Editing Mandelstam variables (section)

==Sum==
Note that
:<math>s + t + u = \left(m_1c^2\right)^2 + \left(m_2c^2\right)^2 + \left(m_3c^2\right)^2 + \left(m_4c^2\right)^2</math>
where ''m''<sub>''i''</sub> is the mass of particle ''i''.<ref>{{cite book |last=Griffiths |first=David |author-link=David J. Griffiths |year=2008 |title=Introduction to Elementary Particles |edition=2nd |publisher=[[Wiley-VCH]] |isbn=978-3-527-40601-2 |page=113}}</ref>

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To prove this, we need to use two facts:
:*The square of a particle's four momentum is the square of its mass,
::<math>p_i^2 = \left(m_ic\right)^2  \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad (1)</math>
:*And conservation of four-momentum,
::<math>p_1 + p_2 = p_3 + p_4</math>
::<math>p_1 = -p_2 + p_3 + p_4 \quad \quad \quad \quad \quad \quad \,\, (2)</math>

So, to begin,
::<math>s /c^2 =(p_1+p_2)^2 =p_1^2 + p_2^2 + 2p_1 \cdot p_2</math>
::<math>t /c^2 =(p_1-p_3)^2=p_1^2 + p_3^2 - 2p_1 \cdot p_3</math>
::<math>u /c^2 =(p_1-p_4)^2=p_1^2 + p_4^2 - 2p_1 \cdot p_4</math>

Then adding the three while inserting squared masses leads to,
::<math>(s+t+u)/c^2=\left(m_1c\right)^2 + \left(m_2c\right)^2 + \left(m_3c\right)^2 + \left(m_4c\right)^2 + 2 p_1^2 + 2p_1 \cdot p_2 - 2p_1 \cdot p_3 - 2p_1 \cdot p_4</math>

Then note that the last four terms add up to zero using conservation of four-momentum,
::<math>2 p_1^2 + 2p_1 \cdot p_2 - 2p_1 \cdot p_3 - 2p_1 \cdot p_4 = 2p_1 \cdot (p_1 + p_2 - p_3 - p_4) = 0</math>

So finally,
:<math>(s+t+u)/c^2 = \left(m_1c\right)^2 + \left(m_2c\right)^2 + \left(m_3c\right)^2 + \left(m_4c\right)^2</math>.
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