Initial CM Frame 4-momentum components: Difference between revisions
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<center><math>\frac {d\vec p}{dt}=0\Rightarrow \frac{d(m\vec v)}{dt}=\frac{ | <center><math>\frac {d\vec p}{dt}=0\Rightarrow \frac{d(m\vec v)}{dt}=\frac{v\ dm}{dt}\Rightarrow \frac{dm}{dt}=0</math></center> | ||
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<center><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></center> | <center><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></center> | ||
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Latest revision as of 18:53, 15 May 2018
[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]
Initial CM Frame 4-momentum components

Starting with the definition for the total relativistic energy:
Since we can assume that the frame of reference is an inertial frame, it moves at a constant velocity, the mass should remain constant.
We can use 4-momenta vectors, i.e. [math]\displaystyle{ {\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,
is invariant.
Using this notation, the sum of two 4-momenta forms a 4-vector as well
The length of this four-vector is an invariant as well
[math]\displaystyle{ \vartriangleleft }[/math] [math]\displaystyle{ \triangle }[/math] [math]\displaystyle{ \vartriangleright }[/math]