# Threshold energy

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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.

## Contents

The threshold energy should not be confused with the threshold displacement energy, which is the minimum energy needed to permanently displace an atom in a crystal to produce a crystal defect in radiation material science.

## 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:

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

$\gamma ={\frac {1}{\sqrt {1-\beta ^{2}}}}={\frac {2m_{p}c^{2}+m_{\pi }c^{2}}{2m_{p}c^{2}}}$ 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 $c^{2}$ terms in our expression will cancel, leaving us with:

$\beta ^{2}=1-\left({\frac {2m_{p}}{2m_{p}+m_{\pi }}}\right)^{2}\approx 0.130$ $\beta \approx 0.360$ $v_{\text{lab}}={\frac {u_{\text{cm}}+V_{\text{cm}}}{1+u_{\text{cm}}V_{\text{cm}}/c^{2}}}$ We know that $V_{cm}$ is equal to the speed of one proton as viewed in the ZMF, so we can re-write with $u_{cm}=V_{cm}$ :

$v_{\text{lab}}={\frac {2u_{\text{cm}}}{1+u_{\text{cm}}^{2}/c^{2}}}\approx 0.64c$ 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 $E_{2}$ and mass $m_{2}$ . 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

$E_{1,{\text{thr}}}={\frac {(m_{a}c^{2}+m_{b}c^{2}+m_{c}c^{2})^{2}-(m_{1}c^{2}+m_{2}c^{2})^{2}}{2m_{2}c^{2}}}$ ## Related Research Articles In nuclear physics, beta decay (β-decay) is a type of radioactive decay in which a beta particle is emitted from an atomic nucleus, transforming the original nuclide to an isobar of that nuclide. For example, beta decay of a neutron transforms it into a proton by the emission of an electron accompanied by an antineutrino; or, conversely a proton is converted into a neutron by the emission of a positron with a neutrino in so-called positron emission. Neither the beta particle nor its associated (anti-)neutrino exist within the nucleus prior to beta decay, but are created in the decay process. By this process, unstable atoms obtain a more stable ratio of protons to neutrons. The probability of a nuclide decaying due to beta and other forms of decay is determined by its nuclear binding energy. The binding energies of all existing nuclides form what is called the nuclear band or valley of stability. For either electron or positron emission to be energetically possible, the energy release or Q value must be positive. An ideal Fermi gas is a state of matter which is an ensemble of many non-interacting fermions. Fermions are particles that obey Fermi–Dirac statistics, like electrons, protons, and neutrons, and, in general, particles with half-integer spin. These statistics determine the energy distribution of fermions in a Fermi gas in thermal equilibrium, and is characterized by their number density, temperature, and the set of available energy states. The model is named after the Italian physicist Enrico Fermi. Surface free energy or interfacial free energy or surface energy quantifies the disruption of intermolecular bonds that occurs when a surface is created. In the physics of solids, surfaces must be intrinsically less energetically favorable than the bulk of a material, otherwise there would be a driving force for surfaces to be created, removing the bulk of the material. The surface energy may therefore be defined as the excess energy at the surface of a material compared to the bulk, or it is the work required to build an area of a particular surface. Another way to view the surface energy is to relate it to the work required to cut a bulk sample, creating two surfaces. There is "excess energy" as a result of the now-incomplete, unrealized bonding at the two surfaces.

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1. Jackson, John. Classical Electrodynamics. Wiley. pp. 533–539. ISBN   978-0-471-30932-1.