Formalism for power spectral density estimation for non-identical and correlated noise using the null channel in Einstein Telescope

Abstract

Several proposed gravitational wave interferometers have a triangular configuration, such as the Einstein Telescope and the Laser Interferometer Space Antenna. For such a configuration one can construct a unique null channel insensitive to gravitational waves from all directions. We expand on earlier work and describe how to use the null channel formalism to estimate the power spectral density for the Einstein Telescope interferometers with non-identical as well as correlated noise sources. The formalism is illustrated with two examples in the context of the Einstein Telescope, with increasing degrees of complexity and realism. By using known mixtures of noises we show the formalism is mathematically correct and internally consistent. Finally we highlight future research needed to use this formalism as an ingredient for a Bayesian estimation framework.

Appendices

Appendix A: Antenna pattern functions for tensorial gravitational waves

We follow the description of the ET’s triangle of Fig. 1 from [11] but interferometer indices 1,2,3 become X,Y,Z.

We introduce the orthogonal triad in the transverse trace less form as $(\mathbf {e_x}, \mathbf {e_y}, \mathbf {e_z})$, which allows us to write the basis polarization tensors:

$$\begin{aligned} \begin{aligned} {\textbf{e}}_+&= \mathbf {e_x} \otimes \mathbf {e_x} - \mathbf {e_y} \otimes \mathbf {e_y} \\ {\textbf{e}}_{\times }&= \mathbf {e_x} \otimes \mathbf {e_y} + \mathbf {e_y} \otimes \mathbf {e_x} \end{aligned} \end{aligned}$$

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with ${\textbf{h}} = h_+{\textbf{e}}_+ + h_+{\textbf{e}}_{\times }$ and $h_+, h_{\times }$ the tensor plus-, respectively cross-polarization components.

The symmetric trace-free tensors representing the three ET interferometers are given by:

$$\begin{aligned} {d}^X&= \frac{1}{2}\mathbf {e_1} \otimes \mathbf {e_1} - \mathbf {e_2} \otimes \mathbf {e_2} \\ {d}^Y&= \frac{1}{2}\mathbf {e_2} \otimes \mathbf {e_2} - \mathbf {e_3} \otimes \mathbf {e_3} \\ {d}^Z&= \frac{1}{2}\mathbf {e_3} \otimes \mathbf {e_3} - \mathbf {e_1} \otimes \mathbf {e_1} \end{aligned}$$

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with the basis of the three arms of the ET’s configuration given by ${\textbf{e}}_1;{\textbf{e}}_2;{\textbf{e}}_3 = \frac{1}{2}(\sqrt{3},-1,0); \frac{1}{2}(\sqrt{3},1,0); (0,1,0)$.

For each interferometer ($I = X,Y,Z$), the interferometer response $h^I(f)$ is given by the product between the detector tensor ${\textbf{d}}^I$ and the tensor ${\textbf{h}}$:

$$\begin{aligned} \begin{aligned} h^I(t)&= d^I_{ij}h^{ij} = d^I_{ij}e_+^{ij}h_+ + d^I_{ij}e_{\times }^{ij}h_{\times } \\&= F_+^I h_+ + F_+^{\times } h_{\times } \end{aligned} \end{aligned}$$

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with $F^I_p$ the antenna pattern function of interferometer I for a GW with polarization p.

According to [11], it is possible to write the unit vector of the basis (xyz) in the radiation frame. We introduce the polarization angle $\psi$ as $\cos \psi = {\textbf{e}}_{\theta } \cdot {\textbf{e}}_x$, so, we can write the antenna pattern function as a function of the sky position $(\theta , \phi )$. For example, the antenna pattern function of the interferometer X is:

$$\begin{aligned} \begin{aligned} F_{+}^X&= \frac{-\sqrt{3}}{4}\Big [\big ( 1 + \cos ^2 \theta \big ) \sin 2\phi \cos 2\psi + 2 \cos \theta \cos 2\phi \sin 2\psi \Big ] \\&= \alpha (\theta ,\phi )\cos 2\psi + \beta (\theta ,\phi )\sin 2\psi \\ F_{\times }^X&= \frac{\sqrt{3}}{4}\Big [\big ( 1 + \cos ^2 \theta \big ) \sin 2\phi \sin 2\psi - 2 \cos \theta \cos 2\phi \cos 2\psi \Big ] \\&= -\alpha (\theta ,\phi )\sin 2\psi + \beta (\theta ,\phi )\cos 2\psi , \end{aligned} \end{aligned}$$

