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Relativistic particle scattering

  1. Oct 1, 2009 #1
    Hello,

    I want to do a scattering calculation with two particles scattering to two (other) particles relativistically.

    Is it correct to use the following conservation laws?
    [tex]
    \vec{p}_1+\vec{p}_2=\vec{p}_3+\vec{p}_4
    [/tex]
    [tex]
    \gamma_1 m_1c^2+\gamma_2 m_2c^2=\gamma_3 m_3c^2+\gamma_4 m_4c^2
    [/tex]
    where the masses m are rest masses and [itex]\gamma[/itex] are the corresponding Lorentz factors .

    Someone on the forum argued that I need to use rest energy [itex]mc^2[/itex] for conservation law instead of [itex]\gamma mc^2[/itex].

    Please clarify.
     
  2. jcsd
  3. Oct 1, 2009 #2
    The total energy of a particle of rest mass m is [tex]E=\gamma mc^{2}[/tex], or you can write the following with kinetic and mass-energy pieces separated: [tex]E^{2}= (pc)^{2}+(mc^{2})^{2}[/tex].

    Then the total energy and momentum are conserved in a fixed inertial reference frame, as you wrote.
     
  4. Oct 1, 2009 #3
    Thanks for the quick response. Actually I know the physics quite well. But some self-announced expert claims I'm wrong.

    So just vote "yes" if you agree and explain if you don't agree :)
     
  5. Oct 1, 2009 #4

    Dale

    Staff: Mentor

    The easiest and most consistent way to do this is to use the four-momentum (http://en.wikipedia.org/wiki/Four-momentum) which is conserved. But yes, you have correctly written the timelike and spacelike components of the four-momentum for a massive particle. The four-momentum and javierR's equations are a little more general and apply to massless particles also.
     
  6. Oct 1, 2009 #5
    True. For mass-less particles I need E=pc.
     
  7. Oct 2, 2009 #6
    Ugh.
    This appears to be you trying to continue an argument in this locked thread:
    https://www.physicsforums.com/showthread.php?t=341514&page=2

    I don't see how what you have written in this thread in anyway proves ZapperZ wrong.
    ZapperZ did not argue that rest energy is conserved. It is of course total energy that is conserved. Much of the argument seems to be a misunderstanding starting with your unconventional and sometimes sloppy terminology. It was a misunderstanding, so just take what you can to learn from the conversation and move on.

    It might be worthwhile to understand the criticisms against using relativistic mass (or at the very least understand that it is not really used by physicists anymore, and so is becoming a bit archaic like "transverse mass" became).
     
  8. Oct 2, 2009 #7
    You are absolutely right with your interpretation and with my initial sloppy interpretation.
    We wrote PM and he explained that my concept of total energy is useless. I was trying to explain to him that I want an energy that obeys the law of conservation of energy. He seems to insist on historically correct quotation of Einstein instead.
    I know he means "E0=m0c^2", but he shouldn't complain when I write "*E*=\gamma m0c^2". Writing "E=m0 c^2" would probably even be incorrect.
     
    Last edited: Oct 2, 2009
  9. Oct 2, 2009 #8

    Dale

    Staff: Mentor

    IMO, this all gets resolved very nicely once the concept of the four-momentum is introduced. You can immediately see the difference between the invariant "rest" mass which is the norm of the four-momentum and the relative total energy and momentum which are the components of the four-momentum. It also helps explain the conservation laws and in what sense the rest mass of a system is also conserved. I would highly recommend the minimal effort required to learn it as it simplifies so many different things.
     
  10. Oct 2, 2009 #9

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This thread is now closed. I intend to make a longer comment in this thread, but I doubt that the thread will be re-opened. Please do not open a new thread on this.
     
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