Polarization profile

Two-dimensional case

Let's assume the polarization and intensity profiles have a gaussian shape:

${\begin{aligned}I_{B}(x,y)&=I_{0,B}\exp \left\{-{\frac {x^{2}}{\sigma _{x,I}^{2}}}-{\frac {y^{2}}{\sigma _{y,I}^{2}}}\right\}\\I_{Y}(x,y)&=I_{0,Y}\exp \left\{-{\frac {x^{2}}{\sigma _{x,I}^{2}}}-{\frac {y^{2}}{\sigma _{y,I}^{2}}}\right\}\\P_{B}(x,y)&=P_{0,B}\exp \left\{-{\frac {x^{2}}{\sigma _{x,P}^{2}}}-{\frac {y^{2}}{\sigma _{y,P}^{2}}}\right\}\\P_{Y}(x,y)&=P_{0,Y}\exp \left\{-{\frac {x^{2}}{\sigma _{x,P}^{2}}}-{\frac {y^{2}}{\sigma _{y,P}^{2}}}\right\}\end{aligned}}$

Since we are interested only in the width of the polarization profile with respect to the intensity one we can use the following relations:

${\begin{aligned}\sigma _{x,I}^{2}&\equiv \sigma _{y,I}^{2}\equiv 1\\\sigma _{x,P}^{2}&\equiv 1/R_{x}\\\sigma _{y,P}^{2}&\equiv 1/R_{y}\end{aligned}}$

Integrating from $-\infty$ to $\infty$ over both dimensions we get for the polarization weighted with intensity of either one or both beams:

${\begin{aligned}{\frac {\iint P(x,y)I(x,y)dxdy}{\iint I(x,y)dxdy}}&={\frac {P_{0}}{{\sqrt {1+R_{x}}}{\sqrt {1+R_{y}}}}}\\{\frac {\iint P(x,y)I_{B}(x,y)I_{Y}(x,y)dxdy}{\iint I_{B}(x,y)I_{Y}(x,y)dxdy}}&={\frac {P_{0}}{{\sqrt {1+{\frac {R_{x}}{2}}}}{\sqrt {1+{\frac {R_{y}}{2}}}}}}\\{\frac {\iint P_{B}(x,y)P_{Y}(x,y)I_{B}(x,y)I_{Y}(x,y)dxdy}{\iint I_{B}(x,y)I_{Y}(x,y)dxdy}}&={\frac {P_{0,B}P_{0,Y}}{{\sqrt {1+{\frac {R_{x,B}}{2}}+{\frac {R_{x,Y}}{2}}}}{\sqrt {1+{\frac {R_{y,B}}{2}}+{\frac {R_{y,Y}}{2}}}}}}\end{aligned}}$

As we normaly measure the average polarization $\langle P\rangle$ given by it is trivial to get the equations for re-weighting factors $k_{SSA}$ and $k_{DSA}$:

${\begin{aligned}\langle P\rangle _{SSA}&=k_{SSA}\times \langle P\rangle \\\langle P_{B}\cdot P_{Y}\rangle _{DSA}&=k_{DSA}\times \langle P_{B}\rangle \cdot \langle P_{Y}\rangle \end{aligned}}$

where

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} k_{SSA} &= \frac{\sqrt{1 + R_x} \sqrt{1 + R_y}}{ \sqrt{1 + \frac{R_x}{2}} \sqrt{1 + \frac{R_y}{2}} }\\ k_{DSA} &= \frac{ \sqrt{1 + R_{x,B}} \sqrt{1 + R_{y,B}} \sqrt{1 + R_{x,Y}} \sqrt{1 + R_{y,Y}} } { \sqrt{1 + \frac{R_{x,B}}{2} + \frac{R_{x,Y}}{2} } \sqrt{1 + \frac{R_{y,B}}{2} + \frac{R_{y,Y}}{2} } }\end{align}}

It is interesting to study the difference between the scale factors $k_{SSA}$ and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle k_{DSA}} . To make things easier we assume the same value for all $R$'s which is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sim 0.2} .

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \frac{k_{DSA}}{k_{SSA,B} k_{SSA,Y}} = 1 + \frac{R^2}{4(1+R)}\end{align}}

where the last term gives a correction on the order of $\lesssim 1\%$. Therefore, with good precision we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \langle P_B\cdot P_Y\rangle_{DSA} &\approx \langle P\rangle_{SSA,B} \langle P\rangle_{SSA,Y} \end{align}}