Difference between revisions of "Polarization profile"

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(Created page with "Polarization and intensity profiles are gaussian <math> \begin{align} 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...")
 
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Polarization and intensity profiles are gaussian
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= Two-dimensional case =
  
<math>
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Let's assume the polarization and intensity profiles have a gaussian shape:
\begin{align}
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 +
<math>\begin{align}
 
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_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\} \\
 
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_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\}
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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{align}</math>
\end{align}
 
</math>
 
  
 +
Since we are interested only in the width of the polarization profile with respect to the intensity one we can use the following relations:
  
<math>
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<math>\begin{align}
\begin{align}
 
 
\sigma^2_{x,I} &\equiv \sigma^2_{y,I} \equiv 1 \\
 
\sigma^2_{x,I} &\equiv \sigma^2_{y,I} \equiv 1 \\
 
\sigma^2_{x,P} &\equiv 1/R_x \\
 
\sigma^2_{x,P} &\equiv 1/R_x \\
\sigma^2_{y,P} &\equiv 1/R_y
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\sigma^2_{y,P} &\equiv 1/R_y\end{align}</math>
\end{align}
 
</math>
 
  
 +
Integrating from <math>-\infty</math> to <math>\infty</math> over both dimensions we get for the polarization weighted with intensity of either one or both beams:
  
<math>
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<math>\begin{align}
\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(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(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} }}
+
\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}</math>
\end{align}
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</math>
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As we normaly measure the average polarization <math>\langle P \rangle</math> given by it is trivial to get the equations for re-weighting factors <math>k_{SSA}</math> and <math>k_{DSA}</math>:
 +
 
 +
<math>\begin{align}
 +
\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{align}</math>
 +
 
 +
where
  
<math>
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<math>\begin{align}
\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_{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}} }
 
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} } }
+
                 { \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}</math>
\end{align}
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</math>
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It is interesting to study the difference between the scale factors <math>k_{SSA}</math> and <math>k_{DSA}</math>. To make things easier we assume the same value for all <math>R</math>'s which is <math>\sim 0.2</math>.
 +
 
 +
<math>\begin{align}
 +
\frac{k_{DSA}}{k_{SSA,B} k_{SSA,Y}} = 1 + \frac{R^2}{4(1+R)}\end{align}</math>
 +
 
 +
where the last term gives a correction on the order of <math>\lesssim 1\%</math>. Therefore, with good precision we have
 +
 
 +
<math>\begin{align}
 +
\langle P_B\cdot P_Y\rangle_{DSA} &\approx \langle P\rangle_{SSA,B} \langle P\rangle_{SSA,Y} \end{align}</math>

Revision as of 14:27, 9 August 2012

Two-dimensional case

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

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

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

As we normaly measure the average polarization given by it is trivial to get the equations for re-weighting factors and :

where

It is interesting to study the difference between the scale factors 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_{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 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 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 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 \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}}