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Energy-momentum Four-vectors

  1. Oct 22, 2005 #1
    Does anyone know how you find the length of the energy-momentum four-vector for a system of particles?
    p_mu=(E/c,p)
    where length is:
    length(p_mu)=-(E/c)^2+(p)^2

    Do you first add the corresponding vector elements then find the length
    OR
    find the length of each particle first then sum the individual lengths.

    Cheers,
    Jimmy
     
  2. jcsd
  3. Oct 22, 2005 #2
    Both values will give you invariants, although the energy-momentum four-vector ([itex]p^\mu[/itex]) of the whole system is equal to the sum of all the individual [itex]p^\mu[/itex], and therefore the length of [itex]p^\mu[/itex] for the system is the length of the sum of all the individual [itex]p^\mu[/itex].
     
  4. Oct 22, 2005 #3

    jtbell

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    Staff: Mentor

    It's exactly analogous to finding the magnitude of the total three-momentum of a system of particles. In that case, you find the total x-momentum, total y-momentum, and total z-momentum of momentum, then use them to find the magnitude of the total-momentum vector.
     
  5. Oct 22, 2005 #4
    If the particles are interacting through when they are seperated (e.g. two charged particles) then the addition of the two 4-vectors is meaningless. Only systems of non-interacting particles anmd systems of particles which interact only through contact forces can be added in a meaningful way. To add the vectors you add components and then take find the magnitude.

    This web page I created will get into great detail regarding this. See
    http://www.geocities.com/physics_world/sr/invariant_mass.htm

    Pete
     
  6. Oct 23, 2005 #5
    Is this method ok?

    Thanks for the clarification...

    What I was trying to do was find the lengths of the four-vectors of this reaction before and after.
    p + p ==> p + p + Z

    Where a proton with 300GeV hits a stationary proton, then producing a particle Z.

    I calculated the length of the Four-vector before the reaction in the stationary proton's frame.

    I then equate this to the length of the four-vector after the collision in the
    center-of-mass frame to extract the rest mass of the Z particle.

    length(p_mu1+p_mu2)=length(p_mu3+p_mu4+p_muZ)

    Question, is there anything wrong with my method?

    I have assumed that after the collision the two protons and the Z particle are at rest, since I want the maximum possible rest mass of Z. Momentum in the COM from is zero so it should be ok?

    Jimmy
     
  7. Oct 24, 2005 #6

    jtbell

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    Staff: Mentor

    Yes, that's a reasonable way to proceed. What you end up with is the largest mass the Z can have, and still be produced under these initial conditions.
     
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