# Two-dimensional case

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

{\displaystyle {\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:

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

\displaystyle \begin{align} \frac{\iint P(x,y) I(x,y) dx dy}{\iint I(x,y) dx dy} &= \frac{P_{0}}{\sqrt{1 + R_x} \sqrt{1 + R_y}} \\ \frac{\iint P(x,y) I_B(x,y) I_Y(x,y) dx dy}{\iint I_B(x,y) I_Y(x,y) dx dy} &= \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) dx dy}{\iint I_B(x,y) I_Y(x,y) dx dy} &= \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{align}

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

{\displaystyle {\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

{\displaystyle {\begin{aligned}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{aligned}}}

It is interesting to study the difference between the scale factors $\displaystyle k_{SSA}$ and ${\displaystyle k_{DSA}}$. To make things easier we assume the same value for all $\displaystyle R$ 's which is ${\displaystyle \sim 0.2}$.

{\displaystyle {\begin{aligned}{\frac {k_{DSA}}{k_{SSA,B}k_{SSA,Y}}}=1+{\frac {R^{2}}{4(1+R)}}\end{aligned}}}

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

{\displaystyle {\begin{aligned}\langle P_{B}\cdot P_{Y}\rangle _{DSA}&\approx \langle P\rangle _{SSA,B}\langle P\rangle _{SSA,Y}\end{aligned}}}

# Time dependent P_SSA

${\displaystyle P_{SSA}=\left(1+{\frac {1}{2}}R(t)\right)P(t)=(1+{\frac {1}{2}}R_{0}+{\frac {1}{2}}R't)(P_{0}+P't)\approx P_{0}+P't+{\frac {1}{2}}R_{0}P_{0}+{\frac {1}{2}}(R_{0}P'+R'P_{0})t=P_{0}(1+{\frac {1}{2}}R_{0})+(P'+{\frac {1}{2}}(R_{0}P'+R'P_{0}))t}$