# Problem with natural units in Cosmology

## Homework Statement

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
Science Advisor
Homework Helper
Gold Member

## Homework Statement

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

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
Science Advisor
Homework Helper
Gold Member
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

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