threshold energy

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

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:

$E\; =\; 2\backslash gamma\; m\_pc^2\; =\; 2\; m\_pc^2+\; m\_\backslash pi\; c^2$

Rearranged to

$\backslash gamma\; =\; \backslash frac\{1\}\{\backslash sqrt\{1-\backslash beta^2\}\}\; =\; \backslash frac\{2\; m\_pc^2+\; m\_\backslash pi\; c^2\}\{2\; m\_pc^2\}$

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:

$\backslash beta^2\; =\; 1-\backslash left(\backslash frac\{2\; m\_p\}\{2\; m\_p+\; m\_\backslash pi\; \}\backslash right)^2\; \backslash approx\; 0.130$

$\backslash beta\; \backslash approx\; 0.360$

Using relativistic velocity additions:

$v\_\backslash text\{lab\}\; =\; \backslash frac\{u\_\backslash text\{cm\}\; +\; V\_\backslash text\{cm\}\}\{1+u\_\backslash text\{cm\}V\_\backslash text\{cm\}/c^2\}$

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

$v\_\backslash text\{lab\}\; =\; \backslash frac\{2\; u\_\backslash text\{cm\}\}\{1+u\_\backslash text\{cm\}^2/c^2\}\; \backslash approx\; 0.64c$

So the energy of the proton must be $E\; =\; \backslash gamma\; m\_p\; c^2\; =\; \backslash frac\{m\_p\; c^2\}\{\backslash sqrt\{1-(v\_\backslash text\{lab\}/c)\; ^2\}\}\; =\; 1221\backslash ,$ MeV.

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

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

:$p^+\; +\; p^+\; \backslash to\; p^+\; +\; p^+\; +\; p^+\; +\; p^-$

If one of the two initial protons is stationary, we find that the impinging proton must be given at least $6m\_pc^2$ 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 $m\_pc^2$ of energy.

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 $E\_\{1,\backslash text\{thr\}\}$ to produce three particles of masses $m\_a$, $m\_b$,

$m\_c$, i.e.

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

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):

$E\_\backslash text\{cm\}\; =\; m\_a\; c^2+\; m\_b\; c^2\; +\; m\_c\; c^2\; =\; \backslash hat\{E\}\_1\; +\; \backslash hat\{E\}\_2\; =\; \backslash gamma\; (E\_1\; -\; \backslash beta\; p\_1\; c)\; +\; \backslash gamma\; m\_2\; c^2$

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

Using $\backslash gamma\; =\; \backslash frac\{E\_1\; +\; m\_2\; c^2\}\{E\_\backslash text\{cm\}\}$, $\backslash beta\; =\; \backslash frac\{p\_1\; c\}\{E\_1\; +\; m\_2\; c^2\}$ and

$p\_1^2\; c^2\; =\; E\_1^2\; -\; m\_1^2\; c^4$ one derives that

$E\_\{1,\backslash text\{thr\}\}\; =\; \backslash frac\{(m\_a+m\_b+m\_c)^2-(m\_1^2+m\_2^2)\}\{2m\_2\}c^2$ {{cite book

|last=Jackson

|first=John

|title= Classical Electrodynamics

|date=14 August 1998

|publisher= Wiley

|pages=533–539

|isbn= 978-0-471-30932-1 }}

{{Reflist}}

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

Category:Energy (physics)

Category:Particle physics

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