Question

The creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a particle of mass $M$ requires an energy $M c^{2} .$ With enough energy, an exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass $m_{p}$, kinetic energy $K$, and momentum magnitude $p$ joins with an originally stationary target proton to form a single product particle of mass $M .$ Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by $$M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{p} c^{2}}}$$ This result shows that when the kinetic energy $K$ of the incident proton is large compared with its rest energy $m_{p} c^{2},$ then $M$ approaches $\left(2 m_{p} K\right)^{1 / 2} / c .$ Therefore, if the energy of the incoming proton is increased by a factor of $9,$ the mass you can create increases only by a factor of $3,$ not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so, in principle, all the initial kinetic energy can be used for particle creation. Show that $$M c^{2}=2 m c^{2}\left(1+\frac{K}{m c^{2}}\right)$$ where $K$ is the kinetic energy of each of the two identical colliding particles. Here, if $K>>m c^{2},$ we have $M$ directly proportional to $K$ as we would desire.

Name: Problem 53, Chapter 38: Physics for Scientists and Engineers with Modern Physics
Uploaded: 2022-11-14T13:38:42-08:00
Duration: 25 min 46 s
Channel: Robert Schnibbe

   The creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a particle of mass $M$ requires an energy $M c^{2} .$ With enough energy, an exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass $m_{p}$, kinetic energy $K$, and momentum magnitude $p$ joins with an originally stationary target proton to form a single product particle of mass $M .$ Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle.
(a) Show that the energy available to create a product particle is given by
$$M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{p} c^{2}}}$$
This result shows that when the kinetic energy $K$ of the incident proton is large compared with its rest energy $m_{p} c^{2},$ then $M$ approaches $\left(2 m_{p} K\right)^{1 / 2} / c .$ Therefore, if the energy of the incoming proton is increased by a factor of $9,$ the mass you can create increases only by a factor of $3,$ not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so, in principle, all the initial kinetic energy can be used for particle creation. Show that
$$M c^{2}=2 m c^{2}\left(1+\frac{K}{m c^{2}}\right)$$
where $K$ is the kinetic energy of each of the two identical colliding particles. Here, if $K>>m c^{2},$ we have $M$ directly proportional to $K$ as we would desire.

Physics for Scientists and Engineers with Modern Physics

Raymond A. Serway,… 10th Edition

Chapter 38, Problem 53

Instant Answer

Solved by Expert Robert Schnibbe

11/14/2022

Step 1

This can be written as: \[E_{1} + E_{2} = E_{f}\] where $E_{1}$ is the energy of the incident proton, $E_{2}$ is the energy of the target proton, and $E_{f}$ is the energy of the product particle. Show more…

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The creation and study of new and very massive elementary particles is an important part of contemporary physics. To create a particle of mass $M$ requires an energy $M c^{2} .$ With enough energy, an exotic particle can be created by allowing a fast-moving proton to collide with a similar target particle. Consider a perfectly inelastic collision between two protons: an incident proton with mass $m_{p}$, kinetic energy $K$, and momentum magnitude $p$ joins with an originally stationary target proton to form a single product particle of mass $M .$ Not all the kinetic energy of the incoming proton is available to create the product particle because conservation of momentum requires that the system as a whole still must have some kinetic energy after the collision. Therefore, only a fraction of the energy of the incident particle is available to create a new particle. (a) Show that the energy available to create a product particle is given by $$M c^{2}=2 m_{p} c^{2} \sqrt{1+\frac{K}{2 m_{p} c^{2}}}$$ This result shows that when the kinetic energy $K$ of the incident proton is large compared with its rest energy $m_{p} c^{2},$ then $M$ approaches $\left(2 m_{p} K\right)^{1 / 2} / c .$ Therefore, if the energy of the incoming proton is increased by a factor of $9,$ the mass you can create increases only by a factor of $3,$ not by a factor of 9 as would be expected. (b) This problem can be alleviated by using colliding beams as is the case in most modern accelerators. Here the total momentum of a pair of interacting particles can be zero. The center of mass can be at rest after the collision, so, in principle, all the initial kinetic energy can be used for particle creation. Show that $$M c^{2}=2 m c^{2}\left(1+\frac{K}{m c^{2}}\right)$$ where $K$ is the kinetic energy of each of the two identical colliding particles. Here, if $K>>m c^{2},$ we have $M$ directly proportional to $K$ as we would desire.

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Key Concepts

Mass–Energy Equivalence

This principle, embodied in the famous relation E = mc², states that mass can be converted into energy and vice versa. It underpins particle creation processes by quantifying the minimum energy required to produce a particle of a given mass. In high-energy physics, it allows us to calculate how much energy is needed to create new particles during collisions.

Conservation of Energy and Momentum

In any isolated system, the total energy and momentum must remain constant throughout any interaction. In collision processes, these conservation laws dictate how the initial kinetic energy is partitioned between creating new mass and the remaining kinetic energy of the system. In perfectly inelastic collisions, some kinetic energy is unavoidably locked into the motion of the resulting particle’s center of mass.

Center-of-Mass Frame in Collisions

The center-of-mass frame is a reference frame in which the total momentum of the system is zero. Analyzing collisions in this frame is crucial because it maximizes the energy available for particle creation, as all kinetic energy can be converted into mass. This is in contrast to fixed-target experiments, where conservation of momentum requires some energy to remain as kinetic energy of the system after the collision.

