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Homework Help: Radius of Surface of Last Scattering

  1. Jan 7, 2010 #1
    1. The problem statement, all variables and given/known data
    Estimate the radius of the last scattering surface, using the age of the Universe. Why might this underestimate the true value?


    2. Relevant equations
    For a matter dominated universe particle horizon distance = 3ct


    3. The attempt at a solution
    I assumed the way to do this would be to get an estimate by putting the age of the universe into the particle horizon distance, but I thought this would give an overestimate as decoupling occurred 400000 years after the big bang and thus the light has been travelling for less time.
    Using the equation and the age of the universe to be 13.5 billion years I get a value of 12000Mpc. The Liddle book tells me the radius is 6000hMpc-1, and I don't think h is anywhere near 0.5. Dont know where i've gone wrong, this clearly does not underestimate....is there a better way of doing this problem?
     
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  3. Jan 10, 2010 #2

    ideasrule

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    For some reason, the book seems to be using distance=ct instead of distance=3ct. It's using light travel time distance instead of actual distance.
     
  4. Jan 10, 2010 #3

    sylas

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    I think you should divide by h in that formula, not multiply. Here's why.

    In a matter dominated universe, the scale factor is
    [tex]a(t) = \left(\frac{3H_0}{2}t \right)^{2/3}[/tex]​
    and the age of the universe is
    [tex]t = \frac{2}{3H_0}[/tex]​
    The distance to the particle horizon is as you have given, 3ct. Substituting for t you get the distance to the particle horizon in terms of H0
    [tex]d = \frac{2c}{H_0}[/tex]​
    The problem is converting units. For some horrible reason, people tend to use h with units of 100km/sec/Mparsec for H0. The speed of light is 300,000 km/sec, which is 3000 (100km/sec). Put that in the equation and you get
    [tex]d = \frac{6000}{h} \; \text{Mpc}[/tex]​

    So it's actually the same formula. The 6000 has units of 100km/sec, and h has units of 100km/sec/Mpc, and you get a result in Mpc. Just make sure you divide by h.

    With h at about 0.74 (current best estimate) you have d = 8100 Mpc.

    Cheers -- sylas

    PS. The real value is about 14000 Mpc. So the big question in your original problem is: how come you have an underestimate? And (for extra credit) how did I get about 14000 as my estimate?
     
    Last edited: Jan 10, 2010
  5. Jan 10, 2010 #4

    Matterwave

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    Regarding your second question. Although it is true decoupling happened ~300k-400k years after the BB and this could mess with your numbers, 300-400k years is puny in astronomical time. It is well within the uncertainty of the age of the Universe ~13.7Billion years.
    (13,700,400,000 vs 13,700,000,000 for example)
     
  6. Jan 12, 2010 #5
    I assume that it is an underestimate because the universe is not exactly matter dominated and thus t is greater than 2/3H0 (this is confusing as in a radiation dominated universe light only travels 2ct in the same time so I would expect th inclusion of the effects of radiation to reduce the radius of the last scattering surface....) but even if you use the believed ''true'' age of the universe you get a value of approx 12000Mpc which is still and underestimate on your value of 14000Mpc....i've no idea how you've got to this....
     
  7. Jan 12, 2010 #6

    sylas

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    I cheated. :biggrin: I used Ned's calculator. However, I have also coded such calculators myself; the method is to solve a differential equation.

    Basically, you've got the main point. The universe isn't matter dominated. Whether age is more, or less than 2/3H0 will depend on what the universe is made of. However, you can't get the distance to the particle horizon simply from the age, because light moving through an expanding universe needs another differential equation.

    You can also think qualitatively like this. The mass of the universe tends to pull it together, or slow down expansion. Hence, any less mass will mean expansion in the past is a bit greater than otherwise, and hence the universe will be older.

    Similarly, if there's any dark energy involved, this works to accelerate expansion rates. Any dark energy will also mean expansion in the past is a bit greater than otherwise, and again the universe will be older.

    We do think there is less than critical mass density, and we also think there's dark energy at work. So the age of the universe, for a given rate of expansion, must be greater. This suggests the distance to the particle horizon may be greater as well, but it's not quite that simple.

    Before I go any further however, I want to make sure I'm not just being confusing.

    Can you tell us what other equations you know that might help? Have you heard of a scale factor, or a co-moving distance?

    Cheers -- sylas
     
  8. Jan 12, 2010 #7
    Your not being confusing at all, very helpful indeed and yes I am aquatinted with the scale factor a(t).
     
  9. Jan 12, 2010 #8
    Is the equation your referring to that could help us further the FRW equation? How would the effect of dark energy manifest itself in that?
     
  10. Jan 12, 2010 #9

    sylas

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    Yes! I am going to use a rather simplified form of the equation.

    There's a simple intuitive derivation of a universe model with matter and no dark energy in the PF library, using Newtonian reasoning rather than relativity. It happens to get the right answer, however, so it is often use to help explain this. See Friedmann Equation.

    I am going to use equations with Ω factors for the different energy terms. These are fractions of critical density. Equations similar to the form I am using are in the appendix of Expanding Confusion by Lineweaver and Davis (arXiv:astro-ph/0310808v2).

