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 $\left(\mathrm{K}^{-}\right)$ and a positive $\mathrm{kaon}\left(\mathrm{K}^{+}\right) :$ $$p+p \rightarrow p+p+\mathbf{K}^{-}+\mathbf{K}^{+}$$ (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 $\mathrm{MeV}$ , 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. See Problem 8.100 , but note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum 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 kimetic 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 $\left(\mathrm{K}^{-}\right)$ and a positive $\mathrm{kaon}\left(\mathrm{K}^{+}\right) :$
$$p+p \rightarrow p+p+\mathbf{K}^{-}+\mathbf{K}^{+}$$
(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 $\mathrm{MeV}$ , 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. See Problem 8.100 , but note that here the Lorentz transformation must be used to relate the velocities in the laboratory frame to those in the zero-total-momentum 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 kimetic 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.)

Key Concepts

Energy and Momentum Conservation

In any particle collision, the total energy and momentum before and after the collision must be conserved. This principle is crucial in high-energy physics for calculating the conditions needed to produce new particles. The conservation laws require that the kinetic energy and rest mass energies add up correctly to meet the threshold for particle creation.

Mass-Energy Equivalence

Mass-energy equivalence, articulated by the famous equation E=mc², means that mass can be converted into energy and vice versa. This concept is fundamental in understanding how kinetic energy can be transformed into the rest mass energy required to create particles, as seen in high-energy collisions.

Threshold Energy in Particle Reactions

Threshold energy refers to the minimum kinetic energy required by the colliding particles to produce the additional particles in a reaction. It accounts for both the energy needed to overcome the rest masses of the new particles and the kinetic energy of the system in the center-of-momentum frame.

Center-of-Momentum Frame

The center-of-momentum (or zero-total-momentum) frame is the reference frame in which the total momentum of the system is zero. Analyzing reactions in this frame simplifies the calculations because the energy available for particle production is maximized and distributed symmetrically between the colliding particles.

Lorentz Transformation

Lorentz transformations are used to relate observations made in different inertial frames, particularly between the laboratory frame and the center-of-momentum frame. These transformations account for the effects of time dilation and length contraction, which are essential when dealing with high-speed (relativistic) particles.

Transcript

00:01 In this problem, we're going to talk about the relativistic energy.

00:04 And what we need to remember is that the relativistic energy is equal to the total energy of a particle is equal to gamma mc squared.

00:14 And this is equal to the rest energy of the particle, which is mc squared, plus the kinetic energy of that particle.

00:22 Meaning that we can write the kinetic energy as gamma minus 1 times mc squared.

00:28 And we need to remember that gamma is 1 over the square root of 1 minus v squared over c squared.

00:39 Besides, we also need to remind ourselves of the relativistic transformation of velocities.

00:51 Transformations of velocities.

00:54 So consider that we have a reference frame.

01:04 And in this reference frame, we have two particles.

01:08 The first one, this red one here, is moving with a speed v.

01:15 The second one, this green one, is moving with a speed capital v.

01:24 I'm going to call this reference frame the s reference ring.

01:28 Now consider that we have a second reference frame in which the first particle is at rest.

01:33 So the red particle has no velocity according to this reference frame.

01:41 And we ask ourselves what is the velocity of the green particle v prime? according to spatial relativity, v prime is equal to capital v plus v divided by 1 plus capital v v over c squared.

02:05 And now we can move on to our exercise.

02:08 So we have a reaction, collision actually, a proton -protein collision.

02:15 That are results in the following.

02:17 We have two proteins.

02:20 One of the protons is at rest, the other one is moving, they collide, and they generate two protons plus two k -ens, one positive and one negative k -a.

02:36 And we need to remember that the mass of the proton is 938 .3m .3m.

02:46 While the mass of the km is 493 .7 meglectron volts, of course, per c squared.

02:59 Okay, so what we need to find in question a is what is the minimum kinetic energy of the moving proton, remember that one of them is moving, the other one is at rest, we need to find what is the minimum kinetic energy of the moving proton such that this reaction can occur.

03:20 Notice that there are two conservation laws that we have to apply here.

03:26 The first one is the conservation of energy, so the initial energy must be able to the final energy, and the conservation of momentum.

03:36 If we work in a reference frame where the initial momentum is non -zero, such as the lab reference frame where the two particles, one of the particles at breast, the other one is not at rest, so pi is different from zero.

03:52 In this reference frame, things are going to get very complicated very soon.

03:58 So we're not going to work in this reference frame.

04:00 Instead, we're going to move on to the center of mass reference frame.

04:08 And then from that reference frame, we can go back to our, we can go back to the lab reference frame afterwards.

04:21 But all the initial calculations we're going to do in the center of mass reference frame.

04:26 In this reference frame, the total momentum of the particles is zero, the total initial momentum.

04:34 So the final initial momentum is also going to be zero.

04:39 This makes the second equation, the conservation momentum, trivial, and makes things a lot easier.

04:47 So notice that in the lab reference frame, we have a proton, moving, another one at rest.

04:57 The proton that is moving has a certain kinetic energy and a certain velocity of v, the second one has neither.

05:07 But in the center of mass reference frame, both protons are moving with the same speed.

05:14 Since they have the same mass, the total momentum is zero, then they must have the same speed just at opposite directions.

05:21 And i'm going to call these speeds, u.

05:24 They're not going to be the same as v because v is the velocity in the lab reference frame and u in the center of mass reference frame.

05:33 Of course, they are connected.

05:36 These two velocities are connected by the fact that v is equal to the sum of the velocities, the two velocity in the center of mass reference frame divided by 1 plus u squared over c squared.

05:52 This comes from this equation here that i talked about when i was talking about transformational velocities.

06:02 So let's keep this equation in mind.

06:04 We're going to need this equation after we work in the center of mass reference rate.

06:11 Okay, so in the center of mass, everything is a lot easier because the initial energy is the kinetic energy, two times the kinetic energy.

06:21 Of each proton, plus the mass, the rest energy of the protons, that's 2m .pc squared, that is equal to, and we want to know what is the minimum kinetic energy, this means that we want the final state with the minimum kinetic energy as possible.

06:41 In the reference frame where the initial momentum is zero, the minimum kinetic energy is allowed to be zero, and this is what makes this calculation so much.

06:54 Simpler.

06:55 Here the final momentum in this reference frame is zero.

06:59 So two, i'm sorry, the final kinetic energy in this reference frame is zero.

07:03 So this is two times the rest energy of one proton plus two times the rest energy of one k .m.

07:11 The protons, rest energy cancel out.

07:15 So the kinetic energy of one proton must be equal to the rest energy of one k...

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