Particle creation, antiproton and 3 protons from 2 protons

Plugging this back into our equation for v_incident, we get v_incident = p_incident/(m-m_rest). This is the velocity needed for the incident proton to have enough energy to produce a proton and an antiproton in the lab frame.In summary, the threshold energy for the creation of a proton and an antiproton in the lab frame is given by the equation E_lab = γ*m_incident*c^2, where γ is the Lorentz factor and m_incident = m - m_rest. This can also be written as a function of the velocity of the incident proton, v_incident = p_incident/(m-m_rest). I
  • #1
Breedlove
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Homework Statement


So basically my instructor did this as an example and I am having trouble figuring out exactly the train of thought. So we've got two protons, in the lab frame, one is at rest and the other is incident on the other. What I want to know is how much energy and how fast the proton has to be moving in order to produce another proton and an antiproton upon the interaction with the rest proton. So I want the threshold energy really.


Homework Equations


Not completely sure. To transform E into another frame, the equation is E[tex]^{1}[/tex] (let's say that is E prime) = [tex]\gamma[/tex] E - [tex]\gamma[/tex] VxPx
first question about that, are Vx and Px the velocity and momentum of the frame? what exactly does everything correspond to here? E is the energy in the rest frame, Eprime is the energy in the moving frame, and I assume that Vx and Px are the velocity and momentum of the frame. (The frame has momentum?) Bah

I've looked online and found a semi walkthrough through the problem, http://galileo.phys.virginia.edu/classes/252/particle_creation.pdf
but I am still confused. What are the equations I should be dealing with?

The Attempt at a Solution



Okay. Taking the situation before the interaction, there's the lab frame, with a proton incident on a rest proton. E(lab)= (m(rest)+m(incident))c^2. This makes sense I think. Is this right? This is the energy in the lab frame prior to the collision. I think the secret to the answer to this problem is in the center of mass frame. In this frame the protons are incident on each other and have the same velocity. The situation has no momentum because they are equal and in opposite directions.

I think what I should do now is transform E to the center of mass frame. In my notes there's an equation E^2=p^2c^2+m(rest)^2c^2, although I'm not sure what this represents. I think it has the "invarient mass" in it. Here is another problem, it would seem to me that mass is always invarient; there isn't a transformation equation for mass that I'm aware of, but it makes sense because if energy is varient, then the relationship between mass and energy would imply that mass is varient.

BLAH. Okay. In my notes I've got m^2c^2=E(lab)^2-p(incident)^2c^2. Is the mass on the left side the invarient mass? What does the invarient mass represent?? How is it related to the two protons? Is it the invarient form of their sum? aughhh.

It occurs to me now that E(lab) is probably wrong, despite my notes. It seems like in the lab frame there is a momentum that affects the E. Hmm. If E=mc^2 then the lab frame is right, and to get the momentum we substitute that guy into this guy
m^2c^2=E(lab)^2-p(incident)^2c^2 and that gives us the momentum. Why or if that is helpful I'm not sure. I think I'm going to stop talking for a bit, think about the problem some more, read some more of the text, and let someone help prod me in the right direction. I'm sorry for this long ramble, I'm just kind of lost I guess. Thanks for whatever help you can give!
 
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  • #2



Dear student,

Thank you for your question. It seems like you are on the right track with your understanding of the problem. Let's break it down step by step.

First, let's define some variables:
- m_rest: the rest mass of the proton (approximately 938 MeV/c^2)
- E_lab: the energy of the system in the lab frame
- E_cm: the energy of the system in the center of mass frame
- p_incident: the momentum of the incident proton in the lab frame
- p_cm: the momentum of both protons in the center of mass frame (since they have the same velocity)

Now, let's start with the equation you mentioned, E^2=p^2c^2+m(rest)^2c^2. This is a special case of the more general equation E^2=p^2c^2+(mc^2)^2, where m is the invariant mass. In this case, m represents the combined mass of the two protons, since they are both present in the system.

Next, we can use the definition of momentum, p=mv, to solve for the momentum of the incident proton in the lab frame: p_incident = m_incident*v_incident. Since we know the mass of the incident proton and we are interested in the threshold energy, we can rearrange this equation to solve for v_incident: v_incident = p_incident/m_incident.

Now, we can use the equation for relativistic energy, E=γmc^2, to calculate the energy of the incident proton in the lab frame: E_lab = γ*m_incident*c^2. Here, γ is the Lorentz factor, which takes into account the relativistic effects of the high speed of the proton.

Next, we can use the conservation of energy and momentum to find the energy in the center of mass frame, E_cm. Since the system has no momentum in this frame, the total energy will be equal to the invariant mass of the two protons, which we can calculate using the equation from earlier: E_cm = m.

Finally, to find the threshold energy, we just need to set E_cm equal to the sum of the rest masses of the two protons, since this is the minimum energy needed to create a proton and an antiproton. This gives us the equation m = m_rest + m_incident. Solving for m_inc
 

1. How is particle creation possible?

Particle creation is possible through the conversion of energy into mass. This process is described by Einstein's famous equation E=mc2, where E represents energy, m represents mass, and c represents the speed of light. In certain conditions, such as high-energy collisions, this energy can be converted into particles, including antiprotons and protons.

2. What is an antiproton?

An antiproton is the antiparticle of a proton, meaning it has the same mass as a proton but with opposite charge. It is created when a high-energy proton collides with a target, producing an antiproton and other particles. Antiprotons are important in scientific research as they can be used to study the properties of antimatter, which is still largely a mystery to scientists.

3. How are 3 protons formed from 2 protons?

In certain nuclear reactions, 2 protons can fuse together to form a deuteron (a nucleus of deuterium, or heavy hydrogen). This deuteron can then collide with another proton, forming a nucleus of helium-3, which contains 2 protons and 1 neutron. Therefore, the end result is 3 protons from the original 2 protons.

4. What applications does particle creation have?

Particle creation has many applications, including in nuclear energy, medical imaging, and scientific research. In nuclear energy, particle creation is used in nuclear reactors to produce heat, which is then converted into electricity. In medical imaging, particle creation is used in positron emission tomography (PET) scans to produce images of the body's internal structures. And in scientific research, particle creation allows scientists to study the fundamental building blocks of matter and the laws of physics.

5. Can particle creation be observed in everyday life?

No, particle creation is not observable in everyday life as it requires extremely high energy conditions that are not present in our daily surroundings. Particle creation can only be observed in particle accelerators or in natural phenomena such as cosmic rays colliding with the Earth's atmosphere.

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