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Does recent expansion change the predicted CMB temperature?

  1. Nov 28, 2009 #1

    BillSaltLake

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    During the interval from t = 380,000 yr (CMB last surface) to the present, if the Universe were exclusively matter at critical density, the scale factor [1+z] would be proportional to t to the 2/3 power. Note that [13.7 Gyr/380,000 yr] raised to the power 2/3 corresponds to [1+z] = 1090. This of course is the correct redshift to take the CMB last surface from 3000K to the present 2.7K.

    However, at t = 380,000 yr, matter probably dominated (matter was about 75% of the total then), while now, matter is only about 25-30% of the total. If the CMB photons have stretched by an extra amount due to recent acceleration/dark energy, wouldn't that cause the predicted value of [1+z] to be larger than 1090 and thus give an incorrect prediction for temperature?

    To put it in numbers, in addition to time dependence, I think the scale factor [1+z] is also proportional to the matter fraction to the negative 1/3 power. Thus the starting (t = 380,000 yr) matter fraction of 0.75 would reduce [1+z] to 0.91 [0.75 to the power 1/3] of the matter-exclusive value of 1090, while the ending matter fraction of about 0.28 would multiply [1+z] by 1/[0.28 to the power 1/3] =1.53. The total correction factor is thus 0.91x1.53 = 1.39, taking the predicted value of [1+z] up to 1515. This would predict a present temperature of about 1.98K--clearly wrong.

    Are photons perhaps not stretched by the recent acceleration? I think such an assumption would cause logical inconsistencies. Please comment on where I went wrong.
     
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  3. Nov 29, 2009 #2

    Chalnoth

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    To really compute this, you'd need to do the full calculation, which includes doing a few integrals numerically. But I think that the basic answer is that this particular calculation isn't changed much by the intervening dark energy content, because the dark energy has only recently become significant.

    I think, if you want to satisfy yourself, you should do the calculation in the following way:

    [tex]t_0 - t_{CMB} = \int_0^{1090} \frac{dz}{H(z)}[/tex]

    Where for [tex]H(z)[/tex] you plug in the Friedmann equation including only matter (dark and normal) and dark energy (cosmological constant works here). See if the value you get matches with the accepted value for this difference.
     
  4. Nov 30, 2009 #3

    BillSaltLake

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    Thank you for the suggestion. Please note however, that if [tex]H = \dot{a}/a = -\dot{z}/[1+z][/tex], then in order to get the differential form of the equation correct (i.e., to get [tex]dt = dt[/tex]), a factor [tex]1+z[/tex] is required:

    [tex]
    t_0 - t_{CMB} = \int_0^{1090} \frac{dz}{[1+z]H(z)}
    [/tex]

    I calculated the approximate integral assuming a linear slope starting from [tex]H=2/(3t)[/tex] at [tex]z = 2[/tex] and going to [tex]H=1/t)[/tex] at [tex]z = 0[/tex] and found the right side to be only [tex]0.8 t_0[/tex]. Note that [tex]t_{CMB}[/tex] on the left side is only 1/30,000 of [tex]t_0[/tex] so we must be very careful using numerical approximations with the equation in the above form. (Nonetheless, the equation integrates perfectly if there were only matter. This of course is just a coincidence. With matter only, [tex]H = 2/(3t)[/tex] and [tex]1+z = (t_0/t)[/tex]2/3.)

    What is really required is evaluation of [tex]a_0/a_{CMB}[/tex]. This should equal 1090, but apparently the calculated ratio is larger than that. We can make an estimate of [tex]a_0/a_{CMB}[/tex] by using the present value of [tex]H = 1/t[/tex] (as observed). Although it is not certain exactly how [tex]H[/tex] was modified in the past by dark energy, a workable approximation would be to use the matter-only expression [tex]a\propto t[/tex]2/3 from time [tex]t_{CMB}[/tex] until [tex] t = 5Gyr[/tex]. Then use [tex]a\propto t[/tex]1 until the present. (This latex is mixing up subscripts & superscripts. I hope it’s readable.)

    This gives an [tex]a_0[/tex] that is larger than the matter-dominated value by factor (13.7/5)1/3 or 1.4x, meaning [tex]a_0/a_{CMB} \cong[/tex] 1500. This is a fairly accurate approximation because it is treated as a perturbation of the matter-only exact solution.

    Are there any papers in which someone has actually compared the calculated value to the observed value 1090?
     
  5. Nov 30, 2009 #4

    Chalnoth

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    Yeah, my calculator's batteries died, so I couldn't check the numerical values easily, and was just working from memory. I thought I might have forgotten a factor of (1+z) to some power.

    I would take that as an experimental input, given by the temperature of the CMB, to very high precision, and then see if other computed values make sense based upon it (e.g. the time, which is not directly observable).
     
  6. Dec 1, 2009 #5

    Chronos

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    The CMB is a measured temperature. Any theoretical model of the universe must produce this result to be taken seriously.
     
  7. Dec 1, 2009 #6

    Chalnoth

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    Whoops, BillSaltLake, I'd also like to point out that you really should do the whole integral, not just a simple approximation. It comes out to the correct value.
     
  8. Dec 1, 2009 #7

    BillSaltLake

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    Unfortunately, the theoretical model that doesn't seem to work here is the standard model. Some numerical integration is required to calculate zCMB using the above integral to verify that z = 1090. Because tCMB is 1/30,000 of t0, this form of the integral is too sensitive to slight errors. A 1% low error would require z to be (more than) infinite, for example.

