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  How does one derive the 2 halo term in two-point correlation function

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This question is in reference to the paper here. In Equation (86) on page 28, the authors have given the two point correlation function \begin{equation*} \xi(\mathbf{x}-\mathbf{x}^{\prime}) = \xi^{1h}(\mathbf{x}-\mathbf{x}^{\prime})+\xi^{2h}(\mathbf{x}-\mathbf{x}^{\prime}) \end{equation*} where the first term comes when we compute the density correlation using the same halo. The superscript $1h$ stands for "1 halo term" which is given by \begin{equation*} \xi^{1h}(\mathbf{x}-\mathbf{x}^{\prime}) =\int dm \frac{m^2 n(m)}{\bar{\rho}^{2}}\int d^3y\ u(\mathbf{y}|m)u(\mathbf{y}+\mathbf{x}-\mathbf{x}^{\prime}|m) \end{equation*} This term is easy to derive once we use formula (83). However, the issue is with the 2halo term which is: \begin{equation*} \begin{aligned} \xi^{2h}(\mathbf{x}-\mathbf{x}^{\prime}) &=\int dm_1 \frac{m_1 n(m_1)}{\bar{\rho}}\int dm_2\frac{m_2n(m_2)}{\bar{\rho}}\int d^3x_1 u(\mathbf{x}-\mathbf{x}_{1}|m_1)\\ &\int d^3x_2 u(\mathbf{x}^{\prime}-\mathbf{x}_{2}|m_2)\xi_{hh}(\mathbf{x}_1 - \mathbf{x}_2|m_1,m_2) \end{aligned} \end{equation*} where $u$ is the normalized density profile i.e $\int d^2x\ u(\mathbf{x}-\mathbf{x}^{\prime})=1$. I have two questions regarding this:

  1. How do I exactly derive the 2halo term (I tried using Equation for $\rho(\mathbf{x})$ as given in Equation (83) but, it does not give much).
  2. Also, what does the term $\xi_{hh}$ physically imply?
This post imported from StackExchange Physics at 2014-07-28 11:15 (UCT), posted by SE-user Debangshu
asked Dec 28, 2012 in Theoretical Physics by DebangshuMukherjee (165 points) [ no revision ]
Very nice question, although unfortunately I haven't the faintest idea of how to answer it.

This post imported from StackExchange Physics at 2014-07-28 11:15 (UCT), posted by SE-user David Z

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