Image Volume

Let $n=\sqrt{1-l^2-m^2}$ be treated as an independent variable. Then one can write a 3D Fourier transform of $V(u,v,w)$ with the conjugate variable for ($u,v,w$) being ($l, m, n$), as

$$F(l,m,n) = \int\int\int{V(u,v,w) e^{2\pi\iota (ul+vm+wn)} du dv dw}.$$ (14.2.2)

Substituting for $V(u,v,w)$ from Eq. 14.1.1 we get

$$F(l,m,n)=~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~$$

$$\int\int\left\{ \int\int\int {I(l^\prime,m^\prime) \over \s... ...e 2}-m^{\prime 2}}-n))}du dv dw \right\} dl^\prime dm^\prime.$$ (14.2.3)

Using the general result

$$\delta(l^\prime-l)=\int e^{-2\pi\iota u(l^\prime-l)} du,$$ (14.2.4)

we get

$$F(l,m,n)=\int\int{ {I(l^\prime,m^\prime) \over \sqrt{1-l^2-m... ...a(\sqrt{1-l^{\prime 2}-m^{\prime 2}}-n) dl^\prime dm^\prime}.$$ (14.2.5)

This equation then provides the connection between the 2D sky brightness distribution given by $I(l,m)$ and the result of 3D Fourier inversion of $V(u,v,w)$ given by $F(l,m,n)$ referred to as the Image volume.

$$F(l,m,n)={I(l,m)\delta(\sqrt{1-l^2-m^2}-n) \over {\sqrt{1-l^2-m^2}}}.$$ (14.2.6)

Figure 14.1: Graphical representation of the geometry of the Image volume and the celestial sphere. The point at which the celestial sphere touches the first plane of the Image volume is the point around which the 2D image inversion approximation is valid. For wider fields, emission at points along the intersection of celestial sphere and the various planes (labeled here as the celestial sphere) needs to be projected to the tangent plane to recover the undistorted 2D image. This is shown for 3 points on the celestial sphere, projected on the tangent plane, along the radial directions.

Figure 14.2: Graphical illustration to compute the distance between the tangent plane and a point in the sky at an angle of $\theta$.

Hereafter, I would use $I(l,m,n)$ to refer to the this Image volume.

In Eq.14.1.1, we have ignored the fringe rotation term $2\pi\iota w$ in the exponent. This is done here only for mathematical (and typing!) convenience. The effect of including this term would be a shift of the Image volume by one unit in the conjugate axis, namely $n$. Hence, the effect of fringe stopping is to make the top most plane of $I(l,m,n)$ tangent to the phase center position on the celestial sphere with the rest of the sphere completely contained inside the Image volume as shown in Fig. 14.1.

Remember that the third variable $n$ of the Image volume is not an independent variable and is constrained to be $n=\sqrt{1-l^2-m^2}$. Eq 14.2.6 then gives the physical interpretation of $I(l,m,n)$. Imagine the celestial sphere defined by $(l,m,n)$ enclosed by the Image volume $I(l,m,n)$, with the top most plane being tangent to the celestial sphere as shown in Fig. 14.1. Eq. 14.2.6 then says that only those parts of the Image volume correspond to the physical emission which lie on the surface of the celestial sphere. Note that since the visibility is written as a function of all the three variables $(u,v,w)$, the transfer function will also be a volume. A little thought will then reveal that $I(l,m,n)$ will be finite away from the surface of the celestial sphere also, but that would correspond to non-physical emission in the Image volume due to the side lobes of the telescope transfer function (referred to by Point spread function (PSF) or Dirty beam in the literature). A 3D deconvolution using the Dirty image- and the Dirty beam-volumes will produce a Clean image-volume. Therefore, after deconvolution, one must perform an extra operation of projecting all points in the image volume along the celestial sphere onto the 2D tangent plane to recover the 2D sky brightness distribution. Fig. 14.2 is the graphical equivalent of the statements in this paragraph.