    A way that is less sensitive to systematic error way is just to integrate da/dt from 380kyr to 13.7Gyr. Then a0/aCMB = 1+z. Functionally, the expansion factor a increases identically to a matter-only Universe at critical density until about t = 5 Gyr, so da/dt = 2a/(3t) then. From 5 to 13.7Gyr, it's closer to da/dt = a/t, yielding 1+z of about 1500. My point is that any recent stretch increases 1+z above 1090 unless there's sum other compensating mechanism.
     
  9. Dec 1, 2009 #8

    Chalnoth

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    It's really not that difficult with any decent numerical integrator. I can supply a simple C++ code if you like, though I'm sure there are easier methods.
     
  10. Dec 2, 2009 #9
    Question: If you add dark energy then doesn't that change the initial conditions in the calculation to compensate? Which is to say that the shouldn't the correct calculation not be "zero dark energy case" + "dark energy" but rather "dark energy" + "a different set of parameters for the early universe"
     
  11. Dec 2, 2009 #10

    Chronos

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    z~1100 is the current accepted value for distance to the surface of last scattering. Is that close enough for comfort?
     
  12. Dec 2, 2009 #11

    Ich

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    Last edited by a moderator: Apr 24, 2017
  13. Dec 2, 2009 #12

    BillSaltLake

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    From the solution referred to in the non-english treatment,

    7ad993fe78b621c43010b8214842b271.png

    where H0 = 1/13.7Gyr, [tex]\Omega_{\Lambda}[/tex] = .73 and [tex]\Omega[/tex]o = .27, we get a0 = 2.732 and aCMB = 0.001506, so a0/aCMB = 1808 = 1+z (without accounting for the CMB light energy density). This not very close to 1090.

    They didn't include CMB energy in the above solution because perhaps this energy doesn't have a simple closed-form solution for a(t). (I think the CMB energy only reduces 1+z by about 9%. I will numerically perform the full integration soon.)

    The point is that there is very little room for extra expansion due to dark energy to jibe with 1+z = 1090. The standard model with a relatively large of dark energy seems to predict a 1+z value that is much higher.
     
    Last edited: Dec 2, 2009
  14. Dec 2, 2009 #13

    Chalnoth

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    I think you did your calculations wrong. I'm getting very different numbers.

    But regardless, doing the calculation in the other direction is vastly more stable. Here's a little online numerical integration utility I found:
    http://people.hofstra.edu/stefan_Waner/RealWorld/integral/integral.html [Broken]

    You can factor [tex]H_0[/tex] out of the equation to make a dimensionless integral, which can then be integrated. You'll have to do the integral in two steps due to the limits of the integrator (I suggest z=0-100 and z=100-1089).

    Once you've done that, input the exact WMAP parameters here:
    http://lambda.gsfc.nasa.gov/product/map/current/params/lcdm_sz_lens_wmap5.cfm [Broken]

    And see what you get for the time since the surface of last scattering (you can safely neglect radiation for getting the time to the surface of last scattering). See how it compares to the WMAP result. Note that the Hubble expansion rate that they give is h = 0.719, which translates to [tex]H_0 = 1/13.6Gyr[/tex].

    You should find that your result for the time to the surface of last scattering is within a few percent of the WMAP result.
     
    Last edited by a moderator: May 4, 2017
  15. Dec 2, 2009 #14

    BillSaltLake

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    My bad; the ratio calculated from above is only a0/aCMB = 1297, not 1808, making agreement with 1090 much more plausible.

    Thanks for the integration link. This will make a lot of things easier because I don't do programming. I'll do the calculation in the next few days. (Maybe this time I won't blabber until I've triple-checked the result.)
     
  16. Dec 6, 2009 #15

    BillSaltLake

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    Fortunately the numbers work out perfectly, and no numerical integration is needed.

    With only matter + radiation, there is an exact solution of t as a function of expansion parameter a. To simplify, the constants can be set to unity with starting eq. (da/dt)2 = 1/a + 1/a2. Solving, t = (2/3)(a+1)3/2 - 2(a+1)1/2 +4/3. (Assumes a=0 at t=0 and da/dt = 0 at t = [tex]\infty[/tex].)

    We then find time tCMB when a(CMB) = 3. This choice corresponds to the time when energy density was 1/3 of the equivalent matter density, which was the case at age 380,000 years. With this dimensionless eq., the solution is tCMB = 2.667 (no units). Multiply this tCMB by the ratio 13.7Gyr/380kyr to get the present time t0 in dimensionless units (t0 = 96100). Then use the eq. backward to solve a0 = 2751. This gives a0/aCMB = 917, which is 1+z if there were no recent acceleration.

    It turns out that the hyperbolic equation (in white background above) is an exact solution for matter + a cosmological constant if radiation is ignored. It predicts an additional 1.189x stretch due to the cosmological constant. Thus 917x1.189 = 1090, which is correct.

    Clearly the slightly slower expansion just after the last scattering surface has reduced the present value of expansion parameter by 1/1.189 (compared to a matter-only expansion), and this is more than a negligible effect. Any model must therefore predict a recent stretch of 1.189x. This recent stretch is obviously used to explain three things: dimness of distant standard candles, the fact that t = 1/Ho (not 2/3Ho), and the observation that T = 2.75 K (not 3.09 K).
     
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