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Fridman's equation help

  1. Jan 18, 2012 #1
    Hi there. My interest in cosmology is purely amateurish, so bear with my ignorance. I have some questions regarding following Fridman's equation:

    [tex]H^{2}=\frac{8\pi G\rho}{3}-\frac{kc^{2}}{a^{2}}+\frac{\Lambda c^{2}}{3}[/tex]

    1. Lambda is usually given in units of s-2, or in units of m-2. When given in s-2 is it already multiplied with c2, so the last term is λ/3, right or wrong?

    2. What are the units of k - normalized spatial curvature? I can't find it anywhere. It is just given as 0,1, or -1. It can't be unitless or equation wouldn't end in 1/s, required units for Hubble parameter. It must also be in m-2, right or wrong?

    3. What is exactly Hubble parameter? Same thing as Hubble constant, or to say, Hubble constant is current value of Hubble parameter?

    Any help would be appreciated, thanks.

  2. jcsd
  3. Jan 18, 2012 #2


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    Yes, that's right.
  4. Jan 19, 2012 #3


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    Honestly, I never pay attention to the units. Instead I just reparameterize everything so that there are no units (except for [itex]H_0[/itex]):

    [tex]H^2 = H_0^2\left({\Omega_m \over a^3} + {\Omega_r \over a^4} + {\Omega_k \over a^2} + \Omega_\Lambda\right)[/tex]

    Here [itex]\Omega_x[/itex] is the current density fraction for component [itex]x[/itex]. In the equation above, I have included matter (normal matter + dark matter), radiation, curvature, and a cosmological constant. These make for a fraction in that the sum of all of the [itex]\Omega_x[/itex] values is equal to one, and they have no dimensions.

    Also useful in some situations is to get rid of the units of [itex]H_0[/itex] by defining:

    [tex]h = {H_0 \over 100km/s/Mpc}[/tex]

    This convention was made before we had an accurate measurement of the Hubble parameter, so a round number not too far away from the very poorly-measured value was used. This is useful in many areas of cosmology where there is a measurement where the precise value of the measurement depends upon the Hubble constant, but doesn't actually measure it.

    But in general, if you want units, you should be able to read them right off of the equation. The units of the Hubble expansion rate are inverse time (or, equivalently, velocity over distance). So just equate the units and solve.
  5. Jan 19, 2012 #4
    Hi Chalnoth, thanks for replying. What set my confusion is fact that lambda term in equation in my first post is somewhere given as λc2/3, and somewhere it is λ/3. Also λ is somewhere given in length-2 (~10-35m-2), and somewhere in time-2 (~10-18s-2). When you multiply 10-35 with c2 you get 10-18. So is this right, or not?

    Also, critical density equation is ρcrit= 3H2/8piG. Does the ρ here apply to all density contributors you mentioned in your post?

    Furthermore, how the curvature density parameter is defined, and is it 0 for flat universe?
  6. Jan 19, 2012 #5


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    Right, so, this is why I prefer dimensionless parameters so much. You don't have to worry about these issues! Units can vary depending upon your convention. There really isn't any meaning to the units beyond the convention that is used. And you can do a lot of calculations correctly (and often more easily!) by instead just using a convention that eliminates units entirely.

    No, the [itex]\rho[/itex] only applies to the radiation and matter densities.

    Well, you can solve for it directly by comparing it to your form:

    [tex] -{k c^2 \over a^2} \to H_0^2 {\Omega_k \over a^2}[/tex]


    [tex]\Omega_k = -{k c^2 \over H_0^2}[/tex]

    But perhaps an easier way to see it is from the definition that the current value of the Hubble parameter is [itex]H_0[/itex]:

    [tex]H(a=1) = H_0[/tex]

    It is really easy to see from the way I wrote the Hubble parameter earlier that:

    [tex]\Omega_m + \Omega_r + \Omega_\Lambda + \Omega_k = 1[/tex]

    So if we measure the matter density, the radiation density, and the dark energy density, the remainder is the curvature.

    (so yes, it's zero for a flat universe)
  7. Jan 19, 2012 #6
    Ok, I see now. Any universe with Ωm,rλ=1 will have flat geometry, right?

    In any case, thank you Chalnoth, you've been very helpful to me.
  8. Jan 19, 2012 #7


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    Exactly! And the way that we measure these different values is by comparing a number of different experiments. Some measurements are more sensitive to one value or another, or are sensitive to a specific combination of the different values.

    For example, supernova observations are very good at measuring the ratio [itex]\Omega_m/\Omega_\Lambda[/itex], while CMB observations are good at measuring the total density, [itex]H_0^2(\Omega_m + \Omega_r + \Omega_\Lambda)[/itex] as well as providing a nearly-exact measurement of the radiation density [itex]H_0^2 \Omega_r[/itex]. So if you combine CMB observations, such as WMAP, with a nearby measurement of the Hubble expansion rate [itex]H_0[/itex], you get a really accurate measurement of how flat the universe is.

    No problem!
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