Difference between revisions of "Initial CM Frame 4-momentum components"

From New IAC Wiki
Jump to navigation Jump to search
 
Line 1: Line 1:
<center><math>\textbf{\underline{Navigation}}</math>
+
<center><math>\underline{\textbf{Navigation}}</math>
 
 
 
[[Initial_Lab_Frame_4-momentum_components|<math>\vartriangleleft </math>]]
 
[[Initial_Lab_Frame_4-momentum_components|<math>\vartriangleleft </math>]]
 
[[VanWasshenova_Thesis#Initial_4-momentum_Components|<math>\triangle </math>]]
 
[[VanWasshenova_Thesis#Initial_4-momentum_Components|<math>\triangle </math>]]
Line 61: Line 60:
  
  
<center><math>\textbf{\underline{Navigation}}</math>
+
<center><math>\underline{\textbf{Navigation}}</math>
  
 
[[Initial_Lab_Frame_4-momentum_components|<math>\vartriangleleft </math>]]
 
[[Initial_Lab_Frame_4-momentum_components|<math>\vartriangleleft </math>]]

Latest revision as of 18:53, 15 May 2018

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

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


Initial CM Frame 4-momentum components

400px-CMcopy.png
Figure 2: Definition of variables in the Center of Mass Frame


Starting with the definition for the total relativistic energy:


[math]E^2\equiv p^2c^2+m^2c^4[/math]


[math]\Longrightarrow {E^2}-p^2c^2=(mc^2)^2[/math]

Since we can assume that the frame of reference is an inertial frame, it moves at a constant velocity, the mass should remain constant.


[math]\frac {d\vec p}{dt}=0\Rightarrow \frac{d(m\vec v)}{dt}=\frac{v\ dm}{dt}\Rightarrow \frac{dm}{dt}=0[/math]


[math] \therefore m=const[/math]


We can use 4-momenta vectors, i.e. [math]{\mathbf P}\equiv \left(\begin{matrix} E\\ p_x \\ p_y \\ p_z \end{matrix} \right)=\left(\begin{matrix} E\\ \vec p \end{matrix} \right)[/math] ,with c=1, to describe the variables in the CM Frame.


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


[math]{\mathbf P_1}\cdot {\mathbf P^1}=P_{\mu}g_{\mu \nu}P^{\nu}=\left(\begin{matrix} E\\ p_x \\ p_y \\ p_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_x & p_y & p_z \end{matrix} \right)[/math]


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


is invariant.


Using this notation, the sum of two 4-momenta forms a 4-vector as well

[math]{\mathbf P_1}+ {\mathbf P_2}= \left( \begin{matrix}E_1+E_2\\\vec p_1 +\vec p_2 \end{matrix} \right)= {\mathbf P}[/math]

The length of this four-vector is an invariant as well

[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]




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

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