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Problem with natural units in Cosmology

  1. Feb 24, 2007 #1
    1. The problem statement, all variables and given/known data
    Reading about the raditation dominated era I saw that the radiation energy density today was given by:

    [tex]
    \rho_r = \frac{\pi^2}{30} g_* T^4 = 8.09 * 10^{-34} g/cm^3
    [/tex]

    where [itex]g_*=3.36[/itex] is the degree of freedom of the radiation (equivalent) and [itex]T=2.75 K[/itex] is the CBR temperature today.

    The problem is that they dont give me the exact expression and so this relation seems to be dimensionally wrong. I suppose that the full relation must be

    [tex]
    \rho_r = \frac{\pi^2}{30} g_* \frac{(kT)^4}{(hc)^3}
    [/tex]

    and this way the relation would be dimensionally correct. But when I put the values on it I get

    [tex]
    \rho_r = 2.88 * 10^{-22} J/cm^3 = 3.20 * 10^{-36} g/cm^3
    [/tex]

    and this is by far wrong. I have spend so much time around this that I am starting to get frustrated!
    Thanks in advance
     
  2. jcsd
  3. Feb 24, 2007 #2

    nrqed

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    If you use [itex] \hbar [/itex] instead of "h" in your equation, it works out (the result is [itex] (2 \pi)^3 [/itex] times larger).

    Patrick
     
  4. Feb 24, 2007 #3
    Humm! You mean:
    [tex]
    \rho_r = \frac{\pi^2}{30} g_* \frac{(kT)^4}{(\hbar c)^3} (2\pi)^3
    [/tex]

    Yes, in fact it works out but does this make any sense? Is this because that in natural units is the [itex]\hbar[/itex] that is equal to 1 instead of [itex]h[/itex]?

    Thanks a lot by the way!!!!!
     
  5. Feb 24, 2007 #4

    nrqed

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    EDIT : No, this is not what I mean You don't have to put in the extra factor of (2 pi)^3. I mean
    [tex]
    \rho_r = \frac{\pi^2}{30} g_* \frac{(kT)^4}{(\hbar c)^3}
    [/tex]
    This result will be (2 pi)^3 times larger than the equation written with only h.


    That's what I mean, yes. Usually, by "natural units", people mean that they set [itex] \hbar [/itex] equal to 1, not "h". Of course, one could also decide to set h to one instead, but this is not what is usually done.

    You are very welcome.

    Patrick
     
    Last edited: Feb 24, 2007
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