Question

1. Consider an electron accelerator able to create electrons with a kinetic energy of 105 mec2. If two such electrons were to experience a head-on collision, what would the energy of the collision look like from the vantage point of either of the electrons?

Answer 1

In the given situation, in the lab frame we have head on collision between two electrons having equal kinetic energy.

This means the velocity of the two electrons is equal in magnitude but opposite in direction. The total kinetic energy of the two electrons is twice that of one of them.

We want to know the energy of collision from the vantage point of either of the electrons i.e. the energy as seen by an observer moving with one of the electrons.

So, we need to consider the energy of electrons in a frame where one of the electrons is at rest.

In the lab frame, if velocity of one electron is  $\vec{v}$ then the velocity of other electron is  $-\vec{v}$ . The energy of one electron, including its rest mass energy is

mo E- T-

where is magnitude of $\vec{v}$ , $m_0$ is electron's rest mass, and c is the speed of light. The kinetic energy of each of these electrons is

Ee - moc? = mo c (7-1); y =

If, for example, the kinetic energy is
100 то с2
, the is about . So, the electrons involved in the collision are moving with velocity very close to the velocity of light.

The question we need to answer is this : if an observer is moving with one of the electrons, so that electron is at rest with respect to this observer, what is the speed of the other electron as seen by this observer.

In the lab frame, let us say that the velocity of one electron is in direction of positive x-axis and the velocity of the other electron is in the negative x direction. If we go to another reference frame which is moving with velocity with respect to the lab frame, in the positive x direction, we find that one of the electrons is at rest. What is the speed of the other electron in this frame?

Since the velocities we are considering are close to the speed of light, we must use the Lorentz transformation of velocities to answer above question. We find that, with respect to the electron at rest, the speed of the other electron is

, 2υ 0 1 + (υ2 /c2)

Energy of the electron moving with speed , including the rest mass energy, is

mo ca E = 1 - (02/c2)

Using above expression for in terms of , we get

E = mo c2 (c+ +04) (C2 - 02)

So, the kinetic energy of collision from the vantage point of one of the electrons is

(c? +0) 20²c E. – moc? = moc? 12 - mo c? = mo 162 v2)

This kinetic energy is approximately $2~\gamma ^2~m_0~c^2$ .

Note that the electron at rest does not contribute to collision energy.