Now let's expand the sky to 2D, well actually 3D.
Starting with the cosine interferometer equation with phase center v_{R_c} , but since we are going to use v as a variable, we use R_c instead of v_{R_c} :
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)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:
Figure :
3D sky (l,m,n) and 3D baseline on Earth (u,v,w)
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) \\Suppose s is near the phase center s_0 (small angular separation), then:
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) \\1.4.1: 2D FT Visibility
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.
1.4.2:
Below interactive figure shows how earth rotation create different kind of ellipse for a baseline.
Feel free to pan around the earth to get a better picture.
Figure :
Baseline vector (u,v) projected onto (l,m) plane.
\text{earth-centered }
\text{coordinates}
\text{telescope }
\text{locations}
\vec{b}:\text{baseline vector}
s: \text{target source}
lm\text{-plane}
\hat s: \color{black}\text{unit vector }
\text{pointing at } s
(u,v,w)
\text{coordinates}
uv\text{-plane}
\vec{b_p}:\text{proj of } \vec{b}
\text{onto } uv\text{-plane}