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Angular size of comoving horizon at last scattering

  1. Feb 29, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the angular size of the comoving horizon at the z=1100 last scattering surface, as projected on to the current (CMB) sky. Assume flat FRW cosmology and no cosmological constant. First calculate angular diameter distance to last scattering, then the particle horizon at last scattering.


    2. Relevant equations

    Particle horizon dH = a(t) ∫0t dt'/at' = a(η)η

    Angular diameter distance dA = R0Sk(χ)/(1 + z) = dL/(1+z)2


    3. The attempt at a solution

    I have used a = 1/(1+z) to go from z = 1100 to a = 9.08 x 10-4

    Looking at the definition of particle horizon, I need to find what t is at z = 1100, so I tried to get this using the first Friedman equation, and rearranging it to get

    da/dt = a √(8∏Gρ/3)

    1/a da = √(8∏Gρ/3) dt

    ∫ 1/a da = √(8∏G/3) ∫ρ(t) dt

    I don't know how to go any further with this though to find t at some a.


    Looking at the angular diameter distance, I have a flat unverse, so Sk(χ) = χ. I think R0 and χ are both just arbitrary measures though, so I have no idea how to do anything useful with that definition!
     
  2. jcsd
  3. Feb 29, 2012 #2

    cepheid

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    Regarding the particle horizon scale:

    You are right that the expression for the co-moving horizon scale is given by[tex]\eta = \int_0^t \frac{dt^\prime}{a(t^\prime)} [/tex]I do not think that it is necessary to solve for t(a) from the Friedmann equation. Instead, you can use the Friedmann equation to carry out a change of variables that allows you to express the integral with respect to a (or z if you want). You can do that, because you have:[tex]\frac{da}{dt} = a\left[\frac{8\pi G}{3}\rho(t)\right]^{1/2} [/tex]If we play fast and loose with notation, we can basically "solve for dt" in terms of da and substitute the expression for dt into the integral. The trick is figuring out what to do with the density. I'm not sure whether you are meant to consider just matter, or matter + radiation. How does the matter density vary with scale factor? Therefore, how can you express [itex]\rho_m(t)[/itex] as a function of the matter density today and the scale factor? Same question for [itex]\rho_r(t)[/itex] if you're meant to take into account radiation as well. So how would then express these in terms of the corresponding density parameters [itex] \Omega_m[/itex] (and [itex]\Omega_r[/itex] if relevant)?
     
  4. Feb 29, 2012 #3

    cepheid

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    Regarding the angular diameter distance: in a flat universe, if [itex]\chi[/itex] is the co-moving radial distance coordinate, then the angular diameter distance to redshift z is given by [itex]d_A(z) = \chi(z) / (1 + z) [/itex]. You could compute [itex]\chi(z)[/itex] using an integral very similar to the one you wrote above for the horizon scale. The difference is that this time you are integrating over the path of a photon travelling between the time when it was emitted, and now (as opposed to before for the horizon scale, where the integral was over a time interval from the beginning of the universe to time t, representing the largest possible distance light could have travelled up to time t).
     
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