# Talent

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Nuclear Talent
Course 7
NT4A

the_nu_wgv

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Members: Don Willcox, Nicole Vassh, Panos Gastis

We would like to find the decoupling temperature, such that: $$\Gamma= G_F^2T^5 + \mu_\nu^2T^3 \sim H = \frac{T^2}{M_{pl}}$$

Let's replace with some more illustrative constants: $$\Gamma = \frac{T^5}{M_W^4} + g_\nu^2\frac{T^3}{m_p^2} \sim H = \frac{T^2}{M_{pl}}$$

For $g_\nu\equiv 0$, we have: $$T_0^3 = \frac{M_W^4}{M_{pl}}$$

We can use this to rewrite the cubic equation, in dimensionless form: $$\left(\frac{T}{T_0}\right)^3 = 1 - g_\nu^2 \frac{M_{pl}^{2/3}M_W^{4/3}}{m_p^2}\frac{T}{T_0}$$

We can see that the last term is of order unity when $g_\nu\sim \frac{m_p}{M_{pl}^{1/3}M_W^{2/3}}\sim 10^{-8}$, allowing us to rewrite the equation as: $$\left(\frac{T}{T_0}\right)^3 = 1 - \left(\frac{g_\nu}{10^{-8}}\right)^2 \frac{T}{T_0}$$