# Invariant mass in special relativity [closed]

I'm following a special relativity course and I'm trying to understand how the invariant mass works. In particular I don't get how the following passages work.

We have a collision between two particles and I get that the invariant mass is the sum squared of the 4-momenta. I also get the second passage, but I don't understand how to go through the last passage, in particular where each piece comes from.

$$M^2 = (p^{\mu}_1 + p^{\mu}_2)^2 = (p^{\mu}_1)^2 + (p^{\mu}_2)^2 + 2 p^{\mu}_1 p_{\mu 2} = m_1^2 + m_2^2 + 2(E_1E_2 - \vec{p}_1 \cdot \vec{p}_2)$$

## closed as off-topic by Aaron Stevens, GiorgioP, ZeroTheHero, Jon Custer, Kyle KanosMar 17 at 11:37

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• Do you know what are the components of the 4-momentum $p_\mu$? – md2perpe Mar 16 at 10:50
• Yes they are (E/c, p_x, p_y, p_z) – Matte Mar 16 at 10:54
• Ok i think i got it, it should be because p²_mu=E²-p²=m² and for the last term from matrix multiplication – Matte Mar 16 at 10:57
• So when can we just use p²_mu=m² – Matte Mar 16 at 10:58
• Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. – Kyle Kanos Mar 17 at 11:37

In general, if you have two four vectors $$A^\mu=(A^0,\vec{A})$$, $$B^\mu=(B^0,\vec{B})$$ then their Minkowski four-product is $$A^\mu B_\mu = A^0B^0-\vec{A}\cdot\vec{B}.$$
In your case particle 1 you have that $$(p_1^\mu)^2 \equiv P_1^\mu P_{1\mu}=E_1^2-\vec{p}_1^2 =m_1^2$$ (I used the shorthand $$\vec{p}^2\equiv\vec{p}\cdot\vec{p}$$). The same holds for particle 2 so that the first two terms should be ok. The last term is $$p_1^\mu p_{2\mu} = E_1E_2-\vec{p}_1\cdot\vec{p}_2$$