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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.
 
==Example==
Look at the reaction of a proton hitting a stationary proton, <math> p + p \to p + p + \pi^0</math>.
 
By going into the center of mass frame, and assuming the outgoing particles have no kinetic energy the conservation of energy equation is:
 
<math> E = 2\gamma m_pc^2 = 2 m_pc^2+ m_\pi c^2</math>
 
<math> \gamma = \frac{1}{\sqrt{1-\beta^2}} = \frac{2 m_pc^2+ m_\pi c^2}{2 m_pc^2} </math>
 
<math> \beta^2 = 1-(\frac{2 m_pc^2}{2 m_pc^2+ m_\pi c^2})^2 \approx 0.13 </math>
 
Using relativistic velocity additions:
 
<math> v_\text{lab} = \frac{u_\text{cm} + v_\text{cm}}{1+u_\text{cm}v_\text{cm}/c^2} \approx 0.64 c</math>
 
So the energy of the proton must be <math> E = \gamma m_p c^2 = \frac{m_p c^2}{\sqrt{1-\beta^2}} = 1221\,</math>MeV.
 
== A more general example ==
 
Let's look at the case where a particle 1 with lab energy <math>E_1</math> (momentum <math>p_1</math>)
and mass <math>m_1</math> impinges on a
target particle 2 at rest in the lab, i.e. with lab energy and mass <math>E_2 = m_2</math>.
The threshold energy <math>E_{1,\text{thr}}</math> to produce three particles of masses <math>m_a</math>, <math>m_b</math>,
<math>m_c</math>, i.e.
 
<math> 1 + 2 \to a + b + c,</math>
 
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):
 
<math> 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 </math> .
 
Here <math> E_\text{cm}</math> is the total energy available in the center  of mass frame.
 
Using <math> \gamma = \frac{E_1 + m_2 c^2}{E_\text{cm}} </math>, <math> \beta = \frac{p_1 c}{E_1 + m_2 c^2} </math>  and
<math>p_1^2 c^2 = E_1^2 - m_1^2 c^4 </math> one derives that
 
<math>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}</math>.
 
== See also ==
 
* [[Threshold displacement energy]]
 
==References==
*http://galileo.phys.virginia.edu/classes/252/particle_creation.html
 
[[Category:Energy (physics)]]
[[Category:Particle physics]]

Revision as of 21:24, 30 May 2013

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+pp+p+π0.

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

E=2γmpc2=2mpc2+mπc2

γ=11β2=2mpc2+mπc22mpc2

β2=1(2mpc22mpc2+mπc2)20.13

Using relativistic velocity additions:

vlab=ucm+vcm1+ucmvcm/c20.64c

So the energy of the proton must be E=γmpc2=mpc21β2=1221MeV.

A more general example

Let's look at the case where a particle 1 with lab energy E1 (momentum p1) and mass m1 impinges on a target particle 2 at rest in the lab, i.e. with lab energy and mass E2=m2. The threshold energy E1,thr to produce three particles of masses ma, mb, mc, i.e.

1+2a+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):

Ecm=mac2+mbc2+mcc2=E^1+E^2=γ(E1βp1c)+γm2c2 .

Here Ecm is the total energy available in the center of mass frame.

Using γ=E1+m2c2Ecm, β=p1cE1+m2c2 and p12c2=E12m12c4 one derives that

E1,thr=(mac2+mbc2+mcc2)2m12c4m22c42m2c2.

See also

References