Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.15

Statement

$14.4.15.$
To study the field of electrons at short distances, they are accelerated to en-
ergies N = 1000 times greater than the rest energy of the electron mec2and
the counter interaction of two such electrons is observed. How many times do you need to increase the energy of an electron to get the same results by ob-
serving the interaction between a moving electron and an initially stationary
electron?

Solution

Center-of-mass energy in the symmetric collider

Each electron has total energy $ E = N m_e c^2$.
When two identical particles collide head‑on with opposite momenta, the total energy in the center‑of‑mass (CM) system is simply the sum of their total energies:

$E_{\text{CM}} = 2E = 2N m_e c^2$

This is the energy available for the interaction.

Center-of-mass energy in the fixed target

Now we have an incident electron with total energy$ E_{\text{lab}}$ and a target electron at rest (energy $m_e c^2)$
The square of the center‑of‑mass energy for this system is calculated using the relativistic invariant s

$s = (E_{\text{lab}} + m_e c^2)^2 - (p_{\text{lab}} c)^2$

where$ p_{\text{lab}} $is the momentum of the incident electron.

Using the relation
$E_{\text{lab}}^2 - (p_{\text{lab}} c)^2 = (m_e c^2)^2$
we expand:

$s = E_{\text{lab}}^2 + 2E_{\text{lab}}m_e c^2 + (m_e c^2)^2 - (p_{\text{lab}} c)^2
= 2(m_e c^2)^2 + 2E_{\text{lab}}m_e c^2$

The center‑of‑mass energy is E_{\text{CM}} = \sqrt{s}.

For high energies $(E_{\text{lab}} \gg m_e c^2) $this approximates to:

$E_{\text{CM}} \approx \sqrt{2m_e c^2 E_{\text{lab}}}$

Equating the center‑of‑mass energies

We want the available energy in the fixed target to be the same as in the symmetric collider:

$\sqrt{2m_e c^2 E_{\text{lab}}} = 2N m_e c^2$

Squaring both sides:

$2m_e c^2 E_{\text{lab}} = 4 N^2 (m_e c^2)^2
\quad\Rightarrow\quad
E_{\text{lab}} = 2 N^2 m_e c^2$

The total energy that the incident electron must have is $E_{\text{lab}} = 2 N^2 m_e c^2$
Since N = 1000, we get

$E_{\text{lab}} = 2 \times (1000)^2\, m_e c^2 = 2 \times 10^6\, m_e c^2$

That is, the electron energy must be 2 million times its rest energy.

Comparing with the energy of each electron in the collider $(N m_e c^2 = 1000\, m_e c^2)$, the increase factor is:

$\frac{E_{\text{lab}}}{N m_e c^2} = \frac{2 N^2}{N} = 2N = 2000$

Therefore, an energy 2000 times greater than that of each electron in the colliding beams is needed.

Answer

$ \boxed{N = 2000}$