(22)

where we have defined $\alpha (\theta ,\phi ) = -\frac{\sqrt{3}}{4} \big ( 1 + \cos ^2 \theta \big ) \sin 2\phi$ and $\beta (\theta ,\phi ) = -\frac{\sqrt{3}}{2} \cos \theta \cos 2\phi$. It is possible, given the equilateral triangle configuration, to calculate the antenna pattern function of the interferometer Y and Z as a rotation of $2\pi /3$. This is given by changing the angle $\phi$ such that $\phi \longrightarrow \phi +\frac{2\pi }{3}$ for Y arm and $\phi \longrightarrow \phi -\frac{2\pi }{3}$ for Z arm.

$$\begin{aligned} \begin{aligned}&F_{p}^Y(\theta , \phi , \psi ) = F_{p}^X(\theta , \phi +\frac{2\pi }{3}, \psi ) \\&F_{p}^Z(\theta , \phi , \psi ) = F_{p}^X(\theta , \phi -\frac{2\pi }{3}, \psi ) \\ \end{aligned} \end{aligned}$$

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with the polarization, $p = +, \times$. We can also write $\sum _p (F^X_p)^2$:

$$\begin{aligned} \begin{aligned} (F_{+}^X)^2 + (F_{\times }^X)^2&= (\alpha \cos 2\psi + \beta \sin 2\psi )^2 \\&\quad + (-\alpha \sin 2\psi + \beta \cos 2\psi )^2 \\&= \alpha ^2 + \beta ^2 \\&= \frac{3}{16} \Big [\big ( 1 + \cos ^2 \theta \big )^2 \sin ^2 2\phi + 4 \cos ^2\theta \cos ^2 2\phi \Big ] \end{aligned} \end{aligned}$$

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We assume that the noise and GWs are not correlated. The PSD of the X channel is given by

$$\begin{aligned} \begin{aligned} \langle X(f) X^*(f^{\prime })\rangle&= \langle n^X(f) n^{X*}(f^{\prime })\rangle + \langle d^X_{ij} h^{ij}(f) (d^X_{ij}h^{ij}(f^{\prime }))^*\rangle \\&=\frac{1}{2}\delta (f-f^{\prime })S_n^X + \left\langle \sum _p F^X_p h_p \left( \sum _{p\prime } F^X_{p\prime } h_{p\prime }\right) ^* \right\rangle \\&= \frac{1}{2}\delta (f-f^{\prime }) S_n^X + \left\langle \sum _p \left( F_p^X h_p\right) ^2 \right\rangle , \\&= \frac{1}{2}\delta (f-f^{\prime }) \left[ S_n^X + \sum _p S_{h_p} \int _{sky} \left( F_p^X\right) ^2 \right] , \\ \end{aligned} \end{aligned}$$

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where $S_X$ is the noise PSD of interferometer X and $S_{h_p}$ the PSD of the GW signal for an isotropic SGWB present with polarization p.

According to Eq. 25, we can calculate the antenna patterns over the sky:

$$\begin{aligned} \begin{aligned} \int _{sky} \left( F_+^X\right) ^2 + \left( F_{\times }^X\right) ^2&= \frac{1}{4\pi ^2}\int _{0}^{\pi } \int _{0}^{2\pi } \int _{0}^{\pi } \sin \theta {\textrm{d}}\theta {\textrm{d}}\phi {\textrm{d}}\psi \Big [\left( F_+^X\right) ^2 + \left( F_{\times }^X\right) ^2 \Big ] \\&= \frac{1}{4\pi ^2} \frac{6\pi ^2}{5}\\&= \frac{3}{10} \end{aligned} \end{aligned}$$

