# Threshold energy

In particle physics, the threshold energy for production of a particle is the minimum kinetic energy a pair of traveling particles must have when they collide. The threshold energy is always greater than or equal to the rest energy of the desired particle. In most cases, since momentum is also conserved, the threshold energy is significantly greater than the rest energy of the desired particle - and thus there will still be considerable kinetic energy in the final particles.

## Example

Look at the reaction of a proton hitting a stationary proton, $p+p\to p+p+\pi ^{0}$.

By going into the center of mass frame, and assuming the outgoing particles have no kinetic energy the conservation of energy equation is:

$E=2\gamma m_{p}c^{2}=2m_{p}c^{2}+m_{\pi }c^{2}$

$\gamma ={\frac {1}{{\sqrt {1-\beta ^{2}}}}}={\frac {2m_{p}c^{2}+m_{\pi }c^{2}}{2m_{p}c^{2}}}$

$\beta ^{2}=1-({\frac {2m_{p}c^{2}}{2m_{p}c^{2}+m_{\pi }c^{2}}})^{2}\approx 0.13$

$v_{{\text{lab}}}={\frac {u_{{\text{cm}}}+v_{{\text{cm}}}}{1+u_{{\text{cm}}}v_{{\text{cm}}}/c^{2}}}\approx 0.64c$

So the energy of the proton must be $E=\gamma m_{p}c^{2}={\frac {m_{p}c^{2}}{{\sqrt {1-\beta ^{2}}}}}=1221\,$MeV.

## A more general example

Let's look at the case where a particle 1 with lab energy $E_{1}$ (momentum $p_{1}$) and mass $m_{1}$ impinges on a target particle 2 at rest in the lab, i.e. with lab energy and mass $E_{2}=m_{2}$. The threshold energy $E_{{1,{\text{thr}}}}$ to produce three particles of masses $m_{a}$, $m_{b}$, $m_{c}$, i.e.

$1+2\to a+b+c,$

is then found by asking these three particles to be at rest in the center of mass frame (symbols with hat indicate quantities in the center of mass frame):

$E_{{\text{cm}}}=m_{a}c^{2}+m_{b}c^{2}+m_{c}c^{2}={\hat {E}}_{1}+{\hat {E}}_{2}=\gamma (E_{1}-\beta p_{1}c)+\gamma m_{2}c^{2}$ .

Here $E_{{\text{cm}}}$ is the total energy available in the center of mass frame.

Using $\gamma ={\frac {E_{1}+m_{2}c^{2}}{E_{{\text{cm}}}}}$, $\beta ={\frac {p_{1}c}{E_{1}+m_{2}c^{2}}}$ and $p_{1}^{2}c^{2}=E_{1}^{2}-m_{1}^{2}c^{4}$ one derives that

$E_{{1,{\text{thr}}}}={\frac {(m_{a}c^{2}+m_{b}c^{2}+m_{c}c^{2})^{2}-m_{1}^{2}c^{4}-m_{2}^{2}c^{4}}{2m_{2}c^{2}}}$.