Head-on collision of an electron and a proton

[Philip Land]

Hey!

Let's say we have an electron and proton colliding head-on.

We will have $|p| \sim E$

Where $p_1=(E_1, \vec{p_1})$ &$p_2=(E_2, \vec{p_2})$

If we want the available energy. We can calculate $\sqrt{s} = \sqrt{(p_1 + p_2)^2}$

We get $$s= p_1^2 + p_2^2 + 2p_1p_2 = m_e^2 + m_p^2 + 2E_1E_2(1-cos(\theta)).$$

My question is why we get $1-cos(\theta)$ and not just $cos(\theta)$ which is one since the angle is zero?

 

[DrClaude]

Let's say we have an electron and proton colliding head-on.

My question is why we get $1-cos(\theta)$ and not just $cos(\theta)$ which is one since the angle is zero?

How do you define the angle?

 

[Orodruin]

The 1 is from the temporal components of the 4-momenta.

 

[mfb]

An angle of zero means the particles fly in the same direction with the same speed.