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I'm reading a cosmology textbook(sorry it's Japanese) and stacked at statement that

  • If the mass of WIMP is enough smaller than 100GeV, the reaction rate is represented as $\langle \sigma_a |\mathbf{\upsilon}|\rangle \simeq \frac{c}{\hbar^4} G_F^2 m^2$
  • If the mass of WIMP is enough larger than 100GeV, the reaction rate is $\langle \sigma_a |\mathbf{\upsilon}|\rangle \simeq \frac{c}{\hbar^4} G_F^2 \frac{m_w^4}{m^2}$

I'm not familiar with particle theory so not sure how the mass dependences appear: $m^2$ v.s. $\frac{m_w^4}{m^2}$. My naive guess is the difference comes from the loop effect(radiative correction) of WIMP/W-boson particles, but I don't know how it affects in details.

Is there any intuitive way to understand the behavior of the mass dimension?

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Sounds like the wings of a Breit-Wigner distribution for virtual weak boson production by the dark matter, with a peak at $m^2 \approx m_W^2$ and a width related to $\Gamma^2 \approx 1/G_F$, and the energy scale set by the mass of the WIMP.

Dimensional analysis: ${G_F}{(\hbar c)^{-3}} m c^2$ has units of $\rm GeV^{-1}$. You've got that squared, multiplied by $(\hbar c)^2$ to give units of area, multiplied again by $c$ so that both sides have units of volume per unit time. To convert to a proper rate you have to multiply again by the number density of dark matter particles, but $\sigma v$ is where the new-particle physics is.

Using $\hbar = c = 1$ units, the Breit-Wigner distribution would have a complete expression like

$$ \left<\sigma |v| \right> \propto \frac {k} {\left( m^2 - m_W^2 \right)^2 + m_W^2/G_F} $$

where the normalization $k$ depends on $m_W$ and $G_F$ in a messy way, with dimensions of $\rm GeV^{+2}$.

I had thought I would do some binomial expansion in the limits of $m \ll m_W$ and $m_W \ll m$, but that's not working for me today. But the Lorentzian has a going-up side and a going-down side, and $m^2$ and $m^{-2}$ are the simplest nonlinear going-up and going-down functions of $m$, and you can hide a lot of handwaving by writing $\simeq$ and making the units right.

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