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The calculation for the Y and Z arms is identical and redundant. We can thus write:

$$\begin{aligned} \begin{aligned} \int _{sky} \left( F_+^X\right) ^2 + \left( F_{\times }^X\right) ^2&= \int _{sky} \left( F_+^Y\right) ^2 + \left( F_{\times }^Y\right) ^2 \\&= \int _{sky} \left( F_+^Z\right) ^2 + \left( F_{\times }^Z\right) ^2= \frac{3}{10} \end{aligned} \end{aligned}$$

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It is also possible to calculate the cross arm integration of the pattern antenna. We can notice first that we can write:

$$\begin{aligned} \left( F_{+}^XF_{+}^Y\right) + \left( F_{\times }^XF_{\times }^Y\right)&= \left( \alpha \cos 2\psi + \beta \sin 2\psi \right) \left( \alpha '\cos 2\psi + \beta '\sin 2\psi \right) \\&\quad + \left( -\alpha \cos 2\psi + \beta \sin 2\psi \right) \left( -\alpha '\cos 2\psi + \beta '\sin 2\psi \right) \\&= \alpha \alpha ' + \beta \beta ' \\&= \frac{3}{16} \Big [\big ( 1 + \cos ^2 \theta \big )^2 \sin 2\phi \sin 2\phi ' \\&\quad + 4 \cos ^2\theta \cos 2\phi \cos 2\phi ' \Big ] \end{aligned}$$

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with $\alpha ' = \alpha (\theta ,\phi +\frac{2\pi }{3})$, $\beta ' = \beta (\theta ,\phi +\frac{2\pi }{3})$ and $\phi ' = \phi +\frac{2\pi }{3}$

we can calculate the antenna patterns over the sky:

$$\begin{aligned} \begin{aligned} \int _{sky} (F_{\times }^XF_{\times }^Y) + (F_{+}^XF_{+}^Y)&= \frac{1}{4\pi ^2}\int _{0}^{\pi } \int _{0}^{2\pi } \int _{0}^{\pi }\sin \theta \textrm{d}\theta \textrm{d}\phi \textrm{d}\psi \Big [(F_{\times }^XF_{\times }^Y) + (F_{+}^XF_{+}^Y) \Big ] \\&= \frac{1}{4\pi ^2} \frac{-3\pi ^2}{5} = -\frac{3}{20} \end{aligned} \end{aligned}$$

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The calculation for the (XZ) and (YZ) arms is also identical and redundant. We can thus write:

$$\begin{aligned} \int _{sky} (F_{\times }^XF_{\times }^Y) + (F_{+}^XF_{+}^Y)&= \int _{sky} (F_{\times }^XF_{\times }^Z)+ (F_{+}^XF_{+}^Z) \\&= \int _{sky} (F_{\times }^YF_{\times }^Z) + (F_{+}^YF_{+}^Z)\\&= -\frac{3}{20} \end{aligned}$$

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We can summarize the contribution of an isotropic SGWB to each channel according to the Eq. 8:

$$\begin{aligned} \begin{aligned} \left<X(f)X^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^X(f) + \frac{3}{10}S_h(f)\right] \\ \left<A(f)A^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^A(f) + \frac{9}{20}S_h(f)\right] \\ \left<E(f)E^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^E(f)+ \frac{9}{20}S_h(f)\right] \\ \left<A(f)E^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^{AE}(f) + 0S_h(f)\right] \\ \end{aligned} \end{aligned}$$

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Appendix B: Antenna pattern functions for non-GR polarization gravitational waves

The non-GR polarization can also be defined from the orthogonal triad $(\mathbf {e_x}, \mathbf {e_y}, \mathbf {e_z})$ [56]. We define two other kind of polarization, the vector (x, y) and scalar (b, l) modes [57, 58].

$$\begin{aligned} \begin{aligned} {\textbf{e}}_x&= \mathbf {e_x} \otimes \mathbf {e_z}+ \mathbf {e_z} \otimes \mathbf {e_x} \\ {\textbf{e}}_y&= \mathbf {e_y} \otimes \mathbf {e_z} + \mathbf {e_z} \otimes \mathbf {e_y} \end{aligned} \end{aligned}$$

