1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Relativistic Collision of Particles

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data

    A particle of mass m whose total energy is twice its rest energy collides with an identical particle at rest. If they stick together, what is the mass of the resulting composite particle? What is its velocity?


    2. Relevant equations

    E = (gamma)mc^2
    p = (gamma)mu

    3. The attempt at a solution

    The total energy of the first particle is twice its rest energy, or

    E_{total} = (gamma)mc^2
    = 2mc^2

    or [tex]\gamma[/tex] = 2.

    Using this, I find that the velocity of particle 1 is:
    u = (sqrt{3})/{2}

    Using this is both momentum and energy conservation equations yields the two comparable equations:

    m_{final} = (gamma 1)m{1} / (gamma final)
    and
    m_{final} = sqrt{3}m{1}c / (gamma final) u{final}

    Solving this, I get:

    u_{final} = c

    and

    m_{final} = 0

    The two answers in relation to each other seem alright, but what is happening here? Is this saying that the particles completely annihilated each other? What is special about the initial conditions that makes this happen?
     
    Last edited: Sep 14, 2007
  2. jcsd
  3. Sep 14, 2007 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Just write the conservation equations in this form


    [tex] E_{tot}_{1}+E_{tot}_{2}=E_{tot}_{\mbox{resulting particle}} [/tex]

    [tex] p_{1}+p_{2}=p_{tot}_{\mbox{resulting particle}} [/tex]

    Then you can solve them quite easily i guess.
     
  4. Aug 18, 2008 #3
    Let v be the velocity of the particle before collision, and v1 be after collision. Similarly, 'm' be the mass before collision, and m1 after collision.

    Now, as correctly stated, v = ([tex]\sqrt{3}[/tex]/2)*c.
    also, [tex]\gamma[/tex](v)=2.
    So, from energy conservation, we get:
    m1*[tex]\gamma[/tex](v1)=2m.

    and, from momentum conservation, we get: v1=v.
    this invariably leads to m1=m.

    i think that this means: the first particle comes at rest, and the second particle travels with exactly the same velocity.
     
  5. Aug 18, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    There are TWO particles before the collision. Energy before the collision is the sum of BOTH their energies. And you don't need to explicitly compute v or gamma, just use conservation and E^2-p^2*c^2=m^2*c^4.
     
  6. Aug 19, 2008 #5
    yup, i missed it. i will try it once again now. thank you for pointing out the mistake.
     
  7. Aug 19, 2008 #6
    I am getting the final velocity to be [tex]\frac{c}{\sqrt{3}}[/tex]
    and combined rest mass as [tex]m\\*\\\sqrt{6}[/tex]

    please correct the answer if i have made a mistake once again.
    thank you.
     
    Last edited: Aug 19, 2008
  8. Aug 19, 2008 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think you have it right.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Relativistic Collision of Particles
  1. Relativistic collision (Replies: 0)

  2. Relativistic collision (Replies: 1)

  3. Relativistic collision (Replies: 2)

  4. Relativistic collision (Replies: 5)

Loading...