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

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

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:

$E = 2\gamma m_pc^2 = 2 m_pc^2+ m_\pi c^2$

Rearranged to give

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

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 $c^2$ terms in our expression will cancel, leaving us with:

$\beta^2 = 1-(\frac{2 m_p}{2 m_p+ m_\pi })^2 \approx 0.130$

$\beta \approx 0.360$

Using relativistic velocity additions:

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

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

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

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

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

A more general example

Consider 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 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):

$E_\text{cm} = m_a c^2+ m_b c^2 + m_c c^2 = \hat{E}_1 + \hat{E}_2 = \gamma (E_1 - \beta p_1 c) + \gamma m_2 c^2$

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

$E_{1,\text{thr}} = \frac{(m_a c^2+ m_b c^2 + m_c c^2)^2 - m_1^2 c^4 - m_2^2 c^4}{2 m_2 c^2}$

See also

• Threshold displacement energy

References

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