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Problem with natural units in Cosmology

  • Thread starter Magister
  • Start date
  • #1
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Homework Statement


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
 

Answers and Replies

  • #2
nrqed
Science Advisor
Homework Helper
Gold Member
3,603
203

Homework Statement


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
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
 
  • #3
83
0
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!!!!!
 
  • #4
nrqed
Science Advisor
Homework Helper
Gold Member
3,603
203
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!!!!!
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:

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