    Added in edit. I have just seen ([post=2529194]msg #2[/post] in thread "light geodesics in FLWR") an excellent pedagogical reference which explains these equations even better. See Distance measures in cosmology, by David Hogg, at arXiv:astro-ph/9905116v4.

    Here's a simple equation using matter and dark energy, and for completeness a curvature term as well, though this term is thought to be very close to zero, for an essentially flat universe.

    [tex]\frac{\dot{a}}{a} = H_0 \sqrt{\Omega_m a^{-3} + \Omega_k a^{-2} + \Omega_\Lambda}[/tex]​

    In this equation, a is the "scale factor" of the universe, defined to be 1 in the present, H0 is the hubble constant in the present, and the Ω terms are constants required to add up to 1. The Ωm is the amount of matter, as a fraction of critical density, and the ΩΛ term is the dark energy fraction, at the time when a=1. The current best estimate has values close to
    • Ωm = 0.27
    • ΩΛ = 0.73
    • Ωk = 1 - Ωm - ΩΛ = 0
    Hence:
    [tex]\begin{align*}
    \frac{\dot{a}}{a} & = H_0 \sqrt{\Omega_m a^{-3} + \Omega_k a^{-2} + \Omega_\Lambda} \\
    \dot{a} = \frac{da}{dt} & = H_0 \sqrt{\Omega_m a^{-1} + \Omega_k + \Omega_\Lambda a^2} \\
    & = H_0 \sqrt{1 + \Omega_m (a^{-1} -1) + \Omega_\Lambda (a^2-1)}
    \end{align*}[/tex]​

    (1) Age of the universe

    (I have edited this to make t=0 the origin of the universe, and t=T the present epoch.)

    Let’s assume t=0 at the origin of the universe. Hence if T is the age of the universe, it is the value of t now. From the scale factor definition, we have a(0) = 0 and a(T) = 1.
    [tex]\begin{align*}
    \frac{da}{dt} &= H_0 \sqrt{1 + \Omega_m (a^{-1} -1) + \Omega_\Lambda (a^2-1)} \\
    T = \int_0^T dt & = \frac{1}{H_0} \int_0^1 \frac{da}{\sqrt{1 + \Omega_m (a^{-1} -1) + \Omega_\Lambda (a^2-1)}}
    \end{align*}[/tex]​

    Caveat: in some cases we cannot satisfy the boundary condition a(0) = 0; this means there is no initial singularity in the model. Ωm = 0, ΩΛ = 1 is the main example. In such cases there is no age to the universe, and I shall apply the boundary condition a(0) = 1, to make 0 the time co-ordinate of the present. But let's not worry about that case just yet. I'll presume a(0) = 0 and a(T) = 1 for the time being.

    (2) Distance to particle horizon

    (Added in edit: I have corrected these formulae some hours after posting to use the time co-ordinate consistently.)

    As for distance to the particle horizon, I can use d to represent the "co-moving" distance to a photon, and this differential equation
    [tex]\frac{dd}{dt} = c/a[/tex]​
    Hence, for a photon we observe in the present, when the time co-ordinate is T, the co-moving distance it has traveled since time t is
    [tex]\begin{align*}
    d(t) & = c \int_t^T \frac{dt}{a} \\
    & = c \int_{a(t)}^1 \frac{da}{\dot{a}a} \\
    & = \frac{c}{H_0} \int_{a(t)}^1 \frac{da}{a \sqrt{1 + \Omega_m (a^{-1} -1) + \Omega_\Lambda (a^2-1)}} \end{align*}[/tex]​

    This gives co-moving distance from our present location to co-moving surface that emitted the photon back at time t. The "proper distance" now is just co-moving distance d times the present scale factor, and so this same value is used as the current "proper" distance to whatever surface emitted the photon. The particle horizon Dph is the largest possible value of this distance, given by starting the photon at t=0. The distance to the surface of last scattering is given by starting the photon when a(t) = 1/1100, because this is the scale change since last scattering.

    [tex]D_{ph} = \frac{c}{H_0} \int_0^1 \frac{da}{a \sqrt{1 + \Omega_m (a^{-1} -1) + \Omega_\Lambda (a^2-1)}} \end{align*}[/tex]​

    If everyone is still on board, I'll go ahead next with some solutions to these equations.

    Cheers -- sylas
     
    Last edited: Jan 13, 2010
  11. Jan 13, 2010 #10
    Yep am still on board and would love some solutions, cheers
     
  12. Jan 17, 2010 #11

    sylas

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    Sorry for the delay.... this post uses a simplified model in which I only consider a slow matter term Ωm ("dust", with ω=0) and a dark energy term ΩΛ ("cosmological constant", with ω=1). (The ω factors represent the "equation of state" for some energy term, which is often used in these types of model.) For numeric answers, I shall use H0 = 72 km/sec/Mparsec, c = 300,000 km/sec, and parsec = 3.26 ly. In this case:
    • 1/H0 is 13.6 Gy
    • c/H0 is 13.6 Gly
    I'll take 1/H0 as exactly 13.6 Gy, which means H0 is actually 71.8981 km/sec/Mparsec.

