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Liquid Density

  1. Jun 13, 2009 #1
    Hypothetically, was there ever a time during inflation of the universe when the density of the universe would have been equal to that of water, and we could have swam through it? If so, how big would the universe have been at that moment?
     
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  3. Jun 13, 2009 #2
    Well if the density started off really,really high and is less than the density of water today which I believe it is, then apparently it did briefly (instantaneously) pass through such a phase....but unless the temperatures were low enough for particles,atoms, etc to form, nothing would have likely been materialized except radiation and maybe a real hot plasma (ionized gas)....likely under tremendous pressure....so nothing could have been "swimming".
     
    Last edited: Jun 14, 2009
  4. Jun 14, 2009 #3

    Chalnoth

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    Oh, certainly! But at that time the temperature would have been pretty darned high too, so it wouldn't be so much swimming through it as being vaporized by the heat.

    So, for some numbers, let's do a bit of math.

    Currently the overall density of our universe right now is about 10^-26 kg/m^3. Now, that includes everything, so it's not trivial to extrapolate that density back in time (the density of different stuff scales differently depending upon its physical properties as the universe expands). By contrast, water has a density of 10^3 kg/m^3, which is obviously quite a bit larger. If we take the simple Lambda-CDM universe, the components of our universe are as follows:

    74%: dark energy [tex]\left(\Omega_\Lambda\right)[/tex]
    26%: normal matter + dark matter [tex]\left(\Omega_m\right)[/tex]
    0.0082%: radiation [tex]\left(\Omega_r\right)[/tex]

    Now, these are the current densities, and they are different going back in time. The overall density scales as:

    [tex]\rho(a) = \rho(1) \left(\Omega_\Lambda + \frac{\Omega_m}{a^3} + \frac{\Omega_r}{a^4}\right)[/tex]

    As you can see, it's not going to be so easy to solve the above equation for the scale factor a. We're going to have to be a bit clever. Fortunately, it turns out that for most scale factors, one single energy density makes up nearly everything. So we just need to test the various energy densities and see which one will be largest.

    What if the energy density when the universe was 10^3 kg/m^3 was dominated by the cosmological constant? Well, clearly that's not possible: the cosmological constant doesn't change with time, so it can't be any more dense than it currently is. That one is out.

    What if the energy density was dominated by normal matter? Well, that would require:
    [tex]10^{3} kg/m^3 = 10^{-26}kg/m^3 \frac{(0.26)}{a^3}[/tex]
    Some quick math gives:
    [tex]a = 1.4 \times 10^{-10}[/tex]

    That's pretty darned small: at that size, the density of radiation will be vastly larger than normal matter. So clearly this isn't right.

    So what if it was dominated by radiation? That would require:
    [tex]10^{3} kg/m^3 = 10^{-26}kg/m^3 \frac{(0.000082)}{a^4}[/tex]

    Which gives us:
    [tex]a = 5 \times 10^{-9}[/tex]

    This is much more reasonable: matter will only be a tiny fraction of the density of radiation at that scale factor, and dark energy will be even smaller. So we can take that as our answer.

    But what does this say about temperature? Well, temperature scales linearly with redshift, so since the temperature of the universe today is about 2.7K, at the above scale factor it would be around 500 million Kelvin, which is much hotter than any star (this is in the X-ray range).
     
    Last edited: Jun 14, 2009
  5. Jun 14, 2009 #4

    DSM

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    The density of water is 10^-3 kg/cm^3
     
  6. Jun 14, 2009 #5

    Chalnoth

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    Ack, duh. I had it correct the first time then "fixed" it.

    Edit: fixed now. Thanks!

    Edit 2: Bah, messed it up again. Now hopefully it's correct...
     
    Last edited: Jun 14, 2009
  7. Jun 15, 2009 #6

    Redbelly98

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    [/URL]

    So that's about

    5×10-9 × 13.7×109 yrs ​

    ≈70 years after the big bang.
     
    Last edited by a moderator: May 4, 2017
  8. Jun 15, 2009 #7

    Chalnoth

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    It's not quite that easy. The expansion rate is nowhere near linear, and at early times the rate was far, far faster than today.

    Basically, to get the time, you have to perform the following integral:

    [tex]\int_0^t dt[/tex]

    To perform the integral, we make use of the Hubble parameter to change variables:

    [tex]H \equiv \frac{1}{a}\frac{da}{dt}[/tex]
    [tex]dt = \frac{da}{aH}[/tex]

    So the above integral becomes:
    [tex]\int_0^a \frac{da}{aH}[/tex]

    Using the Friedmann equation to get H(a):

    [tex]H^2 = \frac{8\pi G}{3} \rho[/tex]

    ...and using the equation in my post above for [tex]\rho[/tex] and numerical integration, I get:
    [tex]t = 592s[/tex]
    or about 10 minutes after t = 0 (t = 0 would have been right around the end of inflation).
     
  9. Jun 15, 2009 #8

    Redbelly98

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    Oops. Thanks for the clarification Chalnoth.
     
  10. Jun 15, 2009 #9
    Thanks guys. Obviously, I know that you can't swim through plasma, but I was speaking metaphorically. I'm far passed the time in my life when I remember how to do integrals and calculate such things, so I thought I'd pose the question to you guys instead. I think it would have been quite funny to see the universe at that time, a big ball of liquid plasma.

    What color would it have been? Would it even have been in the visible wavelengths? Would it have been red in visible light? Or opaque?

    And also, some theories of inflation have the universe expanding at different rates in different parts, right? So I know there was probably no era with perfectly uniform spherical liquid plasma.

    Thanks for humoring me!
     
  11. Jun 15, 2009 #10

    Chalnoth

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    It would have been quite opaque, almost completely uniform, and the light would be in hard X-rays.

    Sort of. But the entire part of inflation that led to our region was expanding at a nearly uniform rate. Those theories that predict different rates of expansion are talking about disconnected regions of the universe beyond our horizon, not different areas within our own universe.
     
  12. Jun 25, 2009 #11
    This can be estimated with a back-of-the-envelope calculation with the classic solution to a flat, matter dominated universe with no cosmological constant (different than the real universe but it gives a decent approximation).

    rho = rho(0) * t(0)^2 / t^2. Where rho and rho(o) are time varying and current matter densities, and t(0) is current age of universe.

    Plugging in current values gives a time = 12 min to equal water density (close to the 10 min answer posted earlier).

    BTW, is there a good way to paste (say Mathcad) formulas into these posts?
     
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