Difference between revisions of "Variables Used in Elastic Scattering"

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Finding the cross terms,
 
Finding the cross terms,
  
<center><math>{\mathbf P_1^*}\cdot {\mathbf P^{'*}}=\left(\begin{matrix} E_1^*\\ p_{1(x)}^* \\ p_{1(y)}^* \\ p_1(z)}^* \end{matrix} \right)\cdot \left( \begin{matrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 &0 & 0 &-1\end{matrix} \right)\cdot \left(\begin{matrix} E^{'*} & p_{1(x)}^{'*} & p_{1(y)}^{'*} & p_{1(z)}^{'*} \end{matrix} \right)=E_1^*E_1^{'*}-\vec p_1^*\cdot \vec p_1^{'*} </math></center>
+
<center><math>{\mathbf P_1^*}\cdot {\mathbf P^{'*}}=\left(\begin{matrix} E_1^*\\ p_{1(x)}^* \\ p_{1(y)}^* \\ p_{1(z)}^* \end{matrix} \right)\cdot \left( \begin{matrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 &0 & 0 &-1\end{matrix} \right)\cdot \left(\begin{matrix} E^{'*} & p_{1(x)}^{'*} & p_{1(y)}^{'*} & p_{1(z)}^{'*} \end{matrix} \right)=E_1^*E_1^{'*}-\vec p_1^*\cdot \vec p_1^{'*} </math></center>
  
 
=Mandelstam Representation=
 
=Mandelstam Representation=
  
 
[[File:Mandelstam.png | 400 px]]
 
[[File:Mandelstam.png | 400 px]]

Revision as of 23:44, 31 January 2016

Lorentz Invariant Quantities

Total 4-Momentums

As was shown earlier the scalar product of a 4-Momentum vector with itself ,

[math]{\mathbf P_1}\cdot {\mathbf P^1}=E_1E_1-\vec p_1\cdot \vec p_1 =m_{1}^2=s[/math]

,

and the length of a 4-Momentum vector composed of 4-Momentum vectors,

[math]{\mathbf P^2}=({\mathbf P_1}+{\mathbf P_2})^2=(E_1+E_2)^2-(\vec p_1 +\vec p_2 )^2=(m_1+m_2)^2=s[/math]

,

are invariant quantities.

It was further shown that

[math]{\mathbf P^*}^2={\mathbf P}^2[/math]


where [math]{\mathbf P^*}=({\mathbf P_1^*}+{\mathbf P_2^*})^2[/math] represents the 4-Momentum Vector in the CM frame


and [math]{\mathbf P}=({\mathbf P_1}+{\mathbf P_2})^2[/math] represents the 4-Momentum Vector in the initial Lab frame

which can be expanded to

[math]{\mathbf P^*}^2={\mathbf P^{'*}}^2={\mathbf P}^2={\mathbf P^'}^2[/math]


where [math]{\mathbf P^'}=({\mathbf P_1^'}+{\mathbf P_2^'})^2[/math] represents the 4-Momentum Vector in the final Lab frame


and [math]{\mathbf P^{'*}}=({\mathbf P_1^{'*}}+{\mathbf P_2^{'*}})^2[/math] represents the 4-Momentum Vector in the final CM frame

New 4-Momentum Quantities

Working in just the CM frame, we can form new 4-Momentum Vectors comprised of 4-Momenta in this frame, with

[math]{\mathbf P_1^*}- {\mathbf P_1^{'*}}= \left( \begin{matrix}E_1^*-E_1^{'*}\\ p_{1(x)}^*-p_{1(x)}^{'*} \\ p_{1(y)}^*-p_{1(y)}^{'*} \\ p_{1(z)}^*-p_{1(z)}^{'*}\end{matrix} \right)={\mathbf P_a^*}[/math]


[math]{\mathbf P_1^*}- {\mathbf P_2^{'*}}= \left( \begin{matrix}E_1^*-E_2^{'*}\\ p_{1(x)}^*-p_{2(x)}^{'*} \\ p_{1(y)}^*-p_{2(y)}^{'*} \\ p_{1(z)}^*-p_{2(z)}^{'*}\end{matrix} \right)={\mathbf P_b^*}[/math]


[math]{\mathbf P_2^*}- {\mathbf P_1^{'*}}= \left( \begin{matrix}E_2^*-E_1{'*}\\ p_{2(x)}^*-p_{1(x)}^{'*} \\ p_{2(y)}^*-p_{1(y)}^{'*} \\ p_{2(z)}^*-p_{1(z)}^{'*}\end{matrix} \right)={\mathbf P_c^*}[/math]


[math]{\mathbf P_2^*}- {\mathbf P_2^{'*}}= \left( \begin{matrix}E_2^*-E_2^{'*}\\ p_{2(x)}^*-p_{2(x)}^{'*} \\ p_{2(y)}^*-p_{2(y)}^{'*} \\ p_{2(z)}^*-p_{2(z)}^{'*}\end{matrix} \right)={\mathbf P_d^*}[/math]