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and

$$\begin{aligned} \begin{aligned} {\textbf{e}}_b&= \mathbf {e_x} \otimes \mathbf {e_x}+ \mathbf {e_y} \otimes \mathbf {e_y} \\ {\textbf{e}}_l&= \mathbf {e_z} \otimes \mathbf {e_z} \end{aligned} \end{aligned}$$

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For interferometer X, the antenna pattern of the non-GR modes are given by,

• Vector modes:

  $$\begin{aligned} \begin{aligned} F^x\left( \theta , \phi , \psi \right)&=\frac{-\sqrt{3}}{2}\sin \theta \left[ \cos \theta \cos \psi \sin 2\phi +\sin \psi \cos 2\phi \right] \\ F^y\left( \theta , \phi , \psi \right)&=\frac{\sqrt{3}}{2}\sin \theta \left[ \cos \theta \sin \psi \sin 2\phi -\cos \psi \cos 2\phi \right] \\ \end{aligned} \end{aligned}$$

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• Scalar modes:

  $$\begin{aligned} \begin{aligned} F^b(\theta , \phi )&=\frac{\sqrt{3}}{4}\sin ^2\theta \sin 2\phi \\ F^l(\theta , \phi )&=\frac{-\sqrt{3}}{4}\sin ^2\theta \sin 2\phi \\ \end{aligned} \end{aligned}$$

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The integration over the sky for differences configuration are:

• Vector modes:

  $$\begin{aligned} \begin{aligned} \int _{sky} \left( F_x^X\right) ^2 + \left( F_y^X\right) ^2&= \int _{sky} \left( F_x^Y\right) ^2 + \left( F_y^Y\right) ^2 \\ {}&= \int _{sky} \left( F_x^Z\right) ^2 + \left( F_y^Z\right) ^2 = \frac{3}{10} \\ \int _{sky} \left( F_y^XF_y^Y\right) + \left( F_x^XF_x^Y\right)&= \int _{sky} \left( F_y^XF_y^Z\right) + \left( F_x^XF_x^Z\right) \\ {}&= \int _{sky} \left( F_y^YF_y^Z\right) + \left( F_x^YF_x^Z\right) = -\frac{-3}{20} \end{aligned} \end{aligned}$$

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• Scalar modes:

  $$\begin{aligned} \begin{aligned} \int _{sky} (F_b^X)^2 + (F_l^X)^2&= \int _{sky} (F_b^Y)^2 + (F_l^Y)^2 \\&= \int _{sky} (F_b^Z)^2 + (F_l^Z)^2= \frac{1}{10} \\ \int _{sky} (F_l^XF_l^Y) + (F_b^XF_b^Y)&= \int _{sky} (F_l^XF_l^Z) + (F_b^XF_b^Z) \\&= \int _{sky} (F_b^YF_b^Z) + (F_l^YF_l^Z)= -\frac{-1}{20} \end{aligned} \end{aligned}$$

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We can summarize the contribution from an isotropic SGWB to each channel for the different polarizations:

• Vector modes:

  $$\begin{aligned} \begin{aligned} \left<X(f)X^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^X(f) + \frac{3}{10}S_h(f)\right] \\ \left<A(f)A^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^A(f) + \frac{9}{20}S_h(f)\right] \\ \left<E(f)E^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^E(f) + \frac{9}{20}S_h(f)\right] \\ \left<A(f)E^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^{AE}(f) + 0S_h(f)\right] \\ \end{aligned} \end{aligned}$$

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• Scalar modes:

  $$\begin{aligned} \begin{aligned} \left<X(f)X^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^X(f) + \frac{1}{10}S_h(f)\right] \\ \left<A(f)A^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^A(f) + \frac{3}{20}S_h(f)\right] \\ \left<E(f)E^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^E(f) + \frac{3}{20}S_h(f)\right] \\ \left<A(f)E^*(f')\right>&= \frac{1}{2}\delta (f-f')\left[ S_n^{AE}(f) + 0S_h(f)\right] \\ \end{aligned} \end{aligned}$$

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