WIMP is a possible dark matter candidates, which self-annihilates.
What is the WIMP annihilation rate, as a function of density? Most sources state this scales as $\rho^2$, (density squared), but they do not state the full equation.
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To get to an expression to use, first note that the "WI" is for "Weakly-Interacting", which tells you that you should expect cross-section similar to those for neutrinos. That means small, so you can treat a WIMP gas as dilute almost without worrying about it's actual density.
So, taking the RMS velocity of on of the particles to be $v_{RMS} = \frac{kT}{2m}$ where $T$ is the temperature of the gas,1 $m$ is the mass of a dark-matter particle and $k$ is the Boltzmann constant, we expect each particle pass "near" other particles at an average rate of $(v_{RMS} \,\sigma)\rho_N$, where "near" is taken to be close enough to interact as parameterized by the cross-section $\sigma$, and $\rho_N$ represent the number density (as opposed to mass density) of the WIMPs. Because that was the interaction rate for a single particle the interaction rate in a unit volume gets another factor of $\rho_N$
Now we merely need to estimate the fraction of interactions that represent annihilations rather than scattering. This is a little harder to get numerically right, because you have to assume some things about the physics here. Or we could just parameterize it for now and more on. I declare that some fraction $f$ of such interactions will be annihilations. In principle, $f$ may (probably does) depend on the energy of the interaction, but perhaps it varies slowly enough to gloss over that.2
At this point the annihilation rate per unit volume3 can be simply written down as $$\begin{align*} R &= f (v_{RMS} \, \sigma) \rho_N^2 \\ &= f\frac{v_{RMS} \, \sigma}{m^2} \rho^2\\ \end{align*}$$ where I wave written $\rho = m \rho_N$ for the mass density. There is also a version that has $\rho_N$ and $kT$, but it is no more interesting.
So, what are the interesting features here? Well, it depends on the RMS velcocity which can be obtained in a rough way from the shape of the halo. It depends on the density squared (either density), which you already remarked on. It's much lower for heavy WIMPs than for light ones.
One more interesting thing we could do is write this in terms of temperature as $$ R = f\frac{kT \sigma}{2m^3} \rho^2 \,,$$ which might shed some light on why dark-matter investigators make distinction between "hot", "warm" and "cold" dark matter.4
Lastly, note that there is no guarantee than WIMPs represent the correct model for dark matter. They are popular right now because they are experimentally tractable, so there is money for chasing them and for writing theories about what to expect from the chases.
1 There is an assumption that the gas has thermalized hidden here, I think that is OK, but I'm not a dark matter guy.
2 If you want to take a crack at $f$, the simplest estimator would involve a weak-universality computation. In essence getting the phase space available to the products.
3 The observationally interesting unit, I think.
4 I believe the current best guess involves at least two different temperature regimes present at once.
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For velocity-averaged cross-section $\sigma v$ and number density $n$:
$dn/dt = -n^2 (\sigma v)$
Replace $\rho=m_\chi$
$d\rho / dt=-\frac{\rho^2 \sigma v}{m_\chi}$
OTH
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