Difference between revisions of "Differential Cross-Section"

From New IAC Wiki
Jump to navigation Jump to search
 
(231 intermediate revisions by the same user not shown)
Line 1: Line 1:
−
<center>[[File:Feynman1stLevel.png]]</center>
+
<center><math>\underline{\textbf{Navigation}}</math></center>
  
 +
<center>
 +
[[Lorentz_Transformation_to_Lab_Frame|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
 +
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
 +
</center>
  
−
<center><math>\mathfrak{M}=\mathfrak{M}_1+\mathfrak{M}_2</math></center>
+
=Differential Cross-Section=
  
 +
<center><math>\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 s}\frac{\mathbf p_{final}}{\mathbf p_{initial}} |\mathfrak{M} |^2</math></center>
  
−
<center><math>-i \mathfrak{M}_1=ie(\mathbf p_1+\mathbf p_2)^{\mu} \left (\frac{-ig_{\mu \nu}}{q^2} \right ) ie \mathbf (p_{1}^'+\mathbf p_2^')^{\nu}</math></center>
 
  
 +
Working in the center of mass frame
  
−
<center><math>-i \mathfrak{M}_1=ie(\mathbf p_1+\mathbf p_2)^{\mu} \left (\frac{-ig_{\mu \nu}}{(\mathbf p_2^'-p_1^')^2} \right ) ie \mathbf (p_{1}^'+\mathbf p_2^')^{\nu}</math></center>
+
<center><math>\mathbf p_{final}=\mathbf p_{initial}</math></center>
  
  
 +
Determining the scattering amplitude in the center of mass frame
  
−
<center><math>-i \mathfrak{M}_1=ie^2\left (\frac{(\mathbf p_1+\mathbf p_2)_{\mu}( (\mathbf p_{1}^'+\mathbf p_2^')^{\mu}}{(\mathbf p_2^'-p_1^')^2} \right ) </math></center>
 
  
 +
<center><math>\mathfrak{M}=e^2 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )</math></center>
  
  
−
<center><math>-i \mathfrak{M}_{e^-e^-}=-i \left ( \frac{e^2(\mathbf p_1^*+\mathbf p_{1}^{'*})_{\mu}(\mathbf p_2^*+\mathbf p_{2}^{'*})^{\mu}}{(\mathbf p_{2}^{'*}-\mathbf p_2^*)^2}- \frac{e^2(\mathbf p_1^*+\mathbf p_{2}^{'*})_{\mu}(\mathbf p_2^*+\mathbf p_{1}^{'*})^{\mu}}{(\mathbf p_{1}^{'*}-\mathbf p_2^*)^2} \right )</math></center>
+
<center><math>\mathfrak{M}^2=e^4 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )\left ( \frac{u-s}{t}+\frac{t-s}{u} \right )</math></center>
  
  
−
<center><math> \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{\mathbf P_1^* \mathbf P_2^*+\mathbf P_{1}^{'*} \mathbf P_{2}^{'*}+\mathbf P_{1}^{'*} \mathbf P_2^*+\mathbf P_1^* \mathbf P_{2}^{'*}}{(\mathbf P_{2}^{'*}-\mathbf P_2^*)^2}- \frac{\mathbf P_1^* \mathbf P_2^*+\mathbf P_{2}^{'*} \mathbf P_{1}^{'*}+\mathbf P_{2}^{'*} \mathbf P_2^*+\mathbf P_1^* \mathbf P_{1}^*}{(\mathbf P_{1}^*-\mathbf P_2^*)^2} \right )</math></center>
 
  
 +
<center><math>\mathfrak{M}^2=e^4 \left ( \frac{(u-s)^2}{t^2}+\frac{(t-s)^2}{u^2} +2\frac{(u-s)}{t}\frac{(t-s)}{u}\right )</math></center>
  
  
−
Using the fact that <math>\mathbf P_1^{'*} \mathbf P_2^{'*}=\mathbf P_1^* \mathbf P_2^* \quad  \quad \mathbf P_1^{'*} \mathbf P_1^{*}=\mathbf P_2^{'*} \mathbf P_2^* \quad  \quad \mathbf P_1^{*}
 
−
\mathbf P_2^{'*}=\mathbf P_2^* \mathbf P_1^{'*}</math>
 
  
 +
<center><math>\mathfrak{M}^2=e^4 \left ( \frac{(u^2-2us+s^2)}{t^2}+\frac{(t^2-2ts+s^2)}{u^2} +2\frac{(ut-st+s^2-us)}{tu}\right )</math></center>
  
  
−
<center><math> \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_{1}^{'*} \mathbf P_2^*}{(\mathbf P_{2}^{'*2}-2\mathbf P_{2}^{'*}\mathbf P_2^{*}+\mathbf P_2^{*2})}- \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_1^* \mathbf P_{1}^{'*}}{(\mathbf P_{1}^{'*2}-2\mathbf P_{1}^{'*}\mathbf P_2^{'*}+\mathbf P_2^{'*2})} \right )</math></center>
 
  
 +
<center><math>\mathfrak{M}^2=e^4 \left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )</math></center>
  
  
−
<center><math> \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_{1}^{'*} \mathbf P_2^*}{(\mathbf P_2^{*2}-2\mathbf P_2^{*}\mathbf P_{2}^{'*}+\mathbf P_{2}^{'*2})}- \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_1^* \mathbf P_{1}^{'*}}{(\mathbf P_2^{'*2}-2\mathbf P_2^{'*}\mathbf P_{1}^{'*}+\mathbf P_{1}^{'*2})} \right )</math></center>
 
  
 +
Using the fine structure constant (<math>with\ c=\hbar=\epsilon_0=1</math>)
 +
<center><math>\alpha \equiv \frac{e^2}{4\pi}</math></center>
  
  
−
<center><math> \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_{1}^{'*} \mathbf P_2^*}{(\mathbf P_2^*-\mathbf P_{2}^{'*})^2}- \frac{2\mathbf P_1^* \mathbf P_2^*+2\mathbf P_1^* \mathbf P_{1}^{'*}}{(\mathbf P_2^{'*}-\mathbf P_{1}^{'*})^2} \right )</math></center>
+
<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4s}\left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )</math></center>
  
  
 +
In the center of mass frame the Mandelstam variables are given by:
  
 +
<center><math>s \equiv 4E^{*2}</math></center>
  
−
<center><math> \mathfrak{M}_{e^-e^-}= e^2\left ( \frac{ (\mathbf P_1^{*2}-2 \mathbf P_1^* \mathbf P_2^{'*}+ \mathbf P_2^{'*2})-(\mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2})}{t}- \frac{(\mathbf P_1^{*2}-2 \mathbf P_1^* \mathbf P_1^{'*}+ \mathbf P_1^{'*2})-( \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2})}{u} \right )</math></center>
 
  
  
 +
<center><math>t \equiv -2p^{*2}(1-\cos{\theta})</math></center>
  
−
<center><math>\mathfrak{M}_{e^-e^-}=e^2 \left (\frac{u-s}{t}+\frac{t-s}{u} \right )</math></center>
+
 
