distances (cosmologically speaking). The curve deviates from that farther out, and how exactly it deviates is where the information about acceleration etc. lies.

    More generally, what curve exactly is found, is revealing of the expansion history, notably if that expansion is accelerating of decelerating - and accelerating expansion was the conclusion reached.

    I have not read the original paper so this is really a broad outline of how the data relates to the conclusion, not an account of the precise method they used.
     
    Last edited: Apr 9, 2015
  4. Apr 9, 2015 #3
    Thank you for your reply! It is really how the curve of velocity against distance deviates that I am wondering about. Does it curve upwards? This would suggest that the redshift, and thus recession velocity, is greater than expected for its distance, and therefore the galaxy appears brighter than what would be expected from its red shift. If it curves downwards, then the galaxies appear dimmer than what would be expected from their redshifts... I think. I can't seem to find a graph of velocity against distance that shows anything past the straight line part of the graph though.
     
  5. Apr 9, 2015 #4

    marcus

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    As I recall the SNe were dimmer than expected, at their measured redshifts

    At each z they turned out to be slightly farther away than would have been expected if Lambda were zero.
     
  6. Apr 9, 2015 #5

    wabbit

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    The supernova cosmology project at http://supernova.lbl.gov has a wealth of information, papers, data, charts and more on this subject (it is my personal impression that their web design might be lagging slightly behind content quality and variety though : ))

    This overview article seems interesting, I had just a quick look but some of the charts may be what you re looking for :
    http://supernova.lbl.gov/PDFs/PhysicsTodayArticle.pdf
     
  7. Apr 9, 2015 #6

    RUTA

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    The way I explain this in my courses on general relativity and modern physics is via Figure 3 of this paper:

    http://users.etown.edu/s/STUCKEYM/GRFessay2012.pdf

    This paper received Honorable Mention in the Gravity Research Foundation 2012 Awards for Essays on Gravitation and was published in International Journal of Modern Physics D21, No. 11, 1242021 (2012). You can read about distance modulus versus redshift used in the plot of Figure 3 in the first paragraph of the paper. [You can also ignore the alternative model pitched therein, you just want to understand LambdaCDM versus Einstein-deSitter cosmology models.] Figure 3 makes it easy to see how the LCDM model fits the large redshift data better than the EdS model. The kinematical consequences of accelerating expansion follow from EdS plus Lambda.
     
  8. Apr 9, 2015 #7
    Here's a graph of apparent (observed) magnitude and redshift. You can see that relative to a matter dominated universe ([itex]\Omega_\Lambda=0[/itex]) a supernova at a given redshift appears dimmer (higher apparent magnitude) in an accelerating universe that contains dark energy [itex](\Omega_M = 0.25, \Omega_\Lambda=0.75)[/itex] because it is more distant (at this given redshift).

    The next attachment is a graph of proper distance at the current time vs. redshift in 3 types of universes: dark energy only, Benchmark (our current best model of the universe), and matter-only. You can see how all three obey the Hubble law [itex]d = \frac{cz}{H_0} [/itex] at low redshift, but there are clear deviations at higher redshifts.
     

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    Last edited: Apr 9, 2015
  9. Apr 13, 2015 #8
    Thank you for all of the brilliant replies, although I still can't quite wrap my head around this statement. The redshift tells us the recession velocity of the Galaxy, so a galaxy with a certain redshift would have the same recession velocity in an accelerating or not accelerating universe. However in an accelerating universe, the Galaxy's speed must have increased much more gradually to its current speed, and therefore it would not have travelled as far a distance as a galaxy in a universe with constant expansion, so I would think that the galaxy in an accelerating universe would appear brighter since it is closer, for a given redshift...

    Apologies if my logic makes no sense!
     
  10. Apr 13, 2015 #9

    wabbit

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    Yes but the curve tells a different story. Broadly speaking, the slope of the redshift luminosity curve (from lots of supernovas) gives the Hubble constant, and its second derivative gives the cosmological constant.

    And accelerating expansion means a given galaxy is (and was when it emitted that light) farther away than it would be in constant expansion, hence more redshifted - or fainter at a given redshift.

    Its current speed is not what the redshift measures, since it measures light emitted a long time ago. What it does measure is how much the universe has expanded since that emission.
     
    Last edited: Apr 13, 2015
  11. Apr 14, 2015 #10

    marcus

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    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, H2 must approach Λ/3 so we give Λ/3 the special name of H2
    *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 Hnow 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. Hnow 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.)
     
    Last edited: Apr 14, 2015
  12. Apr 14, 2015 #11

    marcus

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    Will the "20" one give distances (in billions of lightyears) that are larger or smaller than the one we think is close to right, the "17.3" one?
    Remember that one integral is based on assuming that the longterm Hubble growth rate H is 1/20 per billion years and the other that it is 1/17.3 per billion years.
    To actually compute either of the two integrals for sample S such as 1.5, 2, 2.5, if someone is interested, google "number empire definite integral"
    and put in, for example,
    20*(((20/14.4)^2 - 1)s^3 + 1)^(-1/2)
    change the variable from x to s
    put in the limits 1 and, say, 2.5
    and click "compute".
     
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