Relativistic Collisions Can Produce New Particles

We have mentioned how, using a synchrocyclotron, it ispossible to accelerate protons to relativistic speeds. The rest energy of a proton m p c 2  is 938 MeV, using here the standard highenergy sirhenryjones-museums.orgics energy unit: 1 MeV = 106 eV. The neutron is a bit heavier— m n c 2 =940 MeV. (The electron is 0.51 MeV). Thus to accelerate a proton to relativisticspeeds implies giving it a K.E. of order 1,000 MeV, or 1 GeV.

The standard operating procedure of high energy sirhenryjones-museums.orgicistsis to accelerate particles to relativistic speeds, then smash them into otherparticles to see what happens. Forexample, fast protons will be aimed at protons at rest (hydrogen atoms, inother words—the electron canbe neglected). In a collider, beams ofaccelerated protons have head-on collisions. As we shall see, this greatly increases the center of mass energy (it"snot just doubled) but of course the number of hits goes down a lot.

To see what results from the collision, the resulting debris(usually flying away fast!) must be detected. The first successful detector was the cloud chamber, invented in 1911.If a fast charged particle flies througha supersaturated gas, it ionizes some molecules, they are then nuclei or seedsfor droplet formation, and the path is realized as a string of tiny drops. Thecloud chamber was superseded in the fifties by the bubble chamber, atransparent container filled with a superheated liquid. an energetic particlemoving through the liquid leaves a trail of ionized molecules, which nucleatebubbles. The bubbles grow rapidly to highlightthe path, then rise and go away far faster than the droplets in the cloudchamber. But as accelerators developed, and began searching for less frequentevents, faster and faster turnaround times became essential for the detectors.Clouds and bubbles were replaced by sparks and wires, fine parallel wiresmillimeters apart, in an easily ionized gas, the passing particle generatingsparks between the wires. This improved response time by orders of magnitude. Nowadays, detectors often consist of many thousands of very tiny solid state reversebiased diodes, triggered by the particle, and wired to give precise trajectoryinformation. In fact, majorexperiments have the collision area surrounded by layers of both solidstate detectors and wire grid detectors.

Anyway, back to the first early attempts, and what wasobserved—it turned outthat in p−p  scattering at low but relativistic energies,sometimes more particles came out than went in—particles calledpions, π+,π0,π-were created. The π0 is electrically neutral, the π+has exactly the same amount of charge as the proton. It was found experimentally that totalelectric charge was always conserved in collisions, no matter how many newparticles were spawned, and total baryon number (protons + neutrons) wasconserved.

Possible scenarios include:

p+p→p+p+ π 0 ,

and

p+p→p+n+ π + .

The neutral pion mass is 135 MeV, the charged pions havemass 140 MeV, where we follow standard high energy practice in calling mc2 the “mass”, since this isthe energy equivalent, and hence the energy which, on creation of the particlein a collision, is taken from kinetic energy and stored in mass.

Energy Necessary to Produce a Pion

An incoming proton with 135 MeV of kinetic energy will notbe able to create a neutral pion (rest mass 135 Mev) in a collision with a stationary proton. This is because the incoming proton also hasmomentum, and the collision conserves momentum, so some of the particles afterthe collision must have momentum and hence kinetic energy.

The simplest way to figure out just how much energy theincoming proton needs to create a neutral pion is to go to the center of massframe, where initially two protons are moving towards each other with equal andopposite velocities, there being no total momentum. Obviously, in this frame the least possibleK.E. must be just enough to create the π 0  with allthe final state particles ( p,p, π 0 )  at rest. Thus if the the incoming protons inthe center of mass frame are traveling at ±v,  the total energy, which must equal the restenergies of the final stationary masses, is

E= 2 m p c 2 1− v 2 / c 2 =2 m p c 2 + m π c 2 ,

 we find the twoincoming protons must both be traveling at 0.36c.

Recall that this is the speed in the center of mass frame,and for practical purposes, like designing the accelerator, we need to know theenergy necessary in the “lab” frame—that in which oneof the protons is initially at rest. Thetwo frames obviously have a relative speed of 0.36c, so to get the speed of the incoming proton in the lab frame wemust add a velocity of 0.36c to oneof 0.36c using the relativisticaddition of velocities formula, which gives 0.64c. This implies the incomingproton has a relativistic mass of 1.3 times its rest mass, and thus a K.E.around 280 MeV.

