Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Positron-electron collision can someone check my answer?

  1. May 15, 2010 #1
    1. The problem statement, all variables and given/known data

    a positron and electron travel in opposite directions each with energy 5ooGeV and collide head-on.
    what is the energy in the centre of mass (cm) of the collision?



    3. The attempt at a solution

    E(cm) = E(e+) + E(e-) = 500GeV + 500GeV = 1000GeV

    is this right?

    thanks
     
  2. jcsd
  3. May 15, 2010 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes.
     
  4. May 15, 2010 #3
    okay then thanks for that...
    also could you just look at this as well? i wanted to try it before i put it up

    if the same energy in the cm were to be achieved by colliding a beam of positrins with a target of stationary electrons, what energy of the positron would be required?

    in cm frame

    v = positron vel. in lab frame
    v+' = positron vel. in cm
    v-' = electron vel. in cm
    vcm = cm vel.

    v+' = v-vcm
    v-' = vcm

    [STRIKE]m[/STRIKE][v+'] = [STRIKE]m[/STRIKE][v-']
    so Ecm = 1/2m[v+']2 + 1/2m[v-']2 = m[v+']2

    v+' = v-vcm
    v-' = vcm
    so v - vcm = vcm therefore vcm= v/2

    so Ecm = m[v-vcm]2 = m[v/2]2
    and then v2 = (4Ecm)/m

    Kinetic energy of beam = 1/2mv2 = 1/2m[(4Ecm)/m] = 2Ecm = 2000GeV

    is this right?
    thanks
     
  5. May 15, 2010 #4

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, you're mixing up Newtonian mechanics with relativistic mechanics. You need to be a bit more careful.
     
  6. May 15, 2010 #5
    where is it that i've gone wrong?
    what would you suggest to try and get the right answer?
     
  7. May 15, 2010 #6

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    For one thing, the kinetic energy isn't equal to 1/2 mv2 in relativity. Also, you can't just add and subtract velocities like you did. You have to use the relativistic velocity-addition formula.

    The quantity E2-(pc)2 is an invariant. Use that to relate the quantities in the two frames.
     
  8. May 15, 2010 #7
    ok so i tried this instead

    mass of e+ and e- =>
    m2c4 = (E+mc2)2 - p2c2
    = E2 - p2c2 + 2Emc2 + m2c4
    = 2Emc2 + 2m2c4

    and as mc2<<E
    m2c4 = 2Emc2

    now this is where i'm abit unsure:
    can i say: m2 = (1000GeV)2?
    cause this would give
    (1000GeV)2 = 2E(0.51MeV)
    so E = 1.96x1018eV
     
  9. May 15, 2010 #8

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    How did you get this?
     
  10. May 15, 2010 #9
    (E+mc2)
    is the sum of the energies of the 2 particles. the e- is stationary so i thought that its energy is just mc2 in the lab frame
    p2c2 i didnt put a p(e+) +p(e-) because the momentum of e- is zero so its just the mom. of the e+
     
  11. May 15, 2010 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What about the lefthand side of the equation?
     
  12. May 15, 2010 #11
    as in that i should have put (2m)2c4 on the left hand side?
     
  13. May 15, 2010 #12

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    No, I'm trying to ascertain why you set it to m2c4 in the first place and why later you ask if it should be (1000 GeV)2. The variable m stands for the mass of the electron, right?

    The RHS of your equation is fine. I just want to know what you think that quantity, E2total-(ptotalc)2 should equal and why it should equal that. This is a key point in solving this problem.
     
  14. May 15, 2010 #13
    well i set it to m2c4 cause this is the invariant mass and so i can find this in the lab frame and it will be the same in the cm frame.
    when i found the relation for this i went to the cm frame to use its Ecm to find the energy i need for the beam.

    now while writing this i think i should have done:
    m2c4 = 2Emc2 = 2(1000GeV)(0.52MeV)c2
    m2 = 1x1018eV
    m=1GeV
    and so E=mc2 = (1GeV)c2

    is this right? or am i still failing in my understanding of it?
     
  15. May 15, 2010 #14

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    It's the invariant mass of the two-particle system, not of the electron, so you shouldn't use m, the mass of the electron, on the LHS.
    No, you're mixing up quantities between the two frames. Look at it this way:

    E2cm - p2cm = E2lab - p2lab

    where the lab frame is where the target is stationery and where RHS of the equation is the same as the RHS of your equation. What should be on the left, I think, should be clear to you now.

    Edit: Out of habit, I left out the factors of c, but you should be able to figure out where they should be.
     
  16. May 16, 2010 #15
    ok so on the left the Ecm is 1000GeV,
    is pcm2c2 = Ecm2 - m2c4

    ...wait i dont think that helps...
    my lecturer didnt really go over how to get the momentum. the only thing he metioned ws the 4momentum but i dont think that is what i use here
     
  17. May 16, 2010 #16

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    What is the total momentum in the center-of-mass frame?
     
  18. May 16, 2010 #17
    wait is it just zero?...thats what i got from doing some rearrangement (too much to write out)
     
  19. May 16, 2010 #18

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Yes, the center-of-mass frame is the frame where the total momentum is 0.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook