Difference between revisions of "S-Channel"

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<center><math>\textbf{\underline{Navigation}}</math>
+
<center><math>\underline{\textbf{Navigation}}</math>
  
 
[[Mandelstam_Representation|<math>\vartriangleleft </math>]]
 
[[Mandelstam_Representation|<math>\vartriangleleft </math>]]
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<center><math>s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2</math></center>
 
  
 
+
<center><math>s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2} \equiv \mathbf P_1^{'*2}+2 \mathbf P_1^{'* }\mathbf P_2^{'*}+ \mathbf P_2^{'*2}</math></center>
<center><math>s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2}</math></center>
 
  
  
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This gives,
 
This gives,
<center><math>s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+  m_2^{2}</math></center>
+
<center><math>s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+  m_2^{2} \equiv m_1^{'2}+2 \mathbf P_1^{'*} \mathbf P_2^{'*}+  m_2^{'2}</math></center>
  
  
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<center><math>s \equiv  2m^{2}+2 \mathbf P_1^* \mathbf P_2^*</math></center>
+
<center><math>s \equiv  2m^{2}+2 \mathbf P_1^* \mathbf P_2^* \equiv  2m^{2}+2 \mathbf P_1^{'*} \mathbf P_2^{'*}</math></center>
  
 
Using the relationship
 
Using the relationship
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In the center of mass frame of reference,  
 
In the center of mass frame of reference,  
  
<center><math>E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*= \vec p \ ^*</math></center>
+
<center><math>E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*</math></center>
  
  
<center><math>s_{CM} \equiv 2m^2+2(E_1^{*2}+\vec p \ ^{*2} )</math></center>
+
<center><math>s_{CM} \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} </math></center>
  
  
 
Using the relativistic energy equation
 
Using the relativistic energy equation
  
<center><math>E^2 \equiv \vec p \ ^2+m^2</math></center>
+
<center><math>E^2 \equiv \vec p_1 \ ^2+m^2</math></center>
  
  
<center><math>s_{CM} \equiv 2m^2+2((m^2+\vec p \ ^{*2})+\vec p \ ^{*2})</math></center>
+
<center><math>s_{CM} \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})</math></center>
  
  
<center><math>s_{CM}=4(m^2+\vec p \ ^{*2})=(2E)^{*2}</math></center>
+
<center><math>s_{CM}=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}</math></center>
  
  
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<center><math>\textbf{\underline{Navigation}}</math>
+
<center><math>\underline{\textbf{Navigation}}</math>
  
 
[[Mandelstam_Representation|<math>\vartriangleleft </math>]]
 
[[Mandelstam_Representation|<math>\vartriangleleft </math>]]

Latest revision as of 18:48, 15 May 2018

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

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

s Channel

The s quantity is known as the square of the center of mass energy (invariant mass)

[math]s \equiv \left({\mathbf P_1^*}+ {\mathbf P_2^{*}}\right)^2=\left({\mathbf P_1^{'*}}+ {\mathbf P_2^{'*}}\right)^2[/math]



[math]s \equiv \mathbf P_1^{*2}+2 \mathbf P_1^* \mathbf P_2^*+ \mathbf P_2^{*2} \equiv \mathbf P_1^{'*2}+2 \mathbf P_1^{'* }\mathbf P_2^{'*}+ \mathbf P_2^{'*2}[/math]


As shown earlier, the square of a 4-momentum is


[math]\mathbf P^{2} \equiv m^2[/math]

This gives,

[math]s \equiv m_1^{2}+2 \mathbf P_1^* \mathbf P_2^*+ m_2^{2} \equiv m_1^{'2}+2 \mathbf P_1^{'*} \mathbf P_2^{'*}+ m_2^{'2}[/math]


For the case [math]m_1=m_2=m[/math]


[math]s \equiv 2m^{2}+2 \mathbf P_1^* \mathbf P_2^* \equiv 2m^{2}+2 \mathbf P_1^{'*} \mathbf P_2^{'*}[/math]

Using the relationship


[math]\mathbf P_1 \cdot \mathbf P_2 = E_{1}E_{2}-(\vec p_1 \vec p_2)[/math]


[math]s \equiv 2m^2+2(E_1^*E_2^*-\vec p \ _1^* \vec p \ _2^*)[/math]


In the center of mass frame of reference,

[math]E_1^*=E_2^* \quad and \quad \vec p \ _1^*=-\vec p \ _2^*[/math]


[math]s_{CM} \equiv 2m^2+2E_1^{*2}+2\vec p_1 \ ^{*2} [/math]


Using the relativistic energy equation

[math]E^2 \equiv \vec p_1 \ ^2+m^2[/math]


[math]s_{CM} \equiv 2m^2+2m^2+2\vec p_1 \ ^{*2}+\vec p_1 \ ^{*2})[/math]


[math]s_{CM}=4(m^2+\vec p_1 \ ^{*2})=(2E_1^*)^{2}[/math]




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

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