    We can use the scale factor as a kind of time co-ordinate, representing the time when the universe had that scale factor. Thus a=0 represents the singularity at the beginning of the universe, and we'll let that be time 0. If there's no singularity, then we'll just let time 0 be now.

    Here again are the equations for scale factor a, and for the co-moving distance D(a) traveled by a photon since the scale factor was a. The zero point for time t is arbitrary, but usually the equations work best if t=0 when a=0. If a is always positive, then we just let t=0 when a=1, which is the present. t(a) is just time as a function of a.
    [tex]\begin{align*}
    \frac{da}{dt} & = H_0 \sqrt{1 -\Omega_m - \Omega_\Lambda + \Omega_m a^{-1} + \Omega_\Lambda a^2} \\
    D(a) & = \int_a^1 \frac{c}{a} \frac{dt}{da} da \\
    & = \frac{c}{H_0} \int_a^1 ((1-\Omega_m - \Omega_\Lambda)a^2 + \Omega_m a + \Omega_\Lambda a^4)^{-1/2} da
    \end{align*}[/tex]​

    D(a) can also be understood as the "proper" distance to the co-moving surface which emitted photons when scale factor was a, and which we can just see now. D(0) would be the particle horizon. For anything beyond that horizon, there hasn't been enough time for light to reach us. The "surface of last scattering" is the surface which last scattered the cosmic background radiation. This light has a redshift of about 1100. Hence the distance now to that surface is D(1/1100). I'll calculate also the time since a=1/1100 (or redshift z = 1099), and the corresponding distance.

    The particle horizon, if defined, is at distance D(0), and the surface of last scattering, which is the source of the CMBR, is now at about D(1/1100). I shall use Dph and Dls for these distances, and Tph and Tls for the corresponding times. Hence Tph =t(1)-t(0) and Tls = t(1)-t(1/1100).

    Empty universe, also called the Milne model. Ωm = ΩΛ = 0

    [tex]\begin{align*}
    \frac{da}{dt} & = H_0 \\
    a(t) & = H_0 t \\
    t(a) & = \frac{a}{H_0} \\
    D(a) & = \frac{c}{H_0} \int_a^1 a^{-1} da \\
    & = \frac{-c \ln a}{H_0} \\
    T_{ph} & = \frac{1}{H_0} & = 13.60 \; \text{Gy} \\
    D_{ph} & & \text{(undefined)} \\
    T_{ls} & = \frac{1-1/1100}{H_0} & = 13.59 \; \text{Gy} \\
    D_{ls} & = \frac{c \ln(1100)}{H_0} \approx \frac{7c}{H_0} & = 95.24 \; \text{Gly}
    \end{align*}[/tex]​

    Matter dominated Ωm = 1, ΩΛ = 0

    [tex]\begin{align*}
    \frac{da}{dt} & = H_0 a^{-1/2} \\
    a & = \left( \frac{3 t H_0}{2} \right) ^{2/3} \\
    t(a) & = \frac{2a}{3 H_0} \\
    D(a) & = \frac{c}{H_0} \int_a^1 a^{-1/2} da \\
    & = \frac{c}{H_0} \left[ 2 \sqrt{a} \right]_a^1 \\
    & = \frac{2c(1-\sqrt{a})}{H_0} \\
    T_{ph} & = 9.07 \; \text{Gy}
    D_{ph} & = \frac{2c}{H_0} & = 27.20 \; \text{Gly} \\
    T_{ls} & = 9.06 \; \text{Gy}
    D_{ls} & \approx \frac{1.94 c}{H_0} &= 26.38 \; \text{Gly}
    \end{align*}[/tex]​

    Dark energy dominated Ωm =0, ΩΛ = 1 (inflation)
    [tex]\begin{align*}
    \frac{da}{dt} & = H_0 a \\
    a & = e^{t/ H_0} \\
    t(a) & = H_0 \ln(a) & \text{no singularity, so } \; t(1) = 0 \\
    D(a) & = \frac{c}{H_0} \int_a^1 a^{-2} da \\
    & = \frac{c}{H_0} ( 1/a - 1) \\
    T_{ph} & & \text{undefined} \\
    D_{ph} & & \text{undefined} \\
    T_{ls} & \approx \frac{c}{H_0} \ln(1100) = 95.24 \; \text{Gy} \\
    D_{ls} & \approx \frac{1099 c}{H_0} = 14946.4 \; \text{Gly}
    \end{align*}[/tex]​
    To solve for our current universe, with Ωm = 0.27 and ΩΛ = 0.73, you would have to do some numeric integration... Simpson's rule will be fine. I'll consider that next. But if you want to check my numbers, we have:
    • Tph,0.27,0.73 = 0.9927/H0 = 13.50 Gy
    • Dph,0.27,0.73 = 3.3336c/H0 = 45.34 Gyl
    For the surface of last scattering, the numbers are the same, up to the given accuracy, although another decimal place would start to show a small difference.

    If you compare these numbers with what you can get from Ned's calculator, the results are a bit different. Ned uses a slightly more sophisticated model, which I shall consider next as well.

    Cheers -- sylas
     
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