 +
 
 +
 
 +
<center><math>u \equiv -2p^{*2}(1+\cos{\theta})</math></center>
 +
 
 +
 
 +
Calculating the parts to have common denominators:
 +
 
 +
<center><math>\left(1\right)\qquad \qquad 2=\frac{2p^{*4}\sin^4{\theta}}{p^{*4}\sin^4{\theta}}=\frac{2p^{*4}\left(1-\cos^2{\theta}\right)^2}{p^{*4}\sin^4{\theta}}=\frac{2p^{*4}\left(1-2\cos^2{\theta}+\cos^4{\theta}\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(2\right)\qquad \qquad \frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}}=\frac{8E^{*4}\sin^2{\theta}}{p^{*4}\sin^4}=\frac{8E^{*4}\left(1-\cos^2{\theta}\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(3\right)\qquad \qquad \frac{t^2}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)^2}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\tan^4{\frac{\theta}{2}}=\frac{p^{*4}\left(1-\cos{\theta}\right)^4}{p^{*4}\sin^4{\theta}}=\frac{p^{*4}\left(\cos^4{\theta}-4\cos^3{\theta}+6\cos^2{\theta}-4\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
<center><math>\left(4\right)\qquad \qquad \frac{u^2}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)^2}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\cot^4{\frac{\theta}{2}}=\frac{p^{*4}\left(1+\cos{\theta}\right)^4}{p^{*4}\sin^4{\theta}}=\frac{p^{*4}\left(\cos^4{\theta}+4\cos^3{\theta}+6\cos^2{\theta}+4\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(5\right)\qquad \qquad \frac{s^2}{u^2}=\frac{16E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{E^{*4}\sec^4{\frac{\theta}{2}}}{p^{*4}}=\frac{4E^{*4}}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*4}\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*4}\left(\cos^2{\theta}-2\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(6\right)\qquad \qquad \frac{s^2}{t^2}=\frac{16E^{*4}}{4p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*4}\csc^4{\frac{\theta}{2}}}{p^{*4}}=\frac{4E^{*4}}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*4}\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*4}\left(\cos^2{\theta}+2\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
 
 +
<center><math>\left(7\right)\qquad \qquad \frac{-2s}{t}=\frac{8E^{*2}}{2p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\left(1-\cos{\theta}\right)\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\sin^2{\theta}}=\frac{4E^{*2}p^{*2}sin^2{\theta}\left(1+\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-cos^2{\theta}\right)\left(1+\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}-\cos^2{\theta}-\cos^3{\theta}\right)}{p^{*4}sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(8\right)\qquad \qquad \frac{-2s}{u}=\frac{8E^{*2}}{2p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}sin^2{\theta}}=\frac{4E^{*2}p^{*2}sin^2{\theta}\left(1-\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-cos^2{\theta}\right)\left(1-\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}-\cos^2{\theta}+\cos^3{\theta}\right)}{p^{*4}sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(9\right)\qquad \qquad \frac{-2ts}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1-\cos{\theta}\right)\sec^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(-\cos^3{\theta}+3\cos^2{\theta}-3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
 
 +
<center><math>\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}</math></center>
 +
 
 +
 
 +
Combing like terms further,
 +
 
 +
 
 +
<center><math>\left(1,3/4\right)\rightarrow \qquad \qquad 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}</math></center>
 +
 
 +
 
 +
<center><math>\left(2,6/5\right) \rightarrow \qquad \qquad 16E^{*4}</math></center>
 +
 
 +
 
 +
<center><math>\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}</math></center>
 +
 
 +
 
 +
 
 +
Expressing this as the differential cross-section
 +
 
 +
<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4sp^{*4}\sin^4{\theta}}\left( 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}+16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}+16E^{*4}\right) </math></center>
 +
 
 +
 
 +
 
 +
<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}p^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4E^{*2}p^{*2}+4E^{*2}p^{*2}\cos^2{\theta}+4E^{*4}\right) </math></center>
 +
 
 +
 
 +
In the Ultra-relativistic limit as <math> E \approx p</math>
 +
 
 +
 
 +
<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2\left(3+\cos^2{\theta}\right)^2}{4E^{*2}\sin^4{\theta}} </math></center>
 +
 
 +
 
 +
 
 +
 
 +
In the non-relativistic limit as <math> E \approx m</math>
 +
 
 +
 
 +
<center><math>\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4m^{*2}p^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4m^{*2}p^{*2}+4m^{*2}p^{*2}\cos^2{\theta}+4m^{*4}\right) </math></center>
 +
----
 +
 
 +
<center><math>\underline{\textbf{Navigation}}</math></center>
 +
 
 +
<center>
 +
[[Lorentz_Transformation_to_Lab_Frame|<math>\vartriangleleft </math>]]
 +
[[VanWasshenova_Thesis#Weighted_Isotropic_Distribution_in_Lab_Frame|<math>\triangle </math>]]
 +
[[CED_Verification_of_DC_Angle_Theta_and_Wire_Correspondance|<math>\vartriangleright </math>]]
 +
</center>

Latest revision as of 18:36, 1 January 2019

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

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

Differential Cross-Section

[math]\frac{d\sigma}{d\Omega}=\frac{1}{64\pi ^2 s}\frac{\mathbf p_{final}}{\mathbf p_{initial}} |\mathfrak{M} |^2[/math]


