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Tutorial on rho crit

  1. May 17, 2003 #1

    marcus

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    The critical density of the universe is

    rhocrit = (3c2H02)/8piG

    which is the average energy density required for flatness.
    Many people are put off by formulas this complicated and can't say just by looking at it what the density is in familiar terms---how many BTU per cubic yard or whatever makes sense to you.

    1. why does it matter what it is?
    2. what's a simpler formula for it?
    3. what does it actually work out to be?

    1. It turns out that as far as we can tell our universe is actually flat, and so what you get from the formula (which takes into account the cosmological constant or dark energy) is our best estimate of the density of the world. Cosmologists are in the habit of giving other densities (like matter, radiation, dark matter, etc) as FRACTIONS of rho crit. So if you know rho crit you can interpret what they are saying and if you dont you cant. So it is good to know.

    2. In natural units (c=G=hbar=1) the critical density is simply

    rhocrit = (3/8pi) divided by the square of the Hubble time.

    The Hubble time is 1/H0. It works out to around 8E60 in natural units or about 13.8 billion years. When you see E60 in natural units that is the same order of magnitude as a billion years---same ballpark timescale. The square of the Hubble time is about 64E120.

    The number (3/8pi) is roughly 1/8 so rho crit works out to
    around 1/8 divided by 64E120. It is a very small density which is good because if it were bigger the universe would go crunch.

    If you need more precision in line with presentday accurate measurements of the Hubble parameter, use 8.06E60 for the time.

    rhocrit = (3/8pi) tH-2

    Note that the Hubble time is not generally equal to the age of the universe although in some models it may be fairly close to it. The Hubble parameter is something that is directly measureable from data on observations----whereas the age is something people theorize about and infer from models. Different beasts.

    3. To take a for instance: a density which is 2 joules per cubic mile works out to be E-123 in natural units.
    So if some density happens to be 1.8E-123 in planck, and you want to interpret it in everyday language, you can call it 3.6 joules per cubic mile. This will be considered perverse by metric purists who abhor miles----therefore, so that the purists may rejoice, we say that E-123 is half a joule per cubic kilometer. Then 1.8E-123 comes out to be around 0.9 joule per cubic km. None of this appears to matter much because it is just translation into some arbitrary conventional terms. I find it's more useful to know in planck.

    4. In case you like differential equations the two Friedmann equations are what the Einstein GR equation boils down to assuming a nice homogeneous isotropic universe and the second
    Friedmann says (in the zero curvature case):

    H02 = (8piG/3c2) rho

    that tells what rho has to be in the zero curvature case and
    it's easy to rearrange that equation so as to solve for rho,
    and it gives the definition of rho crit quoted at the beginning.
    This may be why rho crit is so useful----its definition is a disguised
    form of one of the two favorite equations of cosmology.
     
    Last edited: May 17, 2003
  2. jcsd
  3. May 20, 2003 #2
    I think you ought to check that, the ρcrit I have is
    ρ=3H2/8πG
    you can get at it through the Friedmann equations or by equating KE and PE for moving objects and gravity, respectively.
     
  4. May 20, 2003 #3

    marcus

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    if you look back, i nearly always say that I am looking at rho as an energy density

    if you treat it as a mass density, the formula is what you say

    (the formula is missing a c^2 factor because you divide out c^2 to turn energy terms into the "equivalent" mass m = E/c^2)

    it is one possible way to do the accounting

    but if you treat the energy densities as actual energy densities then the formula is as I say.

    The version of the Friedmann eqn that you are using probably has rho as a mass density and therefore is missing a c^2 term

    so when you solve for rho you naturally get a mass density.

    it is legit to go either way as long as you remain alert to which
    form of accounting you are using

    personally, many of the densities in space that interest me are more natural to think of as energy densities

    and the orig. GR eqn were of course written using energy-momentum tensor (getting away from mass idea)

    IMPORTANT SIDE NOTION energy density has the same units as pressure

    if your First Friedman eqn has (rho + 3p) term then it is adding energy density directly to pressure ------mass density will not add like that

    mathematically cleaner to us energy (not mass) density,

    however math formalism is ultimately a matter of esthetics and
    de gustibus non disputandum est
     
  5. May 20, 2003 #4
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