Question

In a collider, electrons and positrons of the same energy were made to collide head on to produce a particle, x, at rest: e+ e- = x. Knowing the mass of x is 1GeV/c^2, calculate the minimum energy of the electron and the positron.

What's the significance of the collision being head on, and how would one calculate this if the collision was at an angle?

Answer 1

The collision being head-on implies that the total momentum after the collision is equal to zero - equivalent to the daughter particle being at rest. The collision being at an angle would imply nonzero final momentum, so the daughter particle would have some kinetic energy which would have to be included in your calculation for the minimum energy necessary.

Based on energy conservation, the initial problem is simple (c = 1):
\[ E_i = E_f \]
\[ (E_e +E_{e^+}) = 2E_0 = M_x c^2\]
So the threshold energy for the electrons and positrons must be \( E_0 = \frac{M_x c^2}{2}\), which we probably could have guessed in the first place.

If we considered the possibility that the particles collided at an angle (which, by the way, is exceptionally unlikely in a particle collider), we would have

Answer 2

So the final particle would have total momentum equal to \(2p_0\cos(\theta/2) \), where \(p_0\) is the momentum of each incoming particle. We would then have the following:
\[E_i^2 = E_f^2\]
\[(2E_0)^2 = (2p_0\cos(\theta/2))^2c^2 + M_x^2 c^4\]
However,
\[E_0^2 = p_0^2c^2 + m^2c^4 \]
where \(m\) is the mass of the electron (or positron). Therefore,
\[(2E_0)^2 = 4\cos^2(\theta/2) \cdot (E_0^2 - m^2c^4) + M_x^2c^4 =4\cos^2(\theta/2)E_0^2 + (M_x^2 - 4m^2\cos^2(\theta/2) ) c^4 \]

leaving us finally with

\[ E_0 = \frac{ \sqrt{M_x^2 - 4m^2\cos^2(\theta/2)} \cdot c^2 }{2[1- \cos(\theta/2) ]} \]

Answer 3

Hopefully I didn't make a typo. Notice that if the angle is 180 degrees (i.e. head on collision) then we get back our original expression, and as the angle goes to 0 degrees (e and positron initially traveling side by side) the expression goes to infinity.

Answer 4

Oh, and lastly - notice that a head on collision will yield the smallest possible threshold energy, so beams should obviously be fired in an attempt for a head-on collision.

Answer 5

oops... I said at the beginning I was going to set c = 1, and then never did it...