Threshold energy

Consider the collision of a mobile proton with a stationary proton so that a ${\displaystyle {\pi }^{0}}$ meson is produced:[1] ${\displaystyle p^{+}+p^{+}\to p^{+}+p^{+}+\pi ^{0}}$

We can calculate the minimum energy that the moving proton must have in order to create a pion. Transforming into the ZMF (Zero Momentum Frame or Center of Mass Frame) and assuming the outgoing particles have no KE (kinetic energy) when viewed in the ZMF, the conservation of energy equation is:

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

Rearranged to

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

By assuming that the outgoing particles have no KE in the ZMF, we have effectively considered an inelastic collision in which the product particles move with a combined momentum equal to that of the incoming proton in the Lab Frame.

Our ${\displaystyle c^{2}}$ terms in our expression will cancel, leaving us with:

${\displaystyle \beta ^{2}=1-\left({\frac {2m_{p}}{2m_{p}+m_{\pi }}}\right)^{2}\approx 0.130}$

${\displaystyle \beta \approx 0.360}$

Using relativistic velocity additions:

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

We know that ${\displaystyle V_{cm}}$ is equal to the speed of one proton as viewed in the ZMF, so we can re-write with ${\displaystyle u_{cm}=V_{cm}}$:

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

So the energy of the proton must be ${\displaystyle E=\gamma m_{p}c^{2}={\frac {m_{p}c^{2}}{\sqrt {1-(v_{\text{lab}}/c)^{2}}}}=1221\,}$ MeV.

Therefore, the minimum kinetic energy for the proton must be ${\displaystyle T=E-{m_{p}c^{2}}\approx 280}$ MeV.

At higher energy, the same collision can produce an antiproton:

${\displaystyle p^{+}+p^{+}\to p^{+}+p^{+}+p^{+}+p^{-}}$

If one of the two initial protons is stationary, we find that the impinging proton must be given at least ${\displaystyle 6m_{p}c^{2}}$ of energy, that is, 5.63 GeV. On the other hand, if both protons are accelerated one towards the other (in a collider) with equal energies, then each needs to be given only ${\displaystyle m_{p}c^{2}}$ of energy.[1]

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

${\displaystyle 1+2\to a+b+c,}$

is then found by assuming that these three particles are at rest in the center of mass frame (symbols with hat indicate quantities in the center of mass frame):

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

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

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

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

• http://galileo.phys.virginia.edu/classes/252/particle_creation.html