Thus to create a pion of rest energy 135 MeV, it isnecessary to give the incoming proton at least 290 MeV of kinetic energy. This is called the “threshold energy” forpion production. This “inefficiency” (more energy than seems necessary)arises because momentum must also be conserved, so, in the lab, there is still considerable K.E.in the final particles.

Antiproton Production

On raising the energy of the incoming proton further, moreparticles are produced, including the “antiproton”—a negativelycharged heavy particle which will annihilate a proton in a flash ofenergy. It turns out experimentally thatan antiproton can only be produced accompanied by a newly created proton,

p+p→p+p+p+ p ¯ .

Notice we could have conserved electric charge with lessenergy with the reaction

p+p→p+p+ π + + p ¯

but this doesn’t happen—so energy, momentum and chargeconservation are not the only constraints in creating new particles. (There"s also angular momentum, but that"s not important here.) 

In fact, what we are seeing here is experimentalconfirmation that the conservation of baryon number, which at the low energiespreviously discussed in the context of pion production just meant that thetotal number of protons plus neutrons stayed fixed, is generalized at highenergies to include antiparticles having negative baryon number, -1 for theantiproton. Thus baryon number conservation becomes parallel to electric chargeconservation.

New particles can always be produced at high enough energiesprovided the total new charge and the total new baryon number are both zero.(Actually there are further conservation laws which become important when moreexotic particles are produced, we may discuss these later.) We should emphasize again that these are experimental results gathered fromexamining millions of collisions between relativistic particles.

A Machine Built to Produce One Particle

One of the first modern accelerators, built at >Berkeley in the fifties,was designed specifically to produce the antiproton, so it was very importantto calculate that antiproton production threshold correctly! This can be done by the same method we usedabove for pion production, but we use a different trick here which is oftenuseful. We have shown that ontransforming the energy and momentum of a particle from one frame to another

E 2 − c 2 p 2 = E ′ 2 − c 2 p ′ 2

Since the Lorentz equations are linear, if we have a systemof particles with total energy E andtotal momentum p in one frame, E" , p"in another, it must again be true that

E 2 − c 2 p → 2 = E ′ 2 − c 2 p ′ → 2 .

We can use this invariance to get lab frame information fromthe center of mass frame. Noting that inthe center of mass (CM) frame the momentum is zero, and in the lab frame themomentum is all in the incoming proton,

E cm 2 = (( m in + m 0 ) c 2 ) 2 − c 2 p in 2  

where here m0is the proton rest mass, m in  is the relativistic mass of the incomingproton: we"re writing m 0 1− v in LAB 2 / c 2 = m in .

At the antiproton production threshold, Ecm = 4m0c2, so

16 m 0 2 c 4 = m in 2 c 4 +2 m in c 2 m 0 c 2 + m 0 2 c 4 − c 2 p in 2 ,

and using

m in 2 c 4 − c 2 p in 2 = m 0 2 c 4 ,

we find

2( m in c 2 )( m 0 c 2 )+2 ( m 0 c 2 ) 2 =16 ( m 0 c 2 ) 2 ,

so

m in c 2 =7 m 0 c 2 .

Therefore to create two extra particles, with total restenergy 2 m 0 c 2 ,  it is necessary for the incoming proton tohave a kinetic energy of 6 m 0 c 2 . TheBerkeley Gevatron had design energy 6.2 GeV. 

Higher Energies

As we go to higher energies, this “inefficiency” gets worse—consider energiessuch that the kinetic energy >> rest energy, and assume the incomingparticle and the target particle have the same rest mass, m 0 ,  with the incoming particle having relativisticmass m in :  

*

Comparing the center of mass energy with the lab energy atthese high energies,

E LAB =( m in + m 0 ) c 2 , E CM 2 = E LAB 2 − p LAB 2 c 2 = m in 2 c 4 +2 m in c 2 m 0 c 2 + m 0 2 c 4 − p LAB 2 c 2 =2 m 0 c 2 ( m in c 2 + m 0 c 2 ).

For m≫ m 0 ,  

E CM 2 ≈2 m 0 c 2 m c 2 ≈2 m 0 c 2 . E LAB

so

E CM ≈ 2 m 0 c 2 .


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E LAB ,

ultimately one must quadruplethe lab energy to double the center of mass energy. And, at higher energies,things get steadily worse——his is whycolliders were built!