Polarization profile

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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{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 ${\displaystyle \langle P\rangle }$ 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{align} \langle P_B\cdot P_Y\rangle_{DSA} &\approx \langle P\rangle_{SSA,B} \langle P\rangle_{SSA,Y} \end{align}

Time dependent P_SSA

$\displaystyle P_{SSA} = \left(1 + \frac12 R(t)\right) P(t) = (1 + \frac12 R_0 + \frac12 R' t)(P_0 + P't) \approx P_0 + P't + \frac12 R_0 P_0 + \frac12 (R_0 P' + R' P_0) t =P_0 (1 + \frac12 R_0) + (P' + \frac12 (R_0 P' + R' P_0) ) t$