Difference between revisions of "Final Lab Frame Moller Electron 4-momentum components in XZ Plane"

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<center><math>\Longrightarrow p^'_{2(z)}</math> should always be positive with respect to the total momentum. This is effectively a ratio of momentum in the z direction versus total momentum. </center>
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<center><math>\Longrightarrow p^'_{2(z)}</math> should always be positive with respect to the total momentum since cosine is an even function.</center>
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<center><math>\cos -x=\cos x</math></center>
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<center>This is effectively a ratio of momentum in the z direction versus total momentum. </center>
  
  
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At <math>\theta=90^{\circ} \quad \cos{90}=0 \Rightarrow p=\sqrt{p_x^2+p_y^2}</math> The motion is in the x-y plane.
 
At <math>\theta=90^{\circ} \quad \cos{90}=0 \Rightarrow p=\sqrt{p_x^2+p_y^2}</math> The motion is in the x-y plane.
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[[VanWasshenova_Thesis#Final_Lab_Frame_Moller_Electron_4-momentum_components|<math>\triangle </math>]]
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Latest revision as of 18:54, 15 May 2018

[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]


Final Lab Frame Moller Electron 4-momentum components in XZ Plane

Xz lab.png
Figure 3: Definition of Moller electron variables in the Lab Frame in the x-z plane.
Using [math]\theta '_2=\arccos \left(\frac{p^'_{2(z)}}{p^'_{2}}\right)[/math]


[math]\Longrightarrow {p^'_{2(z)}=p^'_{2}\cos(\theta '_2)}[/math]




Since,

[math]\frac{p^'_{2(z)}}{p^'_{2}}=cos(\theta '_2)[/math]


[math]\Longrightarrow p^'_{2(z)}[/math] should always be positive with respect to the total momentum since cosine is an even function.


[math]\cos -x=\cos x[/math]


This is effectively a ratio of momentum in the z direction versus total momentum.


Checking on the sign resulting from the cosine function, we are limited to:

[math]0^\circ \le \theta '_2 \le 90^\circ \equiv 0 \le \theta '_2 \le \pi \ Radians[/math]

At [math]\theta=0^{\circ} \quad \cos{0}=1 \Rightarrow p=p_z[/math] and all motion is in the z direction.


At [math]\theta=90^{\circ} \quad \cos{90}=0 \Rightarrow p=\sqrt{p_x^2+p_y^2}[/math] The motion is in the x-y plane.




[math]\underline{\textbf{Navigation}}[/math]

[math]\vartriangleleft [/math] [math]\triangle [/math] [math]\vartriangleright [/math]