Now let's expand the sky to 2D, well actually 3D.

Starting with the cosine interferometer equation with phase center :

R_c = I \cos \left( 2\pi \frac{\vec{b} \cdot \hat{s}}{\lambda}-2\pi \frac{\vec{b} \cdot \hat{s_0}}{\lambda}\right)

Now because we are in a 3D universe, so \vec{b} and \hat{s} are 3D vectors.

\vec{b} lives on a \bold{ (u,v,w) }- grid, which is the original (x,y,z)-grid scaled by \frac{1}{\lambda} , so \frac{\vec{b}}{\lambda} = (u,v,w)

\hat s lives on a \bold{ (l,m,n) }- grid, and \hat s = (l,m,n)

In setting, we assign the phase center at \hat s_0 = (0,0,1) . See the figure below:

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In the figure above:

\hat {s_0} &= (0,0,1)\\ \frac{\vec{b}}{\lambda} &= (u,v,w) \\ \hat s &= (l,m,n) \\

Next we plug those vectors into the cosine interferometer equation:

R_c(\vec{b},\hat{s},\hat{s_0}) &= I \cos \left( 2\pi \frac{\vec{b} \cdot \hat{s}}{\lambda}-2\pi \frac{\vec{b} \cdot \hat{s_0}}{\lambda}\right) \\ &= I \cos \left( 2\pi \begin{bmatrix} u \\ v \\ w \end{bmatrix} \cdot \begin{bmatrix} l \\ m \\ n \end{bmatrix} - 2\pi \begin{bmatrix} u \\ v \\ w \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \right) \\ &= I \cos \left( 2\pi (ul+vm+wn) - 2\pi (0+0+w)\right) \\ &= I \cos \left( 2\pi (ul+vm+w(n-1)) \right) \\ & l^2 + m^ 2 \ll 1 \\ \to & \; n \approx 1 \\ \to & \; R_c(\vec{b},\hat{s},\hat{s_0}) = I \cos \left( 2\pi (ul+vm) \right) \\ &\text{or simply:}\\ R_c&(\vec{b},\hat{s}) = I \cos \left( 2\pi (ul+vm) \right) \\ \\

With the small angular extent assumption, meaning we if only only observe the sources close to the phase center, then we can reduce the 3D equation to 2D:

\vec{b} & \to (u,v) \\ \hat{s} & \to (l,m) \\ R_c(\vec{b},\hat{s}) = R_c(u,v,l,m) = I(l,m) \cos \left( 2\pi (ul+vm) \right)

same logic applies to sine interferometer:

R_s(u,v,l,m) = I(l,m) \sin \left( 2\pi (ul+vm) \right)

So given 2 stations pointing at the phase center s_0 , for a source s that is close to the phase center s_0 , we can calculate the cosine/sin correlator output as:

R_c(u,v,l,m) &= I(l,m) \cos \left( 2\pi (ul+vm) \right) \\ R_s(u,v,l,m) &= I(l,m) \sin \left( 2\pi (ul+vm) \right) \\

Now if we have \infty spatially incoherent sources around the phase center, then we have an integral:

R_c(u,v) &= \int_{l}\int_{m} I(l,m) \cos \left( 2\pi (ul+vm) \right) dldm \\ R_s(u,v) &= \int_{l}\int_{m} I(l,m) \sin \left( 2\pi (ul+vm) \right) dldm \\

And if we put them together in the following way to create the Visibility function:

V(u,v) &= R_c(u,v) -j R_s(u,v) \\ &= \int_{l}\int_{m} I(l,m) [\cos \left( 2\pi (ul+vm) \right) -j \sin \left( 2\pi (ul+vm) \right) ] dldm \\ &= \int_{l}\int_{m} I(l,m) e^{-j2\pi (ul+vm)} dldm \\

It is a 2D Fourier Transform!

\boxed{ V(u,v) = \int_{l}\int_{m} I(l,m) e^{-j2\pi (ul+vm)} dldm }

The above formula is also known as the van Cittert-Zernike Theorem.