Working in the center of mass frame

[math]\mathbf p_{final}=\mathbf p_{initial}[/math]


Determining the scattering amplitude in the center of mass frame


[math]\mathfrak{M}=e^2 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )[/math]


[math]\mathfrak{M}^2=e^4 \left ( \frac{u-s}{t}+\frac{t-s}{u} \right )\left ( \frac{u-s}{t}+\frac{t-s}{u} \right )[/math]


[math]\mathfrak{M}^2=e^4 \left ( \frac{(u-s)^2}{t^2}+\frac{(t-s)^2}{u^2} +2\frac{(u-s)}{t}\frac{(t-s)}{u}\right )[/math]


[math]\mathfrak{M}^2=e^4 \left ( \frac{(u^2-2us+s^2)}{t^2}+\frac{(t^2-2ts+s^2)}{u^2} +2\frac{(ut-st+s^2-us)}{tu}\right )[/math]


[math]\mathfrak{M}^2=e^4 \left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )[/math]


Using the fine structure constant ([math]with\ c=\hbar=\epsilon_0=1[/math])

[math]\alpha \equiv \frac{e^2}{4\pi}[/math]


[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4s}\left ( \frac{(t^2+s^2)}{u^2}+\frac{2s^2}{tu}+2-\frac{2us}{t^2}-\frac{2s}{t}-\frac{2ts}{u^2}-\frac{2s}{u}+\frac{(u^2+s^2)}{t^2}\right )[/math]


In the center of mass frame the Mandelstam variables are given by:

[math]s \equiv 4E^{*2}[/math]


[math]t \equiv -2p^{*2}(1-\cos{\theta})[/math]



[math]u \equiv -2p^{*2}(1+\cos{\theta})[/math]


Calculating the parts to have common denominators:

[math]\left(1\right)\qquad \qquad 2=\frac{2p^{*4}\sin^4{\theta}}{p^{*4}\sin^4{\theta}}=\frac{2p^{*4}\left(1-\cos^2{\theta}\right)^2}{p^{*4}\sin^4{\theta}}=\frac{2p^{*4}\left(1-2\cos^2{\theta}+\cos^4{\theta}\right)}{p^{*4}\sin^4{\theta}}[/math]


[math]\left(2\right)\qquad \qquad \frac{2s^2}{tu}=\frac{32E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{8E^{*4}}{p^{*4}\sin^2{\theta}}=\frac{8E^{*4}\sin^2{\theta}}{p^{*4}\sin^4}=\frac{8E^{*4}\left(1-\cos^2{\theta}\right)}{p^{*4}\sin^4{\theta}}[/math]


[math]\left(3\right)\qquad \qquad \frac{t^2}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)^2}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\tan^4{\frac{\theta}{2}}=\frac{p^{*4}\left(1-\cos{\theta}\right)^4}{p^{*4}\sin^4{\theta}}=\frac{p^{*4}\left(\cos^4{\theta}-4\cos^3{\theta}+6\cos^2{\theta}-4\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}[/math]


[math]\left(4\right)\qquad \qquad \frac{u^2}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)^2}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\cot^4{\frac{\theta}{2}}=\frac{p^{*4}\left(1+\cos{\theta}\right)^4}{p^{*4}\sin^4{\theta}}=\frac{p^{*4}\left(\cos^4{\theta}+4\cos^3{\theta}+6\cos^2{\theta}+4\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}[/math]


[math]\left(5\right)\qquad \qquad \frac{s^2}{u^2}=\frac{16E^{*4}}{4p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{E^{*4}\sec^4{\frac{\theta}{2}}}{p^{*4}}=\frac{4E^{*4}}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*4}\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*4}\left(\cos^2{\theta}-2\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}[/math]


[math]\left(6\right)\qquad \qquad \frac{s^2}{t^2}=\frac{16E^{*4}}{4p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{E^{*4}\csc^4{\frac{\theta}{2}}}{p^{*4}}=\frac{4E^{*4}}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*4}\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*4}\left(\cos^2{\theta}+2\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}[/math]



