- #1
LmdL
- 72
- 1
Hello,
I'm trying to calculate a recurrence relation of the phases of 3 telescopes in a closure phase.
Usually in a stellar interferometer we have 3 telescopes, located in a triangle, measuring intensity of light in 3 points on a far field plane. I found an article, describing how the phase is reconstructed out of the closure phase. However, in this article all 3 telescopes are located in a line, and not on a plane. I'm trying to find a similar derivation as in this article, but for case where 3 telescopes are not necessary on a straight line, but can form a triangle on plane.
Article is here:
https://www.osapublishing.org/ao/abstract.cfm?uri=ao-17-13-2047
Their derivation is as following: we have 3 telescopes at points
(x_0), (x_0-\Delta x), (x_0+n \Delta x)
where n=...-1,0,1,..., i.e. the first two are fixed, with a distance of Δx between them, and the third one is "mapping" the line at points with a "step" of Δx.
Their write a following equation, that satisfy the relation of measured phases around the closed loop of the 3 telescopes:
G\left ( n \Delta x \right ) \equiv \phi \left [ \left ( n+1 \right )\Delta x \right ]-\phi\left ( n \Delta x \right )-\phi \left ( \Delta x \right )
where G is some value, measured for each position of 3 telescopes, and is known.
From this, they write a solution of the above equation as difference equation (recurrence relation) for a phase at any point:
\phi\left ( n \Delta x \right ) = \sum_{k=1}^{n-1}G\left ( k \Delta x \right )+n \phi \left ( \Delta x \right )
I'm trying to find a similar recurrence relation, but for a case where all 3 telescopes are not in line. Actually, from what I know from the stellar interferometry (VLA, VLBI, CHARA, etc.), generally, telescopes are not located at the same line, so there should be equation for a general case of a plane, and not a line.
So, I start from a similar setup, where 3 telescopes are located at points:
(x_0,y_0), (x_0-\Delta x,y_0), (x_0+n \Delta x,y_0+k\Delta x)
i.e. again, first two are fixed in place and the third one is "mapping", but this time a plane, and not a line.
I get the following equation for the phases around a closed loop:
G\left ( n \Delta x, k \Delta y \right) \equiv \phi \left [ \left ( n+1 \right )\Delta x, k \Delta y \right ]-\phi\left ( n \Delta x , k \Delta y \right )-\phi \left ( \Delta x , 0 \right )
Now, what is a recurrence relation solution of this equation? I guess it should be something like:
\phi\left ( n \Delta x , k \Delta y \right ) = \sum_{m=1}^{n-1} \sum_{p=2}^{k}G\left ( m \Delta x , p \Delta y \right )+n \phi \left ( \Delta x ,0 \right )+k \phi \left ( 0, \Delta y \right )
but I'm very unsure in this result.
What wonders me is that in article (1D case), a phase at some point is a linear sum of a phase at some starting point and corrections G along this line. In my guess above, a phase at some point is a linear sum of a phase at some starting point and corrections G at all points in plane, not a line.
I'm very skeptic for my solution above and will be very happy If someone could correct me. Thank you!
I'm trying to calculate a recurrence relation of the phases of 3 telescopes in a closure phase.
Usually in a stellar interferometer we have 3 telescopes, located in a triangle, measuring intensity of light in 3 points on a far field plane. I found an article, describing how the phase is reconstructed out of the closure phase. However, in this article all 3 telescopes are located in a line, and not on a plane. I'm trying to find a similar derivation as in this article, but for case where 3 telescopes are not necessary on a straight line, but can form a triangle on plane.
Article is here:
https://www.osapublishing.org/ao/abstract.cfm?uri=ao-17-13-2047
Their derivation is as following: we have 3 telescopes at points
(x_0), (x_0-\Delta x), (x_0+n \Delta x)
where n=...-1,0,1,..., i.e. the first two are fixed, with a distance of Δx between them, and the third one is "mapping" the line at points with a "step" of Δx.
Their write a following equation, that satisfy the relation of measured phases around the closed loop of the 3 telescopes:
G\left ( n \Delta x \right ) \equiv \phi \left [ \left ( n+1 \right )\Delta x \right ]-\phi\left ( n \Delta x \right )-\phi \left ( \Delta x \right )
where G is some value, measured for each position of 3 telescopes, and is known.
From this, they write a solution of the above equation as difference equation (recurrence relation) for a phase at any point:
\phi\left ( n \Delta x \right ) = \sum_{k=1}^{n-1}G\left ( k \Delta x \right )+n \phi \left ( \Delta x \right )
I'm trying to find a similar recurrence relation, but for a case where all 3 telescopes are not in line. Actually, from what I know from the stellar interferometry (VLA, VLBI, CHARA, etc.), generally, telescopes are not located at the same line, so there should be equation for a general case of a plane, and not a line.
So, I start from a similar setup, where 3 telescopes are located at points:
(x_0,y_0), (x_0-\Delta x,y_0), (x_0+n \Delta x,y_0+k\Delta x)
i.e. again, first two are fixed in place and the third one is "mapping", but this time a plane, and not a line.
I get the following equation for the phases around a closed loop:
G\left ( n \Delta x, k \Delta y \right) \equiv \phi \left [ \left ( n+1 \right )\Delta x, k \Delta y \right ]-\phi\left ( n \Delta x , k \Delta y \right )-\phi \left ( \Delta x , 0 \right )
Now, what is a recurrence relation solution of this equation? I guess it should be something like:
\phi\left ( n \Delta x , k \Delta y \right ) = \sum_{m=1}^{n-1} \sum_{p=2}^{k}G\left ( m \Delta x , p \Delta y \right )+n \phi \left ( \Delta x ,0 \right )+k \phi \left ( 0, \Delta y \right )
but I'm very unsure in this result.
What wonders me is that in article (1D case), a phase at some point is a linear sum of a phase at some starting point and corrections G along this line. In my guess above, a phase at some point is a linear sum of a phase at some starting point and corrections G at all points in plane, not a line.
I'm very skeptic for my solution above and will be very happy If someone could correct me. Thank you!