Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?
share|cite|improve this question
Very nice question, although unfortunately I haven't the faintest idea of how to answer it. – David Z Dec 29 '12 at 5:05

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.