Using the algebraic fact

[math]\left({\mathbf a}- {\mathbf b}\right)^2=\left({\mathbf b}- {\mathbf a}\right)^2[/math]


and the fact that the length of these 4-Momentum Vectors are invariant,

[math]\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_1^*}^2-2{\mathbf P_1^*}\cdot {\mathbf P_1^{'*}}+ {\mathbf P_1^{'*}}\right)= \left( \begin{matrix}E_1^*-E_1^{'*}\\ p_{1(x)}^*-p_{1(x)}^{'*} \\ p_{1(y)}^*-p_{1(y)}^{'*} \\ p_{1(z)}^*-p_{1(z)}^{'*}\end{matrix} \right)^2=\left({\mathbf P_a^*}\right)^2[/math]


[math]\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_1^*}^2-2{\mathbf P_1^*}\cdot {\mathbf P_2^{'*}}+ {\mathbf P_2^{'*}}\right)= \left( \begin{matrix}E_1^*-E_2^{'*}\\ p_{1(x)}^*-p_{2(x)}^{'*} \\ p_{1(y)}^*-p_{2(y)}^{'*} \\ p_{1(z)}^*-p_{2(z)}^{'*}\end{matrix} \right)^2=\left({\mathbf P_b^*}\right)^2[/math]


[math]\left({\mathbf P_2^*}- {\mathbf P_1^{'*}}\right)^2=\left({\mathbf P_2^*}^2-2{\mathbf P_2^*}\cdot {\mathbf P_1^{'*}}+ {\mathbf P_1^{'*}}\right)= \left( \begin{matrix}E_2^*-E_1^{'*}\\ p_{2(x)}^*-p_{1(x)}^{'*} \\ p_{2(y)}^*-p_{1(y)}^{'*} \\ p_{2(z)}^*-p_{1(z)}^{'*}\end{matrix} \right)^2=\left({\mathbf P_c^*}\right)^2[/math]


[math]\left({\mathbf P_2^*}- {\mathbf P_2^{'*}}\right)^2=\left({\mathbf P_2^*}^2-2{\mathbf P_2^*}\cdot {\mathbf P_2^{'*}}+ {\mathbf P_2^{'*}}\right)= \left( \begin{matrix}E_2^*-E_2^{'*}\\ p_{2(x)}^*-p_{2(x)}^{'*} \\ p_{2(y)}^*-p_{2(y)}^{'*} \\ p_{2(z)}^*-p_{2(z)}^{'*}\end{matrix} \right)^2=\left({\mathbf P_d^*}\right)^2[/math]

Using the fact that the scalar product of a 4-momenta with itself is invariant,


[math]{\mathbf P_1}\cdot {\mathbf P^1}=E_1E_1-\vec p_1\cdot \vec p_1 =m_{1}^2[/math]


We can simiplify the expressions

[math]\left({\mathbf P_1^*}- {\mathbf P_1^{'*}}\right)^2=\left( m_1^{*2}-2{\mathbf P_1^*}\cdot {\mathbf P_1^{'*}}+ m_1^{'*2}\right)=\left({\mathbf P_a^*}\right)^2[/math]


[math]\left({\mathbf P_1^*}- {\mathbf P_2^{'*}}\right)^2=\left( m_1^{*2}-2{\mathbf P_1^*}\cdot {\mathbf P_2^{'*}}+ m_2^{'*2}\right)=\left({\mathbf P_b^*}\right)^2[/math]


[math]\left({\mathbf P_2^*}- {\mathbf P_1^{'*}}\right)^2=\left( m_2^{*2}-2{\mathbf P_2^*}\cdot {\mathbf P_1^{'*}}+ m_1^{'*2}\right)=\left({\mathbf P_c^*}\right)^2[/math]


[math]\left({\mathbf P_2^*}- {\mathbf P_2^{'*}}\right)^2=\left( m_2^{*2}-2{\mathbf P_2^*}\cdot {\mathbf P_2^{'*}}+ m_2^{'*2}\right)=\left({\mathbf P_d^*}\right)^2[/math]

Finding the cross terms,

[math]{\mathbf P_1^*}\cdot {\mathbf P^{'*}}=\left(\begin{matrix} E_1^*\\ p_{1(x)}^* \\ p_{1(y)}^* \\ p_{1(z)}^* \end{matrix} \right)\cdot \left( \begin{matrix}1 & 0 & 0 & 0\\0 & -1 & 0 & 0\\0 & 0 & -1 & 0\\0 &0 & 0 &-1\end{matrix} \right)\cdot \left(\begin{matrix} E^{'*} & p_{1(x)}^{'*} & p_{1(y)}^{'*} & p_{1(z)}^{'*} \end{matrix} \right)=E_1^*E_1^{'*}-\vec p_1^*\cdot \vec p_1^{'*} [/math]

Mandelstam Representation

Mandelstam.png