双黑洞旋近事件数计算（中频）

未分类

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$$N(>\varrho_{\rm th})=\iiint_{\varrho(z,\mathcal{M}, \Theta)>\varrho_{\rm th}}d\Omega_{z,\mathcal{M},\Theta}\frac{dN}{dzd\mathcal{M}d\Theta}\\ =\int_{\varrho_{\rm th}}^{\infty}d\varrho\int d\mathcal{M}\int dz\frac{dN}{dzd\mathcal{M}d\varrho}$$
其中做了换元$\varrho=\varrho(\Theta)$

$$\frac{dN}{dzd\mathcal{M}d\varrho}=\frac{dV}{dz}\frac{R(\mathcal{M},z)}{(1+z)}\frac{dp}{d\varrho}(\varrho|z,\mathcal{M})\int\frac{dt_o}{d\lg f_o}d\lg f_o$$
$$\frac{d V}{d z}=\frac{4 \pi c D_{c}^{2}}{H_{0} \sqrt{\Omega_{m}(1+z)^{3}+\Omega_{\Lambda}}}$$
$$R(\mathcal{M},z)=R(z)\frac{dp}{d\mathcal{M}}$$
$$\begin{aligned} \frac{dp}{d\varrho}\left(\varrho \mid z, \mathcal{M}\right) &=\left.\frac{dp}{d\Theta}(\Theta) \frac{\partial \Theta}{\partial \varrho}\right|_{\mathcal{M}, z} \\ &=\frac{dp}{d\Theta}(\Theta) \frac{\Theta_{\varrho}}{\varrho}, \end{aligned}$$
$$P_{\Theta}(\Theta)=\left\{\begin{array}{lr}{5 \Theta(4-\Theta)^{3} / 256} & {\text { if } 0<\Theta<4} \\ {0} & {\text { otherwise }}\end{array}\right.,$$
$$\Theta_{\varrho}=\frac{\varrho}{8} \frac{D_{L}}{r_{\rm det}}\left(\frac{1.2 M_{\odot}}{\mathcal{M}(1+z)}\right)^{5 / 6}$$

$$r_{\rm det}^{2} \equiv \frac{5}{192 \pi}\left(\frac{3G}{20}\right)^{5 / 3} \frac{M_{\odot}^{2}}{c^3}\int_{f_{\min,\rm det}}^{f_{\max,\rm det }} \frac{\left(\pi M_{\odot}\right)^{2}}{\left(\pi f M_{\odot}\right)^{7 / 3} S_{n}(f)} d f$$

$$\frac{d t_{o}}{d \lg f_{o}}=\ln 10 \frac{f_{o}}{\dot{f}_{o}}=\frac{5 \ln 10}{96} \pi^{-8 / 3}\left(\frac{G \mathcal{M}(1+z)}{c^{3}}\right)^{-5 / 3} f_{o}^{-8 / 3}$$
到此为止

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$$N=\frac{5 \ln 10 c^{5} \pi^{-8/3}}{96}\int dz \frac{dV}{dz}R(z)(1+z)^{-8/3} \int f^{-8/3}d\lg f\\ \int (G\mathcal{M})^{-5/3}P(\mathcal{M})d\mathcal{M}$$
$$P(\mathcal{M})=\frac{dp}{d\mathcal{M}}$$

$$\frac{dN}{dz d\mathcal{M}}=\frac{5 \ln 10 c^{5} \pi^{-8/3}}{96} \frac{dV}{dz}R(z)(1+z)^{-8/3}(G\mathcal{M})^{-5/3}P(\mathcal{M})\\ \int f^{-8/3}d\lg f$$

$$\varrho=\varrho(z,\mathcal{M},\Theta)=\sqrt{4\int\frac{|\tilde{h}(f)|^2}{S_{\rm n}(f)}df}$$
$|h(f)|=\frac{1}{D_{L}}\left(\frac{5}{24}\right)^{1 / 2}\left(\frac{\mathcal{M}_{z}^{5}}{\pi^{4}}\right)^{1 / 6}\left(\frac{\Theta}{4}\right) f^{-7 / 6}$

For middle frequency detector, $\zeta(f_{\max})<1$, because $2f_{\max}<f_{\max,\rm det}$

$$f_{\max}=\frac{785\rm Hz}{1+z}\frac{2.8M_{\odot}}{M}$$
$$\rho=8 \Theta \frac{r_{\rm det}}{D_{L}}\left(\frac{\mathcal{M}_{z}}{1.2 M_{\odot}}\right)^{5 / 6}$$
$$r_{\rm det}^{2} \equiv \frac{5}{192 \pi}\left(\frac{3G}{20}\right)^{5 / 3} \frac{M_{\odot}^{2}}{c^3}\int_{f_{\min,\rm det}}^{f_{\max,\rm det }} \frac{\left(\pi M_{\odot}\right)^{2}}{\left(\pi f M_{\odot}\right)^{7 / 3} S_{n}(f)} d f$$

$$N(>\varrho_{\rm th})=\int_{\varrho_{\rm th}}^{\infty}\frac{dN_{\rm tot}}{d\varrho}$$