Description

This routine takes the site geometry described in the LALFrDetector structure, along with a LALDetectorType parameter, and constructs the Cartesian detector location and response tensor needed to fill the LALDetector output.

The detector type is needed because different types of detectors have different response tensors. In each case the response tensor is determined by the unit vectors $\hat{u}_X$ and $\hat{u}_Y$ which are constant in an Earth-fixed rotating reference frame and point in the ``X arm'' and ``Y arm'' directions, respectively; the headings of these directions in a local frame at the detector are stored in the LALFrDetector structure.

The detector types recognized are (all names are prefaced by LALDETECTORTYPE_):

IFODIFF

An interferometer in differential mode. The response tensor is given by $d^{ab}=\frac{1}{2} (u_X^au_X^b-u_Y^au_Y^b)$. Note that this is the preferred form even in the two arms of the detector are not perpendicular (e.g., at the GEO600 site).

IFOXARM

An interferometer in one-armed mode with the X arm active. The response tensor is given by $d^{ab}=\frac{1}{2}u_X^au_X^b$.

IFOYARM

An interferometer in one-armed mode with the Y arm active. The response tensor is given by $d^{ab}=\frac{1}{2}u_Y^au_Y^b$.

IFOCOMM

An interferometer in common mode. The response tensor is given by $d^{ab}=\frac{1}{2} (u_X^au_X^b+u_Y^au_Y^b)$.

CYLBAR

A cylindrical bar detector. In this case the ``X arm'' is actually the symmetry axis of the bar, and the ``Y arm'' is ignored. The response tensor is $d^{ab}=u_X^au_X^b$.

In each of these cases, the basic transformation needed is to express a unit vector $\hat{u}$ in terms of its components in the Earth-fixed basis $\{\hat{e}_1,\hat{e}_2,\hat{e}_3\}$. The altitude angle ${\mathcal{A}}$ and azimuth angle $\zeta$ allow us to express the unit vector $\hat{u}$ corresponding to a direction in terms of an orthonormal basis consisting of a vector $\hat{e}_{\scriptstyle{\rm E}}$ pointing due East within the local horizontal, a vector $\hat{e}_{\scriptstyle{\rm N}}$ pointing due North within the local horizontal, and an upward-pointing vector $\hat{e}_{\scriptstyle{\rm U}}$ normal to the local horizontal plane.^22.1 The relationship is

$$\hat{u} = ( \hat{e}_{\scriptstyle{\rm E}}\sin\zeta + \hat{e... ...cal{A}} + \hat{e}_{\scriptstyle{\rm U}} \sin{\mathcal{A}} .$$ (22.5)

Since the local horizontal is defined as the tangent plane to the reference ellipsoid at the point with the detector's latitude $\beta$ and longitude $\lambda$, the local basis is related to the orthonormal basis $\{\hat{e}_\rho,\hat{e}_\lambda,\hat{e}_z\}$ of a cylindrical coördinate system (related to the Earth-fixed Cartesian coördinates by $x^1=\rho\cos\lambda$, $x^2=\rho\sin\lambda$, $x^3=z$, so that $\hat{e}_\rho$ points away from the Earth's axis, $\hat{e}_\lambda$ points in the direction of increasing longitude, and $\hat{e}_z$ points in the direction of increasing $x^3$) by

$$\hat{e}_{\scriptstyle{\rm E}} = \hat{e}_\lambda$$ (22.6)

$$\hat{e}_{\scriptstyle{\rm N}} = - \hat{e}_\rho \sin\beta + \hat{e}_z \cos\beta$$ (22.7)

$$\hat{e}_{\scriptstyle{\rm U}} = \hat{e}_\rho \cos\beta + \hat{e}_z \sin\beta$$ (22.8)

It is then straightforward to relate the cylindrical basis vectors to those in the Earth-fixed Cartesian system by

$$\hat{e}_\rho = \hat{e}_1\cos\lambda + \hat{e}_2\sin\lambda$$ (22.9)

$$\hat{e}_\lambda = -\hat{e}_1\sin\lambda + \hat{e}_2\cos\lambda$$ (22.10)

$$\hat{e}_z = \hat{e}_3$$ (22.11)

To express $\hat{u}$ in the Cartesian basis, we need $\hat{u}\cdot\hat{e}_1$, $\hat{u}\cdot\hat{e}_2$, and $\hat{u}\cdot\hat{e}_3$. We first observe that

$$\hat{u}\cdot\hat{e}_{\scriptstyle{\rm E}} = \cos{\mathcal{A}} \sin\zeta$$ (22.12)

$$\hat{u}\cdot\hat{e}_{\scriptstyle{\rm N}} = \cos{\mathcal{A}} \cos\zeta$$ (22.13)

$$\hat{u}\cdot\hat{e}_{\scriptstyle{\rm U}} = \sin{\mathcal{A}}$$ (22.14)

then that

$$\hat{u}\cdot\hat{e}_\rho = (\hat{u}\cdot\hat{e}_{\scriptstyle{\rm N}}) (\hat{e}_{\scriptstyl... ...style{\rm N}}) \sin\beta +(\hat{u}\cdot\hat{e}_{\scriptstyle{\rm U}}) \cos\beta$$ (22.15)

$$\hat{u}\cdot\hat{e}_\lambda = \hat{u}\cdot\hat{e}_{\scriptstyle{\rm E}}$$ (22.16)

$$\hat{u}\cdot\hat{e}_z = (\hat{u}\cdot\hat{e}_{\scriptstyle{\rm N}}) (\hat{e}_{\scriptstyl... ...style{\rm N}}) \cos\beta +(\hat{u}\cdot\hat{e}_{\scriptstyle{\rm U}}) \sin\beta$$ (22.17)

and finally that

$$\hat{u}\cdot\hat{e}_1 = (\hat{u}\cdot\hat{e}_\rho) (\hat{e}_\rho\cdot\hat{e}_1) + (\hat{u... ...hat{u}\cdot\hat{e}_\rho) \cos\lambda -(\hat{u}\cdot\hat{e}_\lambda) \sin\lambda$$ (22.18)

$$\hat{u}\cdot\hat{e}_2 = (\hat{u}\cdot\hat{e}_\rho) (\hat{e}_\rho\cdot\hat{e}_2) + (\hat{u... ...hat{u}\cdot\hat{e}_\rho) \sin\lambda +(\hat{u}\cdot\hat{e}_\lambda) \cos\lambda$$ (22.19)

$$\hat{u}\cdot\hat{e}_3 = \hat{u}\cdot\hat{e}_z$$ (22.20)