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In an article its written,
$$\Omega_{\nu} = \frac{\rho_{\nu}}{\rho_{crit}}=\frac{\sum m_{i,\nu}n_{i,\nu}}{\rho_{crit}} = \frac{\sum m_{\nu}}{93.14h^2eV}$$
Now I am trying to derive this for myself but I could not. Can someone help me ?
So the values are,
##\rho_{crit} = 1.053 75 \times 10^{-5}h^2 GeV/c^2~~cm^{-3}##
Total neutrino average number density today : ##n_{\nu} = 339.5~cm^{-3}##
I tried to write it like,
$$\frac{n_{\nu}\sum m_{\nu}}{\rho_{crit}} = \frac{\sum m_{\nu}}{93.14h^2eV}$$
$$\frac{n_{\nu}}{\rho_{cric}} = \frac{339.5cm^{-3}}{1.05375 \times 10^{-5}h^2 GeV/c^2~~cm^{-3}} = \frac{1}{3.103h^2 \times 10^{-8} GeV} = \frac{1}{31.0382916 h^2eV}$$
Which I am missing additional ##1/3##.
I guess its a simple question but I couldn't see the answer.
$$\Omega_{\nu} = \frac{\rho_{\nu}}{\rho_{crit}}=\frac{\sum m_{i,\nu}n_{i,\nu}}{\rho_{crit}} = \frac{\sum m_{\nu}}{93.14h^2eV}$$
Now I am trying to derive this for myself but I could not. Can someone help me ?
So the values are,
##\rho_{crit} = 1.053 75 \times 10^{-5}h^2 GeV/c^2~~cm^{-3}##
Total neutrino average number density today : ##n_{\nu} = 339.5~cm^{-3}##
I tried to write it like,
$$\frac{n_{\nu}\sum m_{\nu}}{\rho_{crit}} = \frac{\sum m_{\nu}}{93.14h^2eV}$$
$$\frac{n_{\nu}}{\rho_{cric}} = \frac{339.5cm^{-3}}{1.05375 \times 10^{-5}h^2 GeV/c^2~~cm^{-3}} = \frac{1}{3.103h^2 \times 10^{-8} GeV} = \frac{1}{31.0382916 h^2eV}$$
Which I am missing additional ##1/3##.
I guess its a simple question but I couldn't see the answer.
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