[math]\left(7\right)\qquad \qquad \frac{-2s}{t}=\frac{8E^{*2}}{2p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\left(1-\cos{\theta}\right)\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1+\cos{\theta}\right)}{p^{*2}\sin^2{\theta}}=\frac{4E^{*2}p^{*2}sin^2{\theta}\left(1+\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-cos^2{\theta}\right)\left(1+\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}-\cos^2{\theta}-\cos^3{\theta}\right)}{p^{*4}sin^4{\theta}}[/math]


[math]\left(8\right)\qquad \qquad \frac{-2s}{u}=\frac{8E^{*2}}{2p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}}{p^{*2}\left(1+\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}\left(1+\cos{\theta}\right)\left(1-\cos{\theta}\right)}=\frac{4E^{*2}\left(1-\cos{\theta}\right)}{p^{*2}sin^2{\theta}}=\frac{4E^{*2}p^{*2}sin^2{\theta}\left(1-\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-cos^2{\theta}\right)\left(1-\cos{\theta}\right)}{p^{*4}sin^4{\theta}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}-\cos^2{\theta}+\cos^3{\theta}\right)}{p^{*4}sin^4{\theta}}[/math]


[math]\left(9\right)\qquad \qquad \frac{-2ts}{u^2}=\frac{4p^{*2}\left(1-\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1-\cos{\theta}\right)\sec^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)}{p^{*4}\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1-\cos{\theta}\right)\left(1-\cos{\theta}\right)^2}{p^{*4}\left(1+\cos{\theta}\right)^2\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(-\cos^3{\theta}+3\cos^2{\theta}-3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}[/math]


[math]\left(10\right)\qquad \qquad \frac{-2us}{t^2}=\frac{4p^{*2}\left(1+\cos{\theta}\right)4E^{*2}}{4p^{*2}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}\left(1+\cos{\theta}\right)\csc^4{\frac{\theta}{2}}}{p^{*2}}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)}{p^{*4}\left(1-\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(1+\cos{\theta}\right)\left(1+\cos{\theta}\right)^2}{p^{*4}\left(1-\cos{\theta}\right)^2\left(1+\cos{\theta}\right)^2}=\frac{4E^{*2}p^{*2}\left(\cos^3{\theta}+3\cos^2{\theta}+3\cos{\theta}+1\right)}{p^{*4}\sin^4{\theta}}[/math]


Combing like terms further,


[math]\left(1,3/4\right)\rightarrow \qquad \qquad 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}[/math]


[math]\left(2,6/5\right) \rightarrow \qquad \qquad 16E^{*4}[/math]


[math]\left(8/7,10/9\right) \rightarrow \qquad \qquad 16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}[/math]


Expressing this as the differential cross-section

[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4sp^{*4}\sin^4{\theta}}\left( 4p^{*4}\cos^4{\theta}+8p^{*4}\cos^2{\theta}+4p^{*4}+16E^{*2}p^{*2}+16E^{*2}p^{*2}\cos^2{\theta}+16E^{*4}\right) [/math]


[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}p^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4E^{*2}p^{*2}+4E^{*2}p^{*2}\cos^2{\theta}+4E^{*4}\right) [/math]


In the Ultra-relativistic limit as [math] E \approx p[/math]


[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4E^{*2}\sin^4{\theta}}\left( \cos^4{\theta}+6\cos^2{\theta}+9\right)=\frac{\alpha ^2\left(3+\cos^2{\theta}\right)^2}{4E^{*2}\sin^4{\theta}} [/math]



In the non-relativistic limit as [math] E \approx m[/math]


[math]\frac{d\sigma}{d\Omega}=\frac{\alpha ^2}{4m^{*2}p^{*4}\sin^4{\theta}}\left( p^{*4}\cos^4{\theta}+2p^{*4}\cos^2{\theta}+p^{*4}+4m^{*2}p^{*2}+4m^{*2}p^{*2}\cos^2{\theta}+4m^{*4}\right) [/math]
[math]\underline{\textbf{Navigation}}[/math]

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

Retrieved from "https://wiki.iac.isu.edu/index.php?title=Differential_Cross-Section&oldid=125977"