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Homework Help: Particle creation, antiproton and 3 protons from 2 protons

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    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.

    2. Relevant 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?

    3. 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 eachother 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!
  2. jcsd
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