In high-energy physics, new particles can be created by...

In high-energy physics, new particles can be created by collisions of fast-moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon ($K^-$) and a positive kaon ($K^+$): $$p + p \rightarrow p + p + K^- + K^+$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and the rest energy of each proton is 938.3 MeV. ($Hint$: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-totalmomentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

In high-energy physics, new particles can be created by collisions of fast-moving projectile particles with stationary particles. Some of the kinetic energy of the incident particle is used to create the mass of the new particle. A proton-proton collision can result in the creation of a negative kaon ($K^-$) and a positive kaon ($K^+$): $$p + p \rightarrow p + p + K^- + K^+$$ (a) Calculate the minimum kinetic energy of the incident proton that will allow this reaction to occur if the second (target) proton is initially at rest. The rest energy of each kaon is 493.7 MeV, and
the rest energy of each proton is 938.3 MeV. ($Hint$: It is useful here to work in the frame in which the total momentum is zero. But note that the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-totalmomentum frame.) (b) How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (c) Suppose that instead the two protons are both in motion with velocities of equal magnitude and opposite direction. Find the minimum combined kinetic energy of the two protons that will allow the reaction to occur. How does this calculated minimum kinetic energy compare with the total rest mass energy of the created kaons? (This example shows that when colliding beams of particles are used instead of a stationary target, the energy requirements for producing new particles are reduced substantially.)

Transcript

00:01 In the first part of this problem, we are going to calculate the minimum kinetic energy of the incident proton that will allow the given reaction.

00:08 So let the minimum kinetic energy is caleb.

00:14 In this problem, we are going to apply the law of conservation of energy.

00:19 So in the center of momentum frame, the initial momentum of protons that are approaching each other should be equal to 0.

00:28 So after the collision, the net momentum will be 0.

00:31 So we can write two times of kinetic energy of proton is equal to two times of rest mass energy of the kiln.

00:53 So this can be written as 2 into gamma cm where the cm is the center of momentum from minus 1 into mp.

01:05 Mp, this is the mass of proton c squared, is equals to 2 mk c squared.

01:15 However, this mk is the mass of con and this is the speed of light.

01:19 From here we can write this gamma cm is equal to 1 plus mk divided by mp.

01:30 And this is equal to 1 .5 to 6.

01:37 Now we can write the equation for the velocity of the velocity.

01:41 Of the proton in the center of momentum frame as v -cm is equal to square root of gamma square cm minus one divided by gamma square cm into c now by inserting values into the square and we can write into square root of 1 .5 to 6 all square minus 1 divided by 1 .526 whole square into c so from here we can write the value for this vcm as vcm is equal to 0 .7 554c now in order to calculate the velocity of the proton in layframe we apply the velocity additional this this can be written as a we lap is equals to vcm plus vcm divided by one plus vcm into vcm divided by c squared now by inserting values into this equation we can write two into vcm which is equals to 0 .7 554c divided by one plus so we cm square which is equal to 0 .7 7554 c whole square divided by c squared so from here we can write the value for this we lap as a v lab is equals to 0 .9619c now let's calculate the gamma factor in the lab which is written as a gamma lab is equals to 1 divided by square root of 1 minus we left square divided by c square.

03:58 So by setting values into this equation we can write one divided by square root of 1 minus will which is equal to 1 .9619c whole square divided by c square.

04:15 So from here we can write the value for this gamma lap as gamma lube is equals to 3 .65 feet...

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