# How are the dark matter (DM) thermalization rates derived in singlet DM model?

This well-cited paper talks about a minimal renormalizable extension to the Standard Model (SM) to incorporate particle dark matter (DM) into it by adding a real scalar field $S$ which (unlike the Higgs doublet $H$) is a singlet under the full SM gauge group. But we have to pay a price, we have to introduce three more free parameters in the theory (in addition to those already present in the Standard model): (i) the mass of the new scalar $m_0$, (ii) dimensionless self coupling of the scalar $\lambda_S$, and (iii) a dimensionless coupling to the Higgs $\lambda$.

The power-counting renormalizable Lagrangian of the model, is therefore, $$\mathscr{L}=\mathscr{L}_{\rm SM}+\frac{1}{2}(\partial_\mu S)^2-\frac{1}{2}m_0^2S^2-\frac{1}{4}\lambda_SS^4-\lambda S^2({\rm H^\dagger H})\tag{1}.$$

Question 1 It is assumed that the DM was in thermal equilibrium at temperatures of the order $m_0$ and above. Is this an ad hoc assumption? Why did the authors have to make this assumption?

Question 2 It is also assumed that the thermalization rate varies as $\sim \lambda^2T$ for $T\gg m_h$ and $\sim \lambda^2T^5m_h^{-4}$ for $T\ll m_h$ where $m_h$ is the Higgs mass. How do they derive these thermalization rates?

All I know is that the dark matter density evolves as the figure below (taken from here.)

• The cross-section of the interaction of dark matter with ordinary matter will determine when it falls out of equilibrium with ordinary matter in the early Universe. This will depend on the value of $\lambda$ in your Lagrangian. Apr 4 '18 at 19:05
• @Virgo My question is "What's the necessity of making the assumption that the DM was in equilibrium around $m_0$" in their paper? Where did they use it?
– SRS
Apr 4 '18 at 19:40
• Given this Lagrangian, why don't the S particles keep on interacting and equilibrating with quarks and leptons via exchange of virtual Higgsons? You could make $\lambda$ arbitrarily weak, but that would be artificial. Apr 8 '18 at 16:00

I'm no expert in this field and I would love to be corrected by others. This is also the answer to question 1. But here is what my heuristic understanding is. It can be argued that if a dark matter species $\xi$ continued to remain in thermal equilibrium until the present, its abundance $$\frac{n_{\xi}}{s}\sim \Big(\frac{m_\xi}{T}\Big)^{3/2}e^{-m_{\xi}/T}$$ would have been negligible and contrary to the observation we wouldn't find any appreciable relic abundance. Therefore, the dark matter species $\xi$ must at some point of time (or temperature) had decoupled from the cosmic thermal bath. To achieve the desired relic density, the $\xi$ should freeze-out at a temperature not much less than the dark matter mass $m_{\xi}$. Therefore, it is reasonable to assume that for temperatures $\sim m_{\xi}$ and above $\xi$ was in equilibrium. The first dashed line in the figure tells that freeze-out occurs at a temperature around $\frac{m_{\xi}}{20}$, and remained in equilibrium around $T\sim m_{\xi}$ and above. If instead, we assume that the dark matter departs from equilibrium at a scale $\frac{m_{\xi}}{30}$, its abundance will decrease compared to what it had been when the departure scale was $\frac{m_{\xi}}{20}$.