4-momenta

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4-momenta

As was previously shown for the space-time 4-vector, a similar 4-vector can be composed of momentum. Using index notation, the energy and momentum components can be combined into a single "4-vector" [math]\mathbf{p^{\mu}},\ \mu=0,\ 1,\ 2,\ 3[/math], that has units of momentum(i.e. E/c is a distance with c=1).

[math]\mathbf{P} \equiv \begin{bmatrix} p^0 \\ p^1 \\ p^2 \\ p^3 \end{bmatrix}= \begin{bmatrix} E \\ p_x \\ p_y \\ p_z \end{bmatrix}[/math]

As shown earlier,

[math]\mathbf R \cdot \mathbf R = x_0^2-(x_1^2+x_2^2+x_3^2)[/math]

Following the 4-vector of space-time for momentum-energy,

[math]\mathbf P \cdot \mathbf P = p_0^2-(p_1^2+p_2^2+p_3^2)[/math]

[math]\mathbf P \cdot \mathbf P = E^2-\vec p\ ^2[/math]

Using the relativistic equation for energy

[math]E^2=\vec p\ ^2+m^2[/math]

[math]\mathbf P \cdot \mathbf P = E^2-E^2+m^2[/math]

[math]\mathbf P \cdot \mathbf P = m^2[/math]

A 4-momenta vector can be composed of different 4-momenta vectors,

[math]\mathbf P \equiv \mathbf P_1 +\mathbf P_2[/math]

This allows us to write

[math]\mathbf P^2 \equiv (\mathbf P_1+\mathbf P_2)^2[/math]

[math]\mathbf (\mathbf P_1 +\mathbf P_2)^2 \equiv \mathbf P_1^2+2 \mathbf P_1 \mathbf P_2+\mathbf P_2^2[/math]

Using

[math]\mathbf (\mathbf P_1 +\mathbf P_2)^2=m^2[/math]

This gives

[math]\mathbf (\mathbf P_1 +\mathbf P_2)^2 \equiv m_1^2+2 \mathbf P_1 \mathbf P_2+m_2^2[/math]

Using the relationship shown for 4-vectors,

[math]\mathbf R_1 \cdot \mathbf R_2 = x_{0_1}x_{0_2}-(x_{1_1}x_{1_2}+x_{2_1}x_{2_2}+x_{3_1}x_{3_2})[/math]

[math]\Rightarrow \mathbf P_1 \cdot \mathbf P_2 = p_{0_1}p_{0_2}-(p_{1_1}p_{1_2}+p_{2_1}p_{2_2}+p_{3_1}p_{3_2})[/math]

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

[math]\mathbf (\mathbf P_1 +\mathbf P_2)^2 \equiv m_1^2+2E_{1}E_{2}-2(\vec p_1 \vec p_2)+m_2^2[/math]

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