Relativistic Kinematics

Relativistic kinematics deals with the motion and energy of objects moving at speeds close to the speed of light, where classical mechanics no longer applies. It involves using the relativistic energy-momentum relation to correctly compute energies and momenta. In high-energy collisions, relativistic effects become prominent, and accurate calculations require accounting for these effects.

Fixed-Target vs. Collider Dynamics

The dynamics of collisions differ significantly between fixed-target setups, where a moving particle collides with a stationary target, and collider setups, where two particles are accelerated toward each other. In fixed-target experiments, conservation of momentum results in a portion of the kinetic energy being tied up in the motion of the center of mass, thus reducing the energy available for creating new particles. In contrast, collider experiments can be arranged so that the center-of-mass is at rest, making the entire kinetic energy available for particle creation.

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Transcript

00:01 This problem is about relativistic collisions where particles hit each other and then their energy is converted into a new particle with a greater mass.

00:12 First, i want to point out that this problem is similar to problems 37 and 52 in this chapter.

00:20 So in problem 37, you had two masses colliding at the same speed, but they had different masses.

00:25 So after the collision, you have a greater mass, but some speed, v.

00:34 So the particle isn't stationary, new particle isn't stationary after the collision.

00:40 So why do you define mass, but you can't find mass without knowing how much of the system's momentum is in speed.

00:50 You have to find the value of fee, find the value of m.

00:53 So in that case, you started with the conservation momentum.

00:56 Law and then you use conservation energy so you had similar terms you substituted those into the conservation momentum expression then use that to solve for v and gamma which is really just another form of v and you would find m and these are numerical values in this case pf the final momentum of the system is not equal zero so in question 52 we instead had a stationary mass, which was an iron nucleus.

01:32 It was, so the p initial was zero.

01:38 So it emitted a photon.

01:44 So the final momentum stays the same as zero, but that means that there was a p nucleus and a p photon.

01:58 So there's a recoil effect.

02:02 The photon is released.

02:03 Photon has its own momentum, iron atom recoils.

02:10 But the problem is, so in that case, you start with conservation of energy.

02:15 So you know that there's energy of the photon.

02:19 You know there's energy of the recoil and rest mass energy.

02:24 But the problem is that energy is coming from the original mass being decreased to a smaller mass.

02:32 The problem is that that doesn't tell you how the energy is.

02:35 Is split up distributed between the recoil and the photon.

02:41 And you need to use momentum conservation, where the momentum of the nucleus is, magnitude is equal to the magnitude of the photon momentum.

03:00 To use that to find the speed v and then for you, i guess, and then use the speed you to find the kinetic energy, which is what that problem is looking for.

03:09 So in general, you use conservation laws to constrain other conservation laws.

03:16 So in this problem, we have two cases.

03:19 One is we have a stationary particle.

03:23 These are all protons, so the mass is the same in all cases.

03:27 So i'm just going to use m without subscripts.

03:32 The second proton is stationary.

03:36 The first one comes in at some high.

03:40 Speed or momentum.

03:42 That means after the collision, there's a new particle, greater mass, but the momentum, final momentum in the system is not equal zero.

03:52 So you need to constrain the conservation of energy equation with conservation of momentum law of relationships.

04:10 So you can find how much of the kinetic energy actually goes into increasing the mass.

04:18 In case two, you have, again, equal and opposite direction protons colliding head -on.

04:26 In that case, they're in the same speed.

04:30 So the magnitudes of p1 and p2 are the same.

04:35 So the final mass is stationary.

04:39 So the final momentum is zero.

04:42 In this case, all of k goes into increasing.

05:00 So the point of this problem is basically to show that case two is much more efficient for creating larger mass particles because in case one, you're losing some of the energy to the final momentum of the particle because it's going to keep going in some direction.

05:20 So all that being said, in case one, it wants us to find the relationship between the final mass, rest mass energy, with the connect energy input.

05:34 So, unfortunately, these collision problems, it's all about choosing which equations to use to try to get to where you want to go.

05:45 So we're starting with conservation of energy.

05:56 So the energy of the first incoming particle is a kinetic energy plus its rest mass energy, and e2 equals just the rest mass energy because it's stationary, so it has no kinetic energy.

06:28 So we also know that in general, e squared equals p .c squared plus mc squared squared.

06:42 So this is the energy momentum equation or relationship.

06:48 So we know from this that e1 squared is p1c squared plus mc squared squared squared.

06:58 So the value of these is that they will help us use conservation of momentum relationship to constrain the momentum term.

07:13 In these energy equations.

07:24 So that would have been p2c, the whole thing squared, but the momentum of the second particle is zero.

07:31 So in case two, we're not going to be able to do that.

07:40 So ef squared is pfc squared plus mz squared squared squared.

07:50 So in this case, there actually is a final momentum of the new mass, whereas in case two, that one is going to be the one that's going to be zero.

08:00 So this one will be zero in case two, but this one up here won't be.

08:11 So one last one equation we have is just since ef equals e1 plus e2, because that's what conservation of energy is.

08:31 That means ef squared equals e1 squared plus 2, e1, e2, plus...

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