Four-momentum conservation: electron-electron collision

Required math: calculus

Required physics: special relativity

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 3; Problem 3.9.

The invariance of the scalar product under Lorentz transformations can be useful in working out energies of particles in collisions. For example, suppose we have one electron colliding with another electron (where the second electron is at rest in the lab) and producing a new electron-positron pair (so after the collision there are 4 particles: 3 electrons and a positron). What is the minimum energy of the incoming electron (in the lab frame) to achieve this?

First, we can look at the problem in the centre of mass (COM) frame. There, the two electrons travel towards each other at the same speed {v} and after the collision all 4 particles are at rest. From conservation of momentum we must have

\displaystyle  \left(\mathbf{p}_{1}+\mathbf{p}_{2}\right)^{2}=-\left(4m\right)^{2} \ \ \ \ \ (1)

where {\mathbf{p}_{i}} is the momentum of electron {i} before the collision, and {m} is the mass of an electron or positron.

Multiplying this out, we get

\displaystyle   p_{1}^{2}+p_{2}^{2}+2\mathbf{p}_{1}\cdot\mathbf{p}_{2} \displaystyle  = \displaystyle  -m^{2}-m^{2}+2\mathbf{p}_{1}\cdot\mathbf{p}_{2}\ \ \ \ \ (2)
\displaystyle  \displaystyle  = \displaystyle  -16m^{2}\ \ \ \ \ (3)
\displaystyle  \mathbf{p}_{1}\cdot\mathbf{p}_{2} \displaystyle  = \displaystyle  -7m^{2} \ \ \ \ \ (4)

Since the scalar product on the LHS is invariant, it must also be true in the lab frame. There, {\mathbf{p}_{1}^{\prime}=\left[m,0\right]} since the first electron is at rest, and {\mathbf{p}_{2}^{\prime}=\left[E,p_{x}\right]}. Therefore

\displaystyle   \mathbf{p}_{1}^{\prime}\cdot\mathbf{p}_{2}^{\prime} \displaystyle  = \displaystyle  -mE\ \ \ \ \ (5)
\displaystyle  \displaystyle  = \displaystyle  \mathbf{p}_{1}\cdot\mathbf{p}_{2}=-7m^{2}\ \ \ \ \ (6)
\displaystyle  E \displaystyle  = \displaystyle  7m \ \ \ \ \ (7)

This gives us the minimum energy of the incoming electron in the lab frame.

To work out the velocity of the incoming electron, we note that

\displaystyle   E \displaystyle  = \displaystyle  \gamma m\ \ \ \ \ (8)
\displaystyle  \gamma \displaystyle  = \displaystyle  7=\frac{1}{\sqrt{1-v^{2}}}\ \ \ \ \ (9)
\displaystyle  1-v^{2} \displaystyle  = \displaystyle  \frac{1}{49}\ \ \ \ \ (10)
\displaystyle  v \displaystyle  = \displaystyle  \frac{4\sqrt{3}}{7} \ \ \ \ \ (11)

6 thoughts on “Four-momentum conservation: electron-electron collision

  1. Pingback: Four-momentum conservation: electron-photon collision | Physics tutorials

  2. Serkan

    I have difficulty understanding first equation. How can you momentum on one side and mass on the other side of the equation (velocity could be 1 but why)?

    1. growescience

      Remember that we’re using the four-momentum here, not the usual Newtonian 3-momentum. The {t} component of the four-momentum is the energy, which has units of mass. In the centre of mass frame the {x} component of the four-momentum adds up to zero, and the {t} component adds up to the total rest mass of the four particles after the collision. (Since all the motion is in the {x} direction, I’ve left off the {y} and {z} components). In full then:

      displaystyle   left(mathbf{p}_{1}+mathbf{p}_{2}right)^{2} displaystyle  = displaystyle  left(left[gamma m,gamma mvright]+left[gamma m,-gamma mvright]right)^{2}=-left(2gamma mright)^{2}
      displaystyle  displaystyle  = displaystyle  left(left[m,0right]+left[m,0right]+left[m,0right]+left[m,0right]right)^{2}-left(4mright)^{2}

      The first line is the total momentum of the electrons before the collision, and the last line is the total momentum of the four particles after the collision, where they are all at rest.

    2. peeterjoot

      Also keep in mind this is all with c = 1. If you want to make the dimensions work out, take each of the m‘s and multiply them by c … this restores dimensions of momentum in that equation.


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