Problem with natural units in Cosmology

[Magister]

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:


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


where g_*=3.36 is the degree of freedom of the radiation (equivalent) and T=2.75 K 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


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


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


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


and this is by far wrong. I have spend so much time around this that I am starting to get frustrated!
Thanks in advance

[nrqed]

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:


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


where g_*=3.36 is the degree of freedom of the radiation (equivalent) and T=2.75 K 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


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


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


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


and this is by far wrong. I have spend so much time around this that I am starting to get frustrated!
Thanks in advance
If you use \hbar instead of "h" in your equation, it works out (the result is (2 \pi)^3 times larger).

Patrick

[Magister]

Humm! You mean:

\rho_r = \frac{\pi^2}{30} g_* \frac{(kT)^4}{(\hbar c)^3} (2\pi)^3


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

Thanks a lot by the way!!!!!

[nrqed]

Humm! You mean:

\rho_r = \frac{\pi^2}{30} g_* \frac{(kT)^4}{(\hbar c)^3} (2\pi)^3


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

Thanks a lot by the way!!!!!
EDIT : No, this is not what I mean You don't have to put in the extra factor of (2 pi)^3. I mean

\rho_r = \frac{\pi^2}{30} g_* \frac{(kT)^4}{(\hbar c)^3}

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 \hbar 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