Difference between revisions of "Altering Phi Angles"

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(Created page with "Using the fact that <center><math>\cos{\phi} \equiv \frac{p_x}{\sqrt{p^2-p_z^2}}</math></center> <center><math>\Longrightarrow \sqrt{p^2-p_z^2}=\frac{p_x}{\cos{\phi}}=constant…")
 
 
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<center><math>\Longrightarrow p_y'=\sqrt{p^2-p_z^2-p_x^{'2}}</math></center>
 
<center><math>\Longrightarrow p_y'=\sqrt{p^2-p_z^2-p_x^{'2}}</math></center>
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Starting with a data file of momentum components constructed using awk as described above
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<center>[[File:Screen_Shot_2016-02-07_at_3.52.37_PM.png | Starting point]]</center>
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A program was written to rotate the phi angle as described above.  The changing x and y components for this distribution can be seen with
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<center>[[File:xy.png]]</center>

Latest revision as of 16:19, 10 March 2016

Using the fact that

[math]\cos{\phi} \equiv \frac{p_x}{\sqrt{p^2-p_z^2}}[/math]


[math]\Longrightarrow \sqrt{p^2-p_z^2}=\frac{p_x}{\cos{\phi}}=constant[/math]


We can simply use the expression

[math]\frac{p_x}{\cos{\phi}}=\frac{p_x'}{cos{\left(\phi+\delta \phi\right)}}[/math]


[math]\Longrightarrow p_x'=\frac{p_x \times \cos{\left(\phi+\delta \phi\right)}}{\cos{\phi}}[/math]


Then, using

[math]\sqrt{p^2-p_z^2}=\sqrt{p_x^2+p_y^2}[/math]


[math]\Longrightarrow p_y'=\sqrt{p^2-p_z^2-p_x^{'2}}[/math]

Starting with a data file of momentum components constructed using awk as described above

Starting point

A program was written to rotate the phi angle as described above. The changing x and y components for this distribution can be seen with

Xy.png