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Cosmology inequality that I'm struggling to make sense of

  1. Nov 14, 2009 #1
    My lecturer has written the following:

    Given [itex] \frac{4 \pi \rho a^3 c^2 }{3}[/itex] and [itex] V = \frac{4 \pi a^3}{3} [/itex], then substituting [itex]\frac{dE}{dt} = -p\frac{dV}{dt}[/itex] we have:

    [itex]\frac{d(\rho a^3)}{da} = -\frac{3pa^2}{c^2} \leq 0 [/itex] (1)

    Ok that part is fine - substitute and use the chain rule:

    So, if we assume [itex]p \geq 0 [/itex] we then see that:

    [itex]d(\rho a^3)} \geq 0 [/itex] if [itex] da < 0[/itex] or [itex]d(\rho a^3)} \leq 0 [/itex] if [itex] da > 0[/itex]

    I understand how he has got this inequality, simply working from the above inequality, however I'm a little unsure of what da actually represents here, it then goes on to say:

    From the first of the inequality, we see that as a decreases, [itex]\rho a^3[/itex] increases. In particular, [itex]\rho a^2[/itex] increases at least as fast as 1/a.

    That is the bit I'm really struggling to get.

    By the way, the he says that a is the scale factor and is a function of t only. He says it gives the "universal expansion rate" - I don't know if he means that it is the universal expansion rate, or just that you kind find the universal expansion rate from it. Either way I fail to see how it's a universal expansion rate since it is just a time dependent function and not a derivative itself.

    So the way I interpret it as follows:

    d(a(t)) = da/dt x dt - the total differential, qualitatively - to me this means the amount a changes when you make an infinitesimal change in t. So for da < 0, this means that as time goes on, the scale factor a is decreasing - the universal expansion is slowing?

    So when he says "when a is decreasing" - he is using the first inequality - because da < 0 is the same as saying a is decreasing. So why does that mean [itex]\rho a^3[/itex] increases?

    Any help is much appreciated guys, thanks.
    Last edited: Nov 14, 2009
  2. jcsd
  3. Nov 15, 2009 #2


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    Well, first, decreasing "a" wouldn't mean a slowing expansion: it'd mean contraction.

    So what this little exercise is doing is analyzing whether the energy in a comoving volume ([tex]\rho a^3[/tex]) is increasing or decreasing with expansion. And whether or not that occurs depends upon the sign of the pressure of the matter that makes up the universe. If there's positive pressure, then the amount of energy decreases with expansion. If there's negative pressure, then it increases.

    If you want to understand why this is, you can think of gravity (which is driving the expansion) as doing work on the matter in space, dependent upon the pressure.

    For example, if we think of a box that holds stuff with positive pressure, and cause said box to shrink, then we have to press in on the sides of the box. This does positive work on the stuff inside, increasing its energy.
  4. Nov 15, 2009 #3

    George Jones

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    It seems that standard units are being used ([itex]c[/itex] hasn't been set to unity), so the units don't appear to work out, i.e., the units of[itex]\rho a^2[/itex] and are different than the units of [itex]1/a[/itex]. I think that it should read "In particular, the rate of increase of [itex]\rho a^2[/itex] is at least as fast as a (natural) fudge factor times the rate of increase of [itex]1/a[/itex]."

    This revised statement is such that: 1) the units of the fudge factor can make the overall units work out; 2) the statement can be demonstrated easily.

    Start with

    [tex]0 \geq \frac{d}{da} \left( \rho a^3 \right)[/tex]

    and multiply both sides by [itex]da/dt[/itex], with [itex]da/dt < 0[/itex], which, as Chalnoth notes, means a contracting universe.

    What happens?
    Last edited: Nov 15, 2009
  5. Nov 17, 2009 #4

    George Jones

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    Then, because [itex]da/dt < 0[/itex], the inequality changes direction and

    [tex]0 \leq \frac{da}{dt} \frac{d}{da} \left( \rho a^3 \right) = \frac{d}{dt} \left( \rho a^3 \right).[/tex]

    Now, since you're interested in the rate of change of [itex]\rho a^2[/itex], write [itex]\rho a^3 =\rho a^2 a[/itex], and apply the product rule to [itex]\rho a^2[/itex] and [